From Eqs.(2.2.2.84) and Eqs.(2.2.2.89.a,b) we obtain

$$
\begin{align*}
& v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \\
& \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \times \\
& \left. \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\} = \\
& \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \exp \left\{ -\frac{m}{\epsilon} \int_0^T dt \left[ \{u(t) + \lambda\}_\eta + \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda, \lambda) \right]^2 \right\} \times \\
& \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \times \\
& \left. \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\} \tag{2.2.2.90.a}
\end{align*}
$$

and
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$
\begin{aligned}
& \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \times \\
& \qquad \left. \sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_1(u_{V(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\} = \\
& \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \exp \left\{ -\frac{m}{\epsilon} \int_{0}^{T} dt \left[ \{u(t) + \lambda\}_{\eta} + \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda, \lambda) \right]^2 \right\} \times \right. \\
& \qquad \left. \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \times \right. \\
& \qquad \left. \sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_1(u_{V(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\}.
\end{aligned}
\tag{2.2.90.b}
$$

Here
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$
\begin{align}
& \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \nonumber \\
& \qquad \left. \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \times \right. \nonumber \\
& \qquad \left. \sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_1(u_{V(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\} = \tag{2.2.90.b} \\
& \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \nonumber \\
& \qquad \left. \exp \left\{ -\frac{m}{\epsilon} \int_{0}^{T} dt \left[ \{u(t) + \lambda\}_{\eta} + \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda, \lambda) \right]^2 \right\} \times \right. \nonumber \\
& \qquad \left. \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \times \right. \nonumber \\
& \qquad \left. \sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_1(u_{V(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\}. \nonumber
\end{align}
$$

Here
$$v(\epsilon, \epsilon) = \epsilon^k \epsilon^{\frac{m}{2}},$$

$$ \frac{1}{\epsilon} V_1(\sqrt{\epsilon} u_{v(\epsilon, \epsilon)}(t), t, \lambda) = \frac{1}{\epsilon} V_1(\sqrt{\epsilon} u_{v(\epsilon, \epsilon)}(t), t, \lambda) = a_3(t, \lambda) \epsilon^{1/2} u_{v(\epsilon, \epsilon)}^3(t) + \dots \\ + a_k(t, \lambda) \epsilon^{(k/2)-1} u_{v(\epsilon, \epsilon)}^k(t) = \hat{V}_1(\sqrt{\epsilon} u_{v(\epsilon, \epsilon)}(t), t, \lambda) \qquad (2.2.91) $$

$$ = \hat{V}_1(\sqrt{\epsilon} u_{v(\epsilon, \epsilon)}(t), t, \lambda) = \sqrt{\epsilon} \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda). $$

$$ \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t)) = a_3(t, \lambda) u_{v(\epsilon, \epsilon)}^3(t) + \dots + a_k(t, \lambda) \epsilon^{(k/2)-3/2} u_{v(\epsilon, \epsilon)}^k(t). $$

Let us rewrite Eqs.(2.2.90.a,b) in the following equivalent form

$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$ \bar{v}_1 dy \left\{ \int_0^T dt \left[ \frac{1}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right] \right\} \times \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \times \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] \qquad (2.2.2.92.a) $$

and
$$v(\epsilon, \epsilon) = \epsilon^k \epsilon^{\frac{m}{2}},$$

$$ \frac{1}{\epsilon} V_1(\sqrt{\epsilon} u_{v(\epsilon, \epsilon)}(t), t, \lambda) = \frac{1}{\epsilon} V_1(\sqrt{\epsilon} u_{v(\epsilon, \epsilon)}(t), t, \lambda) = a_3(t, \lambda) \epsilon^{1/2} u_{v(\epsilon, \epsilon)}^3(t) + \dots \\ + a_k(t, \lambda) \epsilon^{(k/2)-1} u_{v(\epsilon, \epsilon)}^k(t) = \hat{V}_1(\sqrt{\epsilon} u_{v(\epsilon, \epsilon)}(t), t, \lambda) \quad (2.2.91) $$

$$ = \hat{V}_1(\sqrt{\epsilon} u_{v(\epsilon, \epsilon)}(t), t, \lambda) = \sqrt{\epsilon} \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda). $$

$$ \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t)) = a_3(t, \lambda) u_{v(\epsilon, \epsilon)}^3(t) + \dots + a_k(t, \lambda) \epsilon^{(k/2)-3/2} u_{v(\epsilon, \epsilon)}^k(t). $$

Let us rewrite Eqs.(2.2.90.a,b) in the following equivalent form

$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$ \bar{v}_1 dy \left\{ \int_0^T dt \left[ \frac{\int_{u(T)=\frac{\epsilon-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{\epsilon-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon |u(T)|} \right] \times \right. \\ \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right| \times \right. \\ \left. \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] \right\} \quad (2.2.2.92.a) $$

and
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \sin \left[ \left( -\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\} \quad (2.2.2.92.b) $$

Here $\tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon, \epsilon]$ and $\tilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon, \epsilon]$ is the corresponding signed Feynman"measures", i.e.
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \sin \left[ \left( -\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] \right\} \quad (2.2.2.92.b) $$

Here $\tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon, \epsilon]$ and $\tilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon, \epsilon]$ is the corresponding signed Feynman"measures", i.e.
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times $$

$$ [\cos^{+}(S_{\epsilon}(T, x, \lambda, u(t), \epsilon)) + \cos^{-}(S_{\epsilon}(T, x, \lambda, u(t), \epsilon))] \times (\cdot) = $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (2.2.2.93.a) $$

$$ [\cos^{+}(S_{\epsilon}(T, x, \lambda, u(t), \epsilon)) + \cos^{-}(S_{\epsilon}(T, x, \lambda, u(t), \epsilon))] \times $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot), $$

$$ S_{\epsilon}(T, x, \lambda, q(t), \epsilon) = \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{q}_{\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) q_{\epsilon}^2(t) + b_{\epsilon}(t, \lambda) \frac{q_{\nu}(\epsilon, \epsilon)(t)}{\sqrt{\epsilon}} \right]. $$

and
$$
\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) =
$$

$$
\overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}}
\tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon]\Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times
u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}}
$$

$$
[\cos^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \cos^-(S_\epsilon(T, x, \lambda, u(t), \epsilon))] \times (\cdot) =
$$

$$
\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \qquad (2.2.2.93.a)
$$

$$
[\cos^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \cos^-(S_\epsilon(T, x, \lambda, u(t), \epsilon))] \times
$$

$$
\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot),
$$

$$
S_{\epsilon}(T, x, \lambda, q(t), \epsilon) = \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{q}_{\epsilon}(t)}{dt} \right)^{2} - a_{\epsilon}(t, \lambda) q_{\epsilon}^{2}(t) + b_{\epsilon}(t, \lambda) \frac{q_{\nu(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right].
$$

and
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times $$

$$ [\sin^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \sin^-(S_\epsilon(T, x, \lambda, u(t), \epsilon))] \times (\cdot) = $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^+[q(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \qquad (2.2.2.93.b) $$

$$ [\sin^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \sin^-(S_\epsilon(T, x, \lambda, u(t), \epsilon))] \times $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot), $$

$$ S_{\epsilon}(T, x, \lambda, q(t), \epsilon) = \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{q}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) q_{\epsilon}^2(t) + b_{\epsilon}(t, \lambda) \frac{q_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right]. $$

Let us rewrite Eqs.(2.2.2.93.a,b) in the following equivalent form
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times $$

$$ [\sin^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \sin^-(S_\epsilon(T, x, \lambda, u(t), \epsilon))] \times (\cdot) = $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^+[q(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \qquad (2.2.2.93.b) $$

$$ [\sin^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \sin^-(S_\epsilon(T, x, \lambda, u(t), \epsilon))] \times $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot), $$

$$ S_{\epsilon}(T, x, \lambda, q(t), \epsilon) = \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{q}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) q_{\epsilon}^2(t) + b_{\epsilon}(t, \lambda) \frac{q_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right]. $$

Let us rewrite Eqs.(2.2.2.93.a,b) in the following equivalent form
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = $$

$$ [\cos^{+}(S_{\epsilon}(T, x, \lambda, u(t), \epsilon)) + \cos^{-}(S_{\epsilon}(T, x, \lambda, u(t), \epsilon)) + 1 - 1] \times (\cdot) = $$

$$ \begin{aligned} & \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) - \\ & u(0)=\frac{y-\lambda}{\sqrt{\epsilon}} \end{aligned} $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = \qquad (2.2.2.94.a) $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) $$

$$ - \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot) $$

and
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = $$

$$ [\cos^{+}(S_{\epsilon}(T, x, \lambda, u(t), \epsilon)) + \cos^{-}(S_{\epsilon}(T, x, \lambda, u(t), \epsilon)) + 1 - 1] \times (\cdot) = $$

$$ \begin{aligned} & \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) - \\ & u(0)=\frac{y-\lambda}{\sqrt{\epsilon}} \end{aligned} $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) = \qquad (2.2.2.94.a) $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon, \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) $$

$$ - \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times (\cdot) $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot) $$

and
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) = $$

$$ [\sin^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \sin^-(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + 1 - 1] \times (\cdot) = $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{m}}^{(2)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) - $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) = \qquad (2.2.2.94.b) $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{m}}^{(2)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) $$

$$ - \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) = $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot). $$

Here $\tilde{D}_{\text{m}}^{(1)}$ and $\tilde{D}_{\text{m}}^{(2)}$ is the corresponding positive Feynman-Colombeau submeasures, i.e.
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) = $$

$$ [\sin^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \sin^-(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + 1 - 1] \times (\cdot) = $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{m}}^{(2)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) - $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) = \qquad (2.2.2.94.b) $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{m}}^{(2)}[u(t), \epsilon, \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) $$

$$ - \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times (\cdot) = $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot). $$

Here $\tilde{D}_{\text{m}}^{(1)}$ and $\tilde{D}_{\text{m}}^{(2)}$ is the corresponding positive Feynman-Colombeau submeasures, i.e.
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathrm{m}}^{(1)}[u(t), \varepsilon, \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon, \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times $$

$$ [\cos^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \cos^-(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + 1] \times (\cdot) = \qquad (2.2.2.95.a) $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times $$

$$ [\cos^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \cos^-(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + 1] \times $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot) $$

and
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathrm{m}}^{(1)}[u(t), \varepsilon, \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon, \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times $$

$$ [\cos^{+}(S_{\varepsilon}(T, x, \lambda, u(t), \varepsilon)) + \cos^{-}(S_{\varepsilon}(T, x, \lambda, u(t), \varepsilon)) + 1] \times (\cdot) = \qquad (2.2.2.95.a) $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times $$

$$ [\cos^{+}(S_{\varepsilon}(T, x, \lambda, u(t), \varepsilon)) + \cos^{-}(S_{\varepsilon}(T, x, \lambda, u(t), \varepsilon)) + 1] \times $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^{2} \right\} \times (\cdot) $$

and
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathrm{m}}^{(2)}[u(t), \varepsilon, \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon, \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times $$

$$ [\sin^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \sin^-(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + 1] \times (\cdot) = \qquad (2.2.2.95.b) $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times $$

$$ [\sin^+(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + \sin^-(S_\epsilon(T, x, \lambda, u(t), \epsilon)) + 1] \times $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot) $$

From Eq.(2.2.2.92.a,b)-Eq.(2.2.2.95.a,b) we obtain
$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathrm{m}}^{(2)}[u(t), \varepsilon, \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times (\cdot) = $$

$$ \overline{\int_{q(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon, \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times $$

$$ [\sin^+(S_\varepsilon(T, x, \lambda, u(t), \varepsilon)) + \sin^-(S_\varepsilon(T, x, \lambda, u(t), \varepsilon)) + 1] \times (\cdot) = \qquad (2.2.2.95.b) $$

$$ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} D^{+}[q(t), \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times $$

$$ [\sin^+(S_\varepsilon(T, x, \lambda, u(t), \varepsilon)) + \sin^-(S_\varepsilon(T, x, \lambda, u(t), \varepsilon)) + 1] \times $$

$$ \exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times (\cdot) $$

From Eq.(2.2.2.92.a,b)-Eq.(2.2.2.95.a,b) we obtain
$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|}} \times \right. \\ \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda)u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right) \right\} \times$$

$$\cos \left[ \left( \sqrt{\epsilon} \int_{0}^{T} \tilde{V}_{1}(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] =$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{m}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|}} \times \right. \\ \left. \left| \cos \left[ \left( \sqrt{\epsilon} \int_{0}^{T} \tilde{V}_{1}(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\} - \right. \\ \left. \left(2.2.2.96 \cdot a\right) \right\}$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|}} \times \right. \\ \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda)u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right\} \times$$

($\epsilon \approx \frac{1}{\sqrt{\epsilon}}$)
$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon}|u(T)|} \times \right. \\ \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda)u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right| \times \\ \left. \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\} =$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{m}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon}|u(T)|} \times \right. \\ \left. \left| \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right| \right\} - \qquad (2.2.2.96 \text{.a})$$

$$\left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda)u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \\ \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \} -$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon}|u(T)|} \times \right. \\ \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda)u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right| \times \\ \left. \left( \int_{-\infty}^{T} dt \right) \right\}$$
and
and
$$v_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) =$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(2)}[u(t), \varepsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right\} \times$$

$$\sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_{1}(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] =$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{m}}^{(1)}[u(t), \varepsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right\} \times \quad (2.2.2.96.b)$$

$$\sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_{1}(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] -$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right\} \times$$

$$\sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_{1}(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right].$$
$$v_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) =$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(2)}[u(t), \varepsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right\} \times$$

$$\sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_{1}(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] =$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{m}}^{(1)}[u(t), \varepsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right\} \times \quad (2.2.2.96.b)$$

$$\sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_{1}(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] -$$

$$\int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right\} \times$$

$$\sin \left[ \left( -\sqrt{\epsilon} \int_{0}^{T} \tilde{V}_{1}(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right].$$
From Eq.(2.2.2.96.a,b) we obtain

$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \tilde{v}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) - v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta).$$

$$ \tilde{v}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \bar{\int} dy \left\{ \begin{array}{l} \displaystyle \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \\ \displaystyle \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \\ \displaystyle \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \end{array} \right\}, \quad (2.2.2.97.a) $$

$$ v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) = \bar{\int} dy \left\{ \begin{array}{l} \displaystyle \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \\ \displaystyle \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \\ \displaystyle \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \end{array} \right\} $$

and
From Eq.(2.2.2.96.a,b) we obtain

$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \tilde{v}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) - v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta).$$

$$ \tilde{v}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \bar{\int} dy \left\{ \begin{array}{l} \displaystyle \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \\ \displaystyle \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \\ \displaystyle \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \end{array} \right\}, \quad (2.2.2.97.a) $$

$$ v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) = \bar{\int} dy \left\{ \begin{array}{l} \displaystyle \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \\ \displaystyle \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \\ \displaystyle \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \end{array} \right\} $$

and
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \tilde{v}_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) - v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta).$$

$$ \tilde{v}_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \int dy \left\{ \begin{array}{l} \displaystyle \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times} \\ \\ \displaystyle \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \\ \\ \displaystyle \sin \left[ \left( -\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \end{array} \right\}, \quad (2.2.2.97.b) $$

$$ v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) = \int dy \left\{ \begin{array}{l} \displaystyle \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times} \\ \\ \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \\ \\ \sin \left[ \left( -\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \end{array} \right\} $$

