||◦||. The Banach space C'_B = C([0, t], B) consist of continuous functions
ψ(t), ψ : [0, t] → B, with the norm

$$
\|\psi\|_{C_B^t} = \sup_{0 \le \tau \le t} \|\psi(\tau)\|_B. \qquad (2.1.1.10)
$$

If $t' < t$ we denote $\psi'(t) \triangleq \psi(t) \upharpoonright [0, t']$, thus $\psi'(t) \in C([0, t'], \mathbf{B})$.
Suppose that:

$$
(1) \varepsilon_0 > 0, T_0 > 0,
$$

(2) $\forall t, \tilde{\varepsilon}$ such that $0 \le t \le T, (-t) \le \tilde{\varepsilon} \le \varepsilon_0$ there exists a function

$$
S_{t,\tilde{\varepsilon}}[\circ] : \mathbf{C}([0,t'], \mathbf{B}) \times [-t, \varepsilon_0] \to \mathbf{B} \quad (2.1.1.11)
$$

such that $\forall v \in C([0, t'], \mathbf{B})$, $\forall t \in [0, T_0]$ function $q(t, \tilde{\varepsilon}) = \|S_{t, \tilde{\varepsilon}}[v]\|_{\mathbf{B}}$ is continuous on variable $\tilde{\varepsilon}$,

(3) if $\tilde{\varepsilon} \le 0$, then $S_{t,\tilde{\varepsilon}}[v] = S_t[v] \triangleq v(t)$.

Let's consider now finite splitting Δ<sub>N</sub> of the closed interval [0, T<sub>0</sub>] :

$$
\Delta_N = \{0 = t_0 < t_1 < \dots < t_N = T_0\}. \qquad (2.1.1.12)
$$

Note that for every $v_0 \in \mathbf{B}$ there exists an sequence $v_j(t)$, $j = 0, 1, \dots, N-1$ such that $v_{j+1}(t) = S_{t_j, t-t_j}[v_j]$, $0 \le t \le t_{j+1}$.

**Definition 2.1.1.1.** [20].

$$
v_N(t) = S_{t_N, \Delta t_{N-1}} \circ \dots \circ S_{t_j, \Delta t_j} \circ \dots \circ S_{t_0, \Delta t_0}[v_0], \qquad (2.1.1.13)
$$

$$
\Delta t_j = t_{j+1} - t_j, \quad j = 0, 1, \dots, N-1.
$$

**Definition 2.1.1.2.** [20] Suppose that:

(1) $v_0 \in D \subsetneq B$ and

(2) s-lim$_{\delta_{N} \to 0} v_N(t)$ there exists.
We denote this limit by

$$v(t) = \left( \prod_{\tau=0}^{t} S_{\tau, d\tau} \right) [v_0] \qquad (2.1.1.14)$$

and call it chronological exponential with a generator $S_{t, \varepsilon}[v]$.

Let $\mathbf{B}_1 \subsetneq \mathbf{B}_2$ be a complex separable infinite dimensional Banach space with norm $\|\circ\|_1 \ge \|\circ\|_2$ and $C_{\mathbf{B}_j}^t = C([0, t], \mathbf{B}_j), j = 1, 2$. Thus $C_{\mathbf{B}_1}^t \subsetneq C_{\mathbf{B}_2}^t$.

Suppose that for every $t \in [0, T_0]$ there exists continuous operator

$$\Re_t[\circ] : C_{\mathbf{B}_1}^t \to \mathbf{B}_2 \qquad (2.1.1.15)$$

such that for every $v \in C_{\mathbf{B}_1}^t$ function $\|\Re_t[v]\|_{\mathbf{B}_2}$ is continuous on a variable $t \in [0, T_0]$.

Let's consider the Cauchy problem with initial condition $\Phi_0 \in \mathbf{B}_1$:

$$\frac{d}{dt} \Phi(t) = \Re_t[\Phi(t)], \Phi(0) = \Phi_0, \qquad (2.1.1.16)$$

where

$$\frac{d}{dt} \Phi(t) \triangleq \lim_{\Delta t \to 0} \left\| \frac{\Phi(t + \Delta t) - \Phi(t)}{\Delta t} \right\|_2, \qquad (2.1.1.17)$$

i.e. $\forall t$ and $\forall v_t \in \mathbf{B}_2$

$$\frac{d}{dt} \Phi(t) = v_t \Leftrightarrow \lim_{\Delta t \to 0} \left\| v_t - \frac{\Phi(t + \Delta t) - \Phi(t)}{\Delta t} \right\|_2 = 0. \qquad (2.1.1.18)$$
We denote this limit by

$$v(t) = \left( \prod_{\tau=0}^{t} S_{\tau, d\tau} \right) [v_0] \qquad (2.1.1.14)$$

and call it chronological exponential with a generator $S_{t, \varepsilon}[v]$.

Let $\mathbf{B}_1 \subsetneq \mathbf{B}_2$ be a complex separable infinite dimensional Banach space with norm $\|\circ\|_1 \ge \|\circ\|_2$ and $C_{\mathbf{B}_j}^t = C([0, t], \mathbf{B}_j), j = 1, 2$. Thus $C_{\mathbf{B}_1}^t \subsetneq C_{\mathbf{B}_2}^t$. Suppose that for every $t \in [0, T_0]$ there exists continuous operator

$$\Re_t[\circ] : C_{\mathbf{B}_1}^t \to \mathbf{B}_2 \qquad (2.1.1.15)$$

such that for every $v \in C_{\mathbf{B}_1}^t$ function $\|\Re_t[v]\|_{\mathbf{B}_2}$ is continuous on a variable $t \in [0, T_0]$.
Let's consider the Cauchy problem with initial condition $\Phi_0 \in \mathbf{B}_1$:

$$\frac{d}{dt} \Phi(t) = \Re_t[\Phi(t)], \Phi(0) = \Phi_0, \qquad (2.1.1.16)$$

where

$$\frac{d}{dt} \Phi(t) \triangleq \lim_{\Delta t \to 0} \left\| \frac{\Phi(t + \Delta t) - \Phi(t)}{\Delta t} \right\|_2, \qquad (2.1.1.17)$$

i.e. $\forall t$ and $\forall v_t \in \mathbf{B}_2$

$$\frac{d}{dt} \Phi(t) = v_t \Leftrightarrow \lim_{\Delta t \to 0} \left\| v_t - \frac{\Phi(t + \Delta t) - \Phi(t)}{\Delta t} \right\|_2 = 0. \qquad (2.1.1.18)$$
**Definition 2.1.1.3.**[20]. Let be $S_{t,\tilde{\epsilon}}[\circ] : C([0,t], \mathbf{B}) \times [-t, \epsilon_0] \to \mathbf{B}$ the infinitesimal generator of the corresponding $T$-exponent. Suppose that for every $t \in [0, T_0]$ the next condition is satisfied

$$ \lim_{\tilde{\epsilon} \to 0} \left\| \frac{S_{t,\tilde{\epsilon}}[\nu] - \nu(t)}{\tilde{\epsilon}} - \mathfrak{R}_t[\nu] \right\|_2 = 0. \quad (2.1.19) $$

locally uniformly on variable $\nu(t) \in C_{\mathbf{B}_1}^t$. We call operator function $S_{t,\tilde{\epsilon}}[\circ]$ as master operator for Cauchy problem (2.1.16).

**Theorem 2.1.1.2.**[20]. Let be $S_{t,\tilde{\epsilon}}[\circ]$ the master operator for Cauchy problem (2.1.1.16)

and for some $\psi_0 \in \mathbf{B}_1$ there exists $\psi(t) = (\prod_{\tau=0}^t S_{\tau,d\tau})[\psi_0], \psi(t) \in C_1(T_0)$. Then $\psi(t)$ is a solution of the Eq.(2.1.1.16).

**Proof.** Let's consider finite splitting $\Delta_N = \{0 = t_0 < t_1 < \dots < t_N = T_0\}$ of the closed interval $[0, T_0]$ and $\nu_N(t) = S_{t_N, \Delta t_{N-1}} \circ \dots \circ S_{t_j, \Delta t_j} \circ \dots \circ S_{t_0, \Delta t_0}[\psi_0]$. Then for every $t_{j+1} \ge t \ge t_j$, one obtain $\psi_N(t) = S_{t_j, t-t_j}[\psi_N]$. Thus there exists $N_0$ such that

$$ \lim_{\Delta t \to 0} \left\| \frac{\psi_N(t_j + \Delta t) - \psi_N(t_j)}{\Delta t} - \mathfrak{R}_{t_j}[\psi_N] \right\|_2 = 0 \quad (2.1.1.20) $$

uniformly on $N \ge N_0$.

Let $t$ be a fixed point $t \in [0, T_0)$ and let $\{\Delta_N\}_{N=1}^\infty$ be the sequence of the splittings such that for any $N : t \in \Delta_N$.

Now take into account that operator $\mathfrak{R}_t$ is continuous finally we obtain

$$ \lim_{\Delta t \to 0} \left\| \frac{\psi(t + \Delta t) - \psi(t)}{\Delta t} - \mathfrak{R}_t[\psi] \right\|_2 = 0. \quad (2.1.1.21) $$

Let $\mathfrak{R}_t[\psi]$ be operator such that

$$ \mathfrak{R}_t[\psi] = A_t[\psi]\psi(t). \quad (2.1.1.22) $$

where $\forall \psi(\psi \in C_{\mathbf{B}_1}^t)$ operator $A_t[\psi] : \mathbf{B}_1 \to \mathbf{B}_2$ is continuous. Thus $A_t[\psi]$ is a
**Definition 2.1.1.3.** [20]. Let be $S_{t,\tilde{\epsilon}}[\circ] : C([0,t], \mathbf{B}) \times [-t, \epsilon_0] \to \mathbf{B}$ the infinitesimal generator of the corresponding $T$-exponent. Suppose that for every $t \in [0, T_0]$ the next condition is satisfied

$$ \lim_{\tilde{\epsilon} \to 0} \left\| \frac{S_{t,\tilde{\epsilon}}[\nu] - \nu(t)}{\tilde{\epsilon}} - \mathfrak{R}_t[\nu] \right\|_2 = 0. \quad (2.1.19) $$

locally uniformly on variable $\nu(t) \in C_{\mathbf{B}_1}^t$. We call operator function $S_{t,\tilde{\epsilon}}[\circ]$ as master operator for Cauchy problem (2.1.16).

**Theorem 2.1.1.2.** [20]. Let be $S_{t,\tilde{\epsilon}}[\circ]$ the master operator for Cauchy problem (2.1.1.16)

and for some $\psi_0 \in \mathbf{B}_1$ there exists $\psi(t) = (\prod_{\tau=0}^t S_{\tau,d\tau})[\psi_0], \psi(t) \in C_1(T_0)$. Then $\psi(t)$ is a solution of the Eq.(2.1.1.16).

**Proof.** Let's consider finite splitting $\Delta_N = \{0 = t_0 < t_1 < \dots < t_N = T_0\}$ of the closed interval $[0, T_0]$ and $\nu_N(t) = S_{t_N, \Delta t_{N-1}} \circ \dots \circ S_{t_j, \Delta t_j} \circ \dots \circ S_{t_0, \Delta t_0}[\psi_0]$. Then for every $t_{j+1} \ge t \ge t_j$, one obtain $\psi_N(t) = S_{t_j, t-t_j}[\psi_N]$. Thus there exists $N_0$ such that

$$ \lim_{\Delta t \to 0} \left\| \frac{\psi_N(t_j + \Delta t) - \psi_N(t_j)}{\Delta t} - \mathfrak{R}_{t_j}[\psi_N] \right\|_2 = 0 \quad (2.1.1.20) $$

uniformly on $N \ge N_0$.

Let $t$ be a fixed point $t \in [0, T_0)$ and let $\{\Delta_N\}_{N=1}^\infty$ be the sequence of the splittings such that for any $N : t \in \Delta_N$.

Now take into account that operator $\mathfrak{R}_t$ is continuous finally we obtain

$$ \lim_{\Delta t \to 0} \left\| \frac{\psi(t + \Delta t) - \psi(t)}{\Delta t} - \mathfrak{R}_t[\psi] \right\|_2 = 0. \quad (2.1.1.21) $$

Let $\mathfrak{R}_t[\psi]$ be operator such that

$$ \mathfrak{R}_t[\psi] = A_t[\psi]\psi(t). \quad (2.1.1.22) $$

where $\forall \psi(\psi \in C_{\mathbf{B}_1}^t)$ operator $A_t[\psi] : \mathbf{B}_1 \to \mathbf{B}_2$ is continuous. Thus $A_t[\psi]$ is a
infinitesimal generator of the one-parameter semigroup

$$ \exp[\tilde{\epsilon} A_t[\psi]]. \qquad (2.1.1.23) $$

Hence master operator of the Cauchy problem (2.1.1.16) is

$$ S_{t,\tilde{\epsilon}}[\nu] = \exp[\tilde{\epsilon} A_t[\psi]]\nu(t). \qquad (2.1.1.24) $$

Corresponding T-exp is

$$ \psi(t) = \prod_{\tau}^{t} \exp[A_{\tau}[\psi]]d\tau. \qquad (2.1.1.25) $$

**Definition 2.1.1.4.** Let B be a complex separable infinite dimensional Banach space with the norm $||\cdot||$. The $\tilde{C}$-module $C'_{G_B} = C([0,t], G_B)$ consist of continuous functions $(\psi_{\epsilon}(t))_{\epsilon} : [0,t] \to G_B$, with the ultra-pseudo-norm $||\cdot||_{C'_{G_B}}$:

$$ \|(\psi_{\epsilon})_{\epsilon}\|_{C'_{G_B}} = \sup_{0 \le \tau \le t} \|(\psi_{\epsilon}(\tau))_{\epsilon}\|_{G_B}. \qquad (2.1.1.26) $$

If $t' < t$ we denote $(\psi'_{\epsilon}(t))_{\epsilon} \triangleq (\psi_{\epsilon}(t))_{\epsilon} \upharpoonright [0,t']$, thus $(\psi'_{\epsilon}(t))_{\epsilon} \in C([0,t'], G_B)$.

Suppose that:

(1) $\epsilon_0 > 0, T_0 > 0$,

(2) $\forall t, \tilde{\epsilon}$ such that $0 \le t \le T, (-t) \le \tilde{\epsilon} \le \epsilon_0$ there exists Colombeau generalized function

$$ (S_{t,\tilde{\epsilon},\epsilon}[\circ])_{\epsilon \in (0,1]} : C([0,t'], G_B) \times [-t, \epsilon_0] \to G_B \qquad (2.1.1.27) $$
infinitesimal generator of the one-parameter semigroup

$$ \exp[\tilde{\epsilon} A_t[\psi]]. \qquad (2.1.1.23) $$

Hence master operator of the Cauchy problem (2.1.1.16) is

$$ S_{t,\tilde{\epsilon}}[\nu] = \exp[\tilde{\epsilon} A_t[\psi]]\nu(t). \qquad (2.1.1.24) $$

Corresponding T-exp is

$$ \psi(t) = \prod_{\tau}^{t} \exp[A_{\tau}[\psi]]d\tau. \qquad (2.1.1.25) $$

**Definition 2.1.1.4.** Let B be a complex separable infinite dimensional Banach space with the norm $\|\cdot\|$. The $\tilde{C}$-module $C'_{G_B} = C([0,t], G_B)$ consist of continuous functions $(\psi_{\varepsilon}(t))_{\varepsilon \in [0,t]}$, $(\psi_{\tilde{\epsilon}}(t))_{\tilde{\epsilon} \in [0,t]} \to G_B$, with the ultra-pseudo-norm $\|\cdot\|_{C'_{G_B}}$:

$$ \|(\psi_{\varepsilon})_{\varepsilon} \|_{C_{G_B}^t} = \sup_{0 \le \tau \le t} \|(\psi_{\varepsilon}(\tau))_{\varepsilon} \|_{G_B}. \qquad (2.1.1.26) $$

If $t' < t$ we denote $(\psi'_{\varepsilon}(t))_{\varepsilon \in [0,t']}$, thus $(\psi'_{\tilde{\epsilon}}(t))_{\tilde{\epsilon} \in [0,t']}$.

Suppose that:

(1) $\varepsilon_0 > 0, T_0 > 0$,

(2) $\forall t, \tilde{\epsilon}$ such that $0 \le t \le T, (-t) \le \tilde{\epsilon} \le \varepsilon_0$ there exists Colombeau generalized function

$$ (S_{t,\tilde{\epsilon},\varepsilon}[\circ])_{\varepsilon \in (0,1]} : C([0,t'], G_B) \times [-t, \varepsilon_0] \to G_B \qquad (2.1.1.27) $$
such that $\forall (v_\varepsilon)_\varepsilon \in C([0, t'], G_B), \forall t \in [0, T_0]$ function $(q_\varepsilon(t, \tilde{\varepsilon}))_\varepsilon = \|(S_{t, \tilde{\varepsilon}, \varepsilon}[v_\varepsilon])_\varepsilon\|_{G_B}$ is
continuous on variable $\tilde{\varepsilon}$,

(3) if $\tilde{\varepsilon} \le 0$, then $(S_{t,\tilde{\varepsilon},\varepsilon}[v_\varepsilon])_\varepsilon = (S_{t,\varepsilon}[v_\varepsilon])_\varepsilon \triangleq (v_\varepsilon(t))_\varepsilon$.

Let's consider now finite splitting $\Delta_N$ of the closed interval $[0, T_0]$ :

$$
\Delta_N = \{0 = t_0 < t_1 < \dots < t_N = T_0\}. \tag{2.1.1.28}
$$

Note that for every $(v_{0,\varepsilon})_\varepsilon \in G_B$ there exists an sequence $(v_{j,\varepsilon}(t))_\varepsilon, j = 0, 1, \dots, N-1$
such that $(v_{j+1,\varepsilon}(t))_\varepsilon = (S_{t_j, t-t_j,\varepsilon}[v_{j,\varepsilon}])_\varepsilon, 0 \le t \le t_{j+1}$.

**Definition 2.1.1.5.**

$$
(v_{N,\epsilon}(t))_{\epsilon} = (S_{t_N, \Delta t_{N-1}, \epsilon} \circ \dots \circ S_{t_j, \Delta t_j, \epsilon} \circ \dots \circ S_{t_0, \Delta t_0, \epsilon}[v_{0,\epsilon}])_{\epsilon}, \quad (2.1.1.29)
$$

$$
\Delta t_j = t_{j+1} - t_j, j = 0, 1, \dots, N-1.
$$

**Definition 2.1.1.6.** Suppose that:

(1) $(v_{0,\varepsilon})_\varepsilon \in D \subsetneqq G_B$ and

(2) s-lim$_{\delta \to 0}$ $(v_{N,\epsilon}(t))_\epsilon$ there exists.

We denote this limit by

$$
(v_{\varepsilon}(t))_{\varepsilon} = \left( \left( \prod_{\tau=0}^{t} S_{\tau, d\tau, \varepsilon} \right) [v_{0,\varepsilon}] \right)_{\varepsilon} \quad (2.1.1.30)
$$

and call it *chronological exponential with a generator* $(S_{t,\tilde{\varepsilon},\varepsilon}[v_\varepsilon])_\varepsilon$.

Let $\mathbf{B}_1 \subsetneqq \mathbf{B}_2$ be a complex separable infinite dimensional Banach spaces with norm
$\|\cdot\|_1 \ge \|\cdot\|_2$ and let $\mathbf{C}^t_{G_{\mathbf{B}_j}}$ be a $\tilde{\mathbb{C}}$-modules $\mathbf{C}^t_{G_{\mathbf{B}_j}} = \mathbf{C}([0,t], G_{\mathbf{B}_j}), j = 1,2$. Thus
$\mathbf{C}^t_{G_{\mathbf{B}_1}} \subsetneqq \mathbf{C}^t_{G_{\mathbf{B}_2}}$. Suppose that for every $t \in [0,T_0]$ there exists continuous operator

$$
(\mathfrak{R}_{t,\varepsilon}[\circ])_{\varepsilon} : \mathbf{C}_{G_{\mathbf{B}_1}}^{t} \rightarrow \mathbf{G}_{\mathbf{B}_2} \qquad (2.1.1.31)
$$

such that for every $(v_\varepsilon)_\varepsilon \in \mathbf{C}^t_{G_{\mathbf{B}_1}}$ function $\|(\mathfrak{R}_{t,\varepsilon}[v_\varepsilon])_\varepsilon\|_{G_{\mathbf{B}_2}}$ is continuous on a
such that $\forall (v_\epsilon)_\epsilon \in C([0, t'], G_B), \forall t \in [0, T_0]$ function $(q_\epsilon(t, \tilde{\epsilon}))_\epsilon = \|(S_{t, \tilde{\epsilon}, \epsilon}[v_\epsilon])_\epsilon\|_{G_B}$ is
continuous on variable $\tilde{\epsilon}$,

(3) if $\tilde{\epsilon} \le 0$, then $(S_{t,\tilde{\epsilon},\epsilon}[v_\epsilon])_\epsilon = (S_{t,\epsilon}[v_\epsilon])_\epsilon \triangleq (v_\epsilon(t))_\epsilon$.

Let's consider now finite splitting $\Delta_N$ of the closed interval $[0, T_0]$:

$$
\Delta_N = \{0 = t_0 < t_1 < \dots < t_N = T_0\}. \tag{2.1.1.28}
$$

Note that for every $(v_{0,\epsilon})_{\epsilon} \in G_B$ there exists an sequence $(v_{j,\epsilon}(t))_{\epsilon}, j = 0, 1, \dots, N-1$
such that $(v_{j+1,\epsilon}(t))_{\epsilon} = (S_{t_j, t-t_j, \epsilon}[v_{j,\epsilon}])_{\epsilon}, 0 \le t \le t_{j+1}$.

**Definition 2.1.1.5.**

$$
(v_{N,\epsilon}(t))_{\epsilon} = (S_{t_N, \Delta t_{N-1}, \epsilon} \circ \dots \circ S_{t_j, \Delta t_j, \epsilon} \circ \dots \circ S_{t_0, \Delta t_0, \epsilon}[v_{0,\epsilon}])_{\epsilon}, \quad (2.1.1.29)
$$

$$
\Delta t_j = t_{j+1} - t_j, j = 0, 1, \dots, N-1.
$$

**Definition 2.1.1.6.** Suppose that:

(1) $(v_{0,\epsilon})_\epsilon \in D \subsetneq G_B$ and

(2) $s$-lim$_{\delta \to 0}$ $(v_{N,\epsilon}(t))_\epsilon$ there exists.

We denote this limit by

$$
(v_{\epsilon}(t))_{\epsilon} = \left( \left( \prod_{\tau=0}^{t} S_{\tau, d\tau, \epsilon} \right) [v_{0,\epsilon}] \right)_{\epsilon} \quad (2.1.1.30)
$$

and call it *chronological exponential with a generator* $(S_{t,\tilde{\epsilon},\epsilon}[v_\epsilon])_\epsilon$.

Let $\mathbf{B}_1 \subsetneq \mathbf{B}_2$ be a complex separable infinite dimensional Banach spaces with norm
$\|\cdot\|_1 \ge \|\cdot\|_2$ and let $\mathbf{C}^t_{G_{\mathbf{B}_j}}$ be a $\tilde{\mathbb{C}}$-modules $\mathbf{C}^t_{G_{\mathbf{B}_j}} = \mathbf{C}([0,t], G_{\mathbf{B}_j}), j = 1,2$. Thus
$\mathbf{C}^t_{G_{\mathbf{B}_1}} \subsetneq \mathbf{C}^t_{G_{\mathbf{B}_2}}$. Suppose that for every $t \in [0,T_0]$ there exists continuous operator

$$
(\mathfrak{R}_{t,\epsilon}[\circ])_{\epsilon} : \mathbf{C}_{G_{\mathbf{B}_1}}^{t} \rightarrow \mathbf{G}_{\mathbf{B}_2} \qquad (2.1.1.31)
$$

such that for every $(v_\epsilon)_\epsilon \in \mathbf{C}^t_{G_{\mathbf{B}_1}}$ function $\|(\mathfrak{R}_{t,\epsilon}[v_\epsilon])_\epsilon\|_{G_{\mathbf{B}_2}}$ is continuous on a
variable $t \in [0, T_0]$.

Let's consider the Cauchy problem with initial condition $(\Phi_{0,\varepsilon})_{\varepsilon} \in G_{B_1}$:

$$ \frac{d}{dt}(\Phi_{\varepsilon}(t))_{\varepsilon} = (\mathfrak{R}_{t,\varepsilon}[\Phi_{\varepsilon}(t)])_{\varepsilon}, (\Phi_{\varepsilon}(0))_{\varepsilon} = (\Phi_{0,\varepsilon})_{\varepsilon}, \qquad (2.1.1.32) $$

where

$$ \frac{d}{dt}(\Phi_{\varepsilon}(t))_{\varepsilon} \triangleq \lim_{\Delta t \to 0} \left\| \left( \frac{\Phi_{\varepsilon}(t+\Delta t) - \Phi_{\varepsilon}(t)}{\Delta t} \right)_{\varepsilon} \right\|_{G_{B_2}}, \qquad (2.1.1.33) $$

i.e. $\forall t$ and $\forall (v_{t,\varepsilon})_{\varepsilon} \in G_{B_2}$

$$ \frac{d}{dt}(\Phi_{\varepsilon}(t))_{\varepsilon} = (v_{t,\varepsilon})_{\varepsilon} \Leftrightarrow \lim_{\Delta t \to 0} \left\| \left( v_{t,\varepsilon} - \frac{\Phi_{\varepsilon}(t+\Delta t) - \Phi_{\varepsilon}(t)}{\Delta t} \right)_{\varepsilon} \right\|_{G_{B_2}} = 0. \qquad (2.1.1.34) $$

**Definition 2.1.1.7.** Let be $(S_{t,\tilde{\varepsilon},\varepsilon}[\circ])_{\varepsilon} : C([0,t], G_B) \times [-t, \varepsilon_0] \to G_B$ the infinitesimal generator of the corresponding $T$-exponent. Suppose that for every $t \in [0, T_0]$ the next condition is satisfied

$$ \lim_{\tilde{\varepsilon} \to 0} \left\| \left( \frac{S_{t, \tilde{\varepsilon}, \varepsilon}[\nu] - v_{\varepsilon}(t)}{\tilde{\varepsilon}} - \mathfrak{R}_{t, \varepsilon}[\nu_{\varepsilon}] \right)_{\varepsilon} \right\|_{G_{B_2}} = 0. \qquad (2.1.1.35) $$

locally uniformly on variable $(v_{\varepsilon}(t))_{\varepsilon} \in G_{C_{B_1}}$. We call operator function $(S_{t,\tilde{\varepsilon},\varepsilon}[\circ])_{\varepsilon}$ as master operator for Cauchy problem (2.1.1.32).

**Theorem 2.1.1.3.** Let be $(S_{t,\tilde{\varepsilon},\varepsilon}[\circ])_{\varepsilon}$ the master operator for Cauchy problem (2.1.1.32)

and for some $\psi_{0,\varepsilon} \in G_{B_1}$ there exists $(\psi_{\varepsilon}(t))_{\varepsilon} = ((\prod_{\tau=0}^{t} S_{\tau, d\tau, \varepsilon})[\psi_{0,\varepsilon}])_{\varepsilon}$,
variable $t \in [0, T_0]$.

Let's consider the Cauchy problem with initial condition $(\Phi_{0,\varepsilon})_{\varepsilon} \in G_{B_1}$:

$$ \frac{d}{dt}(\Phi_{\varepsilon}(t))_{\varepsilon} = (\mathfrak{R}_{t,\varepsilon}[\Phi_{\varepsilon}(t)])_{\varepsilon}, (\Phi_{\varepsilon}(0))_{\varepsilon} = (\Phi_{0,\varepsilon})_{\varepsilon}, \qquad (2.1.1.32) $$

where

$$ \frac{d}{dt}(\Phi_{\varepsilon}(t))_{\varepsilon} \triangleq \lim_{\Delta t \to 0} \left\| \left( \frac{\Phi_{\varepsilon}(t+\Delta t) - \Phi_{\varepsilon}(t)}{\Delta t} \right)_{\varepsilon} \right\|_{G_{B_2}}, \qquad (2.1.1.33) $$

i.e. $\forall t$ and $\forall (v_{t,\varepsilon})_{\varepsilon} \in G_{B_2}$

$$ \frac{d}{dt}(\Phi_{\varepsilon}(t))_{\varepsilon} = (v_{t,\varepsilon})_{\varepsilon} \Leftrightarrow \lim_{\Delta t \to 0} \left\| \left( v_{t,\varepsilon} - \frac{\Phi_{\varepsilon}(t+\Delta t) - \Phi_{\varepsilon}(t)}{\Delta t} \right)_{\varepsilon} \right\|_{G_{B_2}} = 0. \qquad (2.1.1.34) $$

**Definition 2.1.1.7.** Let be $(S_{t,\tilde{\varepsilon},\varepsilon}[\circ])_{\varepsilon} : C([0,t], G_B) \times [-t, \varepsilon_0] \to G_B$ the infinitesimal generator of the corresponding $T$-exponent. Suppose that for every $t \in [0, T_0]$ the next condition is satisfied

$$ \lim_{\tilde{\varepsilon} \to 0} \left\| \left( \frac{S_{t, \tilde{\varepsilon}, \varepsilon}[\nu] - v_{\varepsilon}(t)}{\tilde{\varepsilon}} - \mathfrak{R}_{t, \varepsilon}[\nu_{\varepsilon}] \right)_{\varepsilon} \right\|_{G_{B_2}} = 0. \qquad (2.1.1.35) $$

locally uniformly on variable $(v_{\varepsilon}(t))_{\varepsilon} \in G_{C_{B_1}}$. We call operator function $(S_{t,\tilde{\varepsilon},\varepsilon}[\circ])_{\varepsilon}$ as master operator for Cauchy problem (2.1.1.32).

**Theorem 2.1.1.3.** Let be $(S_{t,\tilde{\varepsilon},\varepsilon}[\circ])_{\varepsilon}$ the master operator for Cauchy problem (2.1.1.32)

and for some $\psi_{0,\varepsilon} \in G_{B_1}$ there exists $(\psi_{\varepsilon}(t))_{\varepsilon} = ((\prod_{\tau=0}^{t} S_{\tau, d\tau, \varepsilon})[\psi_{0,\varepsilon}])_{\varepsilon}$,
($\psi_{\varepsilon}(t))_{\varepsilon} \in G_{C_1(T_0)}$. Then ($\psi_{\varepsilon}(t))_{\varepsilon}$ is a solution of the Eq.(2.1.1.32).

**Proof.** Let's consider finite splitting $\Delta_N = \{0 = t_0 < t_1 < \dots < t_N = T_0\}$ of the closed interval $[0, T_0]$ and $(v_{N,\varepsilon}(t))_{\varepsilon} = (S_{t_N, \Delta t_{N-1}, \varepsilon} \circ \dots \circ S_{t_j, \Delta t_j, \varepsilon} \circ \dots \circ S_{t_0, \Delta t_0, \varepsilon}[\psi_{0,\varepsilon}])_{\varepsilon}$. Then for every $t_{j+1} \ge t \ge t_j$, one obtain ($\psi_{N,\varepsilon}(t))_{\varepsilon} = (S_{t_j, t-t_j, \varepsilon}[\psi_{N,\varepsilon}])_{\varepsilon}$. Thus there exists $N_0$ such that

$$ \lim_{\Delta t \to 0} \left\| \left( \frac{\psi_{N,\varepsilon}(t_j + \Delta t) - \psi_{N,\varepsilon}(t_j)}{\Delta t} - \mathfrak{R}_{t_j,\varepsilon}[\psi_{N,\varepsilon}] \right)_{\varepsilon} \right\|_{G_{B_2}} = 0 \quad (2.1.1.36) $$

uniformly on $N \ge N_0$.

Let $t$ be a fixed point $t \in [0, T_0)$ and let $\{\Delta_N\}_{N=1}^{\infty}$ be the sequence of the splittings such that for any $N: t \in \Delta_N$.

Now take into account that operator $(\mathfrak{R}_{t,\varepsilon})_{\varepsilon}$ is continuous finally we obtain

$$ \lim_{\Delta t \to 0} \left\| \left( \frac{\psi_{\varepsilon}(t + \Delta t) - \psi_{\varepsilon}(t)}{\Delta t} - \mathfrak{R}_{t,\varepsilon}[\psi_{\varepsilon}] \right)_{\varepsilon} \right\|_{G_{B_2}} = 0. \quad (2.1.1.37) $$

Let $\mathfrak{R}_{t}[\psi]$ be operator such that

$$ (\mathfrak{R}_{t,\varepsilon}[\psi])_{\varepsilon} = (A_{t,\varepsilon}[\psi_{\varepsilon}]\psi_{\varepsilon}(t))_{\varepsilon}. \quad (2.1.1.38) $$

where $\forall (\psi_{\varepsilon})_{\varepsilon}((\psi_{\varepsilon})_{\varepsilon} \in G_{C_{B_1}^t})$ operator $(A_{t,\varepsilon}[\psi_{\varepsilon}])_{\varepsilon}: G_{B_1} \to G_{B_2}$ is continuous. Thus $(A_{t,\varepsilon}[\psi_{\varepsilon}])_{\varepsilon}$ is a infinitesimal generator of the one-parameter semigroup

$$ \exp([\tilde{\varepsilon}A_{t,\varepsilon}[\psi_{\varepsilon}]])_{\varepsilon}. \quad (2.1.1.39) $$

Hence master operator of the Cauchy problem (2.1.1.32) is
($\psi_{\varepsilon}(t))_{\varepsilon} \in G_{C_1(T_0)}$. Then ($\psi_{\varepsilon}(t))_{\varepsilon}$ is a solution of the Eq.(2.1.1.32).