Let us evaluate now path integral $v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta)$
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \tilde{v}_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) - v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta).$$

$$ \tilde{v}_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \sin \left[ \left( -\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\}, \quad (2.2.2.97.b) $$

$$ v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) = \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \times \sin \left[ \left( -\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right\} $$

Let us evaluate now path integral $v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta)$
$$v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$ \bar{\int} dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon |u(T)|} \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times \cos\left[\left(\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt\right)\right] \right. \quad (2.2.2.98.a) $$

$$ \Xi_1(q(t), \lambda, \epsilon, \epsilon) = $$

$$ \left| \cos \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| $$

and path integral $v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta)$

$$ v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) = $$

$$ \bar{\int} dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon |u(T)|} \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times \sin\left[\left(\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt\right)\right] \right. \quad (2.2.2.98.b) $$

$$ \Xi_2(q(t), \lambda, \epsilon, \epsilon) = $$

$$ \left| \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| $$
$$v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$ \bar{\int} dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)|} \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times \right. \\ \left. \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right], \right. \qquad (2.2.2.98.a) $$

$$\Xi_1(q(t), \lambda, \epsilon, \epsilon) =$$

$$ \left| \cos \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| $$

and path integral $v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta)$

$$v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$ \bar{\int} dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)|} \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times \right. \\ \left. \sin \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right], \right. \qquad (2.2.2.98.b) $$

$$\Xi_2(q(t), \lambda, \epsilon, \epsilon) =$$

$$ \left| \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| $$
From Eq.(2.2.2.98.a,b) we obtain
From Eq.(2.2.2.98.a,b) we obtain
$$v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\cos\left(\int_{0}^{T} \sqrt{\epsilon} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt\right) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} =$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times$$

$$\left[ \sum_{n=0}^{M} \frac{(-1)^n \epsilon^n}{(2n)!} \left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n} \right] = \qquad (2.2.2.99.a)$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} +$$

$$\sum_{n=1}^{M} \frac{(-1)^n \epsilon^n}{(2n)!} \times$$

$$\overline{\int dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times}$$
$$v_{1,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\cos\left(\int_{0}^{T} \sqrt{\epsilon} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt\right) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} =$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times$$

$$\left[ \sum_{n=0}^{M} \frac{(-1)^n \epsilon^n}{(2n)!} \left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n} \right] = \qquad (2.2.2.99.a)$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} +$$

$$\sum_{n=1}^{M} \frac{(-1)^n \epsilon^n}{(2n)!} \times$$

$$\overline{\int dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times}$$
and
and
$$v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\sin\left(\int_{0}^{T} \sqrt{\epsilon} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) d t\right) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} =$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times$$

$$\left[ \sum_{n=0}^{M} \frac{(-1)^n \epsilon^{n+1/2}}{(2n+1)!} \left( \int_{0}^{T} \tilde{V}_1(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n+1} + \epsilon^{n+1} O\left( \left( \int_{0}^{T} \tilde{V}_1(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n+2} \right) \right] = \quad (2.2.2.99.b)$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} + \sum_{n=0}^{M} \frac{(-1)^n \epsilon^{n+1/2}}{(2n+1)!} \times$$
$$v_{2,\epsilon}^*(T, x, x_0, \lambda, \epsilon, \eta) =$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\sin\left(\int_{0}^{T} \sqrt{\epsilon} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) d t\right) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} =$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \times$$

$$\left[ \sum_{n=0}^{M} \frac{(-1)^n \epsilon^{n+1/2}}{(2n+1)!} \left( \int_{0}^{T} \tilde{V}_1(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n+1} + \epsilon^{n+1} O\left( \left( \int_{0}^{T} \tilde{V}_1(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n+2} \right) \right] = \qquad (2.2.2.99.b)$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} + \sum_{n=0}^{M} \frac{(-1)^n \epsilon^{n+1/2}}{(2n+1)!} \times$$
Let us evaluate path integral $\tilde{v}_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta)$ and $\tilde{v}_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta)$. Assume now that $p_2 \gg \epsilon^{-1}$, i.e. $p_1 \approx 1$. Then from Eqs.(2.2.2.97.a,b) and Hölder's inequality (2.2.2.33) we obtain:
Let us evaluate path integral $\tilde{v}_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta)$ and $\tilde{v}_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta)$. Assume now that $p_2 \gg \epsilon^{-1}$, i.e. $p_1 \approx 1$. Then from Eqs.(2.2.2.97.a,b) and Hölder's inequality (2.2.2.33) we obtain:
$$|\tilde{V}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta)| \le$$

$$\int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. $$

$$ \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon, \epsilon)}^2(t) + \right. \right. \right. $$

$$ \left| \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{V(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] \right| \le $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \right. $$

$$ \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon, \epsilon)}^2(t) + \right. \right. \right. $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \right. $$

$$ \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon, \epsilon)}^2(t) + \right. \right. \right. $$

$$ \left. \left. u_{V(\epsilon, \epsilon)}(t) \right\} \right\} $$
$$|\tilde{V}_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta)| \le$$

$$\int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\varepsilon}|u(T)| \times \right. $$

$$ \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{V(\varepsilon, \varepsilon)}^2(t) + \right. \right. \right. $$

$$ \left| \cos \left[ \left( \sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{V(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] \right| \le $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\varepsilon}|u(T)| \times \right. \right. $$

$$ \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{V(\varepsilon, \varepsilon)}^2(t) + \right. \right. \right. $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\varepsilon}|u(T)| \times \right. \right. $$

$$ \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{V(\varepsilon, \varepsilon)}^2(t) + \right. \right. \right. $$

$$ u_{V(\varepsilon, \varepsilon)}(t) \ge \frac{1}{\sqrt{\epsilon}} $$
and
and
$$|\tilde{V}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta)| \le$$

$$\int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \right. $$

$$ \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon, \varepsilon)}^2(t) + \right. \right. \right. $$

$$ \left. \left| \sin \left[ \left( -\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon, \epsilon)}(t), t, \lambda) dt \right) \right] \right| \right\} \le $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \right. \right. $$

$$ \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon, \varepsilon)}^2(t) + \right. \right. \right. $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \right. \right. $$

$$ \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon, \varepsilon)}^2(t) + \right. \right. \right. $$

$$ u_{v(\varepsilon, \varepsilon)}(t) \approx \lambda $$
$$|\tilde{V}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta)| \le$$

$$\int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. $$

$$ \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. $$

$$ \left. \left| \sin \left[ \left( -\sqrt{\epsilon} \int_0^T \tilde{V}_1(u_{v(\epsilon,\epsilon)}(t), t, \lambda) dt \right) \right] \right| \right\} \le $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \right. $$

$$ \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \right. $$

$$ \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. $$

$$ \dots \ u_{v(\epsilon, \epsilon)}(t) \quad \dots $$
Therefore from Eqs.(2.2.2.100.a,b) we obtain

$$|\tilde{V}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta)| \le$$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \\ \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon, \epsilon)}^2(t) + \right. \right. \right. \\ \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right]^{1/p_1} \times \quad (2.2.2.101.a) $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \right. \\ \left. \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon, \epsilon)}^2(t) + \right. \right. \right. \right. \\ \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right]^{1/p_2} $$

and
Therefore from Eqs.(2.2.2.100.a,b) we obtain

$$|\tilde{V}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta)| \le$$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \\ \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon, \epsilon)}^2(t) + \right. \right. \right. \\ \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right]^{1/p_1} \times \quad (2.2.2.101.a) $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \right. \\ \left. \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon, \epsilon)}^2(t) + \right. \right. \right. \right. \\ \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right]^{1/p_2} $$

and
$$|\tilde{\mathcal{V}}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta)| \le$$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \right. \\ \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda)u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \right. \\ \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right]^{1/p_1} \times \quad (2.2.2.101.b) $$

$$ \left[ \bar{\int} dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \right. \right. \\ \left. \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda)u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \right. \right. \\ \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right]^{1/p_2}. $$

From Eqs.(2.2.2.101.a,b) we obtain

$$ \tilde{\mathcal{V}}_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \le \mathbf{J}_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \times \Gamma_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta, p_2), \quad (2.2.2.102) $$

$$ \tilde{\mathcal{V}}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \le \mathbf{J}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \times \Gamma_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta, p_2). $$

Here
$$|\tilde{\mathcal{V}}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta)| \le$$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \right. \\ \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda)u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \\ \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right|^{1/p_1} \times \right. \\ \left. (2.2.2.101.b) \right. $$

$$ \left[ \bar{\int} dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \right. \right. \\ \left. \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda)u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \right. \\ \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right|^{1/p_2}. $$

From Eqs. (2.2.2.101.a,b) we obtain

$$ \tilde{\mathcal{V}}_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \le \mathbf{J}_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \times \Gamma_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta, p_2), \quad (2.2.2.102) $$

$$ \tilde{\mathcal{V}}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \le \mathbf{J}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \times \Gamma_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta, p_2). $$

Here
$$ \mathbf{J}_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) = $$

$$
\left[
\begin{aligned}
& \overline{\int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\mathbf{m}}^{(1)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times \right.} \\
& \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda)u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \\
& \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right| \right]^{1/p_1},
\end{aligned}
\right.
\qquad (2.2.2.103.a)
$$

$$ \Gamma_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta, p_2) = $$

$$
\left[
\begin{aligned}
& \overline{\int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\mathbf{m}}^{(1)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times \right.} \\
& \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda)u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \\
& \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right| \right]^{1/p_2}
\end{aligned}
\right.
$$

and
$$
\mathbf{J}_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) =
$$

$$
\left[
\begin{aligned}
& \overline{\int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right.} \\
& \qquad \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right| \right]^{1/p_1},
\end{aligned}
\right.
\tag{2.2.2.103.a}
$$

$$
\Gamma_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta, p_2) =
$$

$$
\left[
\begin{aligned}
& \overline{\int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\mathbf{m}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right.} \\
& \qquad \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right| \right]^{1/p_2}
\end{aligned}
\right.
$$

and
$$
\mathbf{J}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) =
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \\
& \qquad \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right|^{1/p_1},
\end{aligned}
\right.
\qquad (2.2.2.103.b)
$$

$$
\Gamma_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta, p_2) =
$$

$$
\left[
\begin{aligned}
& \bar{\int} dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \epsilon] \Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \\
& \qquad \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right|^{1/p_2}.
\end{aligned}
\right.
$$

From Eq.(2.2.2.103.a,b) by using definition (see Eq.(2.2.2.95)) of the submeasures
$\tilde{D}_{\mathbf{m}}^{(1)}[q(t), \epsilon, \epsilon]$, $\tilde{D}_{\mathbf{m}}^{(2)}[q(t), \epsilon, \epsilon]$ we obtain
$$ \mathbf{J}_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) = $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda)u_{v(\varepsilon,\varepsilon)}^2(t) + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right]^{1/p_1}, \quad (2.2.2.103.b) $$

$$ \Gamma_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta, p_2) = $$

$$ \left[ \bar{\int} dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\mathbf{m}}^{(2)}[u(t), \varepsilon]\Psi_0(y) \sqrt{\sqrt{\epsilon}|u(T)|} \times \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda)u_{v(\varepsilon,\varepsilon)}^2(t) + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right]^{1/p_2}. $$

From Eq.(2.2.2.103.a,b) by using definition (see Eq.(2.2.2.95)) of the submeasures $\tilde{D}_{\mathbf{m}}^{(1)}[q(t), \varepsilon, \epsilon]$, $\tilde{D}_{\mathbf{m}}^{(2)}[q(t), \varepsilon, \epsilon]$ we obtain
$$J_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \le$$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \\ \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda)u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\ \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right| \right] + \qquad (2.2.2.104.a) $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon]\Psi_{0}(y) \sqrt{\epsilon} |u(T)| \times \right. \right. \\ \left. \left. \left| \cos \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda)u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \right. \\ \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right| \right| \right] = \\ \Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) + \Lambda_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) $$

and
$$ \mathbf{J}_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) \le $$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \\
& \qquad \left. \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right] + \\
& \qquad \left. \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon]\Psi_0(y) \sqrt{\epsilon} |u(T)| \times \right. \\
& \qquad \left. \left. \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\epsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right] = \right\}
\end{aligned}
\right] \quad (2.2.2.104.a)
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\epsilon} |u(T)| \times \right. \\
& \qquad \left. \cos \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right\} = \\
& \Theta_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) + \Lambda_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta)
\end{aligned}
\right.
$$

and
$$
\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right| \right] = && (2.2.2.105.a) \\
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right\}
\end{aligned}
\right]
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right\}
\end{aligned}
\right]
$$
$$
\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\text{sm}}^{(1)}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right| \right\} = && (2.2.2.105.a) \\
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \left| \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right\},
\end{aligned}
\right.
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \cos \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right\},
\end{aligned}
\right.
$$
$$
\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right| \right] = && (2.2.2.105.b) \\
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right| \right]
\end{aligned}
\right.
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right\}
\end{aligned}
\right]
$$

and
$$
\Theta_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) =
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\text{sm}}^{(2)}[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \left| \sin \left[ \left( \int_0^T dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right| \right] = && (2.2.2.105.b) \\
&
\end{aligned}
\right.
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{V(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{V(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right\}
\end{aligned}
\right]
$$

and
$$ \Lambda_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) = $$

$$ \left[ \begin{aligned} & \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|}} \times \right. \\ & \left. \quad \cos \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \\ & \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right] \end{aligned} \right] \quad (2.2.2.106.a) $$

$$ \Lambda_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) = $$

$$ \left[ \begin{aligned} & \int dy \left\{ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon} |u(T)|}} \times \\ & \quad \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \\ & \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\varepsilon}} \right] \right) \right] \end{aligned} \right] \quad (2.2.2.106.b) $$

From (2.2.2.102) and (2.2.2.104.a,b) we obtain the inequalities:
$$ \Lambda_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) = $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times \right. \\ \left. \left| \cos \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \right. \\ \left. \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\varepsilon}} \right] \right) \right] \right] \right\} \quad (2.2.2.106.a) $$

$$ \Lambda_{2,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) = $$

$$ \left[ \int dy \left\{ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \times \right. \\ \left. \left( \sin \left[ \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\varepsilon\varepsilon}(t)}{dt} \right)^2 - a_{\varepsilon}(t, \lambda) u_{v(\varepsilon,\varepsilon)}^2(t) + \right. \right. \right. \\ \left. \left. + b_{\varepsilon}(t, \lambda) \frac{u_{v(\varepsilon,\varepsilon)}(t)}{\sqrt{\varepsilon}} \right] \right) \right] \right\} \quad (2.2.2.106.b) $$

From (2.2.2.102) and (2.2.2.104.a,b) we obtain the inequalities:
$$
\begin{align}
|\tilde{\nabla}_{1,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta)| &\le \mathbf{J}_{1,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta) \le \nonumber \\
&\qquad \Theta_{1,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta) + \Lambda_{1,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta), \tag{2.2.2.107} \\
|\tilde{\nabla}_{2,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta)| &\le \mathbf{J}_{2,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta) \le \nonumber \\
&\qquad \Theta_{2,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta) + \Lambda_{2,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta). \nonumber
\end{align}
$$

Therefore

$$
\begin{align*}
\nu_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) &= \tilde{\nu}_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - \nu_{1,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta) \le \\
\mathbf{J}_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - \nu_{1,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta) &\le \tag{2.2.2.108.a} \\
\Theta_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) + \Lambda_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - \nu_{1,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta),
\end{align*}
$$

and

$$
\begin{align*}
\nu_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) &= \tilde{\nu}_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - \nu_{2,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta) \le \\
\mathbf{J}_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - \nu_{2,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta) &\le \tag{2.2.2.108.b} \\
\Theta_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) + \Lambda_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - \nu_{2,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta).
\end{align*}
$$

Substitution Eqs.(2.2.2.99.a,b) and Eqs.(2.2.2.106.a,b) into Eqs.(2.2.2.108.a,b)
gives
$$
\begin{align*}
& |\tilde{\nabla}_{1,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta)| \le \mathbf{J}_{1,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta) \le \\
& \qquad \Theta_{1,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta) + \Lambda_{1,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta), \tag{2.2.2.107} \\
& |\tilde{\nabla}_{2,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta)| \le \mathbf{J}_{2,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta) \le \\
& \qquad \Theta_{2,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta) + \Lambda_{2,\varepsilon}(T,x,x_0,\lambda,\varepsilon,\eta).
\end{align*}
$$

Therefore

$$
\begin{align*}
v_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) &= \tilde{v}_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - v_{1,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta) \le \\
\mathbf{J}_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - v_{1,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta) &\le \tag{2.2.2.108.a} \\
\Theta_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) + \Lambda_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - v_{1,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta),
\end{align*}
$$

and

$$
\begin{align*}
v_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) &= \tilde{v}_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - v_{2,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta) \le \\
\mathbf{J}_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - v_{2,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta) &\le \tag{2.2.2.108.b} \\
\Theta_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) + \Lambda_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) - v_{2,\epsilon}^*(T,x,x_0,\lambda,\epsilon,\eta).
\end{align*}
$$

Substitution Eqs.(2.2.2.99.a,b) and Eqs.(2.2.2.106.a,b) into Eqs.(2.2.2.108.a,b)
gives
$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) \le$$

$$\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) + \Lambda_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) -$$

$$\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \Xi_{1}(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon}\int_{0}^{T} dt \left[ \left\{\sqrt{\epsilon} u(t) + \lambda\right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} +$$

$$\sum_{n=1}^{M} \frac{(-1)^n \epsilon^n}{(2n)!} \times \qquad (2.2.2.109.a)$$

$$\left[ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \Xi_{1}(q(t), \lambda, \epsilon, \epsilon) \times \right.$$

$$\left(\int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt\right)^{2n} \times$$

$$\exp\left\{-\frac{1}{\epsilon}\int_{0}^{T} dt \left[ \left\{\sqrt{\epsilon} u(t) + \lambda\right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\} \right]$$

and
$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) \le$$

$$\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) + \Lambda_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) -$$

$$\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \Xi_{1}(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon}\int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2\right\} +$$

$$\sum_{n=1}^{M} \frac{(-1)^n \epsilon^n}{(2n)!} \times \qquad (2.2.2.109.a)$$

$$\left[ \int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon}|u(T)|} \Xi_{1}(q(t), \lambda, \epsilon, \epsilon) \times \right.$$

$$\left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n} \times$$

$$\exp\left\{-\frac{1}{\epsilon}\int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2\right\}$$

and
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) \le$$

$$\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) + \Lambda_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) -$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon}\int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2\right\} \times$$

$$+ \sum_{n=1}^{M} \frac{(-1)^n \epsilon^{n+1/2}}{(2n+1)!} \times \qquad (2.2.2.109.b)$$

$$\left[ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times \right.$$

$$\left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n+1} \times$$

$$\exp\left\{-\frac{1}{\epsilon}\int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2\right\}].$$

Therefore from Eqs.(2.2.2.109.a,b) we obtain
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) \le$$

$$\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) + \Lambda_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) -$$

$$\overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times$$

$$\exp\left\{-\frac{1}{\epsilon}\int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2\right\} \times$$

$$+ \sum_{n=1}^{M} \frac{(-1)^n \epsilon^{n+1/2}}{(2n+1)!} \times \qquad (2.2.2.109.b)$$

$$\left[ \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times \right.$$

$$\left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n+1} \times$$

$$\exp\left\{-\frac{1}{\epsilon}\int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2\right\}].$$

Therefore from Eqs.(2.2.2.109.a,b) we obtain
$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) \le$$

$$\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) -$$

$$-\sum_{n=1}^{M} \frac{(-1)^n \epsilon^n}{(2n)!} \times$$

$$\left[ \frac{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \Xi_1(q(t), \lambda, \epsilon, \epsilon) \times \right. \quad (2.2.2.110.a)$$

$$\left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n} \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\}$$

and
$$v_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) \le$$

$$\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) -$$

$$-\sum_{n=1}^{M} \frac{(-1)^n \epsilon^n}{(2n)!} \times$$

$$\left[ \frac{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \Xi_{1}(q(t), \lambda, \epsilon, \epsilon) \times \right. \quad (2.2.2.110.a)$$

$$\left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n} \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \{\sqrt{\epsilon} u(t) + \lambda\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\}$$

and
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) \le$$

$$\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) -$$

$$-\sum_{n=1}^{M} \frac{(-1)^n \epsilon^{n+1/2}}{(2n+1)!} \times$$

$$\left[ \frac{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times \right. \quad (2.2.2.110.b)$$

$$\left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n+1} \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\}.$$

Let us calculate now oscillatory path integral $\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta)$:
$$v_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) \le$$

$$\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) -$$

$$-\sum_{n=1}^{M} \frac{(-1)^n \epsilon^{n+1/2}}{(2n+1)!} \times$$

$$\left[ \frac{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy D^+[u(t), \epsilon] \Psi_0(y) \sqrt{\sqrt{\epsilon} |u(T)|} \Xi_2(q(t), \lambda, \epsilon, \epsilon) \times \right. \quad (2.2.2.110.b)$$

$$\left( \int_{0}^{T} \tilde{V}_{1}(q_{v(\epsilon, \epsilon)}(t)) dt \right)^{2n+1} \times$$

$$\exp\left\{-\frac{1}{\epsilon} \int_{0}^{T} dt \left[ \left\{ \sqrt{\epsilon} u(t) + \lambda \right\}_{\eta} - \eta g(t, \sqrt{\epsilon} u(T) + \lambda, \sqrt{\epsilon} u(0) + \lambda) \right]^2 \right\}.$$

Let us calculate now oscillatory path integral $\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta)$:
$$
\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \cos \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right]
\end{aligned}
\right]
\quad (2.2.2.111.a)
$$

and oscillatory path integral $\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta)$.

$$
\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =
$$

$$
\left[
\begin{aligned}
& \int dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \widetilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\
& \qquad \left. \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\
& \qquad \left. \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right] \right) \right] \right]
\end{aligned}
\right]
\quad (2.2.2.111.b)
$$

From Eqs.(2.2.2.110. a,b) we obtain
$$ \Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = $$

$$ \left[ \bar{\int} dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\ \left. \cos \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\ \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right], \quad (2.2.2.111.a) $$

and oscillatory path integral $\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta)$.

$$ \Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = $$

$$ \left[ \bar{\int} dy \left\{ \underbrace{\int}_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \times \right. \\ \left. \sin \left[ \left( \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon,\epsilon)}^2(t) + \right. \right. \right. \\ \left. \left. + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon,\epsilon)}(t)}{\sqrt{\epsilon}} \right) \right] \right] \right]. \quad (2.2.2.111.b) $$

From Eqs.(2.2.2.110. a,b) we obtain
$$
\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =
$$

$$
\frac{1}{2} \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\epsilon} |u(T)|} \times \qquad (2.2.2.112.a)
$$

$$
\left\{ \left[ \exp[i\tilde{S}_0(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)] + \exp[-i\tilde{S}_0(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)] \right] \right\} = \\
\frac{1}{2} \tilde{\Theta}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) + \frac{1}{2} \tilde{\Theta}_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta)
$$

and

$$
\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =
$$

$$
\frac{1}{2i} \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\epsilon} |u(T)|} \times \quad (2.2.2.112.a)
$$

$$
\left\{ \left[ \exp[i\tilde{S}_0(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)] - \exp[-i\tilde{S}_0(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)] \right] \right\} = \\
\frac{1}{2i} \tilde{\Theta}_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) - \frac{1}{2i} \tilde{\Theta}_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta).
$$

Here

$$
\tilde{S}_0(T, x, x_0, \lambda, q(t), \epsilon, \epsilon) =
$$

$$
\int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right] \quad (2.2.2.113)
$$
$$
\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) =
$$

$$
\frac{1}{2} \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|}} \times
$$

(2.2.2.112).a)

$$
\{[\exp[i\tilde{S}_0(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)] + \exp[-i\tilde{S}_0(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)]]\} =
$$

$$
\frac{1}{2} \tilde{\Theta}_{1, \epsilon}(T, x, x_0, \lambda, \epsilon, \eta) + \frac{1}{2} \tilde{\Theta}_{2, \epsilon}(T, x, x_0, \lambda, \epsilon, \eta)
$$

and

$$
\Theta_{2,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta) =
$$

$$
\frac{1}{2i} \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|}} \times
$$

(2.2.2.112).a)

$$
\{[\exp[i\tilde{S}_0(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)] - \exp[-i\tilde{S}_0(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)]]\} =
$$

$$
\frac{1}{2i} \tilde{\Theta}_{1, \epsilon}(T, x, x_0, \lambda, \epsilon, \eta) - \frac{1}{2i} \tilde{\Theta}_{2, \epsilon}(T, x, x_0, \lambda, \epsilon, \eta).
$$

Here

$$
\tilde{S}_0(T, x, x_0, \lambda, q(t), \epsilon, \epsilon) =
$$

$$
\int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{v(\epsilon, \epsilon)}^2(t) + b_{\epsilon}(t, \lambda) \frac{u_{v(\epsilon, \epsilon)}(t)}{\sqrt{\epsilon}} \right]
$$

(2.2.2.113)
and

$$
\Theta_{1,\varepsilon}(T, x, x_0, \lambda, \varepsilon, \eta) = \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\varepsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\varepsilon}}} dy \tilde{D}_{\eta}^{+}[u(t), \varepsilon] \Psi_{0}(y) \sqrt{\sqrt{\varepsilon}|u(T)|} \exp[i\tilde{S}_{0}(T, x, x_0, \lambda, u(t), \varepsilon, \varepsilon)]}, \quad (2.2.2.114.a)
$$

$$
\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{|u(T)|} \exp[-i\tilde{S}_{0}(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)]} \quad (2.2.2.114.b)
$$

Let's calculate now path integral $\tilde{\Theta}_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta)$ by using stationary phase method.

By using replacement $q(t) := \frac{q(t)}{\sqrt{\epsilon}}$ we rewrite (2.2.2.114.a) in the next equivalent form

$$
\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \overline{\int_{u(T)=x-\lambda, u(0)=y-\lambda} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{|u(T)|} \times} \quad (2.2.2.115) \\
\exp\left\{\frac{i}{\epsilon} \int_{0}^{T} dt \left[\frac{m}{2} \left(\frac{d\tilde{u}_{\epsilon}(t)}{dt}\right)^2 - a_{\epsilon}(t, \lambda)u_{\epsilon}^2(t) + b_{\epsilon}(t, \lambda)u_{\epsilon}(t)\right]\right\}.
$$
and

$$
\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{\sqrt{\epsilon} |u(T)|} \exp[i\tilde{S}_{0}(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)]}, \quad (2.2.2.114.a)
$$

$$
\Theta_{2,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \overline{\int_{u(T)=\frac{x-\lambda}{\sqrt{\epsilon}}, u(0)=\frac{y-\lambda}{\sqrt{\epsilon}}} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{|u(T)|} \exp[-i\tilde{S}_{0}(T, x, x_0, \lambda, u(t), \epsilon, \epsilon)]}. \quad (2.2.2.114.b)
$$

Let's calculate now path integral $\tilde{\Theta}_{1,\epsilon}(T,x,x_0,\lambda,\epsilon,\eta)$ by using stationary phase method.

By using replacement $q(t) := \frac{q(t)}{\sqrt{\epsilon}}$ we rewrite (2.2.2.114.a) in the next equivalent form

$$
\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \overline{\int_{u(T)=x-\lambda, u(0)=y-\lambda} dy \tilde{D}_{\eta}^{+}[u(t), \epsilon] \Psi_{0}(y) \sqrt{|u(T)|} \times} \quad (2.2.2.115) \\
\exp\left\{\frac{i}{\epsilon} \int_{0}^{T} dt \left[ \frac{m}{2} \left( \frac{d\tilde{u}_{\epsilon}(t)}{dt} \right)^2 - a_{\epsilon}(t, \lambda) u_{\epsilon}^2(t) + b_{\epsilon}(t, \lambda) u_{\epsilon}(t) \right]\right\}.
$$
We assume now that $a_{\epsilon}(t, \lambda) = \frac{m\omega^2}{2} + \tilde{a}_{\epsilon}(\lambda)$. Then the corresponding master Lagrangian $\tilde{\mathcal{L}}_{\epsilon \approx 0}(\dot{u}, u) = \tilde{\mathcal{L}}(\dot{u}, u) + o(\epsilon)$ is

$$ \tilde{\mathcal{L}}(\dot{u}, u) = \frac{m}{2}\dot{u}^2 - \left(\frac{m\omega^2}{2} + \tilde{a}_{\epsilon}(\lambda)\right)u^2 + b(t, \lambda)u. \qquad (2.2.2.116) $$

The Euler-Lagrange equation of $\tilde{\mathcal{L}}(\dot{u}, u)$ with corresponding boundary conditions is

$$ \frac{d}{dt} \left( \frac{\partial \tilde{\mathcal{L}}}{\partial \dot{u}} \right) - \frac{\partial \tilde{\mathcal{L}}}{\partial u} = 0, \qquad (2.2.2.117) $$

$$ u(0) = -\lambda, u(T) = x - \lambda = x. $$

We assume now that $\frac{m\omega^2}{2} = \frac{m\omega^2}{2} + \tilde{a}_{\epsilon}(\lambda) = \frac{m}{2}\left(\omega^2 + \frac{2\tilde{a}_{\epsilon}(\lambda)}{m}\right) \ge 0$ and rewrite Eq.(2.2.2.116) in the canonical form:

$$ \tilde{\mathcal{L}}(\dot{u}, u) = \frac{m}{2}\dot{u}^2 - \frac{m\omega^2}{2}u^2 + b(t, \lambda)u, \qquad (2.2.2.118) $$

$$ \omega(\lambda) = \left| \sqrt{\left(\omega^2 + \frac{2\tilde{a}_{\epsilon}(\lambda)}{m}\right)} \right|. $$