**Proof.** Let's consider finite splitting $\Delta_N = \{0 = t_0 < t_1 < \dots < t_N = T_0\}$ of the closed interval $[0, T_0]$ and $(v_{N,\varepsilon}(t))_{\varepsilon} = (S_{t_N, \Delta t_{N-1}, \varepsilon} \circ \dots \circ S_{t_j, \Delta t_j, \varepsilon} \circ \dots \circ S_{t_0, \Delta t_0, \varepsilon}[\psi_{0,\varepsilon}])_{\varepsilon}$. Then for every $t_{j+1} \ge t \ge t_j$, one obtain ($\psi_{N,\varepsilon}(t))_{\varepsilon} = (S_{t_j, t-t_j, \varepsilon}[\psi_{N,\varepsilon}])_{\varepsilon}$. Thus there exists $N_0$ such that

$$ \lim_{\Delta t \to 0} \left\| \left( \frac{\psi_{N,\varepsilon}(t_j + \Delta t) - \psi_{N,\varepsilon}(t_j)}{\Delta t} - \mathfrak{R}_{t_j,\varepsilon}[\psi_{N,\varepsilon}] \right)_{\varepsilon} \right\|_{G_{B_2}} = 0 \quad (2.1.1.36) $$

uniformly on $N \ge N_0$.

Let $t$ be a fixed point $t \in [0, T_0)$ and let $\{\Delta_N\}_{N=1}^{\infty}$ be the sequence of the splittings such that for any $N: t \in \Delta_N$.

Now take into account that operator $(\mathfrak{R}_{t,\varepsilon})_{\varepsilon}$ is continuous finally we obtain

$$ \lim_{\Delta t \to 0} \left\| \left( \frac{\psi_{\varepsilon}(t + \Delta t) - \psi_{\varepsilon}(t)}{\Delta t} - \mathfrak{R}_{t,\varepsilon}[\psi_{\varepsilon}] \right)_{\varepsilon} \right\|_{G_{B_2}} = 0. \quad (2.1.1.37) $$

Let $\mathfrak{R}_{t}[\psi]$ be operator such that

$$ (\mathfrak{R}_{t,\varepsilon}[\psi])_{\varepsilon} = (A_{t,\varepsilon}[\psi_{\varepsilon}]\psi_{\varepsilon}(t))_{\varepsilon}. \quad (2.1.1.38) $$

where $\forall (\psi_{\varepsilon})_{\varepsilon}((\psi_{\varepsilon})_{\varepsilon} \in G_{C_{B_1}^t})$ operator $(A_{t,\varepsilon}[\psi_{\varepsilon}])_{\varepsilon}: G_{B_1} \to G_{B_2}$ is continuous. Thus $(A_{t,\varepsilon}[\psi_{\varepsilon}])_{\varepsilon}$ is a infinitesimal generator of the one-parameter semigroup

$$ \exp([\tilde{\mathcal{E}}A_{t,\varepsilon}[\psi_{\varepsilon}]])_{\varepsilon}. \quad (2.1.1.39) $$

Hence master operator of the Cauchy problem (2.1.1.32) is
$$ (S_{t,\bar{\epsilon},\epsilon}[v])_{\epsilon} = \exp([\bar{\epsilon}A_{t,\epsilon}[\psi_{\epsilon}]]v_{\epsilon}(t))_{\epsilon}. \qquad (2.1.1.40) $$

Corresponding T-exp is

$$ (\psi_{\epsilon}(t))_{\epsilon} = \left( \prod_{\tau}^{t} \exp[A_{\tau,\epsilon}[\psi_{\epsilon}]] d\tau \right)_{\epsilon}. \qquad (2.1.1.41) $$

## II.1.2. Generalized Maslov theorem.

Let us consider Cauchy problem:

$$ -i\hbar \frac{d}{dt} \Psi_k(t,x) - h^2 \Delta \Psi_k(t,x) + \sum_{l=1}^{m} V_{k,l}(x) \Psi_l(t,x) + \sum_{l,j=1}^{m} \Psi_l(t,x) \int \rho_{k,l,j}(x,y) |\Psi_j(t,y)|^2 dy = 0, \qquad (2.1.2.1) $$

$$ \Psi_k(0,x) = \Psi_{0,k}(x), $$

$$ k = 1, \dots, m, $$

where
$$ (S_{t,\bar{\epsilon},\epsilon}[v])_{\epsilon} = \exp([\bar{\epsilon}A_{t,\epsilon}[\psi_{\epsilon}]]v_{\epsilon}(t))_{\epsilon}. \qquad (2.1.1.40) $$

Corresponding T-exp is

$$ (\psi_{\epsilon}(t))_{\epsilon} = \left( \prod_{\tau}^{t} \exp[A_{\tau,\epsilon}[\psi_{\epsilon}]] d\tau \right)_{\epsilon}. \qquad (2.1.1.41) $$

## II.1.2. Generalized Maslov theorem.

Let us consider Cauchy problem:

$$ \begin{gathered} -i\hbar \frac{d}{dt} \Psi_k(t,x) - h^2 \Delta \Psi_k(t,x) + \sum_{l=1}^{m} V_{k,l}(x) \Psi_l(t,x) + \\ \sum_{l,j=1}^{m} \Psi_l(t,x) \int \rho_{k,l,j}(x,y) |\Psi_j(t,y)|^2 dy = 0, \end{gathered} \qquad (2.1.2.1) $$

$$ \Psi_k(0,x) = \Psi_{0,k}(x), $$

$$ k = 1, \dots, m, $$

where
$$ \Psi_0 = (\Psi_{0,1}, \dots, \Psi_{0,m}) \in \mathcal{L}_{2,m}(\mathbb{R}^n), $$

$$ V_{k,l}(x) = \overline{V_{l,k}(x)}, \rho_{k,l,j}(x,y) = \overline{\rho_{l,k,j}(x,y)}, \qquad (2.1.2.2) $$

$$ V_{k,l}(x) \in C(\mathbb{R}^n), \rho_{k,l,j}(x,y) \in C(\mathbb{R}^n \times \mathbb{R}^n). $$

We remind that $\mathcal{L}_{2,m}(\mathbb{R}^n) = \{\Psi(x) \|\| \Psi\| < \infty\}$, where

$$ \|\Psi\|^2 = \sum_{j=1}^{m} \int |\Psi_j(x)|^2 dx < \infty \qquad (2.1.2.3) $$

and $W_2^{k,m}(\mathbb{R}^n) = \{\Psi(x) \|\| \Psi\|\|_{W_2^{k,m}} < \infty\}$, where

$$ \|\Psi\|_{W_2^{k,m}} = \sum_{j=1}^{m} \|\Psi_j\|_{W_2^k}. \qquad (2.1.2.4) $$

Let $\Psi(t) = (\Psi_1(t), \dots, \Psi_m(t))$ be a function $\Psi : \mathbb{R} \to \mathcal{L}_{2,m}(\mathbb{R}^n)$. We abbreviate $\Psi(t) \in \mathcal{L}_{2,m}(\mathbb{R}^n)$ iff $\forall t[\Psi(t) \in \mathcal{L}_{2,m}(\mathbb{R}^n)]$.

Let $v \in \mathcal{L}_{2,m}(\mathbb{R}^n)$ and let $A_{ij}[x,v]$ be the continuous matrix-function

$$ A_{ij}[x,v] = V_{ij}(x) + \sum_{j=1}^{m} \int \rho_{k,l,j}(x,y) |v_l(y)|^2 dy. \qquad (2.1.2.5) $$

Let $\hat{A}[v]$ be a linear bounded operator on $\mathcal{L}_{2,m}(\mathbb{R}^n)$, given via formula

$$ (\hat{A}[v]u)_i = \sum_{j=1}^{m} A_{ij}[x,v]u_j. \qquad (2.1.2.6) $$
$$ \Psi_0 = (\Psi_{0,1}, \dots, \Psi_{0,m}) \in \mathcal{L}_{2,m}(\mathbb{R}^n), $$

$$ V_{k,l}(x) = \overline{V_{l,k}(x)}, \rho_{k,l,j}(x,y) = \overline{\rho_{l,k,j}(x,y)}, \qquad (2.1.2.2) $$

$$ V_{k,l}(x) \in C(\mathbb{R}^n), \rho_{k,l,j}(x,y) \in C(\mathbb{R}^n \times \mathbb{R}^n). $$

We remind that $\mathcal{L}_{2,m}(\mathbb{R}^n) = \{\Psi(x) \|\| \Psi\| < \infty\}$, where

$$ \|\Psi\|^2 = \sum_{j=1}^{m} \int |\Psi_j(x)|^2 dx < \infty \qquad (2.1.2.3) $$

and $W_2^{k,m}(\mathbb{R}^n) = \{\Psi(x) \|\| \Psi\|\|_{W_2^{k,m}} < \infty\}$, where

$$ \|\Psi\|_{W_2^{k,m}} = \sum_{j=1}^{m} \|\Psi_j\|_{W_2^k}. \qquad (2.1.2.4) $$

Let $\Psi(t) = (\Psi_1(t), \dots, \Psi_m(t))$ be a function $\Psi : \mathbb{R} \to \mathcal{L}_{2,m}(\mathbb{R}^n)$. We abbreviate $\Psi(t) \in \mathcal{L}_{2,m}(\mathbb{R}^n)$ iff $\forall t[\Psi(t) \in \mathcal{L}_{2,m}(\mathbb{R}^n)]$.

Let $v \in \mathcal{L}_{2,m}(\mathbb{R}^n)$ and let $A_{ij}[x,v]$ be the continuous matrix-function

$$ A_{ij}[x,v] = V_{ij}(x) + \sum_{j=1}^{m} \int \rho_{k,l,j}(x,y) |v_l(y)|^2 dy. \qquad (2.1.2.5) $$

Let $\hat{A}[v]$ be a linear bounded operator on $\mathcal{L}_{2,m}(\mathbb{R}^n)$, given via formula

$$ (\hat{A}[v]u)_i = \sum_{j=1}^{m} A_{ij}[x,v]u_j. \qquad (2.1.2.6) $$
Let's rewrite now Cauchy problem (2.1.2.1) in the form

$$ -i\hbar \frac{d}{dt} \Psi(t) - \hbar^2 \Delta \Psi(t) + \hat{A}[\Psi(t)] \Psi(t) = 0, \qquad (2.1.2.7) $$

$$ \Psi(0) = \Psi_0. $$

**Theorem 2.1.2.1.** Maslov. [20]. Cauchy problem (2.1.2.1) has a solution $\Psi(t)$ such that $\Psi(t) \in \mathcal{L}_{2,m}(\mathbb{R}^n)$ and the function $\Psi(t)$ is given via formula:

$$ \Psi(t) = \left( \prod_{\tau=0}^{t} G_{d\tau} [\Psi(\tau)] \right) [\Psi_0], \qquad (2.1.2.8) $$

where

$$ G_{\varepsilon}[v] = \exp\{i\varepsilon\Delta\} \exp\{i\varepsilon\hat{A}[v]\}, \qquad (2.1.2.9) $$

$$ v \in \mathcal{L}_{2,m}(\mathbb{R}^n). $$

We let: $\delta = \frac{t}{N}$, $N = 1, 2, \dots$ and

$$ \Psi_{N,0} = \Psi_0, $$

$$ \Psi_{N,k+1} = G_{\delta} \Psi_{N,k}, \qquad (2.1.2.10) $$

$$ k = 0, 1, 2, \dots, N-1, $$
Let's rewrite now Cauchy problem (2.1.2.1) in the form

$$-i\hbar \frac{d}{dt} \Psi(t) - \hbar^2 \Delta \Psi(t) + \hat{A}[\Psi(t)] \Psi(t) = 0, \qquad (2.1.2.7)$$

$$\Psi(0) = \Psi_0.$$

**Theorem 2.1.2.1.** Maslov. [20]. Cauchy problem (2.1.2.1) has a solution $\Psi(t)$ such that $\Psi(t) \in \mathcal{L}_{2,m}(\mathbb{R}^n)$ and the function $\Psi(t)$ is given via formula:

$$\Psi(t) = \left( \prod_{\tau=0}^{t} G_{d\tau} [\Psi(\tau)] \right) [\Psi_0], \qquad (2.1.2.8)$$

where

$$G_{\varepsilon}[v] = \exp\{i\varepsilon\Delta\} \exp\{i\varepsilon\hat{A}[v]\}, \qquad (2.1.2.9)$$

$$v \in \mathcal{L}_{2,m}(\mathbb{R}^n).$$

We let: $\delta = \frac{t}{N}$, $N = 1, 2, \dots$ and

$$\Psi_{N,0} = \Psi_0,$$

$$\Psi_{N,k+1} = G_{\delta} \Psi_{N,k}, \qquad (2.1.2.10)$$

$$k = 0, 1, 2, \dots, N-1,$$
where

$$
G_{\delta}u(x) = \exp(i\delta\hbar\Delta)\exp(i\delta\hbar^{-1}A[u])u(x), \qquad (2.1.2.11)
$$

$u(x) \in L_2(\mathbb{R}^n).$

and we set below

$$
\Psi_N \triangleq \Psi_{N,N}. \tag{2.1.2.12}
$$

Suppose that $\Psi_0(x) \in W_2^{2,m}(\mathbb{R}^n)$. Then every solution $\Psi(t,x)$ of the Eq.(2.1.2.7) such that $\Psi(0,x) = \Psi_0$, has the representation:

$$
\Psi(t,x) = s- \lim_{N \to \infty} \Psi_N, \qquad (2.1.2.13)
$$

where the strong limit in (2.1.2.13) exist uniformly on $[0, T]$ for any $T < \infty$ in the
sence of the convergence in $L_{2,m}(\mathbb{R}^n)$ norm.

Proof.

1. We note that

$$
\begin{equation}
\| \exp(i\epsilon\hbar^{-1}A[v]) \| = 1, \tag{2.1.2.14}
\end{equation}
$$

$$
\| \exp(i \epsilon \hbar \Delta) \| = 1,
$$

where we denote
where

$$G_{\delta}u(x) = \exp(i\delta\hbar\Delta)\exp(i\delta\hbar^{-1}A[u])u(x), \qquad (2.1.2.11)$$

$$u(x) \in L_2(\mathbb{R}^n).$$

and we set below

$$\Psi_N \triangleq \Psi_{N,N}. \qquad (2.1.2.12)$$

Suppose that $\Psi_0(x) \in W_2^{2,m}(\mathbb{R}^n)$. Then every solution $\Psi(t,x)$ of the Eq.(2.1.2.7) such that $\Psi(0,x) = \Psi_0$, has the representation:

$$\Psi(t,x) = s- \lim_{N \to \infty} \Psi_N, \qquad (2.1.2.13)$$

where the strong limit in (2.1.2.13) exist uniformly on $[0, T]$ for any $T < \infty$ in the sense of the convergence in $L_{2,m}(\mathbb{R}^n)$ norm.

**Proof.**

1. We note that

$$\|\exp(i\tilde{\epsilon}\hbar^{-1}A[v])\| = 1, \qquad (2.1.2.14)$$

$$\|\exp(i\tilde{\epsilon}\hbar\Delta)\| = 1,$$

where we denote
$$ \|\cdot\| \triangleq \|\cdot\|_{L_{2,m}}. \qquad (2.1.2.15) $$

Hence one obtain

$$ \|\Psi_{N,k}\| = \|\Psi_0\|. \qquad (2.1.2.16) $$

2. Let's now to prove the formula

$$ \begin{gathered} \Psi_{N,k} = \exp(ik\delta\hbar\Delta)\Psi_0 + \\ i\delta\hbar^{-1} \sum_{j=0}^{k-1} \exp[i(k-j-1)\delta\hbar\Delta]A[\Psi_{N,j}]\Psi_{N,j} + O\left(\frac{1}{\hbar N}\right) \end{gathered} \qquad (2.1.2.17) $$

Note that $\|v\| \le r \Rightarrow \|A[v]\| \le C_r$. Hence

$$ \exp(i\delta\hbar^{-1}A[v]) = 1 + i\delta\hbar^{-1}A[\Psi_{N,j}] + O\left(\frac{1}{\hbar^2 N^2}\right) \qquad (2.1.2.18) $$

From Eq.(2.1.2.18) by simple calculation one obtain

$$ \begin{gathered} \Psi_{N,k} = \exp(ik\delta\hbar\Delta)\Psi_{N,k-1} + \\ i\delta\hbar^{-1} \exp[i\delta\hbar\Delta]A[\Psi_{N,k-1}]\Psi_{N,k-1} + O\left(\frac{1}{\hbar^2 N^2}\right), \end{gathered} \qquad (2.1.2.19) $$

where
$$ \|\cdot\| \triangleq \|\cdot\|_{L_{2,m}}. \qquad (2.1.2.15) $$

Hence one obtain

$$ \|\Psi_{N,k}\| = \|\Psi_0\|. \qquad (2.1.2.16) $$

2. Let's now to prove the formula

$$ \begin{gathered} \Psi_{N,k} = \exp(ik\delta\hbar\Delta)\Psi_0 + \\ i\delta\hbar^{-1} \sum_{j=0}^{k-1} \exp[i(k-j-1)\delta\hbar\Delta]A[\Psi_{N,j}]\Psi_{N,j} + O\left(\frac{1}{\hbar N}\right) \end{gathered} \qquad (2.1.2.17) $$

Note that $\|v\| \le r \Rightarrow \|A[v]\| \le C_r$. Hence

$$ \exp(i\delta\hbar^{-1}A[v]) = 1 + i\delta\hbar^{-1}A[\Psi_{N,j}] + O\left(\frac{1}{\hbar^2 N^2}\right) \qquad (2.1.2.18) $$

From Eq.(2.1.2.18) by simple calculation one obtain

$$ \begin{gathered} \Psi_{N,k} = \exp(ik\delta\hbar\Delta)\Psi_{N,k-1} + \\ i\delta\hbar^{-1} \exp[i\delta\hbar\Delta]A[\Psi_{N,k-1}]\Psi_{N,k-1} + O\left(\frac{1}{\hbar^2 N^2}\right), \end{gathered} \qquad (2.1.2.19) $$

where
$$ \|O\left(\frac{1}{\hbar^2 N^2}\right)\|_2 \le \frac{C}{\hbar^2 N^2} \qquad (2.1.2.20) $$

and where constant $C$ does not depend on $k = 1, 2, \dots, N$. From Eq.(2.1.2.19) by simple calculation one obtain Eq.(2.1.2.17).

3. Let's now to prove the inequality:

$$ \|\Psi_{N,l+k} - \Psi_{N,l}\| \le $$

$$ C_0[\exp(C_1 t)](k\hbar^{-1}\delta + \|\exp(ik\delta\hbar\Delta)\Psi_0 - \Psi_0\|), \qquad (2.1.2.21) $$

$$ l + k \le N, $$

where $C_0 = \text{const}$ and $C_1 = \text{const}$.

By using formula (2.1.2.17) one obtain

$$ \Psi_{N,l+k} - \Psi_{N,l} = \exp(ik\delta\hbar\Delta)(\exp(ik\delta\hbar\Delta)\Psi_0 - \Psi_0) + $$

$$ +i\delta\hbar^{-1} \sum_{j=0}^{k-1} \exp[i\delta\hbar(k+l-j-1)\Delta]A\Psi_{N,j} + \qquad (2.1.2.22) $$

$$ +i\delta\hbar^{-1} \sum_{j=0}^{l-1} \exp[i\delta\hbar(l-j-1)\Delta](A\Psi_{N,j+k} - A\Psi_{N,j}) + O\left(\frac{1}{\hbar N}\right) $$

From inequality: $\|A[v]\| \le C_r$ and Eqs.(2.1.2.22) we obtain
$$ \|O(\frac{1}{\hbar^2 N^2})\|_2 \le \frac{C}{\hbar^2 N^2} \qquad (2.1.2.20) $$

and where constant C does not depend on $k = 1, 2, \dots, N$. From Eq.(2.1.2.19) by simple calculation one obtain Eq.(2.1.2.17).

3. Let's now to prove the inequality:

$$ \|\Psi_{N,l+k} - \Psi_{N,l}\| \le $$

$$ C_0[\exp(C_1 t)](kh^{-1}\delta + \|\exp(ik\delta\hbar\Delta)\Psi_0 - \Psi_0\|), \qquad (2.1.2.21) $$

$$ l + k \le N, $$

where $C_0 = \text{const}$ and $C_1 = \text{const}$.

By using formula (2.1.2.17) one obtain

$$ \Psi_{N,l+k} - \Psi_{N,l} = \exp(ik\delta\hbar\Delta)(\exp(ik\delta\hbar\Delta)\Psi_0 - \Psi_0) + $$

$$ +i\delta\hbar^{-1} \sum_{j=0}^{k-1} \exp[i\delta\hbar(k+l-j-1)\Delta]A\Psi_{N,j} + \qquad (2.1.2.22) $$

$$ +i\delta\hbar^{-1} \sum_{j=0}^{l-1} \exp[i\delta\hbar(l-j-1)\Delta](A\Psi_{N,j+k} - A\Psi_{N,j}) + O(\frac{1}{\hbar N}) $$

From inequality: $\|A[v]\| \le C_r$ and Eqs.(2.1.2.22) we obtain
$a_0 \le \epsilon,$

$$
\begin{gather*}
a_1 \le C \left[ \epsilon + k\delta\hbar^{-1} + \hbar^{-1} \sum_{j=0}^{l-1} \delta a_j \right], \\
l = 1, 2, \dots, N, \tag{2.1.2.23}
\end{gather*}
$$

$a_j = \| \Psi_{N,j+k} - \Psi_{N,j} \|,$

$$
\epsilon = \| \exp(ik\delta\hbar\Delta)\Psi_0 - \Psi_0 \|
$$

4. Suppose that $M \in \mathbb{N}, M \ge 1$ and $M = \text{const}$ then for every $j \in \mathbb{N}$ one obtain

$$
\frac{j}{M} - \left[ \frac{j}{M} \right] \le 1. \tag{2.1.2.24}
$$

Hence from (2.1.2.21) we obtain

$$
\Psi_{MN,j} = \Psi_M[\frac{j}{M}] + O\left(\frac{1}{\hbar N}\right), \quad (2.1.2.25)
$$

where

$$
\begin{gather*}
\left\| O\left(\frac{1}{\hbar N}\right) \right\| \le \\
\left( \delta\hbar^{-1} + \left\| \Psi_0 - \exp\left[i\delta\hbar\left(\frac{j}{M} - \left[\frac{j}{M}\right]\right)\Delta\right] \Psi_0 \right\| \right) \le \tag{2.1.2.26}
\end{gather*}
$$

$$
\delta\hbar^{-1} \|\Psi_0\|_{W_2^{2,m}(\mathbb{R}^n)}.
$$
$$a_0 \le \epsilon,$$

$$ a_1 \le C \left[ \epsilon + k\delta\hbar^{-1} + \hbar^{-1} \sum_{j=0}^{l-1} \delta a_j \right], \qquad (2.1.2.23) \\ l = 1, 2, \dots, N, $$

$$a_j = \| \Psi_{N,j+k} - \Psi_{N,j} \|,$$

$$\epsilon = \| \exp(ik\delta\hbar\Delta)\Psi_0 - \Psi_0 \|$$

4. Suppose that $M \in \mathbb{N}, M \ge 1$ and $M = \text{const}$ then for every $j \in \mathbb{N}$ one obtain

$$\frac{j}{M} - \left\lfloor \frac{j}{M} \right\rfloor \le 1. \qquad (2.1.2.24)$$

Hence from (2.1.2.21) we obtain

$$\Psi_{MN,j} = \Psi_M[\frac{j}{M}] + O\left(\frac{1}{\hbar N}\right), \qquad (2.1.2.25)$$

where

$$\begin{gathered} \left\| O\left(\frac{1}{\hbar N}\right) \right\| \le \\ \left( \delta\hbar^{-1} + \left\| \Psi_0 - \exp\left[i\delta\hbar\left(\frac{j}{M} - \left[\frac{j}{M}\right]\right)\Delta\right] \Psi_0 \right\| \right) \le \qquad (2.1.2.26) \\ \delta\hbar^{-1} \|\Psi_0\|_{W_2^{2,m}(\mathbb{R}^n)}. \end{gathered}$$
We denote

$$ \gamma_N(\hbar^{-1}) \triangleq \delta\hbar^{-1} \|\Psi_0\|_{W_2^2(\mathbb{R}^3)}, \qquad (2.1.2.27) $$

$$ \delta = \delta(N). $$

Thus we have two finite splittings $\Delta_N^{(1)} = \{0 = t_0 < t_1^{(1)} < \dots < t_N^{(1)} = t\}$ and $\Delta_{NM}^{(2)} = \{0 = t_0 < t_1^{(2)} < \dots < t_{NM}^{(2)} = t\}$ of the closed interval $[0, t]$ such that $\{t_i^{(1)}\}_{i=0}^N \subseteq \{t_j^{(2)}\}_{j=0}^{NM}$.

Let's rewrite formula (2.1.2.17) by using replacement $N \to NM$ for $k_0 = M \times l$. Thus we obtain

$$ \Psi_{MN,Ml} = \exp(i\delta\hbar\Delta)\Psi_0 + i\delta\hbar^{-1} \sum_{j=0}^{l-1} \exp[i\delta\hbar(l-j-1)\Delta] \times \qquad (2.1.2.28) $$

$$ A\Psi_{MN,Mj} + O(\gamma_N(\hbar^{-1})). $$

Let's denote

$$ b_j = \|\Psi_{N,j} - \Psi_{MN,Mj}\| \qquad (2.1.2.29) $$

By using subtraction Eq.(2.1.2.17) from Eq.(2.1.2.29) we obtain

$$ b_0 \le \alpha_N $$

$$ b_l \le \alpha_N + c\delta\hbar^{-1} \sum_{j=0}^{l-1} b_j, \qquad (2.1.2.30) $$

$$ l = 1, \dots, N, $$
We denote

$$ \gamma_N(\hbar^{-1}) \triangleq \delta\hbar^{-1} \|\Psi_0\|_{W_2^2(\mathbb{R}^3)}, \qquad (2.1.2.27) $$

$$ \delta = \delta(N). $$

Thus we have two finite splittings $\Delta_N^{(1)} = \{0 = t_0 < t_1^{(1)} < \dots < t_N^{(1)} = t\}$ and $\Delta_{NM}^{(2)} = \{0 = t_0 < t_1^{(2)} < \dots < t_{NM}^{(2)} = t\}$ of the closed interval $[0, t]$ such that $\{t_i^{(1)}\}_{i=0}^N \subseteq \{t_j^{(2)}\}_{j=0}^{NM}$.

Let's rewrite formula (2.1.2.17) by using replacement $N \to NM$ for $k_0 = M \times l$. Thus we obtain

$$ \Psi_{MN,Ml} = \exp(i\delta\hbar\Delta)\Psi_0 + i\delta\hbar^{-1} \sum_{j=0}^{l-1} \exp[i\delta\hbar(l-j-1)\Delta] \times \qquad (2.1.2.28) $$

$$ A\Psi_{MN,Mj} + O(\gamma_N(\hbar^{-1})). $$

Let's denote

$$ b_j = \|\Psi_{N,j} - \Psi_{MN,Mj}\| \qquad (2.1.2.29) $$

By using subtraction Eq.(2.1.2.17) from Eq.(2.1.2.29) we obtain

$$ b_0 \le \alpha_N $$

$$ b_l \le \alpha_N + c\delta\hbar^{-1} \sum_{j=0}^{l-1} b_j, \qquad (2.1.2.30) $$

$$ l = 1, \dots, N, $$
where $\alpha_N \le C_1\gamma_N(h^{-1})$. From (2.1.2.30) we obtain

$$b_l \le \alpha_N \exp[C_2 t] \qquad (2.1.2.31)$$

In particular for $l = N$ we obtain:

$$\|\Psi_N(t) - \Psi_{NM}(t)\| \le \frac{t}{N} C_1 \exp(C_2 t) \|\Psi_0\|_{W_2^{2,m}} \qquad (2.1.2.32)$$

and therefore sequence $\Psi_N(t)$ uniformly convergence on interval $[0, T]$ for any $T < \infty$ in the sense of the convergence in $L_{2,m}(\mathbb{R}^n)$ norm.

5. Any function $\Psi_N(t)$ is continuous on $[0, T]$. Therefore $\Psi(t) = s-\lim_{N \to \infty} \Psi_N(t)$ is continuous on $[0, T]$. Hence the Riemann sum

$$S_N(t) = \sum_{k=0}^{N-1} \frac{t}{N} \exp\left\{i \frac{t}{N} (N-k-1) \Delta\right\} \hat{A}[\Psi_{N,k}] \Psi_{N,k} \qquad (2.1.2.33)$$

is convergence in the norm of the $L_{2,m}(\mathbb{R}^n)$ to integral

$$\int_0^t d\tau \exp\{i(t-\tau)\} \hat{A}[\Psi(\tau)] \Psi(\tau). \qquad (2.1.2.34)$$

if $N \to \infty$. Finally from Eq.(2.1.2.17) in the limit $N \to \infty$ we obtain integral equation

$$\Psi(t) = \exp\{it\Delta\}\Psi_0 + \int_0^t d\tau \exp\{i(t-\tau)\} \hat{A}[\Psi(\tau)] \Psi(\tau). \qquad (2.1.2.35)$$

Eq.(2.1.2.35) completed the proof.

Let us consider now Cauchy problem:
where $\alpha_N \le C_1\gamma_N(h^{-1})$. From (2.1.2.30) we obtain

$$b_l \le \alpha_N \exp[C_2 t] \qquad (2.1.2.31)$$

In particular for $l = N$ we obtain:

$$\|\Psi_N(t) - \Psi_{NM}(t)\| \le \frac{t}{N} C_1 \exp(C_2 t) \|\Psi_0\|_{W_2^{2,m}} \qquad (2.1.2.32)$$

and therefore sequence $\Psi_N(t)$ uniformly convergence on interval $[0, T]$ for any $T < \infty$ in the sense of the convergence in $L_{2,m}(\mathbb{R}^n)$ norm.

5. Any function $\Psi_N(t)$ is continuous on $[0, T]$. Therefore $\Psi(t) = s-\lim_{N \to \infty} \Psi_N(t)$ is continuous on $[0, T]$. Hence the Riemann sum

$$S_N(t) = \sum_{k=0}^{N-1} \frac{t}{N} \exp\left\{i \frac{t}{N} (N-k-1) \Delta\right\} \hat{A}[\Psi_{N,k}] \Psi_{N,k} \qquad (2.1.2.33)$$

is convergence in the norm of the $L_{2,m}(\mathbb{R}^n)$ to integral

$$\int_0^t d\tau \exp\{i(t-\tau)\} \hat{A}[\Psi(\tau)] \Psi(\tau). \qquad (2.1.2.34)$$

if $N \to \infty$. Finally from Eq.(2.1.2.17) in the limit $N \to \infty$ we obtain integral equation

$$\Psi(t) = \exp\{it\Delta\}\Psi_0 + \int_0^t d\tau \exp\{i(t-\tau)\} \hat{A}[\Psi(\tau)] \Psi(\tau). \qquad (2.1.2.35)$$

Eq.(2.1.2.35) completed the proof.

Let us consider now Cauchy problem:
$$-i\hbar \frac{d}{dt} (\Psi_{k,\varepsilon}(t,x))_{\varepsilon} - \hbar^2 \Delta (\Psi_{k,\varepsilon}(t,x))_{\varepsilon} + \sum_{l=1}^{m} (V_{k,l,\varepsilon}(x) \Psi_{l,\varepsilon}(t,x))_{\varepsilon} + \\ \sum_{l,j=1}^{m} (\Psi_{l,\varepsilon}(t,x))_{\varepsilon} \int \left[ (\rho_{k,l,j,\varepsilon}(x,y) |\Psi_{j,\varepsilon}(t,y)|^2)_{\varepsilon} \right] dy = 0, \qquad (2.1.2.36)$$

$$(\Psi_{k,\varepsilon}(0,x))_{\varepsilon} = (\Psi_{0,k,\varepsilon}(x))_{\varepsilon},$$

$k = 1, \dots, m,$

where

$$(\Psi_{0,\varepsilon})_{\varepsilon} = ((\Psi_{0,1,\varepsilon})_{\varepsilon}, \dots, (\Psi_{0,m,\varepsilon})_{\varepsilon}) \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)},$$

$$(V_{k,l,\varepsilon}(x))_{\varepsilon} = (\overline{V_{l,k,\varepsilon}(x)})_{\varepsilon},$$

$$(\rho_{k,l,j,\varepsilon}(x,y))_{\varepsilon} = (\overline{\rho_{l,k,j,\varepsilon}(x,y)})_{\varepsilon}, \qquad (2.1.2.37)$$

$$(V_{k,l,\varepsilon}(x))_{\varepsilon} \in G_{C(\mathbb{R}^n)},$$

$$(\rho_{k,l,j,\varepsilon}(x,y))_{\varepsilon} \in G_{C(\mathbb{R}^n \times \mathbb{R}^n)}.$$

Let $(\Psi_{\varepsilon}(t))_{\varepsilon} = ((\Psi_{1,\varepsilon}(t))_{\varepsilon}, \dots, (\Psi_{m,\varepsilon}(t))_{\varepsilon})$ be a function $(\Psi_{\varepsilon})_{\varepsilon}: \mathbb{R} \to G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$. We abbreviate $(\Psi_{\varepsilon}(t))_{\varepsilon} \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$ iff $\forall t[\text{cl}[(\Psi_{\varepsilon}(t))_{\varepsilon}] \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}]$.

Let $(v_{\varepsilon})_{\varepsilon} \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$ and let $(A_{ij,\varepsilon}[x,v_{\varepsilon}])_{\varepsilon}$ be the continuous matrix-function
$$ -i\hbar \frac{d}{dt} (\Psi_{k,\varepsilon}(t,x))_{\varepsilon} - \hbar^2 \Delta (\Psi_{k,\varepsilon}(t,x))_{\varepsilon} + \sum_{l=1}^{m} (V_{k,l,\varepsilon}(x) \Psi_{l,\varepsilon}(t,x))_{\varepsilon} + \\ \sum_{l,j=1}^{m} (\Psi_{l,\varepsilon}(t,x))_{\varepsilon} \int \left[ (\rho_{k,l,j,\varepsilon}(x,y) |\Psi_{j,\varepsilon}(t,y)|^2)_{\varepsilon} \right] dy = 0, \qquad (2.1.2.36) $$

$$ (\Psi_{k,\varepsilon}(0,x))_{\varepsilon} = (\Psi_{0,k,\varepsilon}(x))_{\varepsilon}, $$

$k = 1, \dots, m,$

where

$$ (\Psi_{0,\varepsilon})_{\varepsilon} = ((\Psi_{0,1,\varepsilon})_{\varepsilon}, \dots, (\Psi_{0,m,\varepsilon})_{\varepsilon}) \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}, $$

$$ (V_{k,l,\varepsilon}(x))_{\varepsilon} = (\overline{V_{l,k,\varepsilon}(x)})_{\varepsilon}, $$

$$ (\rho_{k,l,j,\varepsilon}(x,y))_{\varepsilon} = (\overline{\rho_{l,k,j,\varepsilon}(x,y)})_{\varepsilon}, \qquad (2.1.2.37) $$

$$ (V_{k,l,\varepsilon}(x))_{\varepsilon} \in G_{C(\mathbb{R}^n)}, $$

$$ (\rho_{k,l,j,\varepsilon}(x,y))_{\varepsilon} \in G_{C(\mathbb{R}^n \times \mathbb{R}^n)}. $$

Let $(\Psi_{\varepsilon}(t))_{\varepsilon} = ((\Psi_{1,\varepsilon}(t))_{\varepsilon}, \dots, (\Psi_{m,\varepsilon}(t))_{\varepsilon})$ be a function $(\Psi_{\varepsilon})_{\varepsilon}: \mathbb{R} \to G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$. We abbreviate $(\Psi_{\varepsilon}(t))_{\varepsilon} \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$ iff $\forall t[\mathbf{cl}[(\Psi_{\varepsilon}(t))_{\varepsilon}] \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}]$.

Let $(v_{\varepsilon})_{\varepsilon} \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$ and let $(A_{ij,\varepsilon}[x,v_{\varepsilon}])_{\varepsilon}$ be the continuous matrix-function
$$ (A_{ij,\varepsilon}[x, v_{\varepsilon}])_{\varepsilon} = (V_{ij,\varepsilon}(x))_{\varepsilon} + \sum_{j=1}^{m} \int \left[ \left( \rho_{k,l,j,\varepsilon}(x,y) |v_{l,\varepsilon}(y)|^2 \right)_{\varepsilon} \right] dy. \quad (2.1.2.38) $$

Let $(\hat{A}_{\varepsilon}[(v_{\varepsilon})])_{\varepsilon}$ be a linear bounded operator on $G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$, given via formula

$$ \left( \left( \hat{A}_{\varepsilon}[v_{\varepsilon}] u_{\varepsilon} \right)_i \right)_{\varepsilon} = \sum_{j=1}^{m} (A_{ij,\varepsilon}[x, v_{\varepsilon}] u_{j,\varepsilon})_{\varepsilon}. \quad (2.1.2.39) $$

Let's rewrite now Cauchy problem (2.1.2.36) in the form:

$$ -i\hbar \frac{d}{dt} (\Psi_{\varepsilon}(t))_{\varepsilon} - \hbar^2 \Delta (\Psi_{\varepsilon}(t))_{\varepsilon} + \left[ \left( \hat{A}_{\varepsilon}[\Psi_{\varepsilon}(t)] \right)_{\varepsilon} \right] (\Psi_{\varepsilon}(t))_{\varepsilon} = 0, \quad (2.1.2.40) $$

$$ (\Psi_{\varepsilon}(0))_{\varepsilon} = (\Psi_{0,\varepsilon})_{\varepsilon}. $$

**Theorem 2.1.2.2. (Generalized Maslov Theorem).** Generalized Cauchy problem

(2.1.2.40) has a solution $(\Psi_{\varepsilon}(t))_{\varepsilon}$ such that $(\Psi_{\varepsilon}(t))_{\varepsilon} \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$ and the generalized function $(\Psi_{\varepsilon}(t))_{\varepsilon}$ is given via formula:

$$ (\Psi_{\varepsilon}(t))_{\varepsilon} = \left( \left( \prod_{\tau=0}^{t} G_{\tau, d\tau, \varepsilon}[\Psi_{\varepsilon}(\tau)] \right) [\Psi_{0,\varepsilon}] \right)_{\varepsilon}, \quad (2.1.2.41) $$

where

$$ G_{\tau, \varepsilon}[v_{\varepsilon}] = \exp\{i\tau\Delta\} \exp\left\{\left(i\tau\hat{A}_{\varepsilon}[v_{\varepsilon}]\right)_{\varepsilon}\right\}, \quad (2.1.2.42) $$

$$ (v_{\varepsilon})_{\varepsilon} \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}. $$
$$ (A_{ij,\varepsilon}[x, v_{\varepsilon}])_{\varepsilon} = (V_{ij,\varepsilon}(x))_{\varepsilon} + \sum_{j=1}^{m} \int \left[ \left( \rho_{k,l,j,\varepsilon}(x,y) |v_{l,\varepsilon}(y)|^2 \right)_{\varepsilon} \right] dy. \quad (2.1.2.38) $$

Let $(\hat{A}_{\varepsilon}[(v_{\varepsilon})])_{\varepsilon}$ be a linear bounded operator on $G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$, given via formula

$$ \left( \left( \hat{A}_{\varepsilon}[v_{\varepsilon}] u_{\varepsilon} \right)_i \right)_{\varepsilon} = \sum_{j=1}^{m} (A_{ij,\varepsilon}[x, v_{\varepsilon}] u_{j,\varepsilon})_{\varepsilon}. \quad (2.1.2.39) $$

Let's rewrite now Cauchy problem (2.1.2.36) in the form:

$$ -i\hbar \frac{d}{dt} (\Psi_{\varepsilon}(t))_{\varepsilon} - \hbar^2 \Delta (\Psi_{\varepsilon}(t))_{\varepsilon} + \left[ \left( \hat{A}_{\varepsilon}[\Psi_{\varepsilon}(t)] \right)_{\varepsilon} \right] (\Psi_{\varepsilon}(t))_{\varepsilon} = 0, \quad (2.1.2.40) $$

$$ (\Psi_{\varepsilon}(0))_{\varepsilon} = (\Psi_{0,\varepsilon})_{\varepsilon}. $$

**Theorem 2.1.2.2. (Generalized Maslov Theorem).** Generalized Cauchy problem

(2.1.2.40) has a solution $(\Psi_{\varepsilon}(t))_{\varepsilon}$ such that $(\Psi_{\varepsilon}(t))_{\varepsilon} \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}$ and the generalized function $(\Psi_{\varepsilon}(t))_{\varepsilon}$ is given via formula:

$$ (\Psi_{\varepsilon}(t))_{\varepsilon} = \left( \left( \prod_{\tau=0}^{t} G_{\tau, d\tau, \varepsilon} [\Psi_{\varepsilon}(\tau)] \right) [\Psi_{0,\varepsilon}] \right)_{\varepsilon}, \quad (2.1.2.41) $$

where

$$ G_{\varepsilon, \varepsilon}[v_{\varepsilon}] = \exp\{i\varepsilon\Delta\} \exp\left\{\left(i\varepsilon\hat{A}_{\varepsilon}[v_{\varepsilon}]\right)_{\varepsilon}\right\}, \quad (2.1.2.42) $$

$$ (v_{\varepsilon})_{\varepsilon} \in G_{\mathcal{L}_{2,m}(\mathbb{R}^n)}. $$
We let: $\delta = \frac{t}{N}$, $N = 1, 2, \dots$ and

$$ (\Psi_{N,0,\epsilon})_{\epsilon} = (\Psi_{0,\epsilon})_{\epsilon}, $$

$$ (\Psi_{N,k+1,\epsilon})_{\epsilon} = (G_{\delta,\epsilon} \Psi_{N,k,\epsilon})_{\epsilon}, \qquad (2.1.2.43) $$

$$ k = 0, 1, 2, \dots, N-1, $$

where

$$ (G_{\delta,\epsilon} u_{\epsilon}(x))_{\epsilon} = \exp(i\delta\hbar\Delta) \exp((i\delta\hbar^{-1}A_{\epsilon}[u_{\epsilon}])u_{\epsilon}(x))_{\epsilon}, \qquad (2.1.2.44) $$

$$ (u_{\epsilon}(x))_{\epsilon} \in G_{L_2(\mathbb{R}^n)}. $$

and we set below

$$ (\Psi_{N,\epsilon})_{\epsilon} \triangleq (\Psi_{N,N,\epsilon})_{\epsilon}. \qquad (2.1.2.45) $$

Suppose that $\Psi_{0,\epsilon}(x) \in G_{W_2^{2,m}(\mathbb{R}^n)}$. Then every solution $(\Psi_{\epsilon}(t,x))_{\epsilon}$ of the Eq.(2.1.2.40) such that $(\Psi_{\epsilon}(0,x))_{\epsilon} = (\Psi_{0,\epsilon}(x))_{\epsilon}$, has the representation:

$$ (\Psi_{\epsilon}(t,x))_{\epsilon} = s- \lim_{N \to \infty} \Psi_{N}, \qquad (2.1.2.46) $$

where the strong limit in (2.1.2.46) exist uniformly on $[0, T]$ for any $T < \infty$ in the sense of the convergence in ultra-pseudo-norm on $G_{L_{2,m}(\mathbb{R}^n)}$.

**Proof.**
We let: $\delta = \frac{t}{N}$, $N = 1, 2, \dots$ and

$$ (\Psi_{N,0,\epsilon})_{\epsilon} = (\Psi_{0,\epsilon})_{\epsilon}, $$

$$ (\Psi_{N,k+1,\epsilon})_{\epsilon} = (G_{\delta,\epsilon} \Psi_{N,k,\epsilon})_{\epsilon}, \qquad (2.1.2.43) $$

$$ k = 0, 1, 2, \dots, N-1, $$

where

$$ (G_{\delta,\epsilon} u_{\epsilon}(x))_{\epsilon} = \exp(i\delta\hbar\Delta) \exp((i\delta\hbar^{-1}A_{\epsilon}[u_{\epsilon}])u_{\epsilon}(x))_{\epsilon}, \qquad (2.1.2.44) $$

$$ (u_{\epsilon}(x))_{\epsilon} \in G_{L_2(\mathbb{R}^n)}. $$

and we set below

$$ (\Psi_{N,\epsilon})_{\epsilon} \triangleq (\Psi_{N,N,\epsilon})_{\epsilon}. \qquad (2.1.2.45) $$

Suppose that $\Psi_{0,\epsilon}(x) \in G_{W_2^{2,m}(\mathbb{R}^n)}$. Then every solution $(\Psi_{\epsilon}(t,x))_{\epsilon}$ of the Eq.(2.1.2.40) such that $(\Psi_{\epsilon}(0,x))_{\epsilon} = (\Psi_{0,\epsilon}(x))_{\epsilon}$, has the representation:

$$ (\Psi_{\epsilon}(t,x))_{\epsilon} = s- \lim_{N \to \infty} \Psi_{N}, \qquad (2.1.2.46) $$

where the strong limit in (2.1.2.46) exist uniformly on $[0, T]$ for any $T < \infty$ in the sense of the convergence in ultra-pseudo-norm on $G_{L_{2,m}(\mathbb{R}^n)}$.

**Proof.**
1. We note that

$$
\| \exp((i\tilde{\epsilon}\hbar^{-1}A_{\epsilon}[v_{\epsilon}]))_{\epsilon} \|_{\#} = 1, \qquad (2.1.2.47)
$$

$$
\| \exp(i \tilde{\epsilon} \hbar \Delta) \|_{\#} = 1,
$$

where we denote

$$
\| \cdot \|_{\#} \triangleq \| \cdot \|_{G_{L,2,m}}^{\text{op}} . \qquad (2.1.2.48)
$$

Hence one obtain

$$
\|(\Psi_{N,k,\epsilon})_{\epsilon}\|_{G_{L,2,m}} = \|(\Psi_{0,\epsilon})_{\epsilon}\|_{G_{L,2,m}}. \qquad (2.1.2.49)
$$

2. Let's now to prove the formula

$$
(\Psi_{N,k,\epsilon})_{\epsilon} = \exp(ik\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} + \\
i\delta\hbar^{-1} \sum_{j=0}^{k-1} \exp[i(k-j-1)\delta\hbar\Delta](A_{\epsilon}[\Psi_{N,j,\epsilon}])_{\epsilon}(\Psi_{N,j,\epsilon})_{\epsilon} + O\left(\frac{1}{\hbar N}\right) \tag{2.1.2.50}
$$

Note that $\|v\|_{G_{L,2,m}} \le r \Rightarrow \|(A_\epsilon[v_\epsilon])_\epsilon\|_{G_{L,2,m}} \le C_r$. Hence

$$
\exp(i\delta\hbar^{-1}(A_\epsilon[v_\epsilon])_\epsilon) = 1 + i\delta\hbar^{-1}(A_\epsilon[\Psi_{N,j,\epsilon}])_\epsilon + O\left(\frac{1}{\hbar^2 N^2}\right) \quad (2.1.2.51)
$$
1. We note that

$$
\| \exp((i\tilde{\epsilon}\hbar^{-1}A_{\epsilon}[v_{\epsilon}]))_{\epsilon} \|_{\#} = 1, \qquad (2.1.2.47)
$$

$$
\| \exp(i \tilde{\epsilon} \hbar \Delta) \|_{\#} = 1,
$$

where we denote

$$
\| \cdot \|_{\#} \triangleq \| \cdot \|_{G_{L,2,m}}^{\text{op}} . \qquad (2.1.2.48)
$$

Hence one obtain

$$
\|(\Psi_{N,k,\epsilon})_{\epsilon}\|_{G_{L,2,m}} = \|(\Psi_{0,\epsilon})_{\epsilon}\|_{G_{L,2,m}}. \quad (2.1.2.49)
$$

2. Let's now to prove the formula

$$
(\Psi_{N,k,\epsilon})_{\epsilon} = \exp(ik\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} + i\delta\hbar^{-1} \sum_{j=0}^{k-1} \exp[i(k-j-1)\delta\hbar\Delta](A_{\epsilon}[\Psi_{N,j,\epsilon}])_{\epsilon}(\Psi_{N,j,\epsilon})_{\epsilon} + O\left(\frac{1}{\hbar N}\right) \quad (2.1.2.50)
$$

Note that $\|v\|_{G_{L,2,m}} \le r \Rightarrow \|(A_\epsilon[v_\epsilon])_\epsilon\|_{G_{L,2,m}} \le C_r$. Hence

$$
\exp(i\delta\hbar^{-1}(A_\epsilon[v_\epsilon])_\epsilon) = 1 + i\delta\hbar^{-1}(A_\epsilon[\Psi_{N,j,\epsilon}])_\epsilon + O\left(\frac{1}{\hbar^2 N^2}\right) \quad (2.1.2.51)
$$
From Eq.(2.1.2.51) by simple calculation one obtain

$$ (\Psi_{N,k,\epsilon})_{\epsilon} = \exp(ik\delta\hbar\Delta)(\Psi_{N,k-1,\epsilon})_{\epsilon} + i\delta\hbar^{-1}\exp[i\delta\hbar\Delta](A_{\epsilon}[\Psi_{N,k-1,\epsilon}])_{\epsilon}(\Psi_{N,k-1,\epsilon})_{\epsilon} + O\left(\frac{1}{\hbar^2 N^2}\right), \quad (2.1.2.52) $$

where

$$ \left\| O\left(\frac{1}{\hbar^2 N^2}\right) \right\|_{G_{L,2,m}} \le \frac{C}{\hbar^2 N^2} \quad (2.1.2.53) $$

and where constant C does not depend on $k = 1, 2, \dots, N$. From Eq.(2.1.2.52) by simple calculation one obtain Eq.(2.1.2.50).

3. Let's now to prove the inequality:

$$ \|(\Psi_{N,l+k,\epsilon})_{\epsilon} - (\Psi_{N,l,\epsilon})_{\epsilon}\|_{G_{L,2,m}} \le $$

$$ C_0[\exp(C_1 t)] \left( k \hbar^{-1} \delta + \| \exp(ik\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} - (\Psi_{0,\epsilon})_{\epsilon} \|_{G_{L,2,m}} \right), \quad (2.1.2.54) $$

$$ l + k \le N, $$

where $C_0 = \text{const}$ and $C_1 = \text{const}$.

By using formula (2.1.2.50) one obtain
From Eq.(2.1.2.51) by simple calculation one obtain

$$ (\Psi_{N,k,\epsilon})_{\epsilon} = \exp(ik\delta\hbar\Delta)(\Psi_{N,k-1,\epsilon})_{\epsilon} + i\delta\hbar^{-1}\exp[i\delta\hbar\Delta](A_{\epsilon}[\Psi_{N,k-1,\epsilon}])_{\epsilon}(\Psi_{N,k-1,\epsilon})_{\epsilon} + O\left(\frac{1}{\hbar^2 N^2}\right), \quad (2.1.2.52) $$

where

$$ \left\| O\left( \frac{1}{\hbar^2 N^2} \right) \right\|_{G_{L,2,m}} \le \frac{C}{\hbar^2 N^2} \quad (2.1.2.53) $$

and where constant C does not depend on $k = 1, 2, \dots, N$. From Eq.(2.1.2.52) by simple calculation one obtain Eq.(2.1.2.50).

3. Let's now to prove the inequality:

$$ \| (\Psi_{N,l+k,\epsilon})_{\epsilon} - (\Psi_{N,l,\epsilon})_{\epsilon} \|_{G_{L,2,m}} \le $$

$$ C_0[\exp(C_1 t)] \left( k \hbar^{-1} \delta + \| \exp(ik\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} - (\Psi_{0,\epsilon})_{\epsilon} \|_{G_{L,2,m}} \right), \quad (2.1.2.54) $$

$$ l + k \le N, $$

where $C_0 = \text{const}$ and $C_1 = \text{const}$.

By using formula (2.1.2.50) one obtain
$$
\begin{align}
(\Psi_{N,l+k,\epsilon})_{\epsilon} - (\Psi_{N,l,\epsilon})_{\epsilon} ={}& \exp(ik\delta\hbar\Delta)(\exp(ik\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} - (\Psi_{0,\epsilon})_{\epsilon}) + \nonumber \\
& + i\delta\hbar^{-1} \sum_{j=0}^{k-1} \exp[i\delta\hbar(k+l-j-1)\Delta](A_{\epsilon}\Psi_{N,j,\epsilon})_{\epsilon} + \tag{2.1.2.55} \\
& + i\delta\hbar^{-1} \sum_{j=0}^{l-1} \exp[i\delta\hbar(l-j-1)\Delta]((A_{\epsilon}\Psi_{N,j+k,\epsilon})_{\epsilon} - (A_{\epsilon}\Psi_{N,j,\epsilon})_{\epsilon}) + O\left(\frac{1}{\hbar N}\right) \nonumber
\end{align}
$$

From inequality: $\|(A_\epsilon[v_\epsilon])_\epsilon\|_\# \le C_r$ and Eqs.(2.1.2.55) we obtain

$$
\begin{align*}
& a_0 \le \epsilon, \\
& a_1 \le C \left[ \epsilon + k\delta\hbar^{-1} + \hbar^{-1} \sum_{j=0}^{l-1} \delta a_j \right], \\
& l = 1, 2, \dots, N, \tag{2.1.2.56} \\
& a_j = \|(\Psi_{N,j+k,\epsilon})_\epsilon - (\Psi_{N,j,\epsilon})_\epsilon\|_{G_{L,2,m}}, \\
& \epsilon = \|\exp(ik\delta\hbar\Delta)(\Psi_{0,\epsilon})_\epsilon - (\Psi_{0,\epsilon})_\epsilon\|_{G_{L,2,m}}
\end{align*}
$$

4. Suppose that $M \in \mathbb{N}, M \ge 1$ and $M = \text{const}$ then for every $j \in \mathbb{N}$ one obtain

$$
\frac{j}{M} - \left[ \frac{j}{M} \right] \le 1. \qquad (2.1.2.57)
$$

Hence from (2.1.2.21) we obtain
$$ (\Psi_{N,l+k,\epsilon})_{\epsilon} - (\Psi_{N,l,\epsilon})_{\epsilon} = \exp(ik\delta\hbar\Delta)(\exp(ik\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} - (\Psi_{0,\epsilon})_{\epsilon}) + \\ +i\delta\hbar^{-1} \sum_{j=0}^{k-1} \exp[i\delta\hbar(k+l-j-1)\Delta](A_{\epsilon}\Psi_{N,j,\epsilon})_{\epsilon} + \\ +i\delta\hbar^{-1} \sum_{j=0}^{l-1} \exp[i\delta\hbar(l-j-1)\Delta]((A_{\epsilon}\Psi_{N,j+k,\epsilon})_{\epsilon} - (A_{\epsilon}\Psi_{N,j,\epsilon})_{\epsilon}) + O\left(\frac{1}{\hbar N}\right) $$

From inequality: $\|(A_{\epsilon}[v_{\epsilon}])_{\epsilon}\|_{\#} \le C_r$ and Eqs.(2.1.2.55) we obtain

$$ \begin{gathered} a_0 \le \epsilon, \\ a_1 \le C \left[ \epsilon + k\delta\hbar^{-1} + \hbar^{-1} \sum_{j=0}^{l-1} \delta a_j \right], \\ l = 1, 2, \dots, N, \\ a_j = \|(\Psi_{N,j+k,\epsilon})_{\epsilon} - (\Psi_{N,j,\epsilon})_{\epsilon}\|_{G_{L,2,m}}, \\ \epsilon = \|\exp(ik\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} - (\Psi_{0,\epsilon})_{\epsilon}\|_{G_{L,2,m}} \end{gathered} $$

4. Suppose that $M \in \mathbb{N}, M \ge 1$ and $M = \text{const}$ then for every $j \in \mathbb{N}$ one obtain

$$ \frac{j}{M} - \left[ \frac{j}{M} \right] \le 1. \qquad (2.1.2.57) $$

Hence from (2.1.2.21) we obtain
$$ (\Psi_{MN,j,\epsilon})_{\epsilon} = (\Psi_{M,[\frac{j}{M}],\epsilon})_{\epsilon} + O\left(\frac{1}{\hbar N}\right), \qquad (2.1.2.58) $$

where

$$ \begin{gathered} \|O\left(\frac{1}{\hbar N}\right)\|_{G_{L_{2,m}}} \le \\ \left( \delta\hbar^{-1} + \|(\Psi_{0,\epsilon})_{\epsilon} - \exp\left[i\delta\hbar\left(\frac{j}{M} - \left[\frac{j}{M}\right]\right)\Delta\right](\Psi_{0,\epsilon})_{\epsilon} \right) \|_{G_{L_{2,m}}} \le \end{gathered} \qquad (2.1.2.59) $$

$$ \delta\hbar^{-1} \|(\Psi_{0,\epsilon})_{\epsilon} \|_{G_{\mathbb{H}_{2}^{2,m}(\mathbb{R}^{n})}}. $$

We denote

$$ \gamma_N(\hbar^{-1}) \triangleq \delta\hbar^{-1} \|(\Psi_{0,\epsilon})_{\epsilon} \|_{G_{\mathbb{H}_{2}^{2,m}(\mathbb{R}^{n})}}, \qquad (2.1.2.60) $$

$$ \delta = \delta(N). $$

Thus we have two finite splittings $\Delta_N^{(1)} = \{0 = t_0 < t_1^{(1)} < \dots < t_N^{(1)} = t\}$ and $\Delta_{NM}^{(2)} = \{0 = t_0 < t_1^{(2)} < \dots < t_{NM}^{(2)} = t\}$ of the closed interval $[0,t]$ such that $\{t_i^{(1)}\}_{i=0}^N \subsetneq \{t_j^{(2)}\}_{j=0}^{NM}$.

Let's rewrite now formula (2.1.2.50) by using replacement $N \to NM$ for $k_0 = M \times l$. Thus we obtain
$$ (\Psi_{MN,j,\epsilon})_{\epsilon} = (\Psi_{M,[\frac{j}{M}],\epsilon})_{\epsilon} + O\left(\frac{1}{\hbar N}\right), \qquad (2.1.2.58) $$

where

$$ \begin{gathered} \left\| O\left(\frac{1}{\hbar N}\right) \right\|_{G_{L_{2,m}}} \le \\ \left( \delta\hbar^{-1} + \left\| (\Psi_{0,\epsilon})_{\epsilon} - \exp\left[i\delta\hbar\left(\frac{j}{M} - \left[\frac{j}{M}\right]\right)\Delta\right](\Psi_{0,\epsilon})_{\epsilon} \right\|_{G_{L_{2,m}}} \right) \le \end{gathered} \qquad (2.1.2.59) $$

$$ \delta\hbar^{-1} \|(\Psi_{0,\epsilon})_{\epsilon}\|_{G_{\mathbb{H}_{2}^{2,m}(\mathbb{R}^n)}}. $$

We denote

$$ \gamma_N(\hbar^{-1}) \triangleq \delta\hbar^{-1} \|(\Psi_{0,\epsilon})_{\epsilon}\|_{G_{\mathbb{H}_{2}^{2,m}(\mathbb{R}^n)}} \qquad (2.1.2.60) $$

$$ \delta = \delta(N). $$

Thus we have two finite splittings $\Delta_N^{(1)} = \{0 = t_0 < t_1^{(1)} < \dots < t_N^{(1)} = t\}$ and $\Delta_{NM}^{(2)} = \{0 = t_0 < t_1^{(2)} < \dots < t_{NM}^{(2)} = t\}$ of the closed interval $[0,t]$ such that $\{t_i^{(1)}\}_{i=0}^N \subsetneq \{t_j^{(2)}\}_{j=0}^{NM}$.

Let's rewrite now formula (2.1.2.50) by using replacement $N \to NM$ for $k_0 = M \times l$. Thus we obtain
$$ (\Psi_{MN,Ml,\epsilon})_{\epsilon} = \exp(i\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} + i\delta\hbar^{-1} \sum_{j=0}^{l-1} \exp[i\delta\hbar(l-j-1)\Delta] \times (A_{\epsilon}\Psi_{MN,Mj,\epsilon})_{\epsilon} + O(\gamma_N(\hbar^{-1})). \quad (2.1.2.61) $$

Let's denote

$$ b_j = \|(\Psi_{N,j,\epsilon})_{\epsilon} - (\Psi_{MN,Mj,\epsilon})_{\epsilon}\|_{G_{L2,m}} \quad (2.1.2.62) $$

By using subtraction Eq.(2.1.2.50) from Eq.(2.1.2.61) we obtain

$$ b_0 \le \alpha_N $$

$$ b_l \le \alpha_N + c\delta\hbar^{-1} \sum_{j=0}^{l-1} b_j, \quad (2.1.2.63) $$

$$ l = 1, \dots, N, $$

where $\alpha_N \le C_1\gamma_N(\hbar^{-1})$. From (2.1.2.63) we obtain

$$ b_l \le \alpha_N \exp[C_2 t]. \quad (2.1.2.64) $$

In particular for $l = N$ we obtain:

$$ \| (\Psi_{N,\epsilon}(t))_{\epsilon} - (\Psi_{NM,\epsilon}(t))_{\epsilon} \|_{G_{L2,m}} \le \frac{t}{N} C_1 \exp(C_2 t) \| (\Psi_{0,\epsilon})_{\epsilon} \|_{G_{W_2^2,m}} \quad (2.1.2.65) $$

and therefore sequence $(\Psi_{N,\epsilon}(t))_{\epsilon}$ uniformly convergence on interval $[0, T]$ for any
$$ (\Psi_{MN,Ml,\epsilon})_{\epsilon} = \exp(i\delta\hbar\Delta)(\Psi_{0,\epsilon})_{\epsilon} + i\delta\hbar^{-1} \sum_{j=0}^{l-1} \exp[i\delta\hbar(l-j-1)\Delta] \times \quad (2.1.2.61) $$

$$ (A_{\epsilon} \Psi_{MN,Mj,\epsilon})_{\epsilon} + O(\gamma_N(\hbar^{-1})). $$

Let's denote

$$ b_j = \|(\Psi_{N,j,\epsilon})_{\epsilon} - (\Psi_{MN,Mj,\epsilon})_{\epsilon}\|_{G_{L2,m}} \quad (2.1.2.62) $$

By using subtraction Eq.(2.1.2.50) from Eq.(2.1.2.61) we obtain

$$ b_0 \le \alpha_N $$

$$ b_l \le \alpha_N + c\delta\hbar^{-1} \sum_{j=0}^{l-1} b_j, \quad (2.1.2.63) $$

$$ l = 1, \dots, N, $$

where $\alpha_N \le C_1\gamma_N(\hbar^{-1})$. From (2.1.2.63) we obtain

$$ b_l \le \alpha_N \exp[C_2 t]. \quad (2.1.2.64) $$

In particular for $l = N$ we obtain:

$$ \| (\Psi_{N,\epsilon}(t))_{\epsilon} - (\Psi_{NM,\epsilon}(t))_{\epsilon} \|_{G_{L2,m}} \le \frac{t}{N} C_1 \exp(C_2 t) \| (\Psi_{0,\epsilon})_{\epsilon} \|_{G_{W_2^2,m}} \quad (2.1.2.65) $$

and therefore sequence $(\Psi_{N,\epsilon}(t))_{\epsilon}$ uniformly convergence on interval $[0, T]$ for any
$T < \infty$ in the sense of the convergence in ultra-pseudo-seminorm on $G_{L_{2,m}(\mathbb{R}^n)}$.

5. Any function $\Psi_N(t)$ is continuous on $[0, T]$. Therefore $\Psi(t) = s\lim_{N\to\infty}(\Psi_{N,s}(t))_s$ is continuous on $[0, T]$. Hence the Riemann sum

$$ (S_N, \varepsilon(t))_\varepsilon = \sum_{k=0}^{N-1} \frac{t}{N} \exp\left\{i \frac{t}{N} (N-k-1)\Delta\right\} \left(\hat{A}_\varepsilon[\Psi_{N,k,\varepsilon}]\right)_\varepsilon (\Psi_{N,k,\varepsilon})_\varepsilon \quad (2.1.2.66) $$

is convergence in the ultra-pseudo-seminorm of the $G_{L_{2,m}(\mathbb{R}^n)}$ to integral

$$ \int_0^t d\tau \exp\{i(t-\tau)\} \left(\hat{A}_\varepsilon[\Psi_\varepsilon(\tau)]\right)_\varepsilon (\Psi_\varepsilon(\tau))_\varepsilon. \quad (2.1.2.67) $$

Finally from Eq. (II.1.2.50) in the limit $N \to \infty$ we obtain integral equation

$$ (\Psi_{\varepsilon}(t))_{\varepsilon} = \exp\{it\Delta\}(\Psi_{0,\varepsilon})_{\varepsilon} + \int_{0}^{t} d\tau \exp\{i(t-\tau)\} \left[ \left( \hat{A}_{\varepsilon}[\Psi_{\varepsilon}(\tau)] \right)_{\varepsilon} \right] (\Psi_{\varepsilon}(\tau))_{\varepsilon}. \quad (II.1.2.68) $$

Eq. (II.1.2.68) completed the proof.

## II.1.3. Representation of the Maslov-Colombeau T-exponent by using Feynman-Colombeau path Integral.

**Definition II.1.3.1.** We define formal pseudodifferential operator with symbol $A(x,p)$ via formula

$$ A\left(\frac{\partial}{\partial x}, -ih \frac{\partial}{\partial x}\right) v(x) = \frac{1}{2\pi h} \int dp A(x,p) \int dy v(y) \exp\left(-\frac{i}{h}yp\right). \quad (II.1.3.1) $$
$T < \infty$ in the sense of the convergence in ultra-pseudo-seminorm on $G_{L_{2,m}(\mathbb{R}^n)}$.