Thus Eq.(2.2.2.117) takes the form
We assume now that $a_ε(t, λ) = \frac{m\omega^2}{2} + \tilde{a}_ε(λ)$. Then the corresponding master Lagrangian $\tilde{\mathcal{L}}_{ε\approx0}(\dot{u}, u) = \tilde{\mathcal{L}}(\dot{u}, u) + o(ε)$ is

$$ \tilde{\mathcal{L}}(\dot{u}, u) = \frac{m}{2}\dot{u}^2 - \left(\frac{m\omega^2}{2} + \tilde{a}_\epsilon(\lambda)\right)u^2 + b(t, \lambda)u. \qquad (2.2.2.116) $$

The Euler-Lagrange equation of $\tilde{\mathcal{L}}(\dot{u}, u)$ with corresponding boundary conditions is

$$ \begin{gathered} \frac{d}{dt} \left( \frac{\partial \tilde{\mathcal{L}}}{\partial \dot{u}} \right) - \frac{\partial \tilde{\mathcal{L}}}{\partial u} = 0, \qquad (2.2.2.117) \\ u(0) = -\lambda, u(T) = x - \lambda = x. \end{gathered} $$

We assume now that $\frac{m\omega^2}{2} = \frac{m\omega^2}{2} + \tilde{a}_ε(λ) = \frac{m}{2} \left( \omega^2 + \frac{2\tilde{a}_ε(λ)}{m} \right) \ge 0$ and rewrite Eq.(2.2.2.116) in the canonical form:

$$ \begin{gathered} \tilde{\mathcal{L}}(\dot{u}, u) = \frac{m}{2}\dot{u}^2 - \frac{m\omega^2}{2}u^2 + b(t, \lambda)u, \qquad (2.2.2.118) \\ \omega(\lambda) = \left| \sqrt{\left(\omega^2 + \frac{2\tilde{a}_\epsilon(\lambda)}{m}\right)} \right|. \end{gathered} $$

Thus Eq.(2.2.2.117) takes the form
$$m\ddot{u} + m\omega^2(\lambda)u - b(t, \lambda) = 0,$$

$$u(0) = y - \lambda = \tilde{y}, \qquad (2.2.2.119)$$

$$u(T) = x - \lambda = \tilde{x}.$$

Corresponding general solution is

$$u_{cl}(t) = \tilde{u}(t) + \int_{0}^{t} \tilde{u}(t-s)b(s, \lambda)ds, \qquad (2.2.2.120)$$

$$\tilde{u}(t) = c_1 \sin(\omega t) + c_2 \cos(\omega t).$$

Substituting Eq.(2.2.2.120) into boundary conditions (2.2.2.119) we obtain

$$\tilde{u}(0) = c_2 = \tilde{y},$$

$$\tilde{u}(T) = \tilde{x} = c_1 \sin \omega T + \tilde{y} \cos \omega T + \qquad (2.2.2.121)$$

$$\int_{0}^{T} [c_{1} \sin(\omega(t-s)) + \tilde{y} \cos(\omega(t-s))] b(s, \lambda) ds.$$

and
$$m\ddot{u} + m\omega^2(\lambda)u - b(t, \lambda) = 0,$$

$$u(0) = y - \lambda = \tilde{y}, \qquad (2.2.2.119)$$

$$u(T) = x - \lambda = \tilde{x}.$$

Corresponding general solution is

$$u_{cl}(t) = \tilde{u}(t) + \int_{0}^{t} \tilde{u}(t-s)b(s, \lambda)ds, \qquad (2.2.2.120)$$

$$\tilde{u}(t) = c_1 \sin(\omega t) + c_2 \cos(\omega t).$$

Substituting Eq.(2.2.2.120) into boundary conditions (2.2.2.119) we obtain

$$\tilde{u}(0) = c_2 = \tilde{y},$$

$$\tilde{u}(T) = \tilde{x} = c_1 \sin \omega T + \tilde{y} \cos \omega T + \qquad (2.2.2.121)$$

$$\int_{0}^{T} [c_{1} \sin(\omega(t-s)) + \tilde{y} \cos(\omega(t-s))] b(s, \lambda) ds.$$

and
$$c_1 = \frac{\tilde{x} - \tilde{y} \cos \omega T - \tilde{y} \int_0^T \cos(\omega(t-s))b(s, \lambda)ds}{\sin \omega T + \int_0^T \sin(\omega(t-s))b(s, \lambda)ds} \quad (2.2.2.122)$$

Corresponding classical action $\tilde{S}_{cl}$ is [13]:

$$\begin{aligned} \tilde{S}_{cl}(T) = & \\ \frac{m\omega}{2\sin\omega T} \Biggl[ & (\cos\omega T)(\lambda^2 + \bar{x}^2) - 2\tilde{x}\tilde{y} + \frac{2\tilde{x}}{m\omega} \int_0^T g(\lambda, t)\sin(\omega t)dt + \\ & + \frac{2\tilde{y}}{m\omega} \int_0^T g(\lambda, t)\sin\omega(T-t)dt - \\ & - \frac{2}{m^2\omega^2} \int_0^T \int_0^t g(\lambda, t)g(\lambda, s)\sin\omega(T-t)\sin(\omega s)dsdt \Biggr]. \end{aligned} \quad (2.2.2.123)$$

Substitution Eq.(2.2.123) into Eq.(2.2.115) gives

$$\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \int_{-\infty}^{+\infty} dy D\Psi_0(y) \sqrt{|x-\lambda|} \exp\left\{\frac{i}{\epsilon} \int_0^T dt \tilde{\mathcal{L}}(\dot{u}(t), u(t))\right\} = (2.2.2.124)$$
$$c_1 = \frac{\tilde{x} - \tilde{y} \cos \omega T - \tilde{y} \int_0^T \cos(\omega(t-s))b(s, \lambda)ds}{\sin \omega T + \int_0^T \sin(\omega(t-s))b(s, \lambda)ds} \quad (2.2.2.122)$$

Corresponding classical action $\tilde{S}_{cl}$ is [13]:

$$\begin{aligned} \tilde{S}_{cl}(T) = & \\ \frac{m\omega}{2\sin\omega T} \Biggl[ & (\cos\omega T)(\lambda^2 + \bar{x}^2) - 2\tilde{x}\tilde{y} + \frac{2\tilde{x}}{m\omega} \int_0^T g(\lambda, t)\sin(\omega t)dt + \\ & + \frac{2\tilde{y}}{m\omega} \int_0^T g(\lambda, t)\sin\omega(T-t)dt - \\ & - \frac{2}{m^2\omega^2} \int_0^T \int_0^t g(\lambda, t)g(\lambda, s)\sin\omega(T-t)\sin(\omega s)dsdt \Biggr]. \end{aligned} \quad (2.2.2.123)$$

Substitution Eq.(2.2.123) into Eq.(2.2.115) gives

$$\Theta_{1,\epsilon}(T, x, x_0, \lambda, \epsilon, \eta) = \int_{-\infty}^{+\infty} dy D\Psi_0(y) \sqrt{|x-\lambda|} \exp \left\{ \frac{i}{\epsilon} \int_0^T dt \tilde{\mathcal{L}}(\dot{u}(t), u(t)) \right\} = \quad (2.2.2.124)$$
the equation $d\tilde{S}_{cl}/d\tilde{x} = 0$ takes the form

$$2\tilde{x}(\cos \omega T) - 2\tilde{y} + \frac{2}{m\omega} \int_{0}^{T} b(\lambda, t) \sin(\omega t) dt = 0. \quad (2.2.2.12)$$

From Eq.(2.3.2.19) we obtain

$$x = \lambda + \left( \frac{\tilde{y}}{\cos \omega T} - \frac{1}{m\omega \cos \omega T} \int_{0}^{T} g(\lambda, t) \sin(\omega t) dt \right). \quad (2.2.2.12)$$

(2.2.2.1)

(2.2.2.1)

From Theorem 2.2.2.1 and Eqs.(2.2.2.66) we obtain:
**Theorem 2.2.2.2**
the equation $d\tilde{S}_{cl}/d\tilde{x} = 0$ takes the form

$$2\tilde{x}(\cos \omega T) - 2\tilde{y} + \frac{2}{m\omega} \int_{0}^{T} b(\lambda, t) \sin(\omega t) dt = 0. \quad (2.2.2.12)$$

From Eq.(2.3.2.19) we obtain

$$x = \lambda + \left( \frac{\tilde{y}}{\cos \omega T} - \frac{1}{m\omega \cos \omega T} \int_{0}^{T} g(\lambda, t) \sin(\omega t) dt \right). \quad (2.2.2.12)$$

$$(2.2.2.1)$$

$$(2.2.2.1)$$

From Theorem 2.2.2.1 and Eqs.(2.2.2.66) we obtain:
**Theorem 2.2.2.2.**
$$
\begin{align}
& |\langle T, x_0, \lambda; \epsilon, \delta \rangle| \le \hat{U}(T, x_0, \lambda)(C + O(\delta)). \tag{2.2.2.67} \\
& \liminf_{0<h\ll\delta} |\langle T, x_0, \lambda; \epsilon, \delta \rangle| \le \hat{U}(T, x_0, \lambda) \times C. \nonumber
\end{align}
$$

Here

$$
\begin{align}
U(T, x_0, \lambda) &= \left| \frac{1}{m\omega(\lambda)} \int_0^T \hat{b}(t, \lambda) \sin[\omega(\lambda)(T-t)] dt + x_0 \right|, \tag{2.2.2.68} \\
\omega^2(\lambda) &= \frac{m\hat{a}_c(\lambda)}{2}. \notag
\end{align}
$$

From inequality we obtain(2.2.2.67) master Equation for the Feynman-Colombeau
path integral.

**Theorem 2.2.2.2. (Master Equation for the Feynman-Colombeau path integral).**

$$
U(T, x_0, \lambda) = \left| \frac{1}{m\omega(\lambda)} \int_0^T \hat{b}(t, \lambda) \sin[\omega(\lambda)(T-t)] dt + x_0 \right| = 0. \quad (2.2.2.69)
$$

II.3. Quantum anharmonic oscillator with a cubic
potential. Quasiclassical asymptotic of the quantum
trajectories.
$$ 
\begin{gathered}
|\langle T, x_0, \lambda; \epsilon, \delta \rangle| \le \hat{U}(T, x_0, \lambda)(C + O(\delta)). \qquad (2.2.2.67) \\
\liminf_{0<h\ll\delta} |\langle T, x_0, \lambda; \epsilon, \delta \rangle| \le \hat{U}(T, x_0, \lambda) \times C.
\end{gathered}
$$

Here

$$ 
\begin{gathered}
U(T, x_0, \lambda) = \left| \frac{1}{m\omega(\lambda)} \int_0^T \hat{b}(t, \lambda) \sin[\omega(\lambda)(T-t)]dt + x_0 \right|, \qquad (2.2.2.68) \\
\omega^2(\lambda) = \frac{m\hat{a}_c(\lambda)}{2}.
\end{gathered}
$$

From inequality we obtain(2.2.2.67) master Equation for the Feynman-Colombeau path integral.

**Theorem 2.2.2.2. (Master Equation for the Feynman-Colombeau path integral).**

$$ U(T, x_0, \lambda) = \left| \frac{1}{m\omega(\lambda)} \int_0^T \hat{b}(t, \lambda) \sin[\omega(\lambda)(T-t)]dt + x_0 \right| = 0. \qquad (2.2.2.69) $$

## II.3. Quantum anharmonic oscillator with a cubic potential. Quasiclassical asymptotic of the quantum trajectories.
## II.3.1. Quantum harmonic oscillator. Quasiclassical asymptotic of the quantum trajectories.

As a first example we consider quantum harmonic oscillator (as cartooned in fig.2.3.1),

$$V(x) = m \frac{\omega^2}{2} x^2 - bx. \qquad (2.3.1.1)$$

supplemented by an additive sinusoidal driving.

<figure><img src="image_1.png" /><figcaption>Fig. 2.3.1.b = 1, $\omega$ = 3, $m$ = 1.</figcaption></figure>

### Example.1.1.

$$m = 1, \Omega = 10, \omega = 3, b = 1, A = 0.$$
II.3.1.Quantum harmonic oscillator.Quasiclassical
asimptotic of the quantum trajectories.

As a first example we consider quantum harmonic oscillator (as cartooned in fig.2.3.1),

$$
V(x) = m \frac{\omega^2}{2} x^2 - bx. \tag{2.3.1.1}
$$

supplemented by an additive sinusoidal driving.

<figure><img src="image_2.png" /><figcaption>Fig. 2.3.1. b = 1, ω = 3, m = 1.</figcaption></figure>

**Example.1.1.**

$m = 1, \Omega = 10, \omega = 3, b = 1, A = 0.$
<figure><img src="image_2.png" /><figcaption>Fig. 1.1.1. $b = 1, \omega = 3, m = 1$.</figcaption></figure>

<figure><img src="image_3.png" /><figcaption>Fig. 1.1.2.</figcaption></figure>

Example.1.2.

$$m = 1, \Omega = 10, \omega = 3, b = 1, A = 10.$$
<figure><img src="image_3.png" /><figcaption>Fig. 1.1.1. $b = 1, \omega = 3, m = 1$.</figcaption></figure>

<figure><img src="image_4.png" /><figcaption>Fig. 1.1.2.</figcaption></figure>

**Example.1.2.**

$$m = 1, \Omega = 10, \omega = 3, b = 1, A = 10.$$
<figure><img src="image_4.png" /><figcaption>Fig. 1.2.1.b = 1, ω = 3, m = 1.</figcaption></figure>

<figure><img src="image_5.png" /><figcaption>Fig. 1.2.2.</figcaption></figure>

**Example.1.3.**

$m = 1, \Omega = 20, \omega = 5, b = 1, A = 10.$
<figure><img src="image_5.png" /><figcaption>Fig. 1.2.1.b = 1, ω = 3, m = 1.</figcaption></figure>

<figure><img src="image_6.png" /><figcaption>Fig. 1.2.2.</figcaption></figure>

**Example.1.3.**

$m = 1, \Omega = 20, \omega = 5, b = 1, A = 10.$
<figure><img src="image_6.png" /><figcaption>Fig. 1.3.1.</figcaption></figure>

<figure><img src="image_7.png" /><figcaption>Fig. 1.3.2.</figcaption></figure>

### II.3.2. Quantum anharmonic oscillator with a cubic potential. Quasiclassical asymptotic of the generalized quantum trajectories.