5. Any function $\Psi_N(t)$ is continuous on $[0, T]$. Therefore $\Psi(t) = s\lim_{N\to\infty}(\Psi_{N,s}(t))_s$ is continuous on $[0, T]$. Hence the Riemann sum

$$ (S_N, \varepsilon(t))_\varepsilon = \sum_{k=0}^{N-1} \frac{t}{N} \exp\left\{i \frac{t}{N} (N-k-1)\Delta\right\} \left(\hat{A}_\varepsilon[\Psi_{N,k,\varepsilon}]\right)_\varepsilon (\Psi_{N,k,\varepsilon})_\varepsilon \quad (2.1.2.66) $$

is convergence in the ultra-pseudo-seminorm of the $G_{L_{2,m}(\mathbb{R}^n)}$ to integral

$$ \int_0^t d\tau \exp\{i(t-\tau)\} \left(\hat{A}_\varepsilon[\Psi_\varepsilon(\tau)]\right)_\varepsilon (\Psi_\varepsilon(\tau))_\varepsilon. \quad (2.1.2.67) $$

Finally from Eq. (II.1.2.50) in the limit $N \to \infty$ we obtain integral equation

$$ (\Psi_{\varepsilon}(t))_{\varepsilon} = \exp\{it\Delta\}(\Psi_{0,\varepsilon})_{\varepsilon} + \int_{0}^{t} d\tau \exp\{i(t-\tau)\} \left[ \left( \hat{A}_{\varepsilon}[\Psi_{\varepsilon}(\tau)] \right)_{\varepsilon} \right] (\Psi_{\varepsilon}(\tau))_{\varepsilon}. \quad (II.1.2.68) $$

Eq. (II.1.2.68) completed the proof.

## II.1.3. Representation of the Maslov-Colombeau T-exponent by using Feynman-Colombeau path Integral.

**Definition II.1.3.1.** We define formal pseudodifferential operator with symbol $A(x,p)$ via formula

$$ A\left(\frac{\partial}{\partial x}, -ih \frac{\partial}{\partial x}\right) v(x) = \frac{1}{2\pi h} \int dp A(x,p) \int dy v(y) \exp\left(-\frac{i}{h}yp\right). \quad (II.1.3.1) $$
Let's consider formal T-exponent with a generator

$$ G_{t,\bar{\epsilon}} = G_{t,\bar{\epsilon}} \left( \stackrel{2}{x}, -i\epsilon \frac{\partial}{\partial x}, t, [\nu] \right) v(x,t). \qquad (2.1.3.2) $$

Let $\Delta_N = \{0 = t_0 < t_1 < \dots < t_{N+1} = t\}, \Delta t_j = t_{j+1} - t_j, \delta_N = \max_j \Delta t_j, x \in \mathbb{R}^n$.

T-exponent

$$ \Psi(x,t) = \left( \prod_{\tau=0}^{t} G_{\tau, d\tau} \right) [\Psi_0] \qquad (2.1.3.3) $$

is defined via formulae

$$ \Psi(x,t) = s- \lim_{\delta_N \to 0} \Psi_{N+1}(x,t), $$

$$ \Psi_0(x,t) = \Psi_0, $$

$$ \Psi_{k+1}(x,t) = \Psi_k(x,t) \quad \text{iff } 0 \le t \le t_k, \qquad (2.1.3.4) $$

$$ \Psi_{k+1}(x,t) = G_{t_k, t-t_k} \left( \stackrel{2}{x}, -i\epsilon \frac{\partial}{\partial x}, t, [\Psi_k] \right) \Psi_k(x,t) \quad \text{iff } t_k \le t \le t_{k+1} $$

$k = 0, 1, \dots, N$

**Theorem 2.1.3.1.[20]**. Suppose that T-exponent given via Eq.(2.1.3.3) exist and $\Psi(x,t) \in W_2^k(\mathbb{R}^n)$. Then function $\Psi(x,t)$ has the representation
Let's consider formal T-exponent with a generator

$$G_{t,\bar{\epsilon}} = G_{t,\bar{\epsilon}} \left( \int_0^t \bar{x}, -i\epsilon \frac{\partial}{\partial x}, t, [\bar{v}] \right) v(x,t). \quad (2.1.3.2)$$

Let $\Delta_N = \{0 = t_0 < t_1 < \dots < t_{N+1} = t\}, \Delta t_j = t_{j+1} - t_j, \delta_N = \max_j \Delta t_j, x \in \mathbb{R}^n$.

T-exponent

$$\Psi(x,t) = \left( \prod_{\tau=0}^{t} G_{\tau, d\tau} \right) [\Psi_0] \quad (2.1.3.3)$$

is defined via formulae

$$\Psi(x,t) = s- \lim_{\delta_N \to 0} \Psi_{N+1}(x,t),$$

$$\Psi_0(x,t) = \Psi_0,$$

$$\Psi_{k+1}(x,t) = \Psi_k(x,t) \text{ iff } 0 \le t \le t_k, \quad (2.1.3.4)$$

$$\Psi_{k+1}(x,t) = G_{t_k, t-t_k} \left( \int_0^t \bar{x}, -i\epsilon \frac{\partial}{\partial x}, t, [\Psi_k] \right) \Psi_k(x,t) \text{ iff } t_k \le t \le t_{k+1}$$

$$k = 0, 1, \dots, N$$

**Theorem 2.1.3.1.** [20]. Suppose that T-exponent given via Eq.(2.1.3.3) exist and $\Psi(x,t) \in W_2^k(\mathbb{R}^n)$. Then function $\Psi(x,t)$ has the representation
$$
\begin{equation}
\begin{aligned}
\Psi(x,t) = {}& s- \lim_{\delta_N \to 0} \frac{1}{(2\pi h)^{(N+1)n}} \int \dots \int \exp\left[\frac{i}{h}(x_{k+1} - x_k)\xi_k\right] \times \\
& \times \prod_{k=0}^{N} G_{t_k, \Delta t_k}(x_{k+1}, \xi_k; [\Psi]) \Psi_0(x_0) d^{N+1}x d^{N+1}\xi,
\end{aligned}
\tag{2.1.3.5}
\end{equation}
$$

where $x_{N+1} = x$ and $d^{N+1} = dx_0 \dots dx_N$, $d^{N+1}\xi = d\xi_0 \dots d\xi_N$, $x_j \in \mathbb{R}^n$, $\xi_j \in \mathbb{R}^n$.
We abbreviate Eq.(2.1.3.5) in canonical symbolic form

$$
\begin{equation}
\begin{aligned}
\Psi(x,t) = \int_{q(t)=x} D[q(\tau)] \int_{q(t)=x} D[p(\tau)] \Psi_0(q(0)) & \exp \left\{ \frac{i}{h} \int_0^t p(\tau) \dot{q}(\tau) d\tau \right\} \times \\
& \times \prod_{\tau=0}^t G_{\tau,d\tau}(q(\tau), p(\tau); [\Psi]).
\end{aligned}
\tag{2.1.3.6}
\end{equation}
$$

From Theorem 2.1.3.1 and Theorem 2.1.2.1 one obtain directly:

**Theorem 2.1.3.2.** [20]. Suppose that $V(x) \in C(\mathbb{R}^n)$, $a(x,y) \in C(\mathbb{R}^n \times \mathbb{R}^n)$ and $\Psi_0(x) \in W_2^2(\mathbb{R}^n)$. Let $\Psi(x,t)$ be the solution of the Cauchy problem (2.1.2.7).
Then $\Psi(x,t)$ can be expressed via formula:

$$
\begin{align*}
\Psi(x,t) &= \lim_{N \to \infty} (4\pi i h)^{-\frac{(N+1)}{2}} \int \dots \int \Psi_0(x_0) \frac{dx_0 \dots dx_N}{\sqrt{\Delta t_0 \dots \Delta t_N}} \times \\
\exp &\left\{ \frac{i}{\hbar} \sum_{k=0}^{N} \left[ \frac{1}{4} \left( \frac{x_{k+1} - x_k}{\Delta t_k} \right)^2 - V(x_{k+1}) - \int a(x_k, y) \Psi(y, t_k) dy \right] \Delta t_k \right\}, \tag{2.1.3.6'} \\
x_{N+1} &= x.
\end{align*}
$$

Asume that $a(x,y) = 0$. Then for the linear Schrödinger equation (0.1) we obtain
$$
\begin{equation}
\begin{aligned}
\Psi(x,t) = {}& s- \lim_{\delta_N \to 0} \frac{1}{(2\pi h)^{(N+1)n}} \int \dots \int \exp\left[\frac{i}{h}(x_{k+1} - x_k)\xi_k\right] \times \\
& \times \prod_{k=0}^{N} G_{t_k, \Delta t_k}(x_{k+1}, \xi_k; [\Psi]) \Psi_0(x_0) d^{N+1}x d^{N+1}\xi,
\end{aligned}
\tag{2.1.3.5}
\end{equation}
$$

where $x_{N+1} = x$ and $d^{N+1} = dx_0 \dots dx_N$, $d^{N+1}\xi = d\xi_0 \dots d\xi_N$, $x_j \in \mathbb{R}^n$, $\xi_j \in \mathbb{R}^n$.
We abbreviate Eq.(2.1.3.5) in canonical symbolic form

$$
\begin{equation}
\begin{aligned}
\Psi(x,t) = \int_{q(t)=x} D[q(\tau)] \int_{q(t)=x} D[p(\tau)] \Psi_0(q(0)) & \exp \left\{ \frac{i}{h} \int_0^t p(\tau) \dot{q}(\tau) d\tau \right\} \times \\
& \times \prod_{\tau=0}^t G_{\tau,d\tau}(q(\tau), p(\tau); [\Psi]).
\end{aligned}
\tag{2.1.3.6}
\end{equation}
$$

From Theorem 2.1.3.1 and Theorem 2.1.2.1 one obtain directly:

**Theorem 2.1.3.2.**[20]. Suppose that $V(x) \in C(\mathbb{R}^n)$, $a(x,y) \in C(\mathbb{R}^n \times \mathbb{R}^n)$ and $\Psi_0(x) \in W_2^2(\mathbb{R}^n)$. Let $\Psi(x,t)$ be the solution of the Cauchy problem (2.1.2.7).
Then $\Psi(x,t)$ can be expressed via formula:

$$
\begin{align*}
\Psi(x,t) &= \lim_{N \to \infty} (4\pi i h)^{-\frac{(N+1)}{2}} \int \dots \int \Psi_0(x_0) \frac{dx_0 \dots dx_N}{\sqrt{\Delta t_0 \dots \Delta t_N}} \times \\
\exp &\left\{ \frac{i}{\hbar} \sum_{k=0}^{N} \left[ \frac{1}{4} \left( \frac{x_{k+1} - x_k}{\Delta t_k} \right)^2 - V(x_{k+1}) - \int a(x_k, y) \Psi(y, t_k) dy \right] \Delta t_k \right\}, \tag{2.1.3.6'} \\
x_{N+1} &= x.
\end{align*}
$$

Asume that $a(x,y) = 0$. Then for the linear Schrödinger equation (0.1) we obtain
$$
\begin{gather*}
\Psi(x,t) = \lim_{N \to \infty} (4\pi i h)^{-\frac{(N+1)}{2}} \int \dots \int \Psi_0(x_0) \frac{dx_0 \dots dx_N}{\sqrt{\Delta t_0 \dots \Delta t_N}} \times \\
\exp\left\{ \frac{i}{h} \sum_{k=0}^{N} \left[ \frac{1}{4} \left( \frac{x_{k+1} - x_k}{\Delta t_k} \right)^2 - V(x_{k+1}) - \int a(x_k, y) \Psi(y, t_k) dy \right] \Delta t_k \right\}, \quad (2.1.3.6^{\#}) \\
x_{N+1} = x.
\end{gather*}
$$

We abbreviate Eq.(2.1.3.6#) in canonical symbolic form

$$
\begin{equation}
\begin{split}
\Psi(x,t) = \int_{q(t)=x} D[q(t)]\Psi_0(q(0)) & \exp\left\{\frac{i}{h} \int_0^t \left[\frac{1}{4}\dot{q}^2(\tau) - V(q(\tau))\right] d\tau\right\} = \\
& \int_{q(t)=x} D[q(t)]dy\Psi_0(y) \exp\left\{\frac{i}{h} \int_0^t \left[\frac{1}{4}\dot{q}^2(\tau) - V(q(\tau))\right] d\tau\right\}.
\end{split}
\tag{2.1.3.6#}
\end{equation}
$$

**Definition 2.1.3.2.** Assume that $v(x) \in G(\mathbb{R}^n)$, $v(x) = \text{cl}[(v_\epsilon(x))_\epsilon]$, $A(x,p) \in G(\mathbb{R}^n \times \mathbb{R}^n)$,

$A(x,p) = \text{cl}[(A_{\epsilon}(x,p))_{\epsilon}]$. We define formal Colombeau pseudodifferential operator with symbol $A(x,p)$ via formula:

$$
\begin{align}
\left( A_{\epsilon} \left( \frac{2}{x}, -ih \frac{1}{\partial x} \right) v_{\epsilon}(x) \right)_{\epsilon \in (0,1]} & = \tag{2.1.3.7} \\
\frac{1}{2\pi h} \left( \int dp A_{\epsilon}(x,p) \int dy v_{\epsilon}(y) \exp\left(-\frac{i}{h}yp\right) \right)_{\epsilon \in (0,1]} & . \nonumber
\end{align}
$$

Let's consider formal Maslov-Colombeau T-exponent with a generator
$$
\begin{gather*}
\Psi(x,t) = \lim_{N \to \infty} (4\pi i h)^{-\frac{(N+1)}{2}} \int \dots \int \Psi_0(x_0) \frac{dx_0 \dots dx_N}{\sqrt{\Delta t_0 \dots \Delta t_N}} \times \\
\exp\left\{ \frac{i}{h} \sum_{k=0}^{N} \left[ \frac{1}{4} \left( \frac{x_{k+1} - x_k}{\Delta t_k} \right)^2 - V(x_{k+1}) - \int a(x_k, y) \Psi(y, t_k) dy \right] \Delta t_k \right\}, \quad (2.1.3.6^\#) \\
x_{N+1} = x.
\end{gather*}
$$

We abbreviate Eq.(2.1.3.6#) in canonical symbolic form

$$
\begin{equation}
\begin{split}
\Psi(x,t) = \int_{q(t)=x} D[q(t)]\Psi_0(q(0)) & \exp\left\{\frac{i}{h} \int_0^t \left[\frac{1}{4}\dot{q}^2(\tau) - V(q(\tau))\right] d\tau\right\} = \\
& \int_{q(t)=x} D[q(t)]dy\Psi_0(y) \exp\left\{\frac{i}{h} \int_0^t \left[\frac{1}{4}\dot{q}^2(\tau) - V(q(\tau))\right] d\tau\right\}.
\end{split}
\tag{2.1.3.6#}
\end{equation}
$$

**Definition 2.1.3.2.** Assume that $v(x) \in G(\mathbb{R}^n)$, $v(x) = \text{cl}[(v_\epsilon(x))_\epsilon]$, $A(x,p) \in G(\mathbb{R}^n \times \mathbb{R}^n)$,

$A(x,p) = \text{cl}[(A_\epsilon(x,p))_\epsilon]$. We define formal Colombeau pseudodifferential operator with symbol $A(x,p)$ via formula:

$$
\begin{align}
\left( A_{\epsilon} \left( \frac{2}{x}, -ih \frac{1}{\partial x} \right) v_{\epsilon}(x) \right)_{\epsilon \in (0,1]} & = \tag{2.1.3.7} \\
\frac{1}{2\pi h} \left( \int dp A_{\epsilon}(x,p) \int dy v_{\epsilon}(y) \exp\left(-\frac{i}{h}yp\right) \right)_{\epsilon \in (0,1]} & . \nonumber
\end{align}
$$

Let's consider formal Maslov-Colombeau T-exponent with a generator
$$ (G_{t,\bar{z},\varepsilon})_{\varepsilon \in (0,1]} = \left( G_{t,\bar{z},\varepsilon} \left( \frac{\partial}{\partial x}, -ih \frac{1}{\partial x}, t, [v_{\varepsilon}] \right) v_{\varepsilon}(x,t) \right)_{\varepsilon \in (0,1]} . \quad (2.1.3.8) $$

Let $\Delta_N = \{0 = t_0 < t_1 < \dots < t_{N+1} = t\}$, $\Delta t_j = t_{j+1} - t_j$, $\delta_N = \max_j \Delta t_j$, $x \in \mathbb{R}^n$. Maslov-Colombeau T-exponent

$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon \in (0,1]} = \left( \left( \prod_{\tau=0}^{t} G_{\tau,d\tau,\varepsilon} \right) [\Psi_{0,\varepsilon}] \right)_{\varepsilon \in (0,1]} \quad (2.1.3.9) $$

is defined via formulae

$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon} = s- \lim_{\delta_N \to 0} (\Psi_{N+1,\varepsilon}(x,t))_{\varepsilon}, $$

$$ (\Psi_{0,\varepsilon}(x,t))_{\varepsilon} = (\Psi_{0,\varepsilon}(x))_{\varepsilon}, $$

$$ (\Psi_{k+1,\varepsilon}(x,t))_{\varepsilon} = (\Psi_{k,\varepsilon}(x,t))_{\varepsilon} \quad \text{iff } 0 \le t \le t_k, \quad (2.1.3.10) $$

$$ (\Psi_{k+1,\varepsilon}(x,t))_{\varepsilon} = \left( G_{t_k, t-t_k, \varepsilon} \left( \frac{\partial}{\partial x}, -ih \frac{1}{\partial x}, t, [\Psi_{k,\varepsilon}] \right) \Psi_{k,\varepsilon}(x,t) \right)_{\varepsilon} \quad \text{iff } t_k \le t \le t_{k+1} $$

$k = 0, 1, \dots, N$

**Theorem 2.1.3.3.** Suppose that T-exponent given via Eq.(2.1.3.10) exist and

cl[(Ψ_ε(x,t))_ε] ∈ G_{W_2^k(ℝ^n)}. Then Colombeau generalized function (Ψ_ε(x,t))_ε has the representation
$$ (G_{t,\bar{z},\varepsilon})_{\varepsilon \in (0,1]} = \left( G_{t,\bar{z},\varepsilon} \left( \frac{\partial}{\partial x}, -i\hbar \frac{1}{\partial x}, t, [v_{\varepsilon}] \right) v_{\varepsilon}(x,t) \right)_{\varepsilon \in (0,1]} . \quad (2.1.3.8) $$

Let $\Delta_N = \{0 = t_0 < t_1 < \dots < t_{N+1} = t\}$, $\Delta t_j = t_{j+1} - t_j$, $\delta_N = \max_j \Delta t_j$, $x \in \mathbb{R}^n$. Maslov-Colombeau T-exponent

$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon \in (0,1]} = \left( \left( \prod_{\tau=0}^{t} G_{\tau,d\tau,\varepsilon} \right) [\Psi_{0,\varepsilon}] \right)_{\varepsilon \in (0,1]} \quad (2.1.3.9) $$

is defined via formulae

$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon} = s- \lim_{\delta_N \to 0} (\Psi_{N+1,\varepsilon}(x,t))_{\varepsilon}, $$

$$ (\Psi_{0,\varepsilon}(x,t))_{\varepsilon} = (\Psi_{0,\varepsilon}(x))_{\varepsilon}, $$

$$ (\Psi_{k+1,\varepsilon}(x,t))_{\varepsilon} = (\Psi_{k,\varepsilon}(x,t))_{\varepsilon} \quad \text{iff } 0 \le t \le t_k, \quad (2.1.3.10) $$

$$ (\Psi_{k+1,\varepsilon}(x,t))_{\varepsilon} = \left( G_{t_k, t-t_k, \varepsilon} \left( \frac{\partial}{\partial x}, -i\hbar \frac{1}{\partial x}, t, [\Psi_{k,\varepsilon}] \right) \Psi_{k,\varepsilon}(x,t) \right)_{\varepsilon} \quad \text{iff } t_k \le t \le t_{k+1} $$
$$ k = 0, 1, \dots, N $$

**Theorem 2.1.3.3.** Suppose that T-exponent given via Eq.(2.1.3.10) exist and

cl[(Ψ_ε(x,t))_ε] ∈ G_{W_2^k(ℝ^n)}. Then Colombeau generalized function (Ψ_ε(x,t))_ε has the representation
$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon} = s- \lim_{\delta_N \to 0} \left( \frac{1}{(2\pi h)^{(N+1)n}} \int \dots \int \exp\left[\frac{i}{h}(x_{k+1} - x_k)\xi_k\right] \times \prod_{k=0}^{N} G_{t_k, \Delta t_k, \varepsilon}(x_{k+1}, \xi_k; [\Psi_{\varepsilon}]) \Psi_{0,\varepsilon}(x_0) d^{N+1}x d^{N+1}\xi \right)_{\varepsilon} \quad (2.1.3.11) $$

We rewrite Eq.(2.1.3.11) in symbolic form

$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon} = \left( \int_{q(\tau)=x} D[q(\tau)] \int_{\tau} D[p(\tau)] \Psi_0(q(0)) \exp\left\{\frac{i}{h} \int_0^t p(\tau) \dot{q}(\tau) d\tau\right\} \times \prod_{\tau=0}^t G_{\tau, d\tau}(q(\tau), p(\tau); [\Psi], \varepsilon) \right)_{\varepsilon} \quad (2.1.3.12) $$

From Theorem 2.1.2.2 and Theorem 2.1.3.3 one obtain directly:

**Theorem 2.1.3.4.** Suppose that $V(x) \in G_{C(\mathbb{R}^n)}$, $a_{\varepsilon}(x,y) \in G_{C(\mathbb{R}^n \times \mathbb{R}^n)}$ and $\Psi_0(x) \in G_{W_2^2(\mathbb{R}^n)}$. Let $\Psi(x,t)$ be the solution of the Cauchy problem (2.1.2.7). Then $(\Psi_{\varepsilon}(x,t))_{\varepsilon}$ can be expressed via formula:

$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon} = \lim_{N \to \infty} \left( (4\pi i\hbar)^{-\frac{(N+1)}{2}} \int \dots \int \Psi_0(x_0) \frac{dx_0 \dots dx_N}{\sqrt{\Delta t_0 \dots \Delta t_N}} \times \exp\left\{ \frac{i}{\hbar} \sum_{k=0}^{N} \left[ \frac{1}{4} \left( \frac{x_{k+1} - x_k}{\Delta t_k} \right)^2 - V_{\varepsilon}(x_{k+1}) - \int a_{\varepsilon}(x_{k,\varepsilon},y) \Psi(y,t_k) dy \right] \Delta t_k \right\} \right)_{\varepsilon} \quad (2.1.3.13) $$

## II.2. Master Equation for the Feynman-Colombeau path integral.
$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon} = s- \lim_{\delta_N \to 0} \left( \frac{1}{(2\pi h)^{(N+1)n}} \int \dots \int \exp\left[\frac{i}{h}(x_{k+1} - x_k)\xi_k\right] \times \prod_{k=0}^{N} G_{t_k, \Delta t_k, \varepsilon}(x_{k+1}, \xi_k; [\Psi_{\varepsilon}]) \Psi_{0,\varepsilon}(x_0) d^{N+1}x d^{N+1}\xi \right)_{\varepsilon} \quad (2.1.3.11) $$

We rewrite Eq.(2.1.3.11) in symbolic form

$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon} = \left( \int_{q(\tau)=x} D[q(\tau)] \int_{\tau} D[p(\tau)] \Psi_0(q(0)) \exp\left\{\frac{i}{h} \int_0^t p(\tau) \dot{q}(\tau) d\tau\right\} \times \prod_{\tau=0}^t G_{\tau, d\tau}(q(\tau), p(\tau); [\Psi], \varepsilon) \right)_{\varepsilon} \quad (2.1.3.12) $$

From Theorem 2.1.2.2 and Theorem 2.1.3.3 one obtain directly:

**Theorem 2.1.3.4.** Suppose that $V(x) \in G_{C(\mathbb{R}^n)}$, $a_{\varepsilon}(x,y) \in G_{C(\mathbb{R}^n \times \mathbb{R}^n)}$ and $\Psi_0(x) \in G_{W_2^2(\mathbb{R}^n)}$. Let $\Psi(x,t)$ be the solution of the Cauchy problem (2.1.2.7). Then $(\Psi_{\varepsilon}(x,t))_{\varepsilon}$ can be expressed via formula:

$$ (\Psi_{\varepsilon}(x,t))_{\varepsilon} = \lim_{N \to \infty} \left( (4\pi i\hbar)^{-\frac{(N+1)}{2}} \int \dots \int \Psi_0(x_0) \frac{dx_0 \dots dx_N}{\sqrt{\Delta t_0 \dots \Delta t_N}} \times \exp\left\{ \frac{i}{\hbar} \sum_{k=0}^{N} \left[ \frac{1}{4} \left( \frac{x_{k+1} - x_k}{\Delta t_k} \right)^2 - V_{\varepsilon}(x_{k+1}) - \int a_{\varepsilon}(x_{k,\varepsilon}, y) \Psi(y, t_k) dy \right] \Delta t_k \right\} \right)_{\varepsilon} \quad (2.1.3.13) $$

## II.2. Master Equation for the Feynman-Colombeau path integral.
II.2.1. Master Equation for the Feynman-Colombeau path integral corresponding with a Schrödinger equation with non-Hermitian Hamiltonian.

The evolution of a closed quantum system during a time interval $T$ is described by the evolution operator $(\mathbf{U}_{\varepsilon}(T))_{\varepsilon}$. The matrix element of the operator $(\mathbf{U}_{\varepsilon}(T))_{\varepsilon}$ between the states with definite positions is called the propagator and may be expressed in the form of the corresponding Feynman-Colombeau path integral (see

Appendix I.3):

$$ (\mathbf{U}_{\varepsilon}(x, T|x_0, 0))_{\varepsilon \in (0,1]} = $$

$$ \left( \int_{q(T)=x} D_C[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times \exp \left[ \frac{i}{\hbar \varepsilon} \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{\mathbf{q}}_{\mathbb{R},\varepsilon} q(t)}{d\tilde{\mathbf{q}}_{\mathbb{R},\varepsilon} t} \right)^2 - V_{\varepsilon}(q(t)) \right] \right) \right] \right)_{\varepsilon \in (0,1]} \quad (2.1.1) $$

$$ (\Psi_{0,\varepsilon}(x))_{\varepsilon} = \delta(x - x_0). $$

If the system with the same Hamiltonian undergoes a continuous measurement and therefore is considered as being open, interacting with a measuring device or environment, its evolution may be described by the set of partial evolution operators

$(\mathbf{U}_{T,\varepsilon}^{\alpha})_{\varepsilon}$ depending on the output of the measurement:
## II.2.1. Master Equation for the Feynman-Colombeau path integral corresponding with a Schrödinger equation with non-Hermitian Hamiltonian.

The evolution of a closed quantum system during a time interval $T$ is described by the evolution operator $(U_{\varepsilon}(T))_{\varepsilon}$. The matrix element of the operator $(U_{\varepsilon}(T))_{\varepsilon}$ between the states with definite positions is called the propagator and may be expressed in the form of the corresponding Feynman-Colombeau path integral (see

Appendix I.3):

$$ (U_{\varepsilon}(x, T|x_0, 0))_{\varepsilon \in (0,1]} = $$

$$ \left( \int_{q(T)=x} D_C[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times \exp \left[ \frac{i}{\hbar \varepsilon} \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{\varepsilon}_{\mathbb{R},\varepsilon} q(t)}{d\tilde{\varepsilon}_{\mathbb{R},\varepsilon} t} \right)^2 - V_{\varepsilon}(q(t)) \right] \right) \right] \right)_{\varepsilon \in (0,1]} \quad (2.1.1) $$

$$ (\Psi_{0,\varepsilon}(x))_{\varepsilon} = \delta(x - x_0). $$

If the system with the same Hamiltonian undergoes a continuous measurement and therefore is considered as being open, interacting with a measuring device or environment, its evolution may be described by the set of partial evolution operators

$(U_{T,\varepsilon}^a)_{\varepsilon}$ depending on the output of the measurement:
$$ (|\Psi_{\varepsilon}(T, \alpha)\rangle)_{\varepsilon} = (\mathbf{U}_{T, \varepsilon}^{\alpha} |\Psi_{0}\rangle)_{\varepsilon} \qquad (2.1.2) $$

The partial propagators are expressed by restricted path integrals. This means [32]-[35] that the path integral for $(\mathbf{U}_{\varepsilon}(T))_{\varepsilon}$ must be given via formula (2.1.1), but restricted according to the information given by the measurement readout $\alpha$. The information given by may be described by a weight functional $(w_{\alpha,\varepsilon}[q])_{\varepsilon}$ (positive, with values between 0 and 1) so that the partial propagator has to be written as a weighted Feynman-Colombeau path integral:

$$ (\mathbf{U}_{\alpha, \varepsilon}(x, T|x_0, 0))_{\varepsilon \in (0,1]} = $$

$$ \left( \int_{q(T)=x} D_{\mathbb{C}}[q(t), \varepsilon] w_{\alpha, \varepsilon}[q] \Psi_{0, \varepsilon}(q(0)) \times \right. $$

$$ \exp \left[ \frac{i}{h_{\varepsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{\mathcal{R}}_{,\varepsilon} q(t)}{d\tilde{\mathcal{R}}_{,\varepsilon} t} \right)^2 - V_{\varepsilon}(q(t)) \right] \right) \right]_{\varepsilon \in (0,1]} , \qquad (2.1.3) $$

$$ V_{\varepsilon}(q(t)) = V(q_{\varepsilon}(t)), $$

$$ q_{\varepsilon}(t) = \frac{q(t)}{1 + \varepsilon^k q^m(t)}, \quad k \ge 1, m \ge 2, $$

$$ (\Psi_{0,\varepsilon}(x))_{\varepsilon} = \delta(x - x_0). $$

If the readout is unknown (nonselective description), the evolution of the measured system may be presented by the generalized density matrix:

$$ (\rho_{T,\varepsilon})_{\varepsilon} = \int d\mu(\alpha) (\rho_{T,\varepsilon}^{\alpha})_{\varepsilon} = \int d\mu(\alpha) [(\mathbf{U}_{\alpha,\varepsilon}(T))_{\varepsilon}] [(\rho_{0,\varepsilon})_{\varepsilon}] [(\rho_{T,\varepsilon}^{\alpha})^{\dagger}]_{\varepsilon} \qquad (2.1.4) $$

and the generalized unitarity condition
$$ (|\Psi_{\varepsilon}(T, \alpha)\rangle)_{\varepsilon} = (\mathbf{U}_{T, \varepsilon}^{\alpha} |\Psi_{0}\rangle)_{\varepsilon} \qquad (2.1.2) $$

The partial propagators are expressed by restricted path integrals. This means [32]-[35] that the path integral for $(\mathbf{U}_{\varepsilon}(T))_{\varepsilon}$ must be given via formula (2.1.1), but restricted according to the information given by the measurement readout $\alpha$. The information given by may be described by a weight functional $(w_{\alpha,\varepsilon}[q])_{\varepsilon}$ (positive, with values between 0 and 1) so that the partial propagator has to be written as a weighted Feynman-Colombeau path integral:

$$ (\mathbf{U}_{\alpha, \varepsilon}(x, T|x_0, 0))_{\varepsilon \in (0,1]} = $$

$$ \left( \int_{q(T)=x} D_{\mathbb{C}}[q(t), \varepsilon] w_{\alpha, \varepsilon}[q] \Psi_{0, \varepsilon}(q(0)) \times \right. $$

$$ \exp \left[ \frac{i}{h_{\varepsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{\mathbb{R}}_{,\varepsilon} q(t)}{d\tilde{\mathbb{R}}_{,\varepsilon} t} \right)^2 - V_{\varepsilon}(q(t)) \right] \right) \right]_{\varepsilon \in (0,1]} , \qquad (2.1.3) $$

$$ V_{\varepsilon}(q(t)) = V(q_{\varepsilon}(t)), $$

$$ q_{\varepsilon}(t) = \frac{q(t)}{1 + \varepsilon^k q^m(t)}, \quad k \ge 1, m \ge 2, $$

$$ (\Psi_{0,\varepsilon}(x))_{\varepsilon} = \delta(x - x_0). $$

If the readout is unknown (nonselective description), the evolution of the measured system may be presented by the generalized density matrix:

$$ (\rho_{T,\varepsilon})_{\varepsilon} = \int d\mu(\alpha) (\rho_{T,\varepsilon}^{\alpha})_{\varepsilon} = \int d\mu(\alpha) [(\mathbf{U}_{\alpha,\varepsilon}(T))_{\varepsilon}] [(\rho_{0,\varepsilon})_{\varepsilon}] \left[ \left( (\rho_{T,\varepsilon}^{\alpha})^{\dagger} \right)_{\varepsilon} \right] \qquad (2.1.4) $$

and the generalized unitarity condition
$$ \int d\mu(\alpha)[((\mathbf{U}_{\alpha,\varepsilon}(T))^{\dagger})_{\varepsilon}][(\mathbf{U}_{\alpha,\varepsilon}(T))_{\varepsilon}] = 1 \qquad (2.1.5) $$

provides conservation of probabilities.

Let us consider the special case, when the weight functional $(w_{\alpha,\beta,\varepsilon}[q(t),q(0),q(T),\lambda])_{\varepsilon} = (w_{\alpha,\beta,\varepsilon}[q(t),q_0,q_T,\lambda])_{\varepsilon}$ describing an measurement may be taken of the symple Gaussian form such that:

$$ (w_{\alpha,\beta,\varepsilon}[q])_{\varepsilon} = $$

$$ \left( \exp \left[ -v \int_0^T [q(t) - \alpha(t)^2] dt - \frac{v}{\hbar_{\varepsilon}} \int_0^T [q(t) - \beta(t, q_0, q_T, \lambda)]^2 dt - \frac{1}{\hbar_{\varepsilon}} \gamma(q_0, q_T, \lambda) \right] \right)_{\varepsilon} = $$

$$ \left( \exp \left[ -\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} \left\{ \hbar_{\varepsilon} v [q(t) - \alpha(t)]^2 + v [q(t) - \beta(t, q_0, q_T, \lambda)]^2 \right\} dt - \right. \right. \qquad (2.1.6) $$

$$ \left. \left. \frac{1}{\hbar_{\varepsilon}} \gamma(q_0, q_T, \lambda) \right] \right)_{\varepsilon} = $$

$$ \left( \exp \left[ -\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} R_{\alpha,\beta,\varepsilon}(q(t),v,v,t,q_{0},q_{T},\lambda) dt \right] \right)_{\varepsilon}, $$

$$ \gamma(q_0, q_T, \lambda) = v[q_0 - c - \lambda]^2 + v[q_T - c - \lambda]^2, $$

$$ v, v \in \mathbb{R}_{+}, \lambda, c \in \mathbb{R}. $$

The corresponding Feynman-Colombeau path integral is
$$ \int d\mu(\alpha)[((\mathbf{U}_{\alpha,\varepsilon}(T))^{\dagger})_{\varepsilon}][(\mathbf{U}_{\alpha,\varepsilon}(T))_{\varepsilon}] = 1 \qquad (2.1.5) $$

provides conservation of probabilities.

Let us consider the special case, when the weight functional $(w_{\alpha,\beta,\varepsilon}[q(t),q(0),q(T),\lambda])_{\varepsilon} = (w_{\alpha,\beta,\varepsilon}[q(t),q_0,q_T,\lambda])_{\varepsilon}$ describing an measurement may be taken of the symple Gaussian form such that:

$$ (w_{\alpha,\beta,\varepsilon}[q])_{\varepsilon} = $$

$$ \left( \exp \left[ -v \int_0^T [q(t) - \alpha(t)^2] dt - \frac{v}{\hbar_{\varepsilon}} \int_0^T [q(t) - \beta(t, q_0, q_T, \lambda)]^2 dt - \frac{1}{\hbar_{\varepsilon}} \gamma(q_0, q_T, \lambda) \right] \right)_{\varepsilon} = $$

$$ \left( \exp \left[ -\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} \left\{ \hbar_{\varepsilon} v [q(t) - \alpha(t)]^2 + v [q(t) - \beta(t, q_0, q_T, \lambda)]^2 \right\} dt - \right. \right. \qquad (2.1.6) $$

$$ \left. \left. \frac{1}{\hbar_{\varepsilon}} \gamma(q_0, q_T, \lambda) \right] \right)_{\varepsilon} = $$

$$ \left( \exp \left[ -\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} R_{\alpha,\beta,\varepsilon}(q(t),v,v,t,q_{0},q_{T},\lambda) dt \right] \right)_{\varepsilon}, $$

$$ \gamma(q_0, q_T, \lambda) = v[q_0 - c - \lambda]^2 + v[q_T - c - \lambda]^2, $$

$$ v, v \in \mathbb{R}_{+}, \lambda, c \in \mathbb{R}. $$

The corresponding Feynman-Colombeau path integral is
$$ (\mathbf{U}_{\alpha,\varepsilon}(x, T|x_0, 0))_{\varepsilon \in (0,1]} = $$

$$ \left( \int_{q(T)=x} D_{\mathbb{C}}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times \exp \left[ \frac{i}{\hbar_{\varepsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - V(q_{\varepsilon}(t)) \right] - \frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} R_{\alpha,\beta,\varepsilon}(q(t),v,v,t,q_{0},q_{T},\lambda) dt \right) \right]_{\varepsilon} = \right. \tag{2.1.7} $$

$$ \left( \int_{q(T)=x} D_{\mathbb{C}}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times \exp \left[ \frac{i}{\hbar_{\varepsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - V(q_{\varepsilon}(t)) + iR_{\alpha,\beta,\varepsilon}(q(t),v,v,t,q_{0},q_{T},\lambda) \right] \right) \right] \right)_{\varepsilon} $$

$$ q_{\varepsilon}(t) = \frac{q(t)}{1 + \varepsilon^k q^m(t)}, k \ge 1, m \ge 2, $$

$$ (\Psi_{0,\varepsilon}^{2}(x-x_{0}))_{\varepsilon} = \delta(x-x_{0}). $$

Path integral (2.1.7) has the form of a canonical non-weighted path integral but with a non-Hermitian Hamiltonian

$$ (\mathbf{H}_{\alpha,\beta,\varepsilon}(p,q,t))_{\varepsilon} = (\mathbf{H}_{\varepsilon}(p,q,t))_{\varepsilon} - \frac{1}{2i\hbar_{\varepsilon}m_{2}} p^{2} - iR_{\alpha,\beta,\varepsilon}(q,v,v,t,q_{0},q_{T},\lambda) \quad (2.1.8) $$

instead of the original Hamiltonian $(\mathbf{H}_{\varepsilon})_{\varepsilon}$. Therefore, instead of calculating a restricted path integral one may solve the Colombeau-Schrodinger equation with a
$$ (\mathbf{U}_{\alpha,\varepsilon}(x, T|x_0, 0))_{\varepsilon \in (0,1]} = $$

$$ \left( \int_{q(T)=x} D\mathbb{C}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times \exp \left[ \frac{i}{\hbar_{\varepsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - V(q_{\varepsilon}(t)) \right] - \frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} R_{\alpha,\beta,\varepsilon}(q(t),v,v,t,q_{0},q_{T},\lambda) dt \right) \right]_{\varepsilon} = \right. \tag{2.1.7} $$

$$ \left( \int_{q(T)=x} D\mathbb{C}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times \exp \left[ \frac{i}{\hbar_{\varepsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - V(q_{\varepsilon}(t)) + iR_{\alpha,\beta,\varepsilon}(q(t),v,v,t,q_{0},q_{T},\lambda) \right] \right) \right] \right)_{\varepsilon} $$

$$ q_{\varepsilon}(t) = \frac{q(t)}{1 + \varepsilon^k q^m(t)}, k \ge 1, m \ge 2, $$

$$ (\Psi_{0,\varepsilon}^{2}(x-x_{0}))_{\varepsilon} = \delta(x-x_{0}). $$

Path integral (2.1.7) has the form of a canonical non-weighted path integral but with a non-Hermitian Hamiltonian

$$ (\mathbf{H}_{\alpha,\beta,\varepsilon}(p,q,t))_{\varepsilon} = (\mathbf{H}_{\varepsilon}(p,q,t))_{\varepsilon} - \frac{1}{2i\hbar_{\varepsilon}m_{2}} p^{2} - iR_{\alpha,\beta,\varepsilon}(q,v,v,t,q_{0},q_{T},\lambda) \quad (2.1.8) $$

instead of the original Hamiltonian $(\mathbf{H}_{\varepsilon})_{\varepsilon}$. Therefore, instead of calculating a restricted path integral one may solve the Colombeau-Schrodinger equation with a
non- Hermitian Hamiltonian:

$$ \frac{\partial}{\partial t} (\Psi_{\alpha, \beta, \varepsilon}(t))_{\varepsilon} = \left( -\frac{i}{(\hbar_{\varepsilon})_{\varepsilon}} (\mathbf{H}_{\varepsilon})_{\varepsilon} - R_{\alpha, \beta, \varepsilon}(q, v, \nu, t, q_0, q_T, \lambda, c) \right) (\Psi_{\alpha, \beta, \varepsilon}(t))_{\varepsilon}, \qquad (2.1.9) $$

$$ (\Psi_{\alpha, \varepsilon}(0))_{\varepsilon} = (\Psi_{\varepsilon}(x - x_0))_{\varepsilon}. $$

Let us consider now quantum average $\langle (T, x_0, \alpha, \beta; \hbar_{\varepsilon}, \varepsilon) \rangle_{\varepsilon}$:

$$ \begin{aligned} \langle (T, x_0, \alpha; \hbar, \varepsilon) \rangle_{\varepsilon} = & \int dx \left( (\mathbf{U}_{\alpha, \beta, \varepsilon}(x, T|x_0, 0) \mathbf{U}_{\alpha, \varepsilon}^{\dagger}(x, T|x_0, 0)) \right)_{\varepsilon} = \\ & Z^{-1} \int dx \left( |\mathbf{U}_{\alpha, \beta, \varepsilon}(x, T|x_0, 0)|^2 \right)_{\varepsilon} \end{aligned} \qquad (2.1.10) $$

for the case: (i) $\nu = 0$. Thus we set $\nu = 0$ in Eq.(2.1.7). In the sequel we denote for short

$$ R_{\alpha, \beta, \varepsilon}(q, 0, \nu, t, q_0, q_T, \lambda) \triangleq Q_{\beta, \varepsilon}(q, \nu, t, q_0, q_T, \lambda) \triangleq Q_{\beta, \varepsilon}(q(t), \nu) \qquad (2.1.11) $$

From Eq.(2.1.7) and Eq.(2.1.11) $\forall \varepsilon \in (0, 1]$ we obtain
non- Hermitian Hamiltonian:

$$ \frac{\partial}{\partial t} (\Psi_{\alpha, \beta, \varepsilon}(t))_{\varepsilon} = \left( -\frac{i}{(\hbar_{\varepsilon})_{\varepsilon}} (\mathbf{H}_{\varepsilon})_{\varepsilon} - R_{\alpha, \beta, \varepsilon}(q, v, \nu, t, q_0, q_T, \lambda, c) \right) (\Psi_{\alpha, \beta, \varepsilon}(t))_{\varepsilon}, \qquad (2.1.9) $$

$$ (\Psi_{\alpha, \varepsilon}(0))_{\varepsilon} = (\Psi_{\varepsilon}(x - x_0))_{\varepsilon}. $$

Let us consider now quantum average $\langle (T, x_0, \alpha, \beta; \hbar_{\varepsilon}, \varepsilon) \rangle_{\varepsilon}$:

$$ \langle (T, x_0, \alpha; \hbar, \varepsilon) \rangle_{\varepsilon} = \int dx x ((U_{\alpha, \beta, \varepsilon}(x, T|x_0, 0) U_{\alpha, \varepsilon}^{\dagger}(x, T|x_0, 0)))_{\varepsilon} = \qquad (2.1.10) $$

$$ Z^{-1} \int dx x (|U_{\alpha, \beta, \varepsilon}(x, T|x_0, 0)|^2)_{\varepsilon} $$

for the case: (i) $\nu = 0$. Thus we set $\nu = 0$ in Eq.(2.1.7). In the sequel we denote for short

$$ R_{\alpha, \beta, \varepsilon}(q, 0, \nu, t, q_0, q_T, \lambda) \triangleq Q_{\beta, \varepsilon}(q, \nu, t, q_0, q_T, \lambda) \triangleq Q_{\beta, \varepsilon}(q(t), \nu) \qquad (2.1.11) $$

From Eq.(2.1.7) and Eq.(2.1.11) $\forall \varepsilon \in (0, 1]$ we obtain
$$ \langle T, x_0, \alpha, \beta; \hbar_{\varepsilon}, \varepsilon \rangle = $$

$$
\begin{align*}
& \int dx \left| \int_{q(T)=x} |D_C[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times \right. \\
& \qquad \left. \exp\left[ \frac{i}{\hbar_{\varepsilon}} \left( \int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - V(q_{\varepsilon}(t)) \right] \right) \right] \right|^2 = \\
& \int dx \left\{ \left[ \int_{q(T)=x} D^+[q(t), \varepsilon] \Psi_0(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_0^T Q_{\beta,\varepsilon}(q(t),v)dt \right) \times \right. \right. \\
& \qquad \left. \left. \cos\left[ \frac{1}{\hbar_{\varepsilon}} \left( \int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - V(q_{\varepsilon}(t)) \right] \right) \right] \right]^2 + \right. \\
& \qquad \left. \left[ \int_{q(T)=x} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_0^T Q_{\beta,\varepsilon}(q,v)dt \right) \times \right. \right. \\
& \qquad \left. \left. \sin\left[ \frac{1}{\hbar_{\varepsilon}} \left( \int_0^T dt \left[ \frac{m_1}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - V(q_{\varepsilon}(t)) \right] \right) \right] \right]^2 \right\} = \\
& \qquad \qquad \eta_{\varepsilon}^{(1)}(T, \hbar_{\varepsilon}, \alpha, \beta) + \eta_{\varepsilon}^{(2)}(T, \hbar_{\varepsilon}, \alpha, \beta).
\end{align*}
$$

Here
$$ \langle T, x_0, \alpha, \beta; \hbar_{\varepsilon}, \varepsilon \rangle = $$

$$
\begin{align*}
& \int dx \left| \int_{q(T)=x} |D_C[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times \right. \\
& \qquad \left. \exp\left[ \frac{i}{\hbar_{\varepsilon}} \left( \int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - V(q_{\varepsilon}(t)) \right] \right) \right] \right|^2 = \\
& \int dx \left\{ \left[ \int_{q(T)=x} D^+[q(t), \varepsilon] \Psi_0(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_0^T Q_{\beta,\varepsilon}(q(t),v)dt \right) \times \right. \right. \\
& \qquad \left. \left. \cos\left[ \frac{1}{\hbar_{\varepsilon}} \left( \int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - V(q_{\varepsilon}(t)) \right] \right) \right] \right]^2 + \right. \\
& \qquad \left. \left[ \int_{q(T)=x} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_0^T Q_{\beta,\varepsilon}(q,v)dt \right) \times \right. \right. \\
& \qquad \left. \left. \sin\left[ \frac{1}{\hbar_{\varepsilon}} \left( \int_0^T dt \left[ \frac{m_1}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - V(q_{\varepsilon}(t)) \right] \right) \right] \right]^2 \right\} = \\
& \qquad \qquad \eta_{\varepsilon}^{(1)}(T, \hbar_{\varepsilon}, \alpha, \beta) + \eta_{\varepsilon}^{(2)}(T, \hbar_{\varepsilon}, \alpha, \beta).
\end{align*}
$$

Here
$$
\eta_{\epsilon}^{(1)}(T, \hbar_{\epsilon}, \alpha, \beta) =
$$

$$
\begin{align}
& \int dx x \left[ \int_{q(T)=x} D^+[q(t),\epsilon] \Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}} \int_{0}^{T} Q_{\beta,\epsilon}(q,v) dt \right) \times \right. \nonumber \\
& \left. \cos\left[ \frac{1}{\hbar_{\epsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - V(q_{\epsilon}(t)) \right] \right) \right] \right]^2 = \nonumber \\
& \qquad \int dx x [\hat{\eta}_{\epsilon}^{(1)}(T,x,\hbar_{\epsilon},\alpha,\beta)]^2, \tag{2.1.12}
\end{align}
$$

$$
\eta_{\epsilon}^{(2)}(T, \hbar_{\epsilon}, \alpha, \beta) =
$$

$$
\begin{align}
& \int dx \left[ \int_{q(T)=x} D^+[q(t),\epsilon] \Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}} \int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \right. \nonumber \\
& \left. \sin\left[\frac{1}{\hbar_{\epsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - V(q_{\epsilon}(t))\right]\right)\right]^2 = \right. \nonumber \\
& \qquad \left. \int dx [\hat{\eta}_{\epsilon}^{(2)}(T,x,\hbar_{\epsilon},\alpha,\beta)]^2. \right]
\end{align}
$$

Here
$$
\eta_{\varepsilon}^{(1)}(T, \hbar_{\varepsilon}, \alpha, \beta) =
$$

$$
\begin{equation}
\begin{split}
& \int dx \left[ \int_{q(T)=x} D^+[q(t),\varepsilon] \Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_\varepsilon} \int_0^T Q_{\beta,\varepsilon}(q,v)dt\right) \times \right. \\
& \qquad \left. \cos\left[\frac{1}{\hbar_\varepsilon}\left(\int_0^T dt \left[\frac{m}{2}\dot{q}^2_{\tilde{\mathbb{R}},\varepsilon}(t) - V(q_\varepsilon(t))\right]\right)\right]^2 \right] \\
& = \int dx \left[\hat{\eta}_\varepsilon^{(1)}(T,x,\hbar_\varepsilon,\alpha,\beta)\right]^2,
\end{split}
\tag{2.1.12}
\end{equation}
$$

$$
\eta_{\epsilon}^{(2)}(T, \hbar_{\epsilon}, \alpha, \beta) =
$$

$$
\begin{equation}
\begin{split}
& \int dx \left[ \int_{q(T)=x} D^+[q(t),\varepsilon] \Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_\varepsilon} \int_0^T Q_{\beta,\varepsilon}(q,v)dt\right) \times \right. \\
& \qquad \left. \sin\left[\frac{1}{\hbar_\varepsilon}\left(\int_0^T dt \left[\frac{m}{2}\dot{q}^2_{\mathbb{R},\varepsilon}(t) - V(q_\varepsilon(t))\right]\right)\right]^2 \right] \\
& = \\
& \qquad \int dx \left[\hat{\eta}_\varepsilon^{(2)}(T,x,\hbar_\varepsilon,\alpha,\beta)\right]^2.
\end{split}
\tag{2.1.12}
\end{equation}
$$

Here
$$
\hat{\eta}_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon}, \alpha, \beta) =
$$

$$
\int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \quad (2.1.13)
$$

$$
\cos\left[ \frac{1}{\hbar_{\varepsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) \right)^{2} - V(q_{\varepsilon}(t)) \right] \right) \right]
$$

and

$$
\hat{\eta}_{\epsilon}^{(2)}(T, x, x_0, \hbar_{\epsilon}, \alpha, \beta) =
$$

$$
\int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \quad (2.1.14)
$$

$$
\sin\left[ \frac{1}{\hbar_{\epsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^{2}(t) - V(q_{\epsilon}(t)) \right] \right) \right].
$$

Let us now to evaluate the quantities $\hat{\eta}_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon}, \alpha)$ and $\hat{\eta}_{\epsilon}^{(2)}(T, x, x_0, \hbar_{\epsilon}, \alpha)$.

We assume now that: $V(q_{\epsilon}(t)) = V_0(q_{\epsilon}(t)) + V_1(q_{\epsilon}(t))$. Here
$$
\hat{\eta}_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon}, \alpha, \beta) =
$$

$$
\int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \quad (2.1.13)
$$

$$
\cos\left[ \frac{1}{\hbar_{\varepsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) \right)^{2} - V(q_{\varepsilon}(t)) \right] \right) \right]
$$

and

$$
\hat{\eta}_{\epsilon}^{(2)}(T, x, x_0, \hbar_{\epsilon}, \alpha, \beta) =
$$

$$
\int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \quad (2.1.14)
$$

$$
\sin\left[ \frac{1}{\hbar_{\epsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^{2}(t) - V(q_{\epsilon}(t)) \right] \right) \right].
$$

Let us now to evaluate the quantities $\hat{\eta}_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon}, \alpha)$ and $\hat{\eta}_{\epsilon}^{(2)}(T, x, x_0, \hbar_{\epsilon}, \alpha)$.

We assume now that: $V(q_{\epsilon}(t)) = V_0(q_{\epsilon}(t)) + V_1(q_{\epsilon}(t))$. Here
$$V_0(q_e(t),t) = aq_e^2(t) + b(t)(q_e(t)),$$

$$V_1(q_e(t)) = a_3q_e^3(t) + \dots + a_rq_e^r(t),$$

$$a > 0, \tag{2.1.15}$$

$$q_e(t) = \frac{q(t)}{1 + \varepsilon^k q^m(t)},$$

$$k \ge 1, m \ge 2$$

Substitution Eq.(2.1.15) into Eq.(2.1.13) gives
$$V_0(q_e(t),t) = aq_e^2(t) + b(t)(q_e(t)),$$

$$V_1(q_e(t)) = a_3q_e^3(t) + \dots + a_rq_e^r(t),$$

$$a > 0, \tag{2.1.15}$$

$$q_e(t) = \frac{q(t)}{1 + \varepsilon^k q^m(t)},$$

$$k \ge 1, m \ge 2$$

Substitution Eq.(2.1.15) into Eq.(2.1.13) gives
$$
\hat{\eta}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}, \alpha, \beta) =
$$

$$
\begin{align*}
& \int_{q(T)=x} D^+[q(t),\varepsilon]\Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T} Q_{\beta,\varepsilon}(q,v)dt\right) \times \\
& \qquad \cos\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt\left[\frac{m}{2}\dot{q}_{\mathbb{R},\varepsilon}^2(t) - V(q_{\varepsilon}(t))\right]\right)\right] = \\
& \int_{q(T)=x} D^+[q(t),\varepsilon]\Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T} Q_{\beta,\varepsilon}(q,v)dt\right) \times \\
& \qquad \cos\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt\left[\frac{m}{2}\dot{q}_{\mathbb{R},\varepsilon}^2(t) - V_0(q_{\varepsilon}(t)) - V_1(q_{\varepsilon}(t))\right]\right)\right] = \tag{2.1.16}
\end{align*}
$$

$$
\cos\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt\left[\frac{m}{2}\dot{q}_{\mathbb{R},\varepsilon}^2(t) - V_0(q_{\varepsilon}(t))\right]\right)\right] \cos\left(\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T} V_1(q_{\varepsilon}(t))dt\right) +
$$

$$
\int_{q(T)=x} D^+[q(t),\varepsilon]\Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T} Q_{\beta,\varepsilon}(q,v)dt\right) \times
$$

$$
\sin\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt\left[\frac{m}{2}\dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - V_{0}(q_{\varepsilon}(t))\right]\right)\right] \sin\left(\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T} V_{1}(q_{\varepsilon}(t))dt\right) =
$$

$$
\chi_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) + \chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}).
$$
$$
\hat{\eta}_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon}, \alpha, \beta) =
$$

$$
\begin{align*}
& \int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \\
& \qquad \cos\left[\frac{1}{\hbar_{\epsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2}\dot{q}_{\mathbb{R},\epsilon}^2(t) - V(q_{\epsilon}(t))\right]\right)\right] = \\
& \int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \\
& \qquad \cos\left[\frac{1}{\hbar_{\epsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2}\dot{q}_{\mathbb{R},\epsilon}^2(t) - V_0(q_{\epsilon}(t)) - V_1(q_{\epsilon}(t))\right]\right)\right] = \tag{2.1.16}
\end{align*}
$$

$$
\begin{align*}
& \int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \\
& \qquad \cos\left[\frac{1}{\hbar_{\epsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2}\dot{q}_{\mathbb{R},\epsilon}^2(t) - V_0(q_{\epsilon}(t))\right]\right)\right] \cos\left(\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} V_1(q_{\epsilon}(t))dt\right) + \\
& \int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \\
& \qquad \sin\left[\frac{1}{\hbar_{\epsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2}\dot{q}_{\mathbb{R},\epsilon}^2(t) - V_0(q_{\epsilon}(t))\right]\right)\right] \sin\left(\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} V_1(q_{\epsilon}(t))dt\right) = \\
& \qquad \chi_{\epsilon}^{(1)}(T,x,x_0,\hbar_{\epsilon}) + \chi_{\epsilon}^{(2)}(T,x,x_0,\hbar_{\epsilon}).
\end{align*}
$$
Here

$$
\chi_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon}) =
$$

$$
\int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \quad (2.1.17)
$$

$$
\cos\left[ \frac{1}{\hbar_{\epsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^{2}(t) - V_{0}(q_{\epsilon}(t)) \right] \right) \right] \cos\left( \frac{1}{\hbar_{\epsilon}} \int_{0}^{T} V_{1}(q_{\epsilon}(t)) dt \right)
$$

and

$$
\chi_{\epsilon}^{(2)}(T, x, x_0, \hbar_{\epsilon}) =
$$

$$
\int_{q(T)=x} D^+[q(t),\epsilon]\Psi_{0,\epsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}}\int_{0}^{T} Q_{\beta,\epsilon}(q,v)dt\right) \times \quad (2.1.18)
$$

$$
\sin\left[ \frac{1}{\hbar_{\epsilon}} \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^{2}(t) - V_{0}(q_{\epsilon}(t)) \right] \right) \right] \sin\left( \frac{1}{\hbar_{\epsilon}} \int_{0}^{T} V_{1}(q_{\epsilon}(t)) dt \right)
$$

Let us now evaluate the quantities $\chi_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon})$ and $\chi_{\epsilon}^{(2)}(T, x, x_0, \hbar_{\epsilon})$.

Using replacement $q(t) \rightarrow \sqrt{\hbar_{\epsilon}} q(t)$, from Eq.(2.1.17) and Eq.(2.1.18) we obtain:
Here

$$
\begin{gather*}
\chi_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) = \\
\int_{q(T)=x} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(q,v)dt\right) \times \tag{2.1.17} \\
\cos\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - V_{0}(q_{\varepsilon}(t))\right]\right)\right] \cos\left(\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} V_{1}(q_{\varepsilon}(t))dt\right) \nonumber
\end{gather*}
$$

$$
\cos\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - V_{0}(q_{\varepsilon}(t))\right]\right)\right] \cos\left(\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} V_{1}(q_{\varepsilon}(t))dt\right)
$$

and

$$
\begin{gather*}
\chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = \\
\int_{q(T)=x} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(q,v)dt\right) \times \tag{2.1.18} \\
\sin\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - V_{0}(q_{\varepsilon}(t))\right]\right)\right] \sin\left(\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} V_{1}(q_{\varepsilon}(t))dt\right) \nonumber
\end{gather*}
$$

$$
\sin\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - V_{0}(q_{\varepsilon}(t))\right]\right)\right] \sin\left(\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} V_{1}(q_{\varepsilon}(t))dt\right)
$$

Let us now evaluate the quantities $\chi_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon})$ and $\chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon})$.

Using replacement $q(t) \rightarrow \sqrt{\hbar_{\varepsilon}} q(t)$, from Eq.(2.1.17) and Eq.(2.1.18) we obtain:
$$
\chi_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) =
$$

$$
\begin{equation}
\begin{split}
& \int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} D^+[q(t),\varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar \varepsilon} q(0)) \\
& \qquad \exp\left(-\frac{1}{\hbar \varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar \varepsilon} q, v) dt\right) \times \\
& \cos\left[ \frac{1}{\hbar \varepsilon} \left( \int_0^T dt \left[ \frac{m}{2} \hbar \varepsilon \dot{q}_{\sqrt{\hbar \varepsilon},\varepsilon}^2(t) - V_0(\sqrt{\hbar \varepsilon} q_{\varepsilon}(t)) \right] \right) \right] \times \\
& \cos\left[ \frac{1}{\hbar \varepsilon} \left( \int_0^T V_1(\sqrt{\hbar \varepsilon} q_{\varepsilon}(t)) dt \right) \right] =
\end{split}
\tag{2.1.19}
\end{equation}
$$

$$
\begin{equation}
\begin{split}
& \int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} D^+[q(t),\varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar \varepsilon} q(0)) \exp\left(-\frac{1}{\hbar \varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar \varepsilon} q, v) dt\right) \times \\
& \cos\left(\int_0^T dt \left[ \frac{m}{2} \dot{q}_{\sqrt{\hbar \varepsilon},\varepsilon}^2(t) - a q_{\varepsilon,\hbar\varepsilon}^2(t) - b(t) \frac{q_{\varepsilon,\hbar\varepsilon}(t)}{\sqrt{\hbar\varepsilon}} \right] \right) \times \\
& \cos\left[ \left( \int_0^T V_{1,\hbar\varepsilon}(q_{\varepsilon}(t)) dt \right) \right]
\end{split}
\tag{2.1.19}
\end{equation}
$$

and
$$
\chi_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon}) =
$$

$$
\begin{equation}
\begin{split}
& \int_{q(T)=\frac{x}{\sqrt{\hbar \epsilon}}} D^+[q(t),\epsilon] \Psi_{0,\epsilon}(\sqrt{\hbar \epsilon} q(0)) \exp\left(-\frac{1}{\hbar \epsilon} \int_0^T Q_{\beta,\epsilon}(\sqrt{\hbar \epsilon} q, v) dt\right) \times \\
& \cos\left[ \frac{1}{\hbar \epsilon} \left( \int_0^T dt \left[ \frac{m}{2} \hbar \epsilon \dot{q}_{\mathbb{R},\epsilon}^2(t) - V_0(\sqrt{\hbar \epsilon} q_{\epsilon}(t)) \right] \right) \right] \times \\
& \cos\left[ \frac{1}{\hbar \epsilon} \left( \int_0^T V_1(\sqrt{\hbar \epsilon} q_{\epsilon}(t)) dt \right) \right] =
\end{split}
\tag{2.1.19}
\end{equation}
$$

$$
\begin{equation}
\begin{split}
& \int_{q(T)=\frac{x}{\sqrt{\hbar \epsilon}}} D^+[q(t),\epsilon] \Psi_{0,\epsilon}(\sqrt{\hbar \epsilon} q(0)) \exp\left(-\frac{1}{\hbar \epsilon} \int_0^T Q_{\beta,\epsilon}(\sqrt{\hbar \epsilon} q, v) dt\right) \times \\
& \cos\left(\int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - a q_{\epsilon,\hbar\epsilon}^2(t) - b(t) \frac{q_{\epsilon,\hbar\epsilon}(t)}{\sqrt{\hbar\epsilon}} \right] \right) \times \\
& \cos\left[ \left( \int_0^T V_{1,\hbar\epsilon}(q_{\epsilon}(t)) dt \right) \right]
\end{split}
\tag{2.1.20}
\end{equation}
$$

and
$$ \chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \begin{aligned} & \int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar \varepsilon} q(0)) \exp\left(-\frac{1}{\hbar \varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar \varepsilon} q, v) dt\right) \times \\ & \sin\left[\frac{1}{\hbar \varepsilon} \left(\int_0^T dt \left[\frac{m}{2} \hbar \varepsilon \dot{q}_{\varepsilon,\varepsilon}^2(t) - V_0(\sqrt{\hbar \varepsilon} q_{\varepsilon}(t))\right]\right)\right] \times \\ & \sin\left[\frac{1}{\hbar \varepsilon} \left(\int_0^T V_1(\sqrt{\hbar \varepsilon} q_{\varepsilon}(t)) dt\right)\right] = \end{aligned} \quad (2.1.20) $$

$$ \begin{aligned} & \int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar \varepsilon} q(0)) \exp\left(-\frac{1}{\hbar \varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar \varepsilon} q, v) dt\right) \times \\ & \sin\left(\int_0^T dt \left[\frac{m}{2} \dot{q}_{\varepsilon,\varepsilon}^2(t) - a q_{\varepsilon,\varepsilon}^2(t) - b(t) \frac{q_{\varepsilon,\varepsilon}(t)}{\sqrt{\hbar \varepsilon}}\right]\right) \times \\ & \sin\left[\left(\int_0^T V_{1,\varepsilon}(q_{\varepsilon}(t)) dt\right)\right] \end{aligned} $$

Here

$$ \begin{gathered} V_{1,\hbar\varepsilon}(q_{\varepsilon}(t)) = \frac{1}{\hbar\varepsilon} V_1(\sqrt{\hbar\varepsilon} q_{\varepsilon,\varepsilon}(t)) = a_3 \hbar\varepsilon^{1/2} q_{\varepsilon,\hbar\varepsilon}^3(t) + \dots + a_k \hbar\varepsilon^{(r/2)-1} q_{\varepsilon,\hbar\varepsilon}^r(t) \quad (2.1.21) \\ q_{\varepsilon,\hbar\varepsilon}(t) = \frac{q(t)}{1 + \varepsilon^k \hbar\varepsilon^m q^m(t)}. \end{gathered} $$

Let us rewrite the equations Eq.(2.1.19) and Eq.(2.1.20) in the following equivalent
$$ \chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \begin{aligned} & \int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} \int_{0}^{T} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt\right) \times \\ & \sin\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} dt\left[\frac{m}{2} \hbar_{\varepsilon} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t)-V_{0}(\sqrt{\hbar_{\varepsilon}} q_{\varepsilon}(t))\right]\right)\right] \times \\ & \sin\left[\frac{1}{\hbar_{\varepsilon}}\left(\int_{0}^{T} V_{1}(\sqrt{\hbar_{\varepsilon}} q_{\varepsilon}(t)) dt\right)\right] = \end{aligned} \quad (2.1.20) $$

$$ \begin{aligned} & \int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} \int_{0}^{T} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt\right) \times \\ & \sin\left(\int_{0}^{T} dt\left[\frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t)-a q_{\varepsilon, \hbar_{\varepsilon}}^{2}(t)-b(t) \frac{q_{\varepsilon, \hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}}\right]\right) \times \\ & \sin\left[\left(\int_{0}^{T} V_{1, \hbar_{\varepsilon}}(q_{\varepsilon}(t)) dt\right)\right] \end{aligned} $$

Here

$$ V_{1,\hbar_{\varepsilon}}(q_{\varepsilon}(t)) = \frac{1}{\hbar_{\varepsilon}} V_{1}(\sqrt{\hbar_{\varepsilon}} q_{\varepsilon,\varepsilon}(t)) = a_{3}\hbar_{\varepsilon}^{1/2}q_{\varepsilon,\hbar_{\varepsilon}}^{3}(t) + \dots + a_{k}\hbar_{\varepsilon}^{(r/2)-1}q_{\varepsilon,\hbar_{\varepsilon}}^{r}(t) \quad (2.1.21) $$

$$ q_{\varepsilon, \hbar_{\varepsilon}}(t) = \frac{q(t)}{1 + \varepsilon^k \hbar_{\varepsilon}^m q^m(t)}. $$