Let's calculate generalized quantum trajectories (generalized quantum averages) of the one dimensional quantum anharmonic oscillator with a cubic potential, i.e.
<figure><img src="image_7.png" /><figcaption>Fig. 1.3.1.</figcaption></figure>

<figure><img src="image_8.png" /><figcaption>Fig. 1.3.2.</figcaption></figure>

### II.3.2. Quantum anharmonic oscillator with a cubic potential. Quasiclassical asymptotic of the generalized quantum trajectories.

Let's calculate generalized quantum trajectories (generalized quantum averages) of the one dimensional quantum anharmonic oscillator with a cubic potential, i.e.
$$ \langle x_i, t, x_0; \hbar \rangle = \frac{\int_{-L}^{L} x \Psi(x, t, x_0; \hbar) dx}{\int_{-L}^{L} \Psi(x, t, x_0; \hbar) dx}, \qquad (2.3.2.0) $$

$$ \Psi(x, 0, x_0; \hbar) = \frac{1}{\sqrt{2\pi\epsilon}} \exp\left(-\frac{x^2}{\epsilon}\right) \approx \delta(x - x_0), $$ $$ \epsilon \ll \hbar \ll 1. $$

As a second example we consider quantum anharmonic oscillator with a cubic potential (as cartooned in fig.2.3.2),

$$ V(x) = \frac{a}{4}x^3 - \frac{b}{2}x. \qquad (2.3.2.1) $$

supplemented by an additive sinusoidal driving.

<figure><img src="image_8.png" /><figcaption>Fig.2.3.2.a = 0.5, b = 1.</figcaption></figure>

$$ V(x) = \frac{a}{4}x^3 - \frac{b}{2}x, V(x,t) = \frac{m\omega^2}{2}x^2 + ax^3 - bx - [A \sin(\Omega t)]x. $$
$$ \langle x_i, t, x_0; \hbar \rangle = \frac{\int_{-L}^{L} x \Psi(x, t, x_0; \hbar) dx}{\int_{-L}^{L} \Psi(x, t, x_0; \hbar) dx}, \qquad (2.3.2.0) $$

$$ \Psi(x, 0, x_0; \hbar) = \frac{1}{\sqrt{2\pi\epsilon}} \exp\left(-\frac{x^2}{\epsilon}\right) \approx \delta(x - x_0), $$

$$ \epsilon \ll \hbar \ll 1. $$

As a second example we consider quantum anharmonic oscillator with a cubic potential (as cartooned in fig.2.3.2),

$$ V(x) = \frac{a}{4}x^3 - \frac{b}{2}x. \qquad (2.3.2.1) $$

supplemented by an additive sinusoidal driving.

<figure><img src="image_9.png" /><figcaption>Fig.2.3.2.a = 0.5, b = 1.</figcaption></figure>

$$ V(x) = \frac{a}{4}x^3 - \frac{b}{2}x, V(x,t) = \frac{m\omega^2}{2}x^2 + ax^3 - bx - [A \sin(\Omega t)]x. $$
Let us consider transition probability amplitude given via formulae
(2.1.1)-(2.1.2) with

$$
V(x,t) = \frac{m\omega^2}{2}x^2 + ax^3 - bx - [A \sin(\Omega t)]x. \quad (2.3.2.2)
$$

The corresponding classical Lagrangian is

$$
\mathcal{L}(\dot{x}, x) = \frac{m}{2} \dot{x}^2 - \frac{m\omega^2}{2} x^2 - ax^3 + bx + [A \sin(\Omega t)]x. \quad (2.3.2.3)
$$

By substitution $x(\tau) = u(\tau) + \lambda$ we obtain

$$
\begin{align*}
\mathcal{L}(\dot{u}, u) &= \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} (u + \lambda)^2 - a(u + \lambda)^3 + \\
&\qquad b(u + \lambda) + [A \sin(\Omega t)](u + \lambda) = \\
&\qquad \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} u^2 - m\omega^2 \lambda u - \frac{m\omega^2 \lambda^2}{2} - \\
&\qquad -au^3 - 3au^2 \lambda - 3au \lambda^2 - a\lambda^3 + bu + b\lambda + \\
&\qquad [A \sin(\Omega t)]u + [A \sin(\Omega t)]\lambda = \\
&\qquad \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} u^2 - 3a\lambda u^2 - \\
&\qquad -m\omega^2 \lambda u - 3au\lambda^2 + bu + [A \sin(\Omega t)]u - au^3 - \\
&\qquad - \frac{m\omega^2 \lambda^2}{2} - a\lambda^3 + b\lambda + [A \sin(\Omega t)]\lambda.
\end{align*}
\tag{2.3.2.4}
$$
Let us consider transition probability amplitude given via formulae
(2.1.1)-(2.1.2) with

$$
V(x,t) = \frac{m\omega^2}{2}x^2 + ax^3 - bx - [A \sin(\Omega t)]x. \quad (2.3.2.2)
$$

The corresponding classical Lagrangian is

$$
\mathcal{L}(\dot{x}, x) = \frac{m}{2} \dot{x}^2 - \frac{m\omega^2}{2} x^2 - ax^3 + bx + [A \sin(\Omega t)]x. \quad (2.3.2.3)
$$

By substitution $x(\tau) = u(\tau) + \lambda$ we obtain

$$
\begin{align*}
\mathcal{L}(\dot{u}, u) &= \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} (u + \lambda)^2 - a(u + \lambda)^3 + \\
&\qquad b(u + \lambda) + [A \sin(\Omega t)](u + \lambda) = \\
&\qquad \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} u^2 - m\omega^2 \lambda u - \frac{m\omega^2 \lambda^2}{2} - \\
&\qquad -au^3 - 3au^2 \lambda - 3au \lambda^2 - a\lambda^3 + bu + b\lambda + \\
&\qquad [A \sin(\Omega t)]u + [A \sin(\Omega t)]\lambda = \\
&\qquad \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} u^2 - 3a\lambda u^2 - \\
&\qquad -m\omega^2 \lambda u - 3au\lambda^2 + bu + [A \sin(\Omega t)]u - au^3 - \\
&\qquad - \frac{m\omega^2 \lambda^2}{2} - a\lambda^3 + b\lambda + [A \sin(\Omega t)]\lambda.
\end{align*}
\tag{2.3.2.4}
$$
The corresponding master Lagrangian $\tilde{\mathcal{L}}(\dot{u}, u)$ is

$$ 
\begin{aligned}
\tilde{\mathcal{L}}(\dot{u}, u) &= \frac{m}{2}\dot{u}^2 - \left(\frac{m\omega^2}{2} + 3a\lambda\right)u^2 - (m\omega^2\lambda + 3a\lambda^2 - b - A\sin(\Omega t))u = \\
&\qquad \frac{m}{2}\dot{u}^2 - m\left(\frac{\omega^2}{2} + \frac{3a\lambda}{m}\right)u^2 - (m\omega^2\lambda + 3a\lambda^2 - b - A\sin(\Omega t))u.
\end{aligned}
\quad (2.3.2.5) $$

The Euler-Lagrange equation of $\tilde{\mathcal{L}}(\dot{u}, u)$ with corresponding boundary conditions is

$$ 
\begin{aligned}
& \frac{d}{dt}\left(\frac{\partial \tilde{\mathcal{L}}}{\partial \dot{u}}\right) - \frac{\partial \tilde{\mathcal{L}}}{\partial u} = 0, \\
& u(0) = -\lambda, u(T) = x - \lambda = \bar{x}.
\end{aligned}
\quad (2.3.2.6) $$

**Remark 2.3.1.** We assume now that $\frac{\omega^2}{2} + \frac{3a\lambda}{m} \ge 0$ and rewrite Eq.(2.3.2.5) in the form

$$ 
\begin{aligned}
\tilde{\mathcal{L}}(\dot{u}, u) &= \frac{m}{2}\dot{u}^2 - \frac{m\omega^2(\lambda)}{2}u^2 + g(\lambda, t)u, \\
\omega(\lambda) &= \left|\sqrt{2\left(\frac{\omega^2}{2} + \frac{3a\lambda}{m}\right)}\right|, \\
g(\lambda, t) &= -(m\omega^2\lambda + 3a\lambda^2 - b - A\sin(\Omega t)).
\end{aligned}
\quad (2.3.2.7) $$

Thus Eq.(2.3.2.6) takes the form
The corresponding master Lagrangian $\tilde{\mathcal{L}}(\dot{u}, u)$ is

$$ 
\begin{aligned}
\tilde{\mathcal{L}}(\dot{u}, u) &= \frac{m}{2}\dot{u}^2 - \left(\frac{m\omega^2}{2} + 3a\lambda\right)u^2 - (m\omega^2\lambda + 3a\lambda^2 - b - A\sin(\Omega t))u = \\
&\qquad \frac{m}{2}\dot{u}^2 - m\left(\frac{\omega^2}{2} + \frac{3a\lambda}{m}\right)u^2 - (m\omega^2\lambda + 3a\lambda^2 - b - A\sin(\Omega t))u.
\end{aligned}
\qquad (2.3.2.5)
$$

The Euler-Lagrange equation of $\tilde{\mathcal{L}}(\dot{u}, u)$ with corresponding boundary conditions is

$$ 
\begin{aligned}
& \frac{d}{dt}\left(\frac{\partial \tilde{\mathcal{L}}}{\partial \dot{u}}\right) - \frac{\partial \tilde{\mathcal{L}}}{\partial u} = 0, \\
& u(0) = -\lambda, u(T) = x - \lambda = \bar{x}.
\end{aligned}
\qquad (2.3.2.6)
$$

**Remark 2.3.1.** We assume now that $\frac{\omega^2}{2} + \frac{3a\lambda}{m} \ge 0$ and rewrite Eq.(2.3.2.5) in the form

$$ 
\begin{aligned}
\tilde{\mathcal{L}}(\dot{u}, u) &= \frac{m}{2}\dot{u}^2 - \frac{m\omega^2(\lambda)}{2}u^2 + g(\lambda, t)u, \\
\omega(\lambda) &= \left|\sqrt{2\left(\frac{\omega^2}{2} + \frac{3a\lambda}{m}\right)}\right|, \\
g(\lambda, t) &= -(m\omega^2\lambda + 3a\lambda^2 - b - A\sin(\Omega t)).
\end{aligned}
\qquad (2.3.2.7)
$$

Thus Eq.(2.3.2.6) takes the form
$$m\ddot{u} + 2m\dot{\omega}^2(\lambda)u + d(\lambda) - A\sin(\Omega t) = 0, \qquad (2.3.2.8)$$

$$d(\lambda) = m\omega^2\lambda + 3a\lambda^2 - b.$$

Plugging in Eq.(2.3.2.8) this trial solution

$$u(t) = C + D\sin(\Omega t) \qquad (2.3.2.9)$$

we obtain

$$-mD\Omega^2 \sin(\Omega t) + 2m\omega^2(\lambda)(C + D\sin(\Omega t)) + d(\lambda) - A\sin(\Omega t) = 0. \qquad (2.3.2.10)$$

Hence

$$2m\omega^2(\lambda)C + d(\lambda) = 0, \qquad (2.3.2.11)$$

$$-mD\Omega^2 + 2m\omega^2(\lambda)D - A = 0.$$

Solving Eqs.(2.3.2.11) we obtain

$$C = -\frac{d(\lambda)}{2m\omega^2(\lambda)}, \qquad (2.3.2.12)$$

$$D = \frac{A}{(2\omega^2(\lambda) - \Omega^2)m}.$$

Thus general solution of the Eq.(2.3.2.8) is
$$m\ddot{u} + 2m\dot{\omega}^2(\lambda)u + d(\lambda) - A\sin(\Omega t) = 0, \qquad (2.3.2.8)$$

$$d(\lambda) = m\omega^2\lambda + 3a\lambda^2 - b.$$

Plugging in Eq.(2.3.2.8) this trial solution

$$u(t) = C + D\sin(\Omega t) \qquad (2.3.2.9)$$

we obtain

$$-mD\Omega^2 \sin(\Omega t) + 2m\omega^2(\lambda)(C + D\sin(\Omega t)) + d(\lambda) - A\sin(\Omega t) = 0. \qquad (2.3.2.10)$$

Hence

$$2m\omega^2(\lambda)C + d(\lambda) = 0, \qquad (2.3.2.11)$$

$$-mD\Omega^2 + 2m\omega^2(\lambda)D - A = 0.$$

Solving Eqs.(2.3.2.11) we obtain

$$C = -\frac{d(\lambda)}{2m\omega^2(\lambda)}, \qquad (2.3.2.12)$$

$$D = \frac{A}{(2\omega^2(\lambda) - \Omega^2)m}.$$

Thus general solution of the Eq.(2.3.2.8) is
$$ u(t, \lambda) = c_1 \sin \omega t + c_2 \cos \omega t + D \sin(\Omega t) + C. \quad (2.3.2.13) $$

Substituting Eq.(2.3.2.13) into boundary conditions

$$ \begin{aligned} u(0) &= -\lambda, \\ u(T) &= x - \lambda = \bar{x} \end{aligned} \quad (2.3.2.14) $$

we obtain

$$ \begin{aligned} u(0) &= c_2 + C = -\lambda, \\ u(T) &= c_1 \sin \omega T + c_2 \cos \omega T + D \sin(\Omega T) + C = \bar{x}. \end{aligned} \quad (2.3.2.15) $$

Hence

$$ \begin{aligned} c_2 &= -(C + \lambda), \\ c_1 &= \frac{\bar{x} - C - D \sin(\Omega T) + (C + \lambda) \cos \omega T}{\sin \omega T} \end{aligned} \quad (2.3.2.16) $$

and
$$ u(t, \lambda) = c_1 \sin \omega t + c_2 \cos \omega t + D \sin(\Omega t) + C. \quad (2.3.2.13) $$

Substituting Eq.(2.3.2.13) into boundary conditions

$$ \begin{aligned} u(0) &= -\lambda, \\ u(T) &= x - \lambda = \bar{x} \end{aligned} \quad (2.3.2.14) $$

we obtain

$$ \begin{aligned} u(0) &= c_2 + C = -\lambda, \\ u(T) &= c_1 \sin \omega T + c_2 \cos \omega T + D \sin(\Omega T) + C = \bar{x}. \end{aligned} \quad (2.3.2.15) $$

Hence

$$ \begin{aligned} c_2 &= -(C + \lambda), \\ c_1 &= \frac{\bar{x} - C - D \sin(\Omega T) + (C + \lambda) \cos \omega T}{\sin \omega T} \end{aligned} \quad (2.3.2.16) $$

and
$$ u(t, \lambda, \bar{x}) = \frac{\bar{x} - C - D \sin(\Omega T) + (C + \lambda) \cos \omega T}{\sin \omega T} \sin \omega t - (C + \lambda) \cos \omega t + \\ + D \sin(\Omega t) + C, \qquad (2.3.2.17) $$

$$ C = -\frac{d(\lambda)}{2m\omega^2(\lambda)}, \quad D = \frac{A}{(2\omega^2(\lambda) - \Omega^2)m}, $$