Let us rewrite the equations Eq.(2.1.19) and Eq.(2.1.20) in the following equivalent
form

$$
\chi_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) = \\
\int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt\right) \times
$$

$$
[\mathbf{sign}^{+} \mathfrak{R}_1(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-} \mathfrak{R}_1(q_{\varepsilon,h_{\varepsilon}}(t))] \times
$$

$$
\left| \cos\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times \quad (2.1.22)
$$

$$
\cos\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t)) dt\right)\right],
$$

$$
\mathfrak{R}_1(q_{\varepsilon,h_{\varepsilon}}(t)) = \cos\left(\int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - a q_{\varepsilon,h_{\varepsilon}}^2(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right)
$$

and
form

$$
\chi_{\epsilon}^{(1)}(T, x, x_0, \hbar_{\epsilon}) = \\
\int_{q(T)=\frac{x}{\sqrt{\hbar_{\epsilon}}}} D^{+}[q(t), \epsilon] \Psi_{0,\epsilon}(\sqrt{\hbar_{\epsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\epsilon}} \int_{0}^{T} Q_{\beta,\epsilon}(\sqrt{\hbar_{\epsilon}} q, v) dt\right) \times
$$

$$
[\mathbf{sign}^{+} \mathfrak{R}_1(q_{\epsilon,h_{\epsilon}}(t)) - \mathbf{sign}^{-} \mathfrak{R}_1(q_{\epsilon,h_{\epsilon}}(t))] \times
$$

$$
\left| \cos\left( \int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - a q_{\epsilon,h_\epsilon}^2(t) - b(t) \frac{q_{\epsilon,h_\epsilon}(t)}{\sqrt{\hbar_\epsilon}} \right] \right) \right| \times \quad (2.1.22)
$$

$$
\cos\left[\left(\int_{0}^{T} V_{1,h_{\epsilon}}(q_{\epsilon}(t))dt\right)\right],
$$

$$
\mathfrak{R}_1(q_{\epsilon,h_{\epsilon}}(t)) = \cos\left(\int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - a q_{\epsilon,h_{\epsilon}}^2(t) - b(t) \frac{q_{\epsilon,h_{\epsilon}}(t)}{\sqrt{\hbar_{\epsilon}}} \right] \right)
$$

and
$$\chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) =$$

$$\int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} } D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt\right) \times$$

$$[\mathbf{sign}^{+}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t))] \times$$

$$\left| \sin\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times \quad (2.1.23)$$

$$\sin\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t)) dt\right)\right],$$

$$\mathfrak{R}_2(q_{\varepsilon,h_{\varepsilon}}(t)) = \sin\left(\int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - a q_{\varepsilon,h_{\varepsilon}}^2(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right)$$

From Eq.(2.1.22) we obtain:
$$\chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) =$$

$$\int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} } D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt\right) \times$$

$$[\mathbf{sign}^{+}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t))] \times$$

$$\left| \sin\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times \quad (2.1.23)$$

$$\sin\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t)) dt\right)\right],$$

$$\mathfrak{R}_2(q_{\varepsilon,h_{\varepsilon}}(t)) = \sin\left(\int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - a q_{\varepsilon,h_{\varepsilon}}^2(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right)$$

From Eq.(2.1.22) we obtain:
$$
\chi_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) =
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} D^+[q(t),\varepsilon]\Psi_{0,\varepsilon}(\sqrt{\hbar \varepsilon}q(0)) \exp\left(-\frac{1}{\hbar \varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar \varepsilon}q,v)dt\right) \times
$$

$$
\{[\mathbf{sign}^{+}\mathfrak{R}_1(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_1(q_{\varepsilon,h_{\varepsilon}}(t)) + 1] - 1\} \times
$$

$$
\left| \cos\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times
$$

$$
\cos\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right] =
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} \tilde{D}_{1}^{+}[q(t), \varepsilon] \Psi_{0, \varepsilon}(\sqrt{\hbar \varepsilon} q(0)) \times
$$

$$
\left| \cos\left( \int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - a q_{\epsilon,h_\epsilon}^2(t) - b(t) \frac{q_{\epsilon,h_\epsilon}(t)}{\sqrt{\hbar_\epsilon}} \right] \right) \right| \times \quad (2.1.24)
$$

$$
\cos\left[\left(\int_{0}^{T} V_{1,h_{\epsilon}}(q_{\epsilon}(t))dt\right)\right] -
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} D^+[q(t),\varepsilon]\Psi_{0,\varepsilon}(\sqrt{\hbar \varepsilon}q(0)) \exp\left(-\frac{1}{\hbar \varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar \varepsilon}q,v)dt\right) \times
$$

$$
\left| \cos\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^{2}(t) - a q_{\epsilon,h_{\epsilon}}^{2}(t) - b(t) \frac{q_{\epsilon,h_{\epsilon}}(t)}{\sqrt{\hbar_{\epsilon}}} \right] \right) \right| \times
$$

$$
\cos\left[\left(\int_{0}^{T} V_{1,h_{\epsilon}}(q_{\epsilon}(t))dt\right)\right] =
$$
$$
\chi_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) =
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} D^+[q(t),\varepsilon]\Psi_{0,\varepsilon}(\sqrt{\hbar \varepsilon}q(0)) \exp\left(-\frac{1}{\hbar \varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar \varepsilon}q,v)dt\right) \times
$$

$$
\{[\mathbf{sign}^{+}\Re_{1}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\Re_{1}(q_{\varepsilon,h_{\varepsilon}}(t)) + 1] - 1\} \times
$$

$$
\left| \cos\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times
$$

$$
\cos\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right] =
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} \tilde{D}_{1}^{+}[q(t), \varepsilon] \Psi_{0, \varepsilon}(\sqrt{\hbar \varepsilon} q(0)) \times
$$

$$
\left| \cos\left( \int_0^T dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - a q_{\epsilon,h_\epsilon}^2(t) - b(t) \frac{q_{\epsilon,h_\epsilon}(t)}{\sqrt{\hbar_\epsilon}} \right] \right) \right| \times \quad (2.1.24)
$$

$$
\cos\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right] -
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar \varepsilon}}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar \varepsilon} q(0)) \exp\left(-\frac{1}{\hbar \varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar \varepsilon} q, v) dt\right) \times
$$

$$
\left| \cos\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^{2}(t) - a q_{\epsilon,h_{\epsilon}}^{2}(t) - b(t) \frac{q_{\epsilon,h_{\epsilon}}(t)}{\sqrt{\hbar_{\epsilon}}} \right] \right) \right| \times
$$

$$
\cos\left[\left(\int_{0}^{T} V_{1,h_{\epsilon}}(q_{\epsilon}(t))dt\right)\right] =
$$
where

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } \widetilde{D}_{1}^{+}[q(t), \varepsilon](\cdot) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } D^{+}[q(t), \varepsilon][\mathbf{sign}^{+}\mathfrak{R}_{1}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_{1}(q_{\varepsilon,h_{\varepsilon}}(t)) + 1](\cdot), $$

$$ \tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } \widetilde{D}^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt\right) \quad (2.1.25) $$

$$ \left| \cos\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times $$

$$ \cos\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t)) dt\right)\right] $$

and
where

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } \widetilde{D}_{1}^{+}[q(t), \varepsilon](\cdot) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } D^{+}[q(t), \varepsilon][\mathbf{sign}^{+}\mathfrak{R}_{1}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_{1}(q_{\varepsilon,h_{\varepsilon}}(t)) + 1](\cdot), $$

$$ \tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } \widetilde{D}^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp \left( -\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt \right) \quad (2.1.25) $$

$$ \left| \cos \left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times $$

$$ \cos \left[ \left( \int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t)) dt \right) \right] $$

and
$$v_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) =$$

$$ \int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt\right) \times \\ \left| \cos\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,\hbar_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,\hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times \\ \cos\left[ \left( \int_{0}^{T} V_{1,\hbar_{\varepsilon}}(q_{\varepsilon}(t)) dt \right) \right]. \qquad (2.1.26) $$

From Eq.(2.1.23) we obtain:
$$v_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) =$$

$$ \int_{q(T)=\frac{x}{\sqrt{\hbar\varepsilon}}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar\varepsilon}q(0)) \exp\left(-\frac{1}{\hbar\varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar\varepsilon}q, v)dt\right) \times \left|\cos\left(\int_0^T dt \left[\frac{m}{2}\dot{q}_{\sqrt{\hbar\varepsilon},\varepsilon}^2(t) - aq_{\varepsilon,\hbar\varepsilon}^2(t) - b(t)\frac{q_{\varepsilon,\hbar\varepsilon}(t)}{\sqrt{\hbar\varepsilon}}\right]\right)\right| \times \cos\left[\left(\int_0^T V_{1,\hbar\varepsilon}(q_{\varepsilon}(t))dt\right)\right]. \quad (2.1.26) $$

From Eq.(2.1.23) we obtain:
$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } \widetilde{D}_{1}^{+}[q(t), \varepsilon](\cdot) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } D^{+}[q(t), \varepsilon][\mathbf{sign}^{+}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) + 1](\cdot), $$

$$ \chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } D^{+}[q(t), \varepsilon]\Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q(0))\exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T}Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q,v)dt\right) \times $$

$$ \times \{[\mathbf{sign}^{+}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) + 1] - 1\} \times $$

$$ \left| \sin\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - aq_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times $$

$$ \sin\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right] = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } \widetilde{D}_{2}^{+}[q(t), \varepsilon]\Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q(0))\exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T}Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q,v)dt\right) \times \quad (2.1.27) $$

$$ \left| \sin\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - aq_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times $$

$$ \sin\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right] - $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } D^{+}[q(t), \varepsilon]\Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q(0))\exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T}Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q,v)dt\right) \times $$
$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } \widetilde{D}_{1}^{+}[q(t), \varepsilon](\cdot) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } D^{+}[q(t), \varepsilon][\mathbf{sign}^{+}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) + 1](\cdot), $$

$$ \chi_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } D^{+}[q(t), \varepsilon]\Psi_{0,\varepsilon}(\sqrt{h_{\varepsilon}}q(0))\exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T}Q_{\beta,\varepsilon}(\sqrt{h_{\varepsilon}}q,v)dt\right) \times $$

$$ \{[\mathbf{sign}^{+}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) - \mathbf{sign}^{-}\mathfrak{R}_{2}(q_{\varepsilon,h_{\varepsilon}}(t)) + 1] - 1\} \times $$

$$ \left| \sin\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - aq_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{h_{\varepsilon}}} \right] \right) \right| \times $$

$$ \sin\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right] = $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } \widetilde{D}_{2}^{+}[q(t), \varepsilon]\Psi_{0,\varepsilon}(\sqrt{h_{\varepsilon}}q(0))\exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T}Q_{\beta,\varepsilon}(\sqrt{h_{\varepsilon}}q,v)dt\right) \times \quad (2.1.27) $$

$$ \left| \sin\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - aq_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,h_{\varepsilon}}(t)}{\sqrt{h_{\varepsilon}}} \right] \right) \right| \times $$

$$ \sin\left[\left(\int_{0}^{T} V_{1,h_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right] - $$

$$ \int_{q(T)=\frac{x}{\sqrt{h_{\varepsilon}}} } D^{+}[q(t), \varepsilon]\Psi_{0,\varepsilon}(\sqrt{h_{\varepsilon}}q(0))\exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T}Q_{\beta,\varepsilon}(\sqrt{h_{\varepsilon}}q,v)dt\right) \times $$
where

$$
\begin{equation}
\begin{aligned}
\tilde{\chi}_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = & \\
& \int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}}} \tilde{D}_2^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q, v)dt\right) \times \\
& \left|\sin\left(\int_0^T dt \left[\frac{m}{2}\dot{q}_{\tilde{\mathbb{R}},\varepsilon}^2(t) - aq_{\varepsilon,\hbar_{\varepsilon}}^2(t) - b(t)\frac{q_{\varepsilon,\hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}}\right]\right)\right| \times \\
& \sin\left[\left(\int_0^T V_{1,\hbar_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right]
\end{aligned}
\tag{2.1.28}
\end{equation}
$$

and

$$
\begin{equation}
\begin{aligned}
\nu_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = & \\
& \int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q, v)dt\right) \times \\
& \left|\sin\left(\int_{0}^{T} dt \left[\frac{m}{2}\dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - aq_{\varepsilon,\hbar_{\varepsilon}}^{2}(t) - b(t)\frac{q_{\varepsilon,\hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}}\right]\right)\right| \times \\
& \sin\left[\left(\int_{0}^{T} V_{1,\hbar_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right].
\end{aligned}
\tag{2.1.29}
\end{equation}
$$

Let us estimate now the quantities $\tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon})$ and $\tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon})$.
where

$$
\begin{equation}
\begin{aligned}
\tilde{\chi}_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = & \\
& \int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}}} \tilde{D}_2^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q, v)dt\right) \times \\
& \left|\sin\left(\int_0^T dt \left[\frac{m}{2}\dot{q}_{\tilde{\mathbb{R}},\varepsilon}^2(t) - aq_{\varepsilon,\hbar_{\varepsilon}}^2(t) - b(t)\frac{q_{\varepsilon,\hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}}\right]\right)\right| \times \\
& \sin\left[\left(\int_0^T V_{1,\hbar_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right]
\end{aligned}
\tag{2.1.28}
\end{equation}
$$

and

$$
\begin{equation}
\begin{aligned}
\nu_{\varepsilon}^{(2)}(T, x, x_0, \hbar_{\varepsilon}) = & \\
& \int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}}\int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}}q, v)dt\right) \times \\
& \left|\sin\left(\int_{0}^{T} dt \left[\frac{m}{2}\dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - aq_{\varepsilon,\hbar_{\varepsilon}}^{2}(t) - b(t)\frac{q_{\varepsilon,\hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}}\right]\right)\right| \times \\
& \sin\left[\left(\int_{0}^{T} V_{1,\hbar_{\varepsilon}}(q_{\varepsilon}(t))dt\right)\right].
\end{aligned}
\tag{2.1.29}
\end{equation}
$$

Let us estimate now the quantities $\tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon})$ and $\tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon})$.
From Eq.(2.1.25) one obtain directly the inequality:

$$|\tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon})| \le$$

$$\int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}} \tilde{D}^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt\right) \times \quad (2.1.30)$$

$$\left| \cos\left( \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^{2}(t) - a q_{\varepsilon,\hbar_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,\hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times$$

$$\left| \cos \left[ \left( \int_{0}^{T} V_{1,\hbar_{\varepsilon}}(q_{\varepsilon}(t)) dt \right) \right] \right|$$

From Eq.(2.1.30) and Hölder's inequality () for $p_2 = \frac{1}{\sqrt{\hbar_{\varepsilon}}}$, $p_1 = \frac{1}{1 - \sqrt{\hbar_{\varepsilon}}} = 1 + \sqrt{\hbar_{\varepsilon}} + o(\hbar_{\varepsilon}) \approx 1$ we obtain:
From Eq.(2.1.25) one obtain directly the inequality:

$$
\begin{equation} \tag{2.1.30}
\begin{aligned}
& \left| \tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) \right| \le \\
& \qquad \int_{q(T)=\frac{x}{\sqrt{\hbar_{\varepsilon}}}} \tilde{D}^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q(0)) \exp \left( -\frac{1}{\hbar_{\varepsilon}} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar_{\varepsilon}} q, v) dt \right) \times \\
& \qquad \left| \cos \left( \int_0^T dt \left[ \frac{m}{2} \dot{q}_{\tilde{\mathbb{R}},\varepsilon}^2(t) - a q_{\varepsilon,\hbar_{\varepsilon}}^2(t) - b(t) \frac{q_{\varepsilon,\hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) \right| \times \\
& \qquad \left| \cos \left[ \left( \int_0^T V_{1,\hbar_{\varepsilon}}(q_{\varepsilon}(t)) dt \right) \right] \right|
\end{aligned}
\end{equation}
$$

From Eq.(2.1.30) and Hölder's inequality () for $p_2 = \frac{1}{\sqrt{\hbar_{\varepsilon}}}$, $p_1 = \frac{1}{1 - \sqrt{\hbar_{\varepsilon}}} = 1 + \sqrt{\hbar_{\varepsilon}} + o(\hbar_{\varepsilon}) \approx 1$ we obtain:
$$|\tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon})| \le$$

$$ \left[ \int_{q(T)=\frac{x}{\sqrt{\hbar\varepsilon}}} \tilde{D}^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar\varepsilon}q(0)) \exp\left(-\frac{1}{\hbar\varepsilon} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar\varepsilon}q, v) dt\right) \times \left| \cos\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}^{2}_{\tilde{R},\varepsilon}(t) - a q_{\varepsilon,\hbar\varepsilon}^{2}(t) - b(t) \frac{q_{\varepsilon,\hbar\varepsilon}(t)}{\sqrt{\hbar\varepsilon}} \right] \right) \right|^{p_1} \right]^{\frac{1}{p_1}} \times \quad (2.1.31) $$

$$ \left[ \int_{q(T)=\frac{x}{\sqrt{\hbar\varepsilon}}} \tilde{D}^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar\varepsilon}q(0)) \left| \cos \left[ \left( \int_{0}^{T} V_{1,\hbar\varepsilon}(q_{\varepsilon}(t)) dt \right) \right] \right|^{p_2} \right]^{\frac{1}{p_2}} \le $$

Let us
$$|\tilde{\chi}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon})| \le$$

$$ \left[ \int_{q(T)=\frac{x}{\sqrt{\hbar\varepsilon}}} \tilde{D}^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar\varepsilon}q(0)) \exp\left(-\frac{1}{\hbar\varepsilon} \int_{0}^{T} Q_{\beta,\varepsilon}(\sqrt{\hbar\varepsilon}q, v) dt\right) \times \left| \cos\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,\hbar\varepsilon}^{2}(t) - b(t) \frac{q_{\varepsilon,\hbar\varepsilon}(t)}{\sqrt{\hbar\varepsilon}} \right] \right) \right|^{p_1} \right]^{\frac{1}{p_1}} \times \quad (2.1.31) $$

$$ \left[ \int_{q(T)=\frac{x}{\sqrt{\hbar\varepsilon}}} \tilde{D}^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar\varepsilon}q(0)) \left| \cos \left[ \left( \int_{0}^{T} V_{1,\hbar\varepsilon}(q_{\varepsilon}(t)) dt \right) \right] \right|^{p_2} \right]^{\frac{1}{p_2}} \le $$

Let us
$$
\Theta_{\epsilon}(T, x, x_0, \hbar_{\epsilon}) =
$$

$$
\int_{q(T)=\frac{\gamma}{\sqrt{\hbar\epsilon}}} D^+[q(t),\epsilon] \Psi_{0,\epsilon}(\sqrt{\hbar\epsilon}q(0)) \exp\left(-\frac{1}{\hbar\epsilon} \int_0^T Q_{\beta,\epsilon}(\sqrt{\hbar\epsilon}q,v)dt\right) \times \quad (2.1.)
$$

$$
\cos\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^{2}(t) - a q_{\epsilon,h_{\epsilon}}^{2}(t) - b(t) \frac{q_{\epsilon,h_{\epsilon}}(t)}{\sqrt{\hbar_{\epsilon}}} \right] \right)
$$
$$ \Theta_{\varepsilon}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{\hbar\varepsilon}}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(\sqrt{\hbar\varepsilon}q(0)) \exp\left(-\frac{1}{\hbar\varepsilon} \int_0^T Q_{\beta,\varepsilon}(\sqrt{\hbar\varepsilon}q, v)dt\right) \times \quad (2.1.) $$

$$ \cos\left(\int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,\hbar_{\varepsilon}}^{2}(t) - b(t) \frac{q_{\varepsilon,\hbar_{\varepsilon}}(t)}{\sqrt{\hbar_{\varepsilon}}} \right] \right) $$
$$
\hat{\Theta}_{\beta, \varepsilon}(T, x, x_0, \hbar_{\varepsilon}) =
$$

$$
\int |x| dx \int_{q(T)=x} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \exp\left(-\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(q,v) dt\right) \times
$$

$$
\exp\left(\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) q_{\varepsilon,h_{\varepsilon}}(t) \right]\right) =
$$

$$
\int |x| dx \int_{q(T)=x} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times
$$

$$
\exp\left(-\frac{v}{\hbar_{\varepsilon}} \int_{0}^{T} \left[(q(t) - \beta(t, q(0), q(T)))^2\right] dt - \frac{v}{\hbar_{\varepsilon}} q^2(T) - \frac{v}{\hbar_{\varepsilon}} q^2(0)\right) \quad (2.1.)
$$

$$
\exp\left(\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} \mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q_{\varepsilon,h_{\varepsilon}}) dt\right) \approx
$$

$$
\int |x| dx \int_{q(T)=x} \int dy D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(y) \exp\left(\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} \mathcal{L}(\dot{q}, q) dt\right) \times \\
q(0)=y
$$

$$
\exp\left(-\frac{v}{\hbar_{\varepsilon}} \int_{0}^{T} \left[ (q(t) - \beta(t, q(0), q(T)))^2 \right] dt - \frac{v}{\hbar_{\varepsilon}} x^2 - \frac{v}{\hbar_{\varepsilon}} y^2 \right),
$$

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}, q_{\epsilon,h_{\epsilon}}) = \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - a q_{\epsilon,h_{\epsilon}}^2(t) - b(t) q_{\epsilon,h_{\epsilon}}(t) \approx
$$

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}, q) = \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - a q^2(t) - b(t) q(t) \text{ if } \epsilon \to 0.
$$

Let us estimate now path integral
$$
\hat{\Theta}_{\beta, \varepsilon}(T, x, x_0, \hbar_{\varepsilon}) =
$$

$$
\int |x| dx \int_{q(T)=x} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \exp \left( -\frac{1}{\hbar_{\varepsilon}} \int_{0}^{T} Q_{\beta,\varepsilon}(q,v) dt \right) \times
$$

$$
\exp\left(\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) q_{\varepsilon,h_{\varepsilon}}(t) \right]\right) =
$$

$$
\int |x| dx \int_{q(T)=x} D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) \times
$$

$$
\exp\left(-\frac{v}{\hbar_{\varepsilon}} \int_{0}^{T} \left[(q(t) - \beta(t, q(0), q(T)))^2\right] dt - \frac{v}{\hbar_{\varepsilon}} q^2(T) - \frac{v}{\hbar_{\varepsilon}} q^2(0)\right) \quad (2.1.)
$$

$$
\exp\left(\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} \mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q_{\varepsilon,h_{\varepsilon}}) dt\right) \approx
$$

$$
\int |x| dx \int_{q(T)=x} \int dy D^{+}[q(t), \varepsilon] \Psi_{0,\varepsilon}(y) \exp \left( \frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} \mathcal{L}(\dot{q}, q) dt \right) \times \\
q(0)=y
$$

$$
\exp\left(-\frac{v}{\hbar_{\varepsilon}} \int_{0}^{T} \left[(q(t) - \beta(t, q(0), q(T)))^2\right] dt - \frac{v}{\hbar_{\varepsilon}} x^2 - \frac{v}{\hbar_{\varepsilon}} y^2\right),
$$

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q_{\varepsilon,h_{\varepsilon}}) = \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - a q_{\varepsilon,h_{\varepsilon}}^2(t) - b(t) q_{\varepsilon,h_{\varepsilon}}(t) \approx
$$

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q) = \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - a q^2(t) - b(t) q(t) \text{ if } \varepsilon \to 0.
$$

Let us estimate now path integral
$$
\hat{\Theta}_{\beta, \varepsilon}(T, x, x_0, h_{\varepsilon}) =
$$

$$
\int_{\substack{q(T)=x \\ q(0)=y}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) q(T) \exp\left(-\frac{1}{h_\varepsilon} \int_0^T Q_{\beta,\varepsilon}(q,v) dt\right) \times
$$

$$
\exp\left(\frac{i}{h_{\varepsilon}} \int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R}, \varepsilon}^{2}(t)-a q_{\varepsilon, h_{\varepsilon}}^{2}(t)-b(t) q_{\varepsilon, h_{\varepsilon}}(t)\right]\right) =
$$

$$
\int_{\substack{q(T)=x \\ q(0)=y}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) q(T) \times
$$

$$
\exp\left(\frac{i}{h_{\varepsilon}} \int_{0}^{T} \mathcal{L}\left(\dot{q}_{\mathbb{R}, \varepsilon}, q_{\varepsilon, h_{\varepsilon}}\right) d t\right) \times
$$

$$
\exp\left(-\frac{v}{h_{\epsilon}} \int_{0}^{T}\left[(q(t)-c-\beta(t, q(0)-c, q(T)-c)\right]\right)^{2} d t \quad (2.1.)
$$

$$
-\frac{v}{h_{\epsilon}} q^{2}(T) - \frac{v}{h_{\epsilon}} q^{2}(0) \approx
$$

$$
\int_{\substack{q(T)=x \\ q(0)=y}} \int dy D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(y) q(T) \exp\left(\frac{i}{h_\varepsilon} \int_0^T \mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q) dt\right) \times
$$

$$
\exp\left(-\frac{v}{h_{\varepsilon}} \int_{0}^{T} \left[(q(t)-c-\beta(t,q(0)-c,q(T)-c)\right]^{2} d t-
$$

$$
\frac{v}{h_{\varepsilon}}[q(T)-c]^{2}-\frac{v}{h_{\varepsilon}}[q(0)-c]),
$$

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q_{\varepsilon,h_{\varepsilon}}) = \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - a q_{\varepsilon,h_{\varepsilon}}^2(t) - b(t) q_{\varepsilon,h_{\varepsilon}}(t) \approx
$$
$$
\hat{\Theta}_{\beta, \varepsilon}(T, x, x_0, h_{\varepsilon}) =
$$

$$
\int_{\substack{q(T)=x \\ q(0)=y}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) q(T) \exp\left(-\frac{1}{h_\varepsilon} \int_0^T Q_{\beta,\varepsilon}(q,v) dt\right) \times
$$

$$
\exp\left(\frac{i}{h_{\varepsilon}} \int_{0}^{T} dt \left[ \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon,h_{\varepsilon}}^{2}(t) - b(t) q_{\varepsilon,h_{\varepsilon}}(t) \right] \right) =
$$

$$
\int_{\substack{q(T)=x \\ q(0)=y}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q(0)) q(T) \times
$$

$$
\exp\left(\frac{i}{h_{\varepsilon}} \int_{0}^{T} \mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q_{\varepsilon,h_{\varepsilon}}) dt\right) \times
$$

$$
\exp\left(-\frac{v}{h_{\epsilon}} \int_{0}^{T} \left[ (q(t) - c - \beta(t, q(0) - c, q(T) - c)) \right]^2 dt \right) \quad (2.1.)
$$

$$
-\frac{v}{h_{\epsilon}} q^{2}(T) - \frac{v}{h_{\epsilon}} q^{2}(0) \approx
$$

$$
\int_{\substack{q(T)=x \\ q(0)=y}} \int dy D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(y) q(T) \exp\left(\frac{i}{h_\varepsilon} \int_0^T \mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q) dt\right) \times
$$

$$
\exp\left(-\frac{v}{h_{\varepsilon}} \int_{0}^{T} \left[ (q(t) - c - \beta(t, q(0) - c, q(T) - c)) \right]^2 dt - \frac{v}{h_{\varepsilon}} [q(T) - c]^2 - \frac{v}{h_{\varepsilon}} [q(0) - c] \right),
$$

$$
\frac{v}{h_{\varepsilon}} [q(T) - c]^2 - \frac{v}{h_{\varepsilon}} [q(0) - c]
$$

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\varepsilon}, q_{\varepsilon,h_{\varepsilon}}) = \frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^2(t) - a q_{\varepsilon,h_{\varepsilon}}^2(t) - b(t) q_{\varepsilon,h_{\varepsilon}}(t) \approx
$$
We set now for short $b(t) = b = \text{const}$, without loss of generality. Substitution

$$ 
\begin{aligned}
q(t) &= q_1(t) + c, \\
c &= -\frac{b}{2a}
\end{aligned} 
\qquad (2.1.) 
$$

into (2.1.) gives

$$ 
\begin{aligned}
\mathcal{L}(\dot{q}_{1\mathbb{R},\varepsilon}, q) &= \frac{m}{2} \dot{q}_{1\mathbb{R},\varepsilon}^2 - a(q_1 + c)^2 - b(q_1 + c) = \\
&= \frac{m}{2} \dot{q}_{1\mathbb{R},\varepsilon}^2 - aq_1^2 - 2acq_1 - bq_1 - ac^2 - bc =
\end{aligned} 
\qquad (2.1.) 
$$

$$ \mathcal{L}_1(\dot{q}_{1\mathbb{R},\varepsilon}, q_1) = \frac{m}{2} \dot{q}_{1\mathbb{R},\varepsilon}^2 - aq_1^2 + \frac{b^2}{2a}. $$

and
We set now for short $b(t) = b = \text{const}$, without loss of generality. Substitution

$$ 
\begin{aligned}
q(t) &= q_1(t) + c, \\
c &= -\frac{b}{2a}
\end{aligned} 
\qquad (2.1.) 
$$

into (2.1.) gives

$$ 
\begin{aligned}
\mathcal{L}(\dot{q}_{1\mathbb{R},\varepsilon}, q) &= \frac{m}{2} \dot{q}_{1\mathbb{R},\varepsilon}^2 - a(q_1 + c)^2 - b(q_1 + c) = \\
&= \frac{m}{2} \dot{q}_{1\mathbb{R},\varepsilon}^2 - aq_1^2 - 2acq_1 - bq_1 - ac^2 - bc =
\end{aligned} 
\qquad (2.1.) 
$$

$$ \mathcal{L}_1(\dot{q}_{1\mathbb{R},\varepsilon}, q_1) = \frac{m}{2} \dot{q}_{1\mathbb{R},\varepsilon}^2 - aq_1^2 + \frac{b^2}{2a}. $$

and
$$ \Theta_{\beta,\varepsilon}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \exp\left(\frac{i}{\hbar_{\varepsilon}} \frac{b^2 T}{2a}\right) \int_{\substack{q_1(T)=x-c \\ q_1(0)=y-c}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q_1(0) + c)[q_1(T) + c] \times $$

$$ \exp\left(\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} \mathcal{L}_{1}(\dot{q}_{1}, \tilde{\mathbb{R}}, \varepsilon, q_{1}) dt\right) \times \qquad (2.1.) $$

$$ \exp\left(-\frac{v}{\hbar_{\varepsilon}} \int_{0}^{T} [(q_{1}(t) - \beta(t, q_{1}(0), q(T)))]^2 dt - \frac{v}{\hbar_{\varepsilon}} q_{1}^2(T) - \frac{v}{\hbar_{\varepsilon}} q_{1}^2(0)\right) $$

Let us estimate now path integral
$$ \Theta_{\beta, \varepsilon}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \exp\left(\frac{i}{\hbar_{\varepsilon}} \frac{b^2 T}{2a}\right) \int_{\substack{q_1(T)=x-c \\ q_1(0)=y-c}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q_1(0) + c)[q_1(T) + c] \times $$

$$ \exp\left(\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} \mathcal{L}_{1}(\dot{q}_{1}, \tilde{\mathbb{R}}, \varepsilon, q_{1}) dt\right) \times \qquad (2.1.) $$

$$ \exp\left(-\frac{v}{\hbar_{\varepsilon}} \int_{0}^{T} [(q_{1}(t) - \beta(t, q_{1}(0), q(T)))]^2 dt - \frac{v}{\hbar_{\varepsilon}} q_{1}^2(0)\right) $$

Let us estimate now path integral
$$ \Theta'_{\beta, \epsilon}(T, x, x_0, \hbar_{\epsilon}) = $$

$$ \int_{\substack{q_1(T)=x-c \\ q_1(0)=y-c}} D^+[q(t), \epsilon] \Psi_{0,\epsilon}(q_1(0)+c)[q_1(T)+c] \times $$

$$ \exp\left(\frac{i}{\hbar_{\epsilon}} \int_{0}^{T} \mathcal{L}_{1}(\dot{q}_{1}\tilde{\mathbb{R}}, \epsilon, q_{1}) dt\right) \times \qquad (2.1.) $$

$$ \exp\left(-\frac{v}{\hbar_{\epsilon}} \int_{0}^{T} \left[ (q_{1}(t) - \beta(t, q_{1}(0), q(T))) \right]^{2} dt - \frac{v}{\hbar_{\epsilon}} q_{1}^{2}(T) - \frac{v}{\hbar_{\epsilon}} q_{1}^{2}(0) \right) $$

using stationary phase method.

The Euler-Lagrange equation of $\mathcal{L}(\dot{q}_{\tilde{\mathbb{R}},\epsilon}, q)$ with corresponding boundary conditions is

$$ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_{\tilde{\mathbb{R}},\epsilon}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0, \qquad (2.1.) $$

$$ q(0) = y - c, q(T) = x - c. $$

Thus if $\epsilon \to 0$, Euler-Lagrange equation of $\mathcal{L}(\dot{q}_{\tilde{\mathbb{R}},\epsilon}, q)$ is
$$ \Theta'_{\beta, \varepsilon}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \int_{\substack{q_1(T)=x-c \\ q_1(0)=y-c}} D^+[q(t), \varepsilon] \Psi_{0,\varepsilon}(q_1(0)+c)[q_1(T)+c] \times $$

$$ \exp\left(\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} \mathcal{L}_{1}(\dot{q}_{1}\tilde{\mathbb{R}}, \varepsilon, q_{1}) dt\right) \times \qquad (2.1.) $$

$$ \exp\left(-\frac{v}{\hbar_{\varepsilon}} \int_{0}^{T} \left[(q_{1}(t) - \beta(t, q_{1}(0), q(T)))\right]^{2} dt - \frac{v}{\hbar_{\varepsilon}} q_{1}^{2}(T) - \frac{v}{\hbar_{\varepsilon}} q_{1}^{2}(0)\right) $$

using stationary phase method.

The Euler-Lagrange equation of $\mathcal{L}(\dot{q}_{\tilde{\mathbb{R}},\varepsilon}, q)$ with corresponding boundary conditions is

$$ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_{\tilde{\mathbb{R}},\varepsilon}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0, \qquad (2.1.) $$

$$ q(0) = y - c, q(T) = x - c. $$

Thus if $\varepsilon \to 0$, Euler-Lagrange equation of $\mathcal{L}(\dot{q}_{\tilde{\mathbb{R}},\varepsilon}, q)$ is
$$ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0, \qquad (2.1.) $$

$$ q(0) = y - c, q(T) = x - c. $$

Let $q_{cr}(t,y,x)$ be the solution of the Euler-Lagrange equation (2.1.). Substitution

$$ q(t) = \delta(t) + q_{cr}(t,y,x) \qquad (2.1.) $$

$$ \delta(0) = 0, \delta(T) = 0 $$

into (2.1.) gives

$$ \Theta'_{\beta,\epsilon}(T, x_0, \hbar_{\epsilon}) = $$

$$ \int_{\substack{\delta(T)=0 \\ \delta(0)=0}} D^+[\delta(t), \epsilon] x \Psi_{0,\epsilon}^2(y) \exp\left(\frac{i}{\hbar_{\epsilon}} \tilde{S}_{\text{cl}}(T, y-c, x-c)\right) \times $$

$$ \exp\left(\frac{i}{\hbar_{\epsilon}} \int_{0}^{T} \mathcal{L}(\dot{\delta}_{\mathbb{R},\epsilon}, \delta) dt\right) \times \qquad (2.1.) $$

$$ \exp\left( -\frac{v}{\hbar_{\epsilon}} \int_{0}^{T} \left[ (\delta(t) + q_{cr}(t, y-c, x-c) - \beta(t, y-c, x-c))^2 \right] dt - \frac{v}{\hbar_{\epsilon}} (x-c)^2 - \frac{v}{\hbar_{\epsilon}} (y-c)^2 \right). $$

Where
$$ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}} \right) - \frac{\partial \mathcal{L}}{\partial q} = 0, \qquad (2.1.) $$
$$ q(0) = y - c, q(T) = x - c. $$

Let $q_{cr}(t,y,x)$ be the solution of the Euler-Lagrange equation (2.1.). Substitution

$$ q(t) = \delta(t) + q_{cr}(t,y,x) \qquad (2.1.) $$
$$ \delta(0) = 0, \delta(T) = 0 $$

into (2.1.) gives

$$ \Theta'_{\beta,\varepsilon}(T, x_0, \hbar_\varepsilon) = \int_{\substack{\delta(T)=0 \\ \delta(0)=0}} D^+[\delta(t), \varepsilon] x \Psi^2_{0,\varepsilon}(y) \exp\left(\frac{i}{\hbar_\varepsilon} \tilde{S}_{\text{cl}}(T, y-c, x-c)\right) \times \\ \exp\left(\frac{i}{\hbar_\varepsilon} \int_0^T \mathcal{L}(\dot{\delta}_{\mathbb{R},\varepsilon}, \delta) dt\right) \times \qquad (2.1.) $$
$$ \exp\left(-\frac{v}{\hbar_\varepsilon} \int_0^T \left[ (\delta(t) + q_{cr}(t,y-c,x-c) - \beta(t,y-c,x-c))^2 \right] dt \right. \\ \left. - \frac{v}{\hbar_\varepsilon} (x-c)^2 - \frac{v}{\hbar_\varepsilon} (y-c)^2 \right). $$

Where
$$ \tilde{S}_{cl}(T, y-c, x-c) = \qquad (2.1.) $$

We set now

$$ \beta(t,y-c,x-c) = q_{cr}(t,y-c,x-c), \qquad (2.1.) $$

From Eq.(2.1) and Eq.(2.1) we obtain

$$ \hat{\Theta}_{\beta, \varepsilon}(T, x_0, \hbar_{\varepsilon}) = $$

$$ \int |x| dx \int dy \exp \left( \frac{i}{\hbar_{\varepsilon}} \left[ \tilde{S}_{cl}(T, y, x) + ivx^2 + ivy^2 \right] \right) \qquad (2.1.) $$

$$ \int_{\substack{\delta(T)=0 \\ \delta(0)=0}} D^+[\delta(t), \varepsilon] \exp \left( \frac{i}{\hbar_{\varepsilon}} \left[ \int_0^T \mathcal{L}(\dot{\delta}_{\tilde{\mathbb{R}},\varepsilon}, \delta) dt + iv \int_0^T \delta^2(t) dt \right] \right), $$

From Eq.(2.2.1.24) and Hölder's inequality (2.2.2.17) for $p_2 = \frac{1}{\sqrt{\hbar_{\varepsilon}}}$,

$$ p_1 = \frac{1}{1 - \sqrt{\hbar_{\varepsilon}}} = 1 + \sqrt{\hbar_{\varepsilon}} + o(\hbar_{\varepsilon}) \approx 1 \text{ we obtain:} $$
$$ \tilde{S}_{cl}(T, y-c, x-c) = \qquad (2.1.) $$

We set now

$$ \beta(t, y-c, x-c) = q_{cr}(t, y-c, x-c), \qquad (2.1.) $$

From Eq.(2.1) and Eq.(2.1) we obtain

$$ \hat{\Theta}_{\beta, \varepsilon}(T, x_0, \hbar_{\varepsilon}) = $$

$$ \int |x| dx \int dy \exp \left( \frac{i}{\hbar_{\varepsilon}} \left[ \tilde{S}_{cl}(T, y, x) + ivx^2 + ivy^2 \right] \right) \qquad (2.1.) $$

$$ \int_{\substack{\delta(T)=0 \\ \delta(0)=0}} D^+[\delta(t), \varepsilon] \exp \left( \frac{i}{\hbar_{\varepsilon}} \left[ \int_0^T \mathcal{L}(\dot{\delta}_{\tilde{\mathbb{R}}, \varepsilon}, \delta) dt + iv \int_0^T \delta^2(t) dt \right] \right), $$

From Eq.(2.2.1.24) and Hölder's inequality (2.2.2.17) for $p_2 = \frac{1}{\sqrt{\hbar_{\varepsilon}}}$,

$$ p_1 = \frac{1}{1 - \sqrt{\hbar_{\varepsilon}}} = 1 + \sqrt{\hbar_{\varepsilon}} + o(\hbar_{\varepsilon}) \approx 1 \text{ we obtain:} $$
$$|\tilde{V}_{2,\epsilon}(T, x, x_0, \alpha, \hbar)| \le$$

$$\left| \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] \Psi_0(\sqrt{\epsilon} q(0)) \rho_{q,\epsilon}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times \right.$$

$$\sin\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon}^2(t) \right] \right) \times$$

$$\sin\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\epsilon}} \right] \right) \cos\left(\int_0^T V_{\epsilon,\epsilon}(q_{\epsilon}(t)) dt\right) \le$$

$$\int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] \Psi_0(\sqrt{\epsilon} q(0)) |\rho_{q,\epsilon}(\cdot)| \exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times$$

$$\left| \sin\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon}^2(t) \right] \right) \right| \times$$

$$\left| \sin\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\epsilon}} \right] \right) \right| \times \left| \cos\left(\int_0^T V_{\epsilon,\epsilon}(q_{\epsilon}(t)) dt\right) \right| \le$$

(2.2.1.26)

$$\left[ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))|^{1+\sqrt{\epsilon}} |\rho_{q,\epsilon}(\cdot)|^{1+\sqrt{\epsilon}} \times \right.$$

$$\exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times$$

$$\left| \sin\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\epsilon}} \right] \right) \sin\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon}^2(t) \right] \right) \right|^{1+\sqrt{\epsilon}} \times$$
$$|\tilde{V}_{2,\epsilon}(T, x, x_0, \alpha, \hbar)| \le$$

$$\left| \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] \Psi_0(\sqrt{\epsilon} q(0)) \rho_{q,\epsilon}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times \right.$$

$$\sin\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon}^2(t) \right] \right) \times$$

$$\sin\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\epsilon}} \right] \right) \cos\left(\int_0^T V_{\epsilon,\epsilon}(q_{\epsilon}(t)) dt\right) \le$$

$$\int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] \Psi_0(\sqrt{\epsilon} q(0)) |\rho_{q,\epsilon}(\cdot)| \exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times$$

$$\left| \sin\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon}^2(t) \right] \right) \right| \times$$

$$\left| \sin\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\epsilon}} \right] \right) \right| \times \left| \cos\left(\int_0^T V_{\epsilon,\epsilon}(q_{\epsilon}(t)) dt\right) \right| \le$$

(2.2.1.26)

$$\left[ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))|^{1+\sqrt{\epsilon}} |\rho_{q,\epsilon}(\cdot)|^{1+\sqrt{\epsilon}} \times \right.$$

$$\exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times$$

$$\left| \sin\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\epsilon}} \right] \right) \sin\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon}^2(t) \right] \right) \right|^{1+\sqrt{\epsilon}} \times$$
Note that $|\rho_q(\cdot)|^{1+\sqrt{\epsilon}} = 1$. Let us calculate now the oscillatory path integral

$$ \theta_{\epsilon}(T, x, x_0, \alpha, \hbar) = $$

$$ \begin{aligned} & \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))| \exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times \\ & \left| \cos\left(\int_0^T dt \left[ \frac{m}{2} \left(\frac{dq(t)}{dt}\right)^2 - aq_{\epsilon}^2(t) \right] \right) \right| \times \\ & \left| \cos\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\hbar}} \right] \right) \right| = \tag{2.2.1.27} \end{aligned} $$

$$ \begin{aligned} & \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))| \exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times \\ & \mathbf{sign}[S_0^{(1)}(T, x, x_0, q_{\epsilon}(t), \epsilon)] \times \mathbf{sign}[S_0^{(2)}(T, x, x_0, q_{\epsilon}(t), \epsilon)] \times \\ & \cos\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\hbar}} \right] \right) \cos\left(\int_0^T dt \left[ \frac{m}{2} \left(\frac{dq(t)}{dt}\right)^2 - aq_{\epsilon}^2(t) \right] \right). \end{aligned} $$

Here

$$ \begin{aligned} S_0^{(1)}(T, x, x_0, q(t), \hbar) &= \cos\left(\int_0^T dt \left[ \frac{m}{2} \left(\frac{dq(t)}{dt}\right)^2 - aq_{\epsilon, \epsilon}^2(t) \right]\right), \\ S_0^{(2)}(T, x, x_0, q(t), \hbar) &= \cos\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon, \epsilon}(t)}{\sqrt{\hbar}} \right]\right). \end{aligned} \tag{2.2.1.28} $$
Note that $|\rho_q(\cdot)|^{1+\sqrt{\epsilon}} = 1$. Let us calculate now the oscillatory path integral

$$ \theta_{\epsilon}(T, x, x_0, \alpha, \hbar) = $$

$$ \begin{aligned} & \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))| \exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times \\ & \left| \cos\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - aq_{\epsilon}^2(t) \right] \right) \right| \times \\ & \left| \cos\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\hbar}} \right] \right) \right| = \tag{2.2.1.27} \end{aligned} $$

$$ \begin{aligned} & \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))| \exp\left(-\frac{1}{\hbar} \int_0^T \left(\sqrt{\hbar} q(t) - \alpha(t)\right)^2 dt\right) \times \\ & \mathbf{sign}[S_0^{(1)}(T, x, x_0, q_{\epsilon}(t), \epsilon)] \times \mathbf{sign}[S_0^{(2)}(T, x, x_0, q_{\epsilon}(t), \epsilon)] \times \\ & \cos\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\hbar}} \right] \right) \cos\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - aq_{\epsilon}^2(t) \right] \right). \end{aligned} $$

Here

$$ \begin{aligned} S_0^{(1)}(T, x, x_0, q(t), \hbar) &= \cos\left(\int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - aq_{\epsilon, \epsilon}^2(t) \right]\right), \\ S_0^{(2)}(T, x, x_0, q(t), \hbar) &= \cos\left(\int_0^T dt \left[ b(t) \frac{q_{\epsilon, \epsilon}(t)}{\sqrt{\hbar}} \right]\right). \end{aligned} \tag{2.2.1.28} $$
From Eq.(2.2.1.28) we obtain

$$ \theta_{\epsilon}(T, x, x_0, \alpha, \hbar) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))| \tilde{p}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar} q(t) - \alpha(t))^2 dt\right) \times \\ \left\{ \left[ \exp[i\tilde{S}_0^{(1)}(T, x, x_0, q_{\epsilon,\epsilon}(t), \epsilon)] + \exp[-i\tilde{S}_0^{(1)}(T, x, x_0, q_{\epsilon,\epsilon}(t), \epsilon)] \right] \times \right. \\ \left. \left[ \exp[i\tilde{S}_0^{(2)}(T, x, x_0, q_{\epsilon,\epsilon}(t), \epsilon)] + \exp[-i\tilde{S}_0^{(2)}(T, x, x_0, q_{\epsilon,\epsilon}(t), \epsilon)] \right] \right\} = \qquad (2.2.1.29) $$

$$ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))| \tilde{p}_{q,\epsilon,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar} q(t) - \alpha(t))^2 dt\right) \times \\ \left\{ \exp[i(\tilde{S}_0^{(1)}(\cdot) + \tilde{S}_0^{(2)}(\cdot))] + \exp[i(-\tilde{S}_0^{(1)}(\cdot) + S_0^{(2)}(\cdot))] + \right. \\ \left. \exp[i(\tilde{S}_0^{(1)}(\cdot) - \tilde{S}_0^{(2)}(\cdot))] + \exp[i(-\tilde{S}_0^{(1)}(\cdot) - \tilde{S}_0^{(2)}(\cdot))] \right\}. $$

Here
From Eq.(2.2.1.28) we obtain

$$ \theta_{\epsilon}(T, x, x_0, \alpha, \hbar) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))| \tilde{p}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar} q(t) - \alpha(t))^2 dt\right) \times \\ \left\{ \left[ \exp[i\tilde{S}_0^{(1)}(T, x, x_0, q_{\epsilon,\epsilon}(t), \epsilon)] + \exp[-i\tilde{S}_0^{(1)}(T, x, x_0, q_{\epsilon,\epsilon}(t), \epsilon)] \right] \times \right. \\ \left. \left[ \exp[i\tilde{S}_0^{(2)}(T, x, x_0, q_{\epsilon,\epsilon}(t), \epsilon)] + \exp[-i\tilde{S}_0^{(2)}(T, x, x_0, q_{\epsilon,\epsilon}(t), \epsilon)] \right] \right\} = \quad (2.2.1.29) $$

$$ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\epsilon} q(0))| \tilde{p}_{q,\epsilon,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar} q(t) - \alpha(t))^2 dt\right) \times \\ \left\{ \exp[i(\tilde{S}_0^{(1)}(\cdot) + \tilde{S}_0^{(2)}(\cdot))] + \exp[i(-\tilde{S}_0^{(1)}(\cdot) + S_0^{(2)}(\cdot))] + \right. \\ \left. \exp[i(\tilde{S}_0^{(1)}(\cdot) - \tilde{S}_0^{(2)}(\cdot))] + \exp[i(-\tilde{S}_0^{(1)}(\cdot) - \tilde{S}_0^{(2)}(\cdot))] \right\}. $$

Here
$$
\begin{align}
\tilde{P}_{q,\epsilon,\epsilon}^{(1)}(\cdot) &= \mathbf{sign}[S_0^{(1)}(T,x,x_0,q_{\epsilon,\epsilon}(t),\hbar)]\mathbf{sign}[S_0^{(2)}(T,x,x_0,q_{\epsilon}(t),\hbar)], \nonumber \\
\tilde{S}_0^{(1)}(T,x,x_0,q_{\epsilon}(t),\epsilon) &= \int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon,\epsilon}^2(t) \right], \tag{2.2.1.30} \\
\tilde{S}_0^{(2)}(T,x,x_0,q_{\epsilon}(t),\epsilon) &= \int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\hbar}} \right]. \nonumber
\end{align}
$$

From Eq.(2.2.1.29) we obtain

$$
\begin{align}
\theta_{\epsilon}(T, x, x_0, \hbar) = & \nonumber \\
& \theta_{\epsilon}^{(1)}(T, x, x_0, \alpha, \hbar) + \theta_{\epsilon}^{(2)}(T, x, x_0, \alpha, \hbar) + \tag{2.2.1.30} \\
& + \theta_{\epsilon}^{(3)}(T, x, x_0, \alpha, \hbar) + \theta_{\epsilon}^{(4)}(T, x, x_0, \alpha, \hbar). \nonumber
\end{align}
$$

Here
$$
\begin{align}
\tilde{P}_{q,\epsilon,\epsilon}^{(1)}(\cdot) &= \mathbf{sign}[S_0^{(1)}(T,x,x_0,q_{\epsilon,\epsilon}(t),\hbar)]\mathbf{sign}[S_0^{(2)}(T,x,x_0,q_{\epsilon}(t),\hbar)], \nonumber \\
\tilde{S}_0^{(1)}(T,x,x_0,q_{\epsilon}(t),\epsilon) &= \int_0^T dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon,\epsilon}^2(t) \right], \tag{2.2.1.30} \\
\tilde{S}_0^{(2)}(T,x,x_0,q_{\epsilon}(t),\epsilon) &= \int_0^T dt \left[ b(t) \frac{q_{\epsilon}(t)}{\sqrt{\hbar}} \right]. \nonumber
\end{align}
$$

From Eq.(2.2.1.29) we obtain

$$
\begin{align}
\theta_{\epsilon}(T, x, x_0, \hbar) = & \nonumber \\
& \theta_{\epsilon}^{(1)}(T, x, x_0, \alpha, \hbar) + \theta_{\epsilon}^{(2)}(T, x, x_0, \alpha, \hbar) + \tag{2.2.1.30} \\
& + \theta_{\epsilon}^{(3)}(T, x, x_0, \alpha, \hbar) + \theta_{\epsilon}^{(4)}(T, x, x_0, \alpha, \hbar). \nonumber
\end{align}
$$

Here
$$
\theta_{\epsilon}^{(1)}(T, x, x_0, \alpha, \hbar) =
$$

$$
\begin{align*}
& \int_{q(T)=\frac{x}{\sqrt{\hbar}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t)-\alpha(t))^2 dt\right) \times \\
& \qquad \exp\left[i(\tilde{S}_0^{(1)}(\cdot) + \tilde{S}_0^{(2)}(\cdot))\right],
\end{align*}
$$

$$
\theta_{\epsilon}^{(2)}(T, x, x_0, \alpha, \hbar) =
$$

$$
\begin{equation}
\begin{split}
& \int_{q(T)=\frac{x}{\sqrt{\hbar}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t)-\alpha(t))^2 dt\right) \times \\
& \qquad \exp\left[-i(\tilde{S}_0^{(1)}(\cdot) - \tilde{S}_0^{(2)}(\cdot))\right],
\end{split}
\tag{2.2.1.31}
\end{equation}
$$

$$
\theta_{\epsilon}^{(3)}(T, x, x_0, \epsilon) =
$$

$$
\begin{equation}
\begin{split}
& \int_{q(T)=\frac{x}{\sqrt{\hbar}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t)-\alpha(t))^2 dt\right) \times \\
& \qquad \exp\left[i(\tilde{S}_0^{(1)}(\cdot) - \tilde{S}_0^{(2)}(\cdot))\right],
\end{split}
\tag{2.2.1.32}
\end{equation}
$$

$$
\theta_{\epsilon}^{(4)}(T, x, x_0, \alpha, \hbar) =
$$

$$
\begin{equation}
\begin{split}
& \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t)-\alpha(t))^2 dt\right) \times \\
& \qquad \exp\left[-i(\tilde{S}_0^{(1)}(\cdot) + \tilde{S}_0^{(2)}(\cdot))\right].
\end{split}
\tag{2.2.1.33}
\end{equation}
$$
$$
\theta_{\varepsilon}^{(1)}(T, x, x_0, \alpha, \hbar) =
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\varepsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t)-\alpha(t))^2 dt\right) \times \\
\exp\left[i(\tilde{S}_0^{(1)}(\cdot) + \tilde{S}_0^{(2)}(\cdot))\right],
$$

$$
\theta_{\epsilon}^{(2)}(T, x, x_0, \alpha, \hbar) =
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\varepsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t)-\alpha(t))^2 dt\right) \times \\
\exp\left[-i(\tilde{S}_0^{(1)}(\cdot) - \tilde{S}_0^{(2)}(\cdot))\right], \qquad (2.2.1.31)
$$

$$
\theta_{\epsilon}^{(3)}(T, x, x_0, \epsilon) =
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\hbar}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t)-\alpha(t))^2 dt\right) \times \\
\exp\left[i(\tilde{S}_0^{(1)}(\cdot) - \tilde{S}_0^{(2)}(\cdot))\right],
$$

$$
\theta_{\epsilon}^{(4)}(T, x, x_0, \alpha, \hbar) =
$$

$$
\int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t)-\alpha(t))^2 dt\right) \times \\
\exp\left[-i(\tilde{S}_0^{(1)}(\cdot) + \tilde{S}_0^{(2)}(\cdot))\right].
$$
Let's calculate now path integral $\theta_{\epsilon}^{(1)}(T, x, x_0, \alpha, \hbar)$:

$$ \theta_{\epsilon}^{(1)}(T, x, x_0, \epsilon) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t) - \alpha(t))^2 dt\right) \times $$

$$ \exp\left[i\left(\tilde{S}_0^{(1)}(\cdot) + \tilde{S}_0^{(2)}(\cdot)\right)\right] = $$

$$ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{\rho}_{q,\epsilon}^{(1)}(\cdot) \times \qquad (2.1.32) $$

$$ \exp\left\{i \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon}^2(t) + b(t) \frac{q_{\epsilon}(t)}{\sqrt{\hbar}} \right] \right\} \times $$

$$ \exp\left(-\frac{1}{\hbar} \int_{0}^{T} (\sqrt{\hbar} q(t) - \alpha(t))^2 dt\right) $$

by using saddle point method. Using replacement $q_{\epsilon}(t) := \frac{q_{\epsilon, \hbar_{\epsilon}}(t)}{\sqrt{\hbar_{\epsilon}}}$ we rewrite
(2.2.1.32) in the next equivalent form
Let's calculate now path integral $\theta_{\epsilon}^{(1)}(T, x, x_0, \alpha, \hbar)$:

$$ \theta_{\epsilon}^{(1)}(T, x, x_0, \epsilon) = $$

$$ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{p}_{q,\epsilon}^{(1)}(\cdot) \exp\left(-\frac{1}{\hbar} \int_0^T (\sqrt{\hbar}q(t) - \alpha(t))^2 dt\right) \times $$

$$ \exp\left[i\left(\tilde{S}_0^{(1)}(\cdot) + \tilde{S}_0^{(2)}(\cdot)\right)\right] = $$

$$ \int_{q(T)=\frac{x}{\sqrt{\epsilon}}} D[q(t)] |\Psi_0(\sqrt{\hbar}q(0))| \tilde{p}_{q,\epsilon}^{(1)}(\cdot) \times \qquad (2.1.32) $$

$$ \exp\left\{i \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{dq(t)}{dt} \right)^2 - a q_{\epsilon}^2(t) + b(t) \frac{q_{\epsilon}(t)}{\sqrt{\hbar}} \right] \right\} \times $$

$$ \exp\left(-\frac{1}{\hbar} \int_{0}^{T} (\sqrt{\hbar}q(t) - \alpha(t))^2 dt\right) $$

by using saddle point method. Using replacement $q_{\epsilon}(t) := \frac{q_{\epsilon, \hbar_{\epsilon}}(t)}{\sqrt{\hbar_{\epsilon}}}$ we rewrite
(2.2.1.32) in the next equivalent form
$$ \theta_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \int_{q(T)=x} D^{+}[q(t), \varepsilon] |\Psi_{0,\varepsilon}(q(0))| \times $$

$$ \exp\left\{\frac{i}{\hbar} \int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R}, \varepsilon}^{2}(t) - a q_{\varepsilon}^{2}(t) + b(t) q_{\varepsilon}(t)\right]\right\} \times $$

$$ \exp\left(-v\hbar\varepsilon \int_{0}^{T} (q_{\varepsilon}(t) - \alpha(t))^{2} dt\right) \exp\left[-\frac{1}{\hbar\varepsilon}(\gamma_{T}^{2}q(0) + \theta_{T}^{2}q(T))\right] = \tag{2.1.33} $$

$$ \int_{q(T)=x} D^{+}[q(t), \varepsilon] |\Psi_{0,\varepsilon}(q(0))| \times $$

$$ \exp\left\{\frac{i}{\hbar} \int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - a q_{\varepsilon}^{2}(t) + b(t) q_{\varepsilon}(t) + i\hbar\varepsilon(q_{\varepsilon}(t) - \alpha(t))^{2}\right]\right\} = $$

$$ \int_{q(T)=x} D^{+}[q(t), \varepsilon] |\Psi_{0,\varepsilon}(q(0))| \exp\left[-\frac{1}{\hbar\varepsilon}(\gamma_{T}^{2}q(0) + \theta_{T}^{2}q(T))\right] $$

$$ \exp\left\{\frac{i}{\hbar} \int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R},\varepsilon}^{2}(t) - (a - \hbar\varepsilon i)q_{\varepsilon}^{2}(t) + (b(t) - 2i\hbar\varepsilon\alpha(t))q_{\varepsilon}(t) + i\hbar\varepsilon\alpha^{2}(t)\right]\right\}. $$

We assume now that

$$ \Psi(x, 0) = \frac{1}{\sqrt{2\pi\hbar\varepsilon}} \exp\left(-\frac{x^2}{2\hbar\varepsilon}\right), \qquad (2.1.34) $$

Substitution Eq.(2.1.34) into Eq.(2.1.33) gives
$$ \theta_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}) = $$

$$ \int_{q(T)=x} D^{+}[q(t), \varepsilon] |\Psi_{0,\varepsilon}(q(0))| \times $$

$$ \exp\left\{\frac{i}{\hbar} \int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R}, \varepsilon}^{2}(t) - a q_{\varepsilon}^{2}(t) + b(t) q_{\varepsilon}(t)\right]\right\} \times $$

$$ \exp\left(-v\hbar\varepsilon \int_{0}^{T} (q_{\varepsilon}(t) - \alpha(t))^{2} dt\right) \exp\left[-\frac{1}{\hbar\varepsilon}(\gamma_{T}^{2}q(0) + \theta_{T}^{2}q(T))\right] = \tag{2.1.33} $$

$$ \int_{q(T)=x} D^{+}[q(t), \varepsilon] |\Psi_{0,\varepsilon}(q(0))| \times $$

$$ \exp\left\{\frac{i}{\hbar} \int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R}, \varepsilon}^{2}(t) - a q_{\varepsilon}^{2}(t) + b(t) q_{\varepsilon}(t) + i \hbar \varepsilon (q_{\varepsilon}(t) - \alpha(t))^{2}\right]\right\} = $$

$$ \int_{q(T)=x} D^{+}[q(t), \varepsilon] |\Psi_{0,\varepsilon}(q(0))| \exp\left[-\frac{1}{\hbar\varepsilon}(\gamma_{T}^{2}q(0) + \theta_{T}^{2}q(T))\right] $$

$$ \exp\left\{\frac{i}{\hbar} \int_{0}^{T} dt \left[\frac{m}{2} \dot{q}_{\mathbb{R}, \varepsilon}^{2}(t) - (a - \hbar\varepsilon i)q_{\varepsilon}^{2}(t) + (b(t) - 2i\hbar\varepsilon\alpha(t))q_{\varepsilon}(t) + i\hbar\varepsilon\alpha^{2}(t)\right]\right\}. $$

We assume now that

$$ \Psi(x, 0) = \frac{1}{\sqrt{2\pi\hbar\varepsilon}} \exp\left(-\frac{x^2}{2\hbar\varepsilon}\right), \qquad (2.1.34) $$

Substitution Eq.(2.1.34) into Eq.(2.1.33) gives
$$
\theta_{\varepsilon}^{(1)}(T, x, \alpha, \hbar_{\varepsilon}) =
$$

$$
\frac{1}{\sqrt[4]{2\pi\hbar\epsilon}} \int_{\substack{q(T)=x \\ q(0)=y}} dy D^+[q(t), \epsilon] \exp\left(-\frac{(y-x_0)^2}{2\hbar\epsilon}\right) \exp\left[-\frac{1}{\hbar\epsilon}(\gamma_T^2 q^2(0) + \theta_T^2 q^2(T))\right] \times
$$

$$
\exp\left\{\frac{i}{\hbar} \int_{0}^{T} dt \left[\frac{m}{2} q_{\tilde{\mathbb{R}}, \epsilon}^{2}(t) - \hat{a} q_{\epsilon}^{2}(t) + \hat{b}(t) q_{\epsilon}(t)\right]\right\} =
$$

$$
\frac{1}{\sqrt[4]{2\pi\hbar\epsilon}} \int_{\substack{q(T)=x \\ q(0)=y}} dy D^+[q(t), \epsilon] \exp\left(-\frac{(y-x_0)^2}{2\hbar\epsilon}\right) \exp\left[-\frac{1}{\hbar\epsilon}(\gamma_T^2 y^2 + \theta_T^2 x^2)\right] \times \quad (2.1.35)
$$

$$
\exp\left\{\frac{i}{\hbar\epsilon} \int_{0}^{T} \mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}(t), q_{\epsilon}(t)) dt\right\},
$$

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}(t), q_{\epsilon}(t)) = \frac{m}{2} q_{\tilde{\mathbb{R}},\epsilon}^2(t) - \hat{a} q_{\epsilon}^2(t) + \hat{b}(t) q_{\epsilon}(t) + i\hbar\epsilon\alpha^2(t),
$$

$$
\hat{a} = a - \hbar \epsilon i,
$$

$$
\hat{b}(t) = b(t) - 2i\hbar\epsilon\alpha(t).
$$

We rewrite $\mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}(t), q_{\epsilon}(t))$ in the next equivalent form:

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}(t), q_{\epsilon}(t)) = \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - \frac{m\omega^2}{2} q_{\epsilon}^2(t) + \hat{b}(t)q_{\epsilon}(t) + i\hbar\epsilon\alpha^2(t), \quad (2.1.36)
$$

$$
\frac{m\omega^2}{2} = \hat{a}.
$$
$$
\theta_{\varepsilon}^{(1)}(T, x, \alpha, \hbar_{\varepsilon}) =
$$

$$
\frac{1}{\sqrt[4]{2\pi\hbar\epsilon}} \int_{\substack{q(T)=x \\ q(0)=y}} dy D^+[q(t), \epsilon] \exp\left(-\frac{(y-x_0)^2}{2\hbar\epsilon}\right) \exp\left[-\frac{1}{\hbar\epsilon}(\gamma_T^2 q^2(0) + \theta_T^2 q^2(T))\right] \times
$$

$$
\exp\left\{\frac{i}{\hbar} \int_{0}^{T} dt \left[\frac{m}{2} q_{\tilde{\mathbb{R}}, \epsilon}^{2}(t) - \hat{a} q_{\epsilon}^{2}(t) + \hat{b}(t) q_{\epsilon}(t)\right]\right\} =
$$

$$
\frac{1}{\sqrt[4]{2\pi\hbar\epsilon}} \int_{\substack{q(T)=x \\ q(0)=y}} dy D^+[q(t), \epsilon] \exp\left(-\frac{(y-x_0)^2}{2\hbar\epsilon}\right) \exp\left[-\frac{1}{\hbar\epsilon}(\gamma_T^2 y^2 + \theta_T^2 x^2)\right] \times \quad (2.1.35)
$$

$$
\exp\left\{\frac{i}{\hbar\epsilon} \int_{0}^{T} \mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}(t), q_{\epsilon}(t)) dt\right\},
$$