Note that

$$ u(T, \lambda, \bar{x}) = \frac{\bar{x} - C - D \sin(\Omega T) + (C + \lambda) \cos \omega T}{\sin \omega T} \sin \omega T - (C + \lambda) \cos \omega T + \\ + D \sin(\Omega T) + C = \qquad (2.3.17') $$

$$ \bar{x} - C - D \sin(\Omega T) + (C + \lambda) \cos \omega T - (C + \lambda) \cos \omega T + D \sin(\Omega T) + C = \bar{x}, $$

$$ u(T, \lambda, \bar{x}) = \bar{x}. $$

Corresponding master action $\tilde{S}_{cl}$ is [13]:

$$ \begin{aligned} \tilde{S}_{cl} = \frac{m\omega}{2\sin\omega T} \Biggl[ & (\cos\omega T)(\lambda^2 + \bar{x}^2) + 2\lambda\bar{x} + \frac{2\bar{x}}{m\omega} \int_0^T g(\lambda, t) \sin(\omega t) dt - \\ & - \frac{2\lambda}{m\omega} \int_0^T g(\lambda, t) \sin\omega(T-t) dt - \\ & - \frac{2}{m^2\omega^2} \int_0^T \int_0^t g(\lambda, t) g(\lambda, s) \sin\omega(T-t) \sin(\omega s) ds dt \Biggr]. \end{aligned} \qquad (2.3.2.18) $$

Thus the equation $d\tilde{S}_{cl}/d\bar{x} = 0$ takes the form
$$ u(t, \lambda, \bar{x}) = \frac{\bar{x} - C - D \sin(\Omega T) + (C + \lambda) \cos \omega T}{\sin \omega T} \sin \omega t - (C + \lambda) \cos \omega t + \\ + D \sin(\Omega t) + C, \qquad (2.3.2.17) $$

$$ C = -\frac{d(\lambda)}{2m\omega^2(\lambda)}, \quad D = \frac{A}{(2\omega^2(\lambda) - \Omega^2)m}, $$

Note that

$$ u(T, \lambda, \bar{x}) = \frac{\bar{x} - C - D \sin(\Omega T) + (C + \lambda) \cos \omega T}{\sin \omega T} \sin \omega T - (C + \lambda) \cos \omega T + \\ + D \sin(\Omega T) + C = \qquad (2.3.17') $$

$$ \bar{x} - C - D \sin(\Omega T) + (C + \lambda) \cos \omega T - (C + \lambda) \cos \omega T + D \sin(\Omega T) + C = \bar{x}, $$

$$ u(T, \lambda, \bar{x}) = \bar{x}. $$

Corresponding master action $\tilde{S}_{cl}$ is [13]:

$$ \begin{aligned} \tilde{S}_{cl} = \frac{m\omega}{2\sin\omega T} \Biggl[ & (\cos\omega T)(\lambda^2 + \bar{x}^2) + 2\lambda\bar{x} + \frac{2\bar{x}}{m\omega} \int_0^T g(\lambda, t) \sin(\omega t) dt - \\ & - \frac{2\lambda}{m\omega} \int_0^T g(\lambda, t) \sin\omega(T-t) dt - \\ & - \frac{2}{m^2\omega^2} \int_0^T \int_0^t g(\lambda, s) g(\lambda, s) \sin\omega(T-t) \sin(\omega s) ds dt \Biggr]. \end{aligned} \qquad (2.3.2.18) $$

Thus the equation $d\tilde{S}_{cl}/d\bar{x} = 0$ takes the form
$$2\bar{x}(\cos \omega T) + 2\lambda + \frac{2}{m\omega} \int_{0}^{T} g(\lambda, t) \sin(\omega t) dt = 0. \quad (2.3.2.19)$$

From Eq.(2.3.2.19) we obtain

$$\bar{x} = - \left( \frac{\lambda}{\cos \omega T} + \frac{1}{m\omega \cos \omega T} \int_{0}^{T} g(\lambda, t) \sin(\omega t) dt \right) \quad (2.3.2.20)$$

Substituting Eq.(2.3.2.20) into Eq.(2.3.2.17') gives master equation of the form

$$u(T, \lambda) = 0,$$

$$\begin{aligned} u(T, \lambda) = & -\frac{1}{\sin \omega T} \left( \frac{\lambda}{\cos \omega T} + \frac{1}{m\omega \cos \omega T} \int_0^T g(\lambda, t) \sin(\omega t) dt \right) \sin \omega T + \\ & \frac{-C - D \sin(\Omega T) + (C + \lambda) \cos \omega T}{\sin \omega T} \sin \omega T - (C + \lambda) \cos \omega T + \\ & + D \sin(\Omega T) + C = \\ & - \left( \frac{\lambda}{\cos \omega T} + \frac{1}{m\omega \cos \omega T} \int_0^T g(\lambda, t) \sin(\omega t) dt \right) = 0, \end{aligned} \quad (2.3.2.22)$$

$$g(\lambda, t) = -(m\omega^2\lambda + 3a\lambda^2 - b - A \sin(\Omega t)) = -d(\lambda) + A \sin(\Omega t),$$

$$C = -\frac{d(\lambda)}{2m\omega^2(\lambda)}, \quad D = \frac{A}{(2\omega^2(\lambda) - \Omega^2)m},$$

$$\omega^2(\lambda) = 2\left(\frac{\omega^2}{2} + \frac{3a\lambda}{m}\right) \geq 0,$$

$$d(\lambda) = m\omega^2\lambda + 3a\lambda^2 - b.$$
$$2\bar{x}(\cos \omega T) + 2\lambda + \frac{2}{m\omega} \int_{0}^{T} g(\lambda, t) \sin(\omega t) dt = 0. \quad (2.3.2.19)$$

From Eq.(2.3.2.19) we obtain

$$\bar{x} = - \left( \frac{\lambda}{\cos \omega T} + \frac{1}{m\omega \cos \omega T} \int_{0}^{T} g(\lambda, t) \sin(\omega t) dt \right) \quad (2.3.2.20)$$

Substituting Eq.(2.3.2.20) into Eq.(2.3.2.17') gives master equation of the form

$$u(T, \lambda) = 0,$$

$$\begin{aligned} u(T, \lambda) = & -\frac{1}{\sin \omega T} \left( \frac{\lambda}{\cos \omega T} + \frac{1}{m\omega \cos \omega T} \int_0^T g(\lambda, t) \sin(\omega t) dt \right) \sin \omega T + \\ & \frac{-C - D \sin(\Omega T) + (C + \lambda) \cos \omega T}{\sin \omega T} \sin \omega T - (C + \lambda) \cos \omega T + \\ & + D \sin(\Omega T) + C = \\ & - \left( \frac{\lambda}{\cos \omega T} + \frac{1}{m\omega \cos \omega T} \int_0^T g(\lambda, t) \sin(\omega t) dt \right) = 0, \end{aligned} \quad (2.3.2.22)$$

$$g(\lambda, t) = -(m\omega^2\lambda + 3a\lambda^2 - b - A \sin(\Omega t)) = -d(\lambda) + A \sin(\Omega t),$$

$$C = -\frac{d(\lambda)}{2m\omega^2(\lambda)}, \quad D = \frac{A}{(2\omega^2(\lambda) - \Omega^2)m},$$

$$\omega^2(\lambda) = 2\left(\frac{\omega^2}{2} + \frac{3a\lambda}{m}\right) \geq 0,$$

$$d(\lambda) = m\omega^2\lambda + 3a\lambda^2 - b.$$
Finally we obtain master equation of the form

$$
\begin{align}
& \lambda + \frac{1}{m\bar{\omega}} \int_0^T g(\lambda, t) \sin(\bar{\omega}t) dt = 0, \tag{2.3.2.23} \\
& \cos[\bar{\omega}(\lambda)T] \neq 0, \bar{\omega}(\lambda) \neq 0. \nonumber
\end{align}
$$

By simple calculation we obtain

$$
\begin{align*}
& \int_0^T g(\lambda, t) \sin(\bar{\omega}t) dt = \int_0^T [-d(\lambda) + A \sin(\Omega t)] \sin(\bar{\omega}t) dt = \\
& \qquad -d(\lambda) \int_0^T \sin(\bar{\omega}t) dt + A \int_0^T \sin(\bar{\omega}t) \sin(\Omega t) dt = \\
& \int_0^T d(\lambda) \sin(\bar{\omega}t) dt + \frac{A}{2} \left( \int_0^T \cos[(\bar{\omega} - \Omega)t] dt - \int_0^T \cos[(\bar{\omega} + \Omega)t] dt \right) = \\
& \qquad -d(\lambda) \left( \left. \frac{\cos(\bar{\omega}t)}{\bar{\omega}} \right|_0^T \right) + \frac{A}{2} \left( \left. \frac{\sin[(\bar{\omega} - \Omega)t]}{\bar{\omega} - \Omega} \right|_0^T \right) - \tag{2.3.24} \\
& \qquad \frac{A}{2} \left( \left. \frac{\sin[(\bar{\omega} + \Omega)t]}{\bar{\omega} + \Omega} \right|_0^T \right) = \\
& \qquad -d(\lambda) \left( -\frac{\cos(\bar{\omega}t)}{\bar{\omega}} + \frac{1}{\bar{\omega}} \right) + \frac{A}{2} \frac{\sin[(\bar{\omega} - \Omega)T]}{\bar{\omega} - \Omega} - \frac{A}{2} \frac{\sin[(\bar{\omega} + \Omega)T]}{\bar{\omega} + \Omega} = \\
& \qquad \left( \frac{d(\lambda) \cos(\bar{\omega}t)}{\bar{\omega}} - \frac{d(\lambda)}{\bar{\omega}} \right) + \frac{A}{2} \frac{\sin[(\bar{\omega} - \Omega)T]}{\bar{\omega} - \Omega} - \frac{A}{2} \frac{\sin[(\bar{\omega} + \Omega)T]}{\bar{\omega} + \Omega}.
\end{align*}
$$

Substitution Eq.(2.3.24) into Eq.(2.3.23) gives
Finally we obtain master equation of the form

$$
\begin{align}
& \lambda + \frac{1}{m\bar{\omega}} \int_0^T g(\lambda, t) \sin(\bar{\omega}t) dt = 0, \tag{2.3.2.23} \\
& \cos[\bar{\omega}(\lambda)T] \neq 0, \bar{\omega}(\lambda) \neq 0. \nonumber
\end{align}
$$

By simple calculation we obtain

$$
\begin{align*}
& \int_0^T g(\lambda, t) \sin(\bar{\omega}t) dt = \int_0^T [-d(\lambda) + A \sin(\Omega t)] \sin(\bar{\omega}t) dt = \\
& \qquad -d(\lambda) \int_0^T \sin(\bar{\omega}t) dt + A \int_0^T \sin(\bar{\omega}t) \sin(\Omega t) dt = \\
& \int_0^T d(\lambda) \sin(\bar{\omega}t) dt + \frac{A}{2} \left( \int_0^T \cos[(\bar{\omega} - \Omega)t] dt - \int_0^T \cos[(\bar{\omega} + \Omega)t] dt \right) = \\
& \qquad -d(\lambda) \left( \left. \frac{\cos(\bar{\omega}t)}{\bar{\omega}} \right|_0^T \right) + \frac{A}{2} \left( \left. \frac{\sin[(\bar{\omega} - \Omega)t]}{\bar{\omega} - \Omega} \right|_0^T \right) - \tag{2.3.24} \\
& \qquad \frac{A}{2} \left( \left. \frac{\sin[(\bar{\omega} + \Omega)t]}{\bar{\omega} + \Omega} \right|_0^T \right) = \\
& \qquad -d(\lambda) \left( -\frac{\cos(\bar{\omega}t)}{\bar{\omega}} + \frac{1}{\bar{\omega}} \right) + \frac{A}{2} \frac{\sin[(\bar{\omega} - \Omega)T]}{\bar{\omega} - \Omega} - \frac{A}{2} \frac{\sin[(\bar{\omega} + \Omega)T]}{\bar{\omega} + \Omega} = \\
& \qquad \left( \frac{d(\lambda) \cos(\bar{\omega}t)}{\bar{\omega}} - \frac{d(\lambda)}{\bar{\omega}} \right) + \frac{A}{2} \frac{\sin[(\bar{\omega} - \Omega)T]}{\bar{\omega} - \Omega} - \frac{A}{2} \frac{\sin[(\bar{\omega} + \Omega)T]}{\bar{\omega} + \Omega}.
\end{align*}
$$

Substitution Eq.(2.3.24) into Eq.(2.3.23) gives
$$ \lambda + \frac{1}{m\omega(\lambda)} \times \left[ \frac{d(\lambda)\cos(\omega T)}{\omega(\lambda)} - \frac{d(\lambda)}{\omega(\lambda)} - \frac{A\sin[(\omega(\lambda) - \Omega)T]}{2(\omega(\lambda) - \Omega)} + \frac{A\sin[(\omega(\lambda) + \Omega)T]}{2(\omega(\lambda) + \Omega)} \right] = 0. \qquad (2.3.2.25) $$

Let us consider now quantum harmonic oscillator, i.e. in Eqs. (2.3.22) $a = 0$ and therefore

$$ d(\lambda, a = 0) = m\omega^2\lambda - b, \qquad (2.3.2.26) $$

$$ \omega(\lambda) = \omega. $$
$$ \lambda + \frac{1}{m\omega(\lambda)} \times \left[ \frac{d(\lambda)\cos(\omega T)}{\omega(\lambda)} - \frac{d(\lambda)}{\omega(\lambda)} - \frac{A\sin[(\omega(\lambda) - \Omega)T]}{2(\omega(\lambda) - \Omega)} + \frac{A\sin[(\omega(\lambda) + \Omega)T]}{2(\omega(\lambda) + \Omega)} \right] = 0. \qquad (2.3.2.25) $$

Let us consider now quantum harmonic oscillator, i.e. in Eqs. (2.3.22) $a = 0$ and therefore

$$ d(\lambda, a = 0) = m\omega^2\lambda - b, \qquad (2.3.2.26) $$

$$ \omega(\lambda) = \omega. $$
$$ \lambda + \frac{1}{m\omega} \times $$

$$ \left[ \frac{(m\omega^2\lambda - b)\cos(\omega T)}{\omega} - \frac{m\omega^2\lambda - b}{\omega} \right] = 0. $$

$$ \lambda + \frac{(m\omega^2\lambda - b)\cos(\omega T)}{m\omega^2} - \frac{m\omega^2\lambda - b}{m\omega^2} = 0, $$

$$ \lambda + \lambda \cos(\omega T) - \frac{b \cos(\omega T)}{m\omega^2} - \lambda + \frac{b}{m\omega^2} = 0, \quad (2.3.2.27) $$

$$ \frac{\lambda}{\cos \omega T} + \lambda \cos(\omega T) - \frac{b \cos(\omega T)}{m\omega^2} + \frac{b}{m\omega^2} = 0, $$

$$ \lambda = \frac{b}{m\omega^2} - \frac{b}{m\omega^2 \cos(\omega T)} = $$

$$ \frac{b}{m\omega^2}\left(1 - \frac{1}{\cos(\omega T)}\right). $$

### Example.2.1.

$$ m = 10, \Omega = 10, \omega = 30, a = -200, b = 1, A = 1 $$
$$ \lambda + \frac{1}{m\omega} \times $$

$$ \left[ \frac{(m\omega^2\lambda - b)\cos(\omega T)}{\omega} - \frac{m\omega^2\lambda - b}{\omega} \right] = 0. $$

$$ \lambda + \frac{(m\omega^2\lambda - b)\cos(\omega T)}{m\omega^2} - \frac{m\omega^2\lambda - b}{m\omega^2} = 0, $$

$$ \lambda + \lambda \cos(\omega T) - \frac{b \cos(\omega T)}{m\omega^2} - \lambda + \frac{b}{m\omega^2} = 0, \quad (2.3.2.27) $$

$$ \frac{\lambda}{\cos \omega T} + \lambda \cos(\omega T) - \frac{b \cos(\omega T)}{m\omega^2} + \frac{b}{m\omega^2} = 0, $$

$$ \lambda = \frac{b}{m\omega^2} - \frac{b}{m\omega^2 \cos(\omega T)} = $$

$$ \frac{b}{m\omega^2}\left(1 - \frac{1}{\cos(\omega T)}\right). $$

### Example.2.1.

$$ m = 10, \Omega = 10, \omega = 30, a = -200, b = 1, A = 1 $$
<figure><img src="image_9.png" /><figcaption>Fig. 2.1.1.</figcaption></figure>