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}(t), q_{\epsilon}(t)) = \frac{m}{2} q_{\tilde{\mathbb{R}},\epsilon}^2(t) - \hat{a} q_{\epsilon}^2(t) + \hat{b}(t) q_{\epsilon}(t) + i\hbar\epsilon\alpha^2(t),
$$

$$
\hat{a} = a - \hbar \epsilon i,
$$

$$
\hat{b}(t) = b(t) - 2i\hbar\epsilon\alpha(t).
$$

We rewrite $\mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}(t), q_{\epsilon}(t))$ in the next equivalent form:

$$
\mathcal{L}(\dot{q}_{\mathbb{R},\epsilon}(t), q_{\epsilon}(t)) = \frac{m}{2} \dot{q}_{\mathbb{R},\epsilon}^2(t) - \frac{m\omega^2}{2} q_{\epsilon}^2(t) + \hat{b}(t)q_{\epsilon}(t) + i\hbar\epsilon\alpha^2(t), \quad (2.1.36)
$$

$$
\frac{m\omega^2}{2} = \hat{a}.
$$
The Euler-Lagrange equation of $\mathcal{L}_\epsilon(\dot{q}, q_\epsilon)$ with corresponding boundary conditions is

$$ \frac{d}{dt} \left( \frac{\partial \mathcal{L}_\epsilon}{\partial \dot{q}_{\mathcal{R},\epsilon}} \right) - \frac{\partial \mathcal{L}_\epsilon}{\partial q} = 0, \qquad (2.1.37) $$

$$ q(0) = y, q(T) = x. $$

Corresponding complex action $S(y, x, T)$ is:

$$ S(y,x,T) = \frac{m\bar{\omega}}{2\sin\bar{\omega}T} \left[ \begin{aligned} & (\cos\bar{\omega}T)(y^2 + x^2) - 2xy + \frac{2x}{m\bar{\omega}} \int_0^T \hat{b}(t)\sin(\bar{\omega}t)dt + \\ & \qquad + \frac{2y}{m\bar{\omega}} \int_0^T \hat{b}(t)\sin\bar{\omega}(T-t)dt - \\ & \qquad - \frac{2}{m^2\bar{\omega}^2} \int_0^T \int_0^t \hat{b}(t)b(s)\sin\bar{\omega}(T-t)\sin(\bar{\omega}s)dsdt + i\hbar_\epsilon \int_0^T \alpha^2(t)dt \end{aligned} \right] = \frac{m\bar{\omega}}{2\sin\bar{\omega}T} [(\cos\bar{\omega}T)(y^2 + x^2) - 2xy + 2w_2x + 2w_1y + \Theta_T] \quad (2.1.38) $$

$$ \Theta_T = -\frac{2}{m^2\bar{\omega}^2} \int_0^T \int_0^t \hat{b}(t)b(s)\sin\bar{\omega}(T-t)\sin(\bar{\omega}s)dsdt + i\hbar_\epsilon \int_0^T \alpha^2(t)dt, $$

$$ w_1 = \frac{1}{m\bar{\omega}} \int_0^T \hat{b}(t) \sin \bar{\omega} (T-t) dt, $$

$$ w_2 = \frac{1}{m\bar{\omega}} \int_0^T \hat{b}(t) \sin(\bar{\omega}t) dt $$

$$ y = \frac{1}{\cos \bar{\omega} T} \left( x - \frac{1}{m\bar{\omega}} \int_{0}^{T} \hat{b}(t) \sin \bar{\omega} (T-t) dt \right), w_T = \frac{1}{m\bar{\omega}} \int_{0}^{T} \hat{b}(t) \sin \bar{\omega} (T-t) dt $$

$$ y = \frac{1}{\cos \bar{\omega} T} (x - w_T), \quad \frac{m\bar{\omega}}{2\sin \bar{\omega} T} = \alpha_T $$
The Euler-Lagrange equation of $\mathcal{L}_\epsilon(\dot{q}, q_\epsilon)$ with corresponding boundary conditions is

$$ \frac{d}{dt} \left( \frac{\partial \mathcal{L}_\epsilon}{\partial \dot{q}_{\mathcal{R},\epsilon}} \right) - \frac{\partial \mathcal{L}_\epsilon}{\partial q} = 0, \qquad (2.1.37) $$

$$ q(0) = y, q(T) = x. $$

Corresponding complex action $S(y, x, T)$ is:

$$ S(y,x,T) = \frac{m\bar{\omega}}{2\sin\bar{\omega}T} \left[ \begin{aligned} & (\cos\bar{\omega}T)(y^2 + x^2) - 2xy + \frac{2x}{m\bar{\omega}} \int_0^T \hat{b}(t) \sin(\bar{\omega}t)dt + \\ & \qquad + \frac{2y}{m\bar{\omega}} \int_0^T \hat{b}(t) \sin\bar{\omega}(T-t)dt - \\ & \qquad - \frac{2}{m^2\bar{\omega}^2} \int_0^T \int_0^t \hat{b}(t)b(s) \sin\bar{\omega}(T-t)\sin(\bar{\omega}s)dsdt + i\hbar_\epsilon \int_0^T \alpha^2(t)dt \end{aligned} \right] = \frac{m\bar{\omega}}{2\sin\bar{\omega}T} [(\cos\bar{\omega}T)(y^2 + x^2) - 2xy + 2w_2x + 2w_1y + \Theta_T] \quad (2.1.38) $$

$$ \Theta_T = -\frac{2}{m^2\bar{\omega}^2} \int_0^T \int_0^t \hat{b}(t)b(s) \sin\bar{\omega}(T-t)\sin(\bar{\omega}s)dsdt + i\hbar_\epsilon \int_0^T \alpha^2(t)dt, $$

$$ w_1 = \frac{1}{m\bar{\omega}} \int_0^T \hat{b}(t) \sin\bar{\omega}(T-t)dt, $$

$$ w_2 = \frac{1}{m\bar{\omega}} \int_0^T \hat{b}(t) \sin(\bar{\omega}t)dt $$

$$ y = \frac{1}{\cos\bar{\omega}T} \left( x - \frac{1}{m\bar{\omega}} \int_0^T \hat{b}(t) \sin\bar{\omega}(T-t)dt \right), w_T = \frac{1}{m\bar{\omega}} \int_0^T \hat{b}(t) \sin\bar{\omega}(T-t)dt $$

$$ y = \frac{1}{\cos\bar{\omega}T}(x - w_T), \quad \frac{m\bar{\omega}}{2\sin\bar{\omega}T} = \alpha_T $$
$$S'(v, x, T) = i\alpha_T[(\cos \omega T)(v^2 + x^2) - 2xy + 2xw_T + 2yw_T] - \gamma_T^2 y^2$$

$$\frac{\partial S'}{\partial y} = i\alpha_T[2y \cos \omega T - 2x + 2w_T] - 2y\gamma_T^2 = 0$$

$$2y \cos \omega T - 2x + 2w_T - 2y \frac{\gamma_T^2}{i\alpha_T} = 0, \quad 2y \cos \omega T - 2x + 2w_T + 2y \frac{i\gamma_T^2}{\alpha_T} = 0$$

$$y \left[ \cos \omega T + \frac{i\gamma_T^2}{\alpha_T} \right] = x - w_T,$$

$$y = \frac{x}{\beta_T} - \frac{w_T}{\beta_T}, \quad \beta_T = \cos \omega T + \frac{i\gamma_T^2}{\alpha_T},$$

$$S'(v, x, T) = i\alpha_T[(\cos \omega T)(v^2 + x^2) - 2xy + 2xw_T + 2yw_T] - \gamma_T^2 y^2 =$$

$$i\alpha_T \left[ (\cos \omega T) \left[ \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 + x^2 \right] - 2x \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) + 2xw_T + 2w_T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) \right] - \gamma_T^2 \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 =$$

$$i\alpha_T \cos \omega T \left[ \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 + x^2 \right] - 2x i\alpha_T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) + 2x i\alpha_T w_T + 2i\alpha_T w_T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) - \gamma_T^2 \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 =$$

$$i\alpha_T \cos \omega T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 + x^2 i\alpha_T \cos \omega T - 2x i\alpha_T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) + 2x i\alpha_T w_T + 2i\alpha_T w_T \left( \frac{x}{\beta_T} - \gamma_T^2 \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 \right) =$$

$$\left(\frac{x}{\beta_T} - \frac{w_T}{\beta_T}\right)^2 \left(i\alpha_T \cos \omega T - \gamma_T^2\right) + x^2 i\alpha_T \cos \omega T - 2i\alpha_T w_T \left(\frac{x}{\beta_T} - \frac{w_T}{\beta_T}\right) (x - w_T) + 2x i\alpha_T w_T \beta_T^{-2} \left(i\alpha_T \cos \omega T - \gamma_T^2\right) (x - w_T)^2 + x^2 i\alpha_T \cos \omega T - 2i\alpha_T w_T \beta_T^{-1} (x - w_T)^2 + 2x i\alpha_T w_T = \\ (x - w_T)^2 \underbrace{\left[\beta_T^{-2} \left(i\alpha_T \cos \omega T - \gamma_T^2\right) - 2i\alpha_T w_T \beta_T^{-1}\right]}_{\delta_T} + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T = \\ (x - w_T)^2 \delta_T + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T = \delta_T (x^2 - 2xw_T + w_T^2) + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T = \\ \delta_T x^2 - 2x \delta_T w_T + \delta_T w_T^2 + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T =$$

$$\delta_T x^2 - 2x \delta_T w_T + \delta_T w_T^2 + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T =$$

$$S(x) = x^2(\delta_T + i\alpha_T \cos \omega T) - 2xw_T(\delta_T - i\alpha_T) + \delta_T w_T^2 - \theta_T^2 x^2 =$$

$$x^2[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2] - 2xw_T(\delta_T - i\alpha_T) + \delta_T w_T^2 =$$

$$\frac{\partial S(x)}{\partial x} = 2x(\delta_T + i\alpha_T \cos \omega T) - 2w_T(\delta_T - i\alpha_T) - 2x\theta_T^2 = 0$$

$$x[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2] = (\delta_T w_T - i\alpha_T w_T), x_{cr} = \frac{w_T(\delta_T - i\alpha_T)}{[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2]}$$
$$S'(v, x, T) = i\alpha_T[(\cos \omega T)(v^2 + x^2) - 2xy + 2xw_T + 2yw_T] - \gamma_T^2 y^2$$

$$\frac{\partial S'}{\partial y} = i\alpha_T[2y \cos \omega T - 2x + 2w_T] - 2y\gamma_T^2 = 0$$

$$2y \cos \omega T - 2x + 2w_T - 2y \frac{\gamma_T^2}{i\alpha_T} = 0, \quad 2y \cos \omega T - 2x + 2w_T + 2y \frac{i\gamma_T^2}{\alpha_T} = 0$$

$$y \left[ \cos \omega T + \frac{i\gamma_T^2}{\alpha_T} \right] = x - w_T,$$

$$y = \frac{x}{\beta_T} - \frac{w_T}{\beta_T}, \quad \beta_T = \cos \omega T + \frac{i\gamma_T^2}{\alpha_T},$$

$$S'(v, x, T) = i\alpha_T[(\cos \omega T)(v^2 + x^2) - 2xy + 2xw_T + 2yw_T] - \gamma_T^2 y^2 =$$

$$i\alpha_T \left[ (\cos \omega T) \left[ \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 + x^2 \right] - 2x \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) + 2xw_T + 2w_T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) \right] - \gamma_T^2 \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 =$$

$$i\alpha_T \cos \omega T \left[ \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 + x^2 \right] - 2x i\alpha_T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) + 2x i\alpha_T w_T + 2i\alpha_T w_T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) - \gamma_T^2 \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 =$$

$$i\alpha_T \cos \omega T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 + x^2 i\alpha_T \cos \omega T - 2x i\alpha_T \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right) + 2x i\alpha_T w_T + 2i\alpha_T w_T \left( \frac{x}{\beta_T} - \gamma_T^2 \left( \frac{x}{\beta_T} - \frac{w_T}{\beta_T} \right)^2 \right) =$$

$$\left(\frac{x}{\beta_T} - \frac{w_T}{\beta_T}\right)^2 \left(i\alpha_T \cos \omega T - \gamma_T^2\right) + x^2 i\alpha_T \cos \omega T - 2i\alpha_T w_T \left(\frac{x}{\beta_T} - \frac{w_T}{\beta_T}\right) (x - w_T) + 2x i\alpha_T w_T \beta_T^{-2} \left(i\alpha_T \cos \omega T - \gamma_T^2\right) (x - w_T)^2 + x^2 i\alpha_T \cos \omega T - 2i\alpha_T w_T \beta_T^{-1} (x - w_T)^2 + 2x i\alpha_T w_T = \\ (x - w_T)^2 \underbrace{\left[\beta_T^{-2} \left(i\alpha_T \cos \omega T - \gamma_T^2\right) - 2i\alpha_T w_T \beta_T^{-1}\right]}_{\delta_T} + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T = \\ (x - w_T)^2 \delta_T + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T = \delta_T (x^2 - 2xw_T + w_T^2) + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T = \\ \delta_T x^2 - 2x \delta_T w_T + \delta_T w_T^2 + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T =$$

$$\delta_T x^2 - 2x \delta_T w_T + \delta_T w_T^2 + x^2 i\alpha_T \cos \omega T + 2x i\alpha_T w_T =$$

$$S(x) = x^2(\delta_T + i\alpha_T \cos \omega T) - 2xw_T(\delta_T - i\alpha_T) + \delta_T w_T^2 - \theta_T^2 x^2 =$$

$$x^2[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2] - 2xw_T(\delta_T - i\alpha_T) + \delta_T w_T^2 =$$

$$\frac{\partial S(x)}{\partial x} = 2x(\delta_T + i\alpha_T \cos \omega T) - 2w_T(\delta_T - i\alpha_T) - 2x\theta_T^2 = 0$$

$$x[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2] = (\delta_T w_T - i\alpha_T w_T), x_{cr} = \frac{w_T(\delta_T - i\alpha_T)}{[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2]}$$
$$S(x_{cr}) = \frac{w_T^2(\delta_T - i\alpha_T)^2}{[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2]} - \frac{2w_T^2(\delta_T - i\alpha_T)^2}{[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2]} = \frac{w_T^2(\delta_T - i\alpha_T)^2}{[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2]}$$

Substitution Eq.(2.1.38) into Eq.(2.1.35) gives

$$ 
\begin{align} 
\theta_{\varepsilon}^{(1)}(T, x, x_0, \alpha, \hbar_{\varepsilon}) &= \nonumber \\ 
&\quad \frac{1}{\sqrt[4]{2\pi\hbar_{\varepsilon}}} \int_{\substack{q(T)=x \\ q(0)=y}} dy D^{+}[q(t), \varepsilon] \exp\left(-\frac{y^2}{2\hbar_{\varepsilon}}\right) \exp\left[-\frac{1}{\hbar_{\varepsilon}}(\gamma_T^2 q^2(0) + \theta_T^2 q^2(T))\right] \times \tag{2.1.39} \\ 
&\quad \exp\left\{\frac{i}{\hbar_{\varepsilon}} \int_{0}^{T} \mathcal{L}(\dot{q}_{\varepsilon}(t), q_{\varepsilon}(t)) dt\right\} = \nonumber \\ 
&\quad = \tilde{\theta}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}). \nonumber 
\end{align} 
$$

Here
$$S(x_{cr}) = \frac{w_T^2(\delta_T - i\alpha_T)^2}{[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2]} - \frac{2w_T^2(\delta_T - i\alpha_T)^2}{[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2]} = \frac{w_T^2(\delta_T - i\alpha_T)^2}{[(\delta_T + i\alpha_T \cos \omega T) - \theta_T^2]}$$

Substitution Eq.(2.1.38) into Eq.(2.1.35) gives

$$\begin{align}
\theta_{\varepsilon}^{(1)}(T, x, x_0, \alpha, \hbar_{\varepsilon}) &= \nonumber \\
&\quad \frac{1}{\sqrt[4]{2\pi\hbar_{\varepsilon}}} \int_{\substack{q(T)=x \\ q(0)=y}} dy D^{+}[q(t), \varepsilon] \exp\left(-\frac{y^2}{2\hbar_{\varepsilon}}\right) \exp\left[-\frac{1}{\hbar_{\varepsilon}}(\gamma_T^2 q^2(0) + \theta_T^2 q^2(T))\right] \times \tag{2.1.39} \\
&\quad \exp\left\{\frac{i}{\hbar_{\varepsilon}} \int_0^T \mathcal{L}(\dot{q}_{\varepsilon}(t), q_{\varepsilon}(t)) dt\right\} = \nonumber \\
&\quad = \tilde{\theta}_{\varepsilon}^{(1)}(T, x, x_0, \hbar_{\varepsilon}). \nonumber
\end{align}$$

Here
$$
\tilde{\theta}_{\epsilon}^{(1)}(T, x, \alpha, \hbar_{\epsilon}) =
$$

$$
\frac{1}{\sqrt[4]{2\pi\hbar\epsilon}} \sqrt{\frac{m\bar{\omega}}{2\pi i\hbar\epsilon \sin\omega T}} \int_{-\infty}^{\infty} dy \exp\left(-\frac{y^2}{2\hbar\epsilon}\right) \times
$$

$$
\exp\left\{ \frac{1}{\hbar_{\epsilon}} \left[ i \frac{m\bar{\omega}}{2\sin\bar{\omega}T} S(y,x,T) - (\gamma_T^2 y^2 + \theta_T^2 x^2) \right] \right\} =
$$

$$
\frac{1}{\sqrt{2\pi\hbar\epsilon}} \sqrt{\frac{\alpha_T}{2\pi i\hbar\epsilon}} \int_{-\infty}^{\infty} dy \exp\left\{\frac{1}{\hbar\epsilon}\left[i\alpha_T S(y,x,T) - (\bar{\gamma}_T^2 y^2 + \theta_T^2 x^2)\right]\right\} =
$$

$$
\frac{1}{\sqrt{2\pi\hbar\epsilon}} \sqrt{\frac{\alpha_T}{2\pi i\hbar\epsilon}} \int_{-\infty}^{\infty} dy \exp\left\{\frac{1}{\hbar\epsilon}\tilde{S}(y,x,T)\right\}, \qquad (2.1.40)
$$

$$
\tilde{S}(y, x, T) = i\alpha_T S(y, x, T) - (\bar{\gamma}_T^2 y^2 + \theta_T^2 x^2),
$$

$$
S(y, x, T) = (\cos \omega T)(y^2 + x^2) - 2xy + 2w_2x + 2w_1y + \Theta_T,
$$

$$
\bar{\gamma}_T^2 = 2\gamma_T^2,
$$

$$
\alpha_T = \frac{m\bar{\omega}}{2\sin\bar{\omega}T}
$$

$\sin \omega T \neq 0.$

Let's calculate now integral $\tilde{\theta}_{\epsilon}^{(1)}(T, x, \alpha, \hbar_{\epsilon})$ by using saddle point method.

$$
\cos \omega T + \frac{i \gamma_T^2}{\alpha_T}
$$
$$
\tilde{\theta}_{\epsilon}^{(1)}(T, x, \alpha, \hbar_{\epsilon}) =
$$

$$
\frac{1}{\sqrt[4]{2\pi\hbar\epsilon}} \sqrt{\frac{m\omega}{2\pi i\hbar\epsilon \sin\omega T}} \int_{-\infty}^{\infty} dy \exp\left(-\frac{y^2}{2\hbar\epsilon}\right) \times
$$

$$
\exp\left\{ \frac{1}{\hbar_{\epsilon}} \left[ i \frac{m\omega}{2\sin\omega T} S(y, x, T) - (\gamma_T^2 y^2 + \theta_T^2 x^2) \right] \right\} =
$$

$$
\frac{1}{\sqrt[4]{2\pi\hbar\epsilon}} \sqrt{\frac{\alpha_T}{2\pi i\hbar\epsilon}} \int_{-\infty}^{\infty} dy \exp\left\{\frac{1}{\hbar\epsilon}\left[i\alpha_T S(y,x,T) - (\bar{\gamma}_T^2 y^2 + \theta_T^2 x^2)\right]\right\} =
$$

$$
\frac{1}{\sqrt[4]{2\pi\hbar\epsilon}} \sqrt{\frac{\alpha_T}{2\pi i\hbar\epsilon}} \int_{-\infty}^{\infty} dy \exp\left\{\frac{1}{\hbar\epsilon}\tilde{S}(y,x,T)\right\}, \qquad (2.1.40)
$$

$$
\tilde{S}(y, x, T) = i\alpha_T S(y, x, T) - (\bar{\gamma}_T^2 y^2 + \theta_T^2 x^2),
$$

$$
S(y, x, T) = (\cos \omega T)(y^2 + x^2) - 2xy + 2w_2x + 2w_1y + \Theta_T,
$$

$$
\bar{\gamma}_T^2 = 2\gamma_T^2,
$$

$$
\alpha_T = \frac{m\omega}{2\sin\omega T}
$$

$\sin \omega T \neq 0.$

Let's calculate now integral $\tilde{\theta}_{\epsilon}^{(1)}(T, x, \alpha, \hbar_{\epsilon})$ by using saddle point method.

$$
\cos \omega T + \frac{i \gamma_T^2}{\alpha_T}
$$
$$ \frac{\partial \tilde{S}}{\partial y} = i\alpha_T[2y\cos\omega T - 2x + 2w_1] - 2\bar{\gamma}_T^2y = 0, $$

$$ y(i\alpha_T \cos \omega T - \bar{\gamma}_T^2) = i\alpha_T(x - w_1), \qquad (2.1.41) $$

$$ y\beta_T = i\alpha_T(x - w_1), $$

$$ \beta_T = i\alpha_T \cos \omega T - \bar{\gamma}_T^2. $$

Therefore saddle point $y_{cr}(x, T)$ is

$$ y_{cr}(x, T) = \frac{i\alpha_T(x - w_1)}{i\alpha_T \cos \omega T - \bar{\gamma}_T^2} = i\alpha_T \beta_T^{-1}(x - w_1), $$

$$ \beta_T = i\alpha_T \cos \omega T - \bar{\gamma}_T^2, \qquad (2.1.42) $$

$$ y_{cr}(x, T) = i\alpha_T \beta_T^{-1} \left( x - \frac{1}{m\omega} \int_0^T \hat{b}(t) \sin \omega (T-t) dt \right), $$

$$ \cos \omega T \neq 0. $$

Hence

$$ y_{cr}(x, T) = i\alpha_T \beta_T^{-1} (x - w_1) $$

$$ y_{cr}^2(x, T) = -\alpha_T^2 \beta_T^{-2} (x - w_1)^2 $$
$$ \frac{\partial \tilde{S}}{\partial y} = i\alpha_T[2y\cos\omega T - 2x + 2w_1] - 2\bar{\gamma}_T^2y = 0, $$

$$ y(i\alpha_T \cos \omega T - \bar{\gamma}_T^2) = i\alpha_T(x - w_1), \qquad (2.1.41) $$

$$ y\beta_T = i\alpha_T(x - w_1), $$

$$ \beta_T = i\alpha_T \cos \omega T - \bar{\gamma}_T^2. $$

Therefore saddle point $y_{cr}(x, T)$ is

$$ y_{cr}(x, T) = \frac{i\alpha_T(x - w_1)}{i\alpha_T \cos \omega T - \bar{\gamma}_T^2} = i\alpha_T \beta_T^{-1}(x - w_1), $$

$$ \beta_T = i\alpha_T \cos \omega T - \bar{\gamma}_T^2, \qquad (2.1.42) $$

$$ y_{cr}(x, T) = i\alpha_T \beta_T^{-1} \left( x - \frac{1}{m\omega} \int_0^T \hat{b}(t) \sin \omega (T-t) dt \right), $$

$$ \cos \omega T \neq 0. $$

Hence

$$ y_{cr}(x, T) = i\alpha_T \beta_T^{-1} (x - w_1) $$

$$ y_{cr}^2(x, T) = -\alpha_T^2 \beta_T^{-2} (x - w_1)^2 $$
$$
\tilde{S}(y_{cr}, x, T) = i\alpha_T S(y_{cr}, x, T) - (\bar{\gamma}_T^2 y_{cr}^2 + \theta_T^2 x^2) =
$$

$$
i\alpha_T[(\cos \omega T)(y_{cr}^2 + x^2) - 2xy_{cr} + 2w_2x + 2w_1y_{cr} + \Theta_T] - (\bar{\gamma}_T^2 y_{cr}^2 + \theta_T^2 x^2) =
$$

$$
i\alpha_T(\cos \omega T)y_{cr}^2 + i\alpha_T(\cos \omega T)x^2 - 2i\alpha_T xy_{cr} + 2i\alpha_T w_2 x + 2i\alpha_T w_1 y_{cr} + i\alpha_T \Theta_T -
$$

$$
-\bar{\gamma}_T^2 y_{cr}^2 - \theta_T^2 x^2 =
$$

$$
(i\alpha_T \cos \omega T - \bar{\gamma}_T^2)y_{cr}^2 + (i\alpha_T \cos \omega T - \theta_T^2)x^2 - 2i\alpha_T y_{cr}(x - w_1) + 2i\alpha_T w_2 x + i\alpha_T \Theta_T =
$$

$$
-\alpha_T^2 \beta_T^{-2} \delta_T^{(1)} (x - w_1)^2 + 2\alpha_T^2 \beta_T^{-1} (x - w_1)^2 + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T = \quad (2.1.)
$$

$$
(x - w_1)^2 (2\alpha_T^2 \beta_T^{-1} - \alpha_T^2 \beta_T^{-2} \delta_T^{(1)}) + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T =
$$

$$
(x - w_1)^2 \eta_T + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T,
$$

$$
\eta_T = 2\alpha_T^2 \beta_T^{-1} - \alpha_T^2 \beta_T^{-2} \delta_T^{(1)} = \alpha_T^2 \beta_T^{-1} (2 - \beta_T^{-1} \delta_T^{(1)}) = \alpha_T^2 \beta_T^{-1},
$$

$$
\delta_T^{(1)} = (i\alpha_T \cos \omega T - \bar{\gamma}_T^2) = \beta_T,
$$

$$
\delta_T^{(2)} = i\alpha_T \cos \omega T - \theta_T^2
$$

Thus

$$
\delta_T^{(2)} = i\alpha_T \cos \omega T - \theta_T^2
$$
$$ \tilde{S}(y_{cr}, x, T) = i\alpha_T S(y_{cr}, x, T) - (\bar{\gamma}_T^2 y_{cr}^2 + \theta_T^2 x^2) = $$

$$ i\alpha_T[(\cos \omega T)(y_{cr}^2 + x^2) - 2xy_{cr} + 2w_2x + 2w_1y_{cr} + \Theta_T] - (\bar{\gamma}_T^2 y_{cr}^2 + \theta_T^2 x^2) = $$

$$ i\alpha_T(\cos \omega T)y_{cr}^2 + i\alpha_T(\cos \omega T)x^2 - 2i\alpha_T xy_{cr} + 2i\alpha_T w_2 x + 2i\alpha_T w_1 y_{cr} + i\alpha_T \Theta_T - $$

$$ -\bar{\gamma}_T^2 y_{cr}^2 - \theta_T^2 x^2 = $$

$$ (i\alpha_T \cos \omega T - \bar{\gamma}_T^2)y_{cr}^2 + (i\alpha_T \cos \omega T - \theta_T^2)x^2 - 2i\alpha_T y_{cr}(x - w_1) + 2i\alpha_T w_2 x + i\alpha_T \Theta_T = $$

$$ -\alpha_T^2 \beta_T^{-2} \delta_T^{(1)} (x - w_1)^2 + 2\alpha_T^2 \beta_T^{-1} (x - w_1)^2 + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T = \quad (2.1.) $$

$$ (x - w_1)^2 (2\alpha_T^2 \beta_T^{-1} - \alpha_T^2 \beta_T^{-2} \delta_T^{(1)}) + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T = $$

$$ (x - w_1)^2 \eta_T + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T, $$

$$ \eta_T = 2\alpha_T^2 \beta_T^{-1} - \alpha_T^2 \beta_T^{-2} \delta_T^{(1)} = \alpha_T^2 \beta_T^{-1} (2 - \beta_T^{-1} \delta_T^{(1)}) = \alpha_T^2 \beta_T^{-1}, $$

$$ \delta_T^{(1)} = (i\alpha_T \cos \omega T - \bar{\gamma}_T^2) = \beta_T, $$

$$ \delta_T^{(2)} = i\alpha_T \cos \omega T - \theta_T^2 $$

Thus

$$ \delta_T^{(2)} = i\alpha_T \cos \omega T - \theta_T^2 $$
$$
\tilde{S}(y_{cr}, x, T) =
$$

$$
\eta_T (x - w_1)^2 + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T =
$$

$$
\eta_T x^2 - 2\eta_T w_1 x + \eta_T w_1^2 + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T =
$$

$$
(\eta_T + \delta_T^{(2)})x^2 + 2(i\alpha_T w_2 - \eta_T w_1)x + \eta_T w_1^2 + i\alpha_T \Theta_T =
$$

$$
\vartheta_T x^2 + \sigma_T x + \eta_T w_1^2 + \Theta_T^{(1)},
$$

$$
\vartheta_T = \eta_T + \delta_T^{(2)} = \eta_T + i\alpha_T \cos \omega T - \theta_T^2, \quad (2.1.)
$$

$$
\sigma_T = 2(i\alpha_T w_2 - \eta_T w_1),
$$

$$
\Theta_T^{(1)} = i\alpha_T \Theta_T,
$$

$$
\eta_T = \alpha_T^2 \beta_T^{-1} = \frac{\alpha_T^2}{i \alpha_T \cos \omega T - \bar{\gamma}_T^2}.
$$

$$
\eta_T = 2\alpha_T^2 \beta_T^{-1} - \alpha_T^2 \beta_T^{-2} \delta_T^{(1)} = \alpha_T^2 \beta_T^{-1} (2 - \beta_T^{-1} \delta_T^{(1)}) =
$$

$$
\alpha_T^2 \beta_T^{-1} (2 - \beta_T^{-1} \beta_T) = \alpha_T^2 \beta_T^{-1}.
$$

From Eq.(2.1.42) and Eq.(2.1.40) we obtain

$$
\tilde{\theta}_{\epsilon}^{(1)}(T, x, \alpha, \hbar_{\epsilon}) =
$$

$$
\frac{1}{\sqrt{2\pi\hbar\epsilon}} \sqrt{\frac{\alpha_T}{2\pi i \tilde{S}_{yy}(x)}} \exp\left\{\frac{1}{\hbar\epsilon} \tilde{S}(y_{cr}, x, T)\right\}. \qquad (2.1.)
$$
$$
\tilde{S}(y_{cr}, x, T) =
$$

$$
\eta_T (x - w_1)^2 + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T =
$$

$$
\eta_T x^2 - 2\eta_T w_1 x + \eta_T w_1^2 + \delta_T^{(2)} x^2 + 2i\alpha_T w_2 x + i\alpha_T \Theta_T =
$$

$$
(\eta_T + \delta_T^{(2)})x^2 + 2(i\alpha_T w_2 - \eta_T w_1)x + \eta_T w_1^2 + i\alpha_T \Theta_T =
$$

$$
\vartheta_T x^2 + \sigma_T x + \eta_T w_1^2 + \Theta_T^{(1)},
$$

$$
\vartheta_T = \eta_T + \delta_T^{(2)} = \eta_T + i\alpha_T \cos \omega T - \theta_T^2, \quad (2.1.)
$$

$$
\sigma_T = 2(i\alpha_T w_2 - \eta_T w_1),
$$

$$
\Theta_T^{(1)} = i\alpha_T \Theta_T,
$$

$$
\eta_T = \alpha_T^2 \beta_T^{-1} = \frac{\alpha_T^2}{i \alpha_T \cos \omega T - \bar{\gamma}_T^2}.
$$

$$
\eta_T = 2\alpha_T^2 \beta_T^{-1} - \alpha_T^2 \beta_T^{-2} \delta_T^{(1)} = \alpha_T^2 \beta_T^{-1} (2 - \beta_T^{-1} \delta_T^{(1)}) =
$$

$$
\alpha_T^2 \beta_T^{-1} (2 - \beta_T^{-1} \beta_T) = \alpha_T^2 \beta_T^{-1}.
$$

From Eq.(2.1.42) and Eq.(2.1.40) we obtain

$$
\tilde{\theta}_{\epsilon}^{(1)}(T, x, \alpha, \hbar_{\epsilon}) =
$$

$$
\frac{1}{\sqrt{2\pi\hbar\epsilon}} \sqrt{\frac{\alpha_T}{2\pi i \tilde{S}_{yy}(x)}} \exp\left\{\frac{1}{\hbar\epsilon} \tilde{S}(y_{cr}, x, T)\right\}. \qquad (2.1.)
$$
Let's consider integral

$$
\begin{align*}
U(T, \alpha, \hbar_{\varepsilon}) &= \int dx|x| |\tilde{\theta}_{\varepsilon}^{(1)}(T, x, \alpha, \hbar_{\varepsilon})|^2 = \\
&\qquad \frac{1}{\sqrt{2\pi\hbar_{\varepsilon}}} \sqrt{\frac{|\alpha_T|}{2\pi |\tilde{S}_{yy}(x)|}} \times \\
&\qquad \int dx|x| \exp\left[\frac{1}{\hbar_{\varepsilon}} \operatorname{Re}[\tilde{S}(y_{cr}(x, T), x)]\right] = \tag{2.1.44} \\
&\qquad \frac{1}{\sqrt{2\pi\hbar_{\varepsilon}}} \sqrt{\frac{|\alpha_T|}{2\pi |\tilde{S}_{yy}(x)|}} \int dx|x| \exp\left[\frac{1}{\hbar_{\varepsilon}} [\vartheta'_{T}x^2 + \sigma'_{T}x + \eta'_{T}w'^2_1 + \Theta'_{T}]\right], \\
&\qquad \vartheta'_{T} = \operatorname{Re}\vartheta_{T}, \sigma'_{T} = \operatorname{Re}\sigma_{T}, \Theta'_{T} = \operatorname{Re}\Theta^{(2)}_{T}.
\end{align*}
$$

$$
\begin{align}
\frac{\partial \tilde{S}(y_{cr}(x, T), x)}{\partial x} &= 2\vartheta'_{T}x + \sigma'_{T} = 0, \tag{2.1.44} \\
x_{cr} &= -\frac{\sigma'_{T}}{2\vartheta'_{T}}. \nonumber
\end{align}
$$

$$
\begin{align}
\tilde{S}(y_{cr}(x_{cr}, T), x_{cr}) &= \vartheta'_{T} x_{cr}^{2} + \sigma'_{T} x_{cr} + \eta'_{T} w_{1}^{\prime 2} + \Theta'_{T} = \nonumber \\
&\quad \frac{\sigma'_{T}^{2}}{4\vartheta'_{T}} - \frac{\sigma'_{T}^{2}}{2\vartheta'_{T}} + \eta'_{T} w_{1}^{\prime 2} + \Theta'_{T} = -\frac{\sigma'_{T}^{2}}{4\vartheta'_{T}} + \eta'_{T} w_{1}^{\prime 2} + \Theta'_{T}. \tag{2.1.}
\end{align}
$$