<figure><img src="image_10.png" /><figcaption>Fig. 2.1.2.a.</figcaption></figure>
<figure><img src="image_10.png" /><figcaption>Fig. 2.1.1.</figcaption></figure>

<figure><img src="image_11.png" /><figcaption>Fig. 2.1.2.a.</figcaption></figure>
<figure><img src="image_11.png" /><figcaption>Fig. 2.1.2.b.</figcaption></figure>

<figure><img src="image_12.png" /><figcaption>Fig. 2.1.3.a $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>
<figure><img src="image_12.png" /><figcaption>Fig. 2.1.2.b.</figcaption></figure>

<figure><img src="image_13.png" /><figcaption>Fig. 2.1.3.a $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>
<figure><img src="image_13.png" /><figcaption>Fig. 2.1.3.b $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

**Example.2.2.**

$$m = 10, \Omega = 10, \omega = 30, a = -200, b = 1, A = 10$$

<figure><img src="image_14.png" /><figcaption>Fig. 2.2.1.</figcaption></figure>
<figure><img src="image_14.png" /><figcaption>Fig. 2.1.3.b $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

Example.2.2.

$$m = 10, \Omega = 10, \omega = 30, a = -200, b = 1, A = 10$$

<figure><img src="image_15.png" /><figcaption>Fig. 2.2.1.</figcaption></figure>
<figure><img src="image_15.png" /><figcaption>Fig. 2.2.2.a.</figcaption></figure>

<figure><img src="image_16.png" /><figcaption>Fig. 2.2.2.b.</figcaption></figure>
<figure><img src="image_16.png" /><figcaption>Fig. 2.2.2.a.</figcaption></figure>

<figure><img src="image_17.png" /><figcaption>Fig. 2.2.2.b.</figcaption></figure>
<figure><img src="image_17.png" /><figcaption>Fig. 2.2.3.a. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

<figure><img src="image_18.png" /><figcaption>Fig. 2.2.3.b. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

Example.2.3.

$$m = 10, \Omega = 10, \omega = 40, a = -200, b = 1, A = 10$$
<figure><img src="image_18.png" /><figcaption>Fig. 2.2.3.a. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

<figure><img src="image_19.png" /><figcaption>Fig. 2.2.3.b. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

Example.2.3.

$$m = 10, \Omega = 10, \omega = 40, a = -200, b = 1, A = 10$$
<figure><img src="image_19.png" /><figcaption>Fig. 2.3.1.</figcaption></figure>

<figure><img src="image_20.png" /><figcaption>Fig. 2.3.2.</figcaption></figure>
<figure><img src="image_20.png" /><figcaption>Fig. 2.3.1.</figcaption></figure>

<figure><img src="image_21.png" /><figcaption>Fig. 2.3.2.</figcaption></figure>
<figure><img src="image_21.png" /><figcaption>Fig. 2.3.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

**Example.2.4.**

$$m = 10, \Omega = 20, \omega = 40, a = -200, b = 1, A = 10$$

<figure><img src="image_22.png" /><figcaption>Fig. 2.4.1.</figcaption></figure>
<figure><img src="image_22.png" /><figcaption>Fig. 2.3.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

**Example.2.4.**

$$m = 10, \Omega = 20, \omega = 40, a = -200, b = 1, A = 10$$

<figure><img src="image_23.png" /><figcaption>Fig. 2.4.1.</figcaption></figure>
<figure><img src="image_23.png" /><figcaption>Fig. 2.4.2.</figcaption></figure>

<figure><img src="image_24.png" /><figcaption>Fig. 2.4.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

**Example.2.5.**

$$m = 10, \Omega = 25, \omega = 40, a = -200, b = 1, A = 10$$
<figure><img src="image_24.png" /><figcaption>Fig. 2.4.2.</figcaption></figure>

<figure><img src="image_25.png" /><figcaption>Fig.2.4.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

**Example.2.5.**

$$m = 10, \Omega = 25, \omega = 40, a = -200, b = 1, A = 10$$
<figure><img src="image_25.png" /><figcaption>Fig. 2.5.1.</figcaption></figure>

<figure><img src="image_26.png" /><figcaption>Fig. 2.5.2.</figcaption></figure>
<figure><img src="image_26.png" /><figcaption>Fig. 2.5.1.</figcaption></figure>

<figure><img src="image_27.png" /><figcaption>Fig. 2.5.2.</figcaption></figure>
<figure><img src="image_27.png" /><figcaption>Fig. 2.5.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

## II.4. Quantum anharmonic oscillator with a double well potential.

As a second example we consider quantum anharmonic oscillator with a double well potential (as cartooned in fig.2.4.1),

$$V(x) = \frac{a}{4}x^4 - \frac{b}{2}x^2, \quad a, b > 0 \qquad (2.4.1)$$

supplemented by an additive sinusoidal driving.
<figure><img src="image_28.png" /><figcaption>Fig. 2.5.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

## II.4. Quantum anharmonic oscillator with a double well potential.

As a second example we consider quantum anharmonic oscillator with a double well potential (as cartooned in fig.2.4.1),

$$V(x) = \frac{a}{4}x^4 - \frac{b}{2}x^2, \quad a, b > 0 \qquad (2.4.1)$$

supplemented by an additive sinusoidal driving.
<figure><img src="image_28.png" /><figcaption>Fig.2.4.1.a = 0.5, b = 1</figcaption></figure>

Let us consider transition probability amplitude given via formulae (2.1.1)-

(2.1.2) with

$$V(x) = \frac{m\omega^2}{2}x^2 - ax^4 + bx^3 - [A \sin(\Omega t)]x. \quad (2.4.2)$$

The corresponding classical Lagrangian is

$$\mathcal{L}(\dot{x}, x) = \frac{m}{2}\dot{x}^2 - \frac{m\omega^2}{2}x^2 - bx^3 + ax^4 + [A \sin(\Omega t)]x. \quad (2.4.3)$$

By substitution $x = u + \lambda$ we obtain
<figure><img src="image_29.png" /><figcaption>Fig.2.4.1.a = 0.5, b = 1</figcaption></figure>

Let us consider transition probability amplitude given via formulae (2.1.1)-

(2.1.2) with

$$V(x) = \frac{m\omega^2}{2}x^2 - ax^4 + bx^3 - [A \sin(\Omega t)]x. \quad (2.4.2)$$

The corresponding classical Lagrangian is

$$\mathcal{L}(\dot{x}, x) = \frac{m}{2}\dot{x}^2 - \frac{m\omega^2}{2}x^2 - bx^3 + ax^4 + [A \sin(\Omega t)]x. \quad (2.4.3)$$

By substitution $x = u + \lambda$ we obtain
$$ 
\begin{aligned}
\mathcal{L}(\dot{u}, u) &= \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} (u + \lambda)^2 - b(u + \lambda)^3 + a(u + \lambda)^4 + \\
& \qquad [A \sin(\Omega t)](u + \lambda) =
\end{aligned}
$$

$$ \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} u^2 - m\omega^2\lambda u - 3b\lambda u^2 - 3b\lambda^2 u + 6a\lambda^2 u^2 + 4a\lambda^3 u + \dots = \quad (2.4.4) $$

$$ \frac{m}{2} \dot{u}^2 - \left( \frac{m\omega^2}{2} + 3b\lambda - 6a\lambda^2 \right) u^2 - (m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3)u + [A \sin(\Omega t)]u + au^4 - 4a\lambda u^3 + a\lambda^4 + [A \sin(\Omega t)]\lambda + \dots $$

The corresponding master Lagrangian $\tilde{\mathcal{L}}(\dot{u}, u)$ is

$$ 
\begin{aligned}
\tilde{\mathcal{L}}(\dot{u}, u) &= \frac{m}{2} \dot{u}^2 - \left( \frac{m\omega^2}{2} + 3b\lambda - 6a\lambda^2 \right) u^2 - (m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3 - A \sin(\Omega t)) u = \\
& \qquad \frac{m}{2} \dot{u}^2 - m \left( \frac{\omega^2}{2} + \frac{3b\lambda}{m} - \frac{6a\lambda^2}{m} \right) u^2 - (m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3 - A \sin(\Omega t)) u.
\end{aligned}
\quad (2.4.5) $$

The Euler-Lagrange equation of $\tilde{\mathcal{L}}(\dot{u}, u)$ with corresponding boundary conditions is

$$ 
\begin{aligned}
& \frac{d}{dt} \left( \frac{\partial \tilde{\mathcal{L}}}{\partial \dot{u}} \right) - \frac{\partial \tilde{\mathcal{L}}}{\partial u} = 0, \\
& u(0) = -\lambda, u(T) = x.
\end{aligned}
\quad (2.4.6) $$

**Remark 2.4.1.** We assume now that $2\omega^2(\lambda) = \frac{\omega^2}{2} + \frac{3b\lambda}{m} - \frac{6a\lambda^2}{m} \ge 0$ and rewrite Eq.(2.4.5) of the form
$$ \mathcal{L}(\dot{u}, u) = \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} (u + \lambda)^2 - b(u + \lambda)^3 + a(u + \lambda)^4 + $$

$$ [A \sin(\Omega t)](u + \lambda) = $$

$$ \frac{m}{2} \dot{u}^2 - \frac{m\omega^2}{2} u^2 - m\omega^2\lambda u - 3b\lambda u^2 - 3b\lambda^2 u + 6a\lambda^2 u^2 + 4a\lambda^3 u + \dots = \quad (2.4.4) $$

$$ \frac{m}{2} \dot{u}^2 - \left( \frac{m\omega^2}{2} + 3b\lambda - 6a\lambda^2 \right) u^2 - (m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3) u + $$

$$ [A \sin(\Omega t)]u + au^4 - 4a\lambda u^3 + a\lambda^4 + [A \sin(\Omega t)]\lambda + \dots $$

The corresponding master Lagrangian $\tilde{\mathcal{L}}(\dot{u}, u)$ is

$$ \tilde{\mathcal{L}}(\dot{u}, u) = \frac{m}{2} \dot{u}^2 - \left( \frac{m\omega^2}{2} + 3b\lambda - 6a\lambda^2 \right) u^2 - (m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3 - A \sin(\Omega t)) u = \quad (2.4.5) $$

$$ \frac{m}{2} \dot{u}^2 - m \left( \frac{\omega^2}{2} + \frac{3b\lambda}{m} - \frac{6a\lambda^2}{m} \right) u^2 - (m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3 - A \sin(\Omega t)) u. $$

The Euler-Lagrange equation of $\tilde{\mathcal{L}}(\dot{u}, u)$ with corresponding boundary conditions is

$$ \frac{d}{dt} \left( \frac{\partial \tilde{\mathcal{L}}}{\partial \dot{u}} \right) - \frac{\partial \tilde{\mathcal{L}}}{\partial u} = 0, \qquad (2.4.6) $$

$$ u(0) = -\lambda, u(T) = x. $$

**Remark 2.4.1.** We assume now that $2\omega^2(\lambda) = \frac{\omega^2}{2} + \frac{3b\lambda}{m} - \frac{6a\lambda^2}{m} \ge 0$ and rewrite Eq.(2.4.5) of the form
$$
\begin{gather*}
\tilde{\mathcal{L}}(\dot{u}, u) = \frac{m}{2}\dot{u}^2 - \frac{m\omega^2(\lambda)}{2}u^2 + g(\lambda, t)u, \\
\omega(\lambda) = \sqrt{2\left(\frac{\omega^2}{2} + \frac{3b\lambda}{m} - \frac{6a\lambda^2}{m}\right)}, \tag{2.4.7}
\end{gather*}
$$

$$
g(\lambda, t) = -[m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3 - A\sin(\Omega t)].
$$

Finally we obtain

$$
\begin{gather*}
\lambda + \frac{1}{m\omega(\lambda)} \times \\
\left[ \frac{d(\lambda) \cos(\omega T)}{\omega(\lambda)} - \frac{d(\lambda)}{\omega(\lambda)} + \frac{A \sin[(\omega(\lambda) - \Omega)T]}{2(\omega(\lambda) - \Omega)} - \frac{A \sin[(\omega(\lambda) + \Omega)T]}{2(\omega(\lambda) + \Omega)} \right] = 0, \tag{2.4.8} \\
\omega(\lambda) = \sqrt{2\left(\frac{\omega^2}{2} + \frac{3b\lambda}{m} - \frac{6a\lambda^2}{m}\right)}, \\
d(\lambda) = m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3.
\end{gather*}
$$

**Example.2.1.**

$m = 10, \Omega = 20, \omega = 10, a = -20, b = 10, A = 10$
$$
\begin{gather*}
\tilde{\mathcal{L}}(\dot{u}, u) = \frac{m}{2}\dot{u}^2 - \frac{m\omega^2(\lambda)}{2}u^2 + g(\lambda, t)u, \\
\omega(\lambda) = \sqrt{2\left(\frac{\omega^2}{2} + \frac{3b\lambda}{m} - \frac{6a\lambda^2}{m}\right)}, \tag{2.4.7}
\end{gather*}
$$

$$
g(\lambda, t) = -[m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3 - A \sin(\Omega t)].
$$

Finally we obtain

$$
\begin{gather*}
\lambda + \frac{1}{m\omega(\lambda)} \times \\
\left[ \frac{d(\lambda) \cos(\omega T)}{\omega(\lambda)} - \frac{d(\lambda)}{\omega(\lambda)} + \frac{A \sin[(\omega(\lambda) - \Omega)T]}{2(\omega(\lambda) - \Omega)} - \frac{A \sin[(\omega(\lambda) + \Omega)T]}{2(\omega(\lambda) + \Omega)} \right] = 0, \tag{2.4.8} \\
\omega(\lambda) = \sqrt{2\left(\frac{\omega^2}{2} + \frac{3b\lambda}{m} - \frac{6a\lambda^2}{m}\right)},
\end{gather*}
$$

$$
d(\lambda) = m\omega^2\lambda + 3b\lambda^2 - 4a\lambda^3.
$$

**Example.2.1.**

$m = 10, \Omega = 20, \omega = 10, a = -20, b = 10, A = 10$
<figure><img src="image_29.png" /></figure>

2.1.1.

<figure><img src="image_30.png" /></figure>

2.1.2.
<figure><img src="image_30.png" /></figure>

2.1.1.

<figure><img src="image_31.png" /></figure>

2.1.2.
<figure><img src="image_31.png" /><figcaption>Fig. 2.1.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

Example.2.2.

$$m = 100, \Omega = 20, \omega = 10, a = -20, b = 50, A = 10$$

<figure><img src="image_32.png" /></figure>

2.1.1.
<figure><img src="image_32.png" /><figcaption>Fig. 2.1.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

Example.2.2.

$$m = 100, \Omega = 20, \omega = 10, a = -20, b = 50, A = 10$$

<figure><img src="image_33.png" /></figure>

2.1.1.
<figure><img src="image_33.png" /><figcaption>2.1.2.</figcaption></figure>

<figure><img src="image_34.png" /><figcaption>Fig. 2.1.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

Example.2.3.

$$m = 100, \Omega = 20, \omega = 10, a = 20, b = 50, A = 100$$
<figure><img src="image_34.png" /><figcaption>2.1.2.</figcaption></figure>

<figure><img src="image_35.png" /><figcaption>Fig. 2.1.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

Example.2.3.

$$m = 100, \Omega = 20, \omega = 10, a = 20, b = 50, A = 100$$
<figure><img src="image_35.png" /></figure>

2.1.1.

<figure><img src="image_36.png" /></figure>

2.1.2.
<figure><img src="image_36.png" /></figure>

2.1.1.

<figure><img src="image_37.png" /></figure>

2.1.2.
<figure><img src="image_37.png" /><figcaption>Fig. 2.1.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

## II.5. Quantum trajectories of the one dimensional quantum anharmonic oscillator with a cubic potential.

Let's calculate canonical quantum trajectories (canonical quantum averages) of the one dimensional quantum anharmonic oscillator with a cubic potential, i.e.

$$ \langle x_i, t, x_0; \hbar \rangle = \frac{\int_{-\infty}^{\infty} x |\Psi(x, t, x_0; \hbar)|^2 dx}{\int_{-\infty}^{\infty} |\Psi(x, t, x_0; \hbar)|^2 dx}, \qquad (2.5.1) $$
$$ \Psi(x, 0, x_0; \hbar) = \frac{1}{\sqrt{2\pi\epsilon}} \exp\left(-\frac{x^2}{\epsilon}\right), $$
$$ \Psi^2(x, 0, x_0; \hbar) \approx \delta(x - x_0), $$
$$ \epsilon \ll \hbar \ll 1. $$
<figure><img src="image_38.png" /><figcaption>Fig. 2.1.3. $\omega(t) \triangleq \omega[\lambda(t)]$.</figcaption></figure>

## II.5. Quantum trajectories of the one dimensional quantum anharmonic oscillator with a cubic potential.

Let's calculate canonical quantum trajectories (canonical quantum averages) of the one dimensional quantum anharmonic oscillator with a cubic potential, i.e.

$$ \langle x_i, t, x_0; \hbar \rangle = \frac{\int_{-\infty}^{\infty} x |\Psi(x, t, x_0; \hbar)|^2 dx}{\int_{-\infty}^{\infty} |\Psi(x, t, x_0; \hbar)|^2 dx}, $$ 
$$ \Psi(x, 0, x_0; \hbar) = \frac{1}{\sqrt{2\pi\epsilon}} \exp\left(-\frac{x^2}{\epsilon}\right), \qquad (2.5.1) $$ 
$$ \Psi^2(x, 0, x_0; \hbar) \approx \delta(x - x_0), $$ 
$$ \epsilon \ll \hbar \ll 1. $$
Corresponding master action $\tilde{S}_{cl}$ is :

$$
\begin{equation}
\begin{aligned}
\tilde{S}_{cl}(\bar{x}, y, \lambda, T) ={}& \frac{m\bar{\omega}}{2\sin\bar{\omega}T} \left[ (\cos\bar{\omega}T)(y^2 + \bar{x}^2) - 2\bar{x}y + \frac{2\bar{x}}{m\bar{\omega}} \int_0^T g(\lambda, t)\sin(\bar{\omega}t)dt - \right. \\
& \left. \qquad + \frac{2y}{m\bar{\omega}} \int_0^T g(\lambda, t)\sin\bar{\omega}(T-t)dt - \right. \\
& \left. \qquad - \frac{2}{m^2\bar{\omega}^2} \int_0^T \int_0^t g(\lambda, t)g(\lambda, s)\sin\bar{\omega}(T-t)\sin(\bar{\omega}s)dsdt \right],
\end{aligned}
\tag{2.5.2}
\end{equation}
$$

$\bar{x} = x - \lambda.$

**Remark.** We remain that

$$
\int_{x(T)=x} (x(\tau) - \lambda)\Psi(x(0)) \exp[\Phi(x(\tau))] Dx(\tau) = \\
\left[ \int_{\substack{x(T)=x \\ x(0)=y}} (x(\tau) - \lambda)\Psi(x(0)) = y \right] \exp[\Phi(x(\tau))] Dx(\tau) \quad dy = \tag{2.5.2'}
$$

$$
\left[ \begin{array}{l} \displaystyle \int_{\substack{u(T)=x-\lambda \\ u(0)=y-\lambda}} (u(\tau))\Psi(u(0)+\lambda)\exp[\Phi(u(\tau)+\lambda)]Du(\tau) \, dy. \\ \\ \hline \end{array} \right]
$$

Here $x(\tau) - \lambda = u(\tau)$, $x(\tau) = u(\tau) + \lambda$.
Corresponding master action $\tilde{S}_{cl}$ is :

$$
\begin{equation}
\begin{aligned}
\tilde{S}_{cl}(\bar{x}, y, \lambda, T) ={}& \frac{m\bar{\omega}}{2\sin\bar{\omega}T} \left[ (\cos\bar{\omega}T)(y^2 + \bar{x}^2) - 2\bar{x}y + \frac{2\bar{x}}{m\bar{\omega}} \int_0^T g(\lambda, t)\sin(\bar{\omega}t)dt - \right. \\
& \left. \qquad + \frac{2y}{m\bar{\omega}} \int_0^T g(\lambda, t)\sin\bar{\omega}(T-t)dt - \right. \\
& \left. \qquad - \frac{2}{m^2\bar{\omega}^2} \int_0^T \int_0^t g(\lambda, t)g(\lambda, s)\sin\bar{\omega}(T-t)\sin(\bar{\omega}s)dsdt \right],
\end{aligned}
\tag{2.5.2}
\end{equation}
$$

$\bar{x} = x - \lambda.$

**Remark.** We remain that

$$
\int_{x(T)=x} (x(\tau) - \lambda)\Psi(x(0)) \exp[\Phi(x(\tau))] Dx(\tau) = \\
\left[ \int_{\substack{x(T)=x \\ x(0)=y}} (x(\tau) - \lambda)\Psi(x(0)) = y \right] \exp[\Phi(x(\tau))] Dx(\tau) \quad dy = \tag{2.5.2'}
$$

$$
\left[ \begin{array}{l} \displaystyle \int_{\substack{u(T)=x-\lambda \\ u(0)=y-\lambda}} (u(\tau))\Psi(u(0)+\lambda)\exp[\Phi(u(\tau)+\lambda)]Du(\tau) \, dy. \\ \\ \hline \end{array} \right]
$$

Here $x(\tau) - \lambda = u(\tau)$, $x(\tau) = u(\tau) + \lambda$.
Therefore from Eq.(2.5.2') and Maslov formula (0.1.5) we obtain:

$$
\begin{align*}
\Psi(\bar{x}, \lambda, T) &= \frac{\sqrt{m\bar{\omega}}}{\sqrt[4]{\epsilon} \sqrt{2\pi i\hbar} |\sin\omega T|^{1/2}} \int_{-\infty}^{\infty} dy \exp\left(-\frac{(y+\lambda)^2}{\epsilon}\right) \times \\
&\quad \times \exp\left[\frac{i}{\hbar} \tilde{S}_{cl}(\bar{x}, y, \lambda, T) + \frac{i\pi\gamma}{2}\right] = \\
&\quad \frac{\sqrt{m\bar{\omega}}}{\sqrt[4]{\epsilon} \sqrt{2\pi i\hbar} |\sin\omega T|^{1/2}} \int_{-\infty}^{\infty} dy \exp\left(-\frac{(y+\lambda)^2}{\epsilon}\right) \times \\
&\quad \exp\left\{\frac{im\bar{\omega}}{2\hbar \sin\omega T} \times \right. \\
&\quad \left[ (\cos\omega T)(y^2 + \bar{x}^2) - 2\bar{x}y + \frac{2\bar{x}}{m\bar{\omega}} \int_0^T g(\lambda, t) \sin(\omega t) dt + \right. \\
&\quad \left. \frac{2y}{m\bar{\omega}} \int_0^T g(\lambda, t) \sin\omega(T-t) dt - \right. \\
&\quad \left. - \frac{2}{m^2\bar{\omega}^2} \int_0^T \int_0^t g(\lambda, t) g(\lambda, s) \sin\omega(T-t) \sin(\omega s) ds dt \right] + \frac{i\pi\gamma}{2} \right\},
\end{align*}
\tag{2.5.3}
$$

Let's to calculate integral (2.5.3) using stationary phase method.

$$
\begin{align*}
& q(t) - \lambda = u(t) \\
& q(0) - \lambda = u(0), q(0) = u(0) + \lambda \\
& q(T) - \lambda = u(T), q(T) = u(T) + \lambda \\
& q(T) = x, u(T) = x - \lambda \\
& q(0) = y, u(0) = y - \lambda \\
& \int_{-\infty}^{\infty} dy \iint_{q(T)=x} [q(T) - \lambda] \exp(-[q(0)]^2) = \int_{-\infty}^{\infty} dy \iint_{u(T)+\lambda=x} u(T) \exp(-[u(0)+\lambda]^2) = \\
& \int_{-\infty}^{\infty} dy \iint_{\substack{u(T)=x-\lambda \\ u(0)=y-\lambda}} u(T) \exp(y^2), \quad \frac{-[\sqrt{\hbar} u'(0) + \lambda]^2}{\hbar} = -\left[u'(0) + \frac{\lambda}{\sqrt{\hbar}}\right]^2
\end{align*}
$$
Therefore from Eq.(2.5.2') and Maslov formula (0.1.5) we obtain:

$$
\begin{align*}
\Psi(\bar{x}, \lambda, T) &= \frac{\sqrt{m\bar{\omega}}}{\sqrt[4]{\epsilon} \sqrt{2\pi i\hbar} |\sin \omega T|^{1/2}} \int_{-\infty}^{\infty} dy \exp\left(-\frac{(y+\lambda)^2}{\epsilon}\right) \times \\
&\quad \times \exp\left[\frac{i}{\hbar} \tilde{S}_{cl}(\bar{x}, y, \lambda, T) + \frac{i\pi\gamma}{2}\right] = \\
&\quad \frac{\sqrt{m\bar{\omega}}}{\sqrt[4]{\epsilon} \sqrt{2\pi i\hbar} |\sin \omega T|^{1/2}} \int_{-\infty}^{\infty} dy \exp\left(-\frac{(y+\lambda)^2}{\epsilon}\right) \times \\
&\quad \exp\left\{\frac{im\bar{\omega}}{2\hbar \sin \omega T} \times \right. \\
&\quad \left[ (\cos \omega T)(y^2 + \bar{x}^2) - 2\bar{x}y + \frac{2\bar{x}}{m\bar{\omega}} \int_0^T g(\lambda, t) \sin(\omega t) dt + \right. \\
&\quad \left. \frac{2y}{m\bar{\omega}} \int_0^T g(\lambda, t) \sin \omega (T-t) dt - \right. \\
&\quad \left. - \frac{2}{m^2\bar{\omega}^2} \int_0^T \int_0^t g(\lambda, t) g(\lambda, s) \sin \omega (T-t) \sin(\omega s) ds dt \right] + \frac{i\pi\gamma}{2} \right\},
\end{align*}
$$

Let's to calculate integral (2.5.3) using stationary phase method.

$$
\begin{align*}
& q(t) - \lambda = u(t) \\
& q(0) - \lambda = u(0), q(0) = u(0) + \lambda \\
& q(T) - \lambda = u(T), q(T) = u(T) + \lambda \\
& q(T) = x, u(T) = x - \lambda \\
& q(0) = y, u(0) = y - \lambda \\
& \int_{-\infty}^{\infty} dy \int_{q(T)=x} [q(T) - \lambda] \exp(-[q(0)]^2) = \int_{-\infty}^{\infty} dy \int_{u(T)+\lambda=x} u(T) \exp(-[u(0)+\lambda]^2) = \\
& \int_{-\infty}^{\infty} dy \int_{\substack{u(T)=x-\lambda \\ u(0)=y-\lambda}} u(T) \exp(y^2), \quad \frac{-[\sqrt{\hbar} u'(0) + \lambda]^2}{\hbar} = -\left[u'(0) + \frac{\lambda}{\sqrt{\hbar}}\right]^2
\end{align*}
$$
$$ \frac{a+b}{2} - a = \frac{b}{2} - \frac{a}{2} $$

$$ (x-a)^2 + (x-b)^2 $$

$$ 2x - a - b = 0 $$

$$ \frac{\partial \tilde{S}_{cl}}{\partial y} = 2y \cos \omega T - 2\bar{x} + \frac{2}{m\omega} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt = 0 \quad (2.5.4) $$

Therefore

$$ \begin{align} y_{cr}(\bar{x}, \lambda, T) &= \frac{\bar{x}}{\cos \omega T} - \frac{1}{m\omega \cos \omega T} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt \tag{2.5.5} \\ & \qquad - \frac{1}{\cos \omega T} \left( \bar{x} - \frac{1}{m\omega} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt \right). \notag \end{align} $$

Simple calculation gives
$$ \frac{a+b}{2} - a = \frac{b}{2} - \frac{a}{2} $$

$$ (x-a)^2 + (x-b)^2 $$

$$ 2x - a - b = 0 $$

$$ \frac{\partial \tilde{S}_{cl}}{\partial y} = 2y \cos \omega T - 2\bar{x} + \frac{2}{m\omega} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt = 0 \quad (2.5.4) $$

Therefore

$$ \begin{aligned} y_{cr}(\bar{x}, \lambda, T) = & \frac{\bar{x}}{\cos \omega T} - \frac{1}{m\omega \cos \omega T} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt = \\ & \frac{1}{\cos \omega T} \left( \bar{x} - \frac{1}{m\omega} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt \right). \end{aligned} \quad (2.5.5) $$

Simple calculation gives
$$ \int_{0}^{T} g(\lambda, t) \sin[\omega(T-t)]dt = \int_{0}^{T} [-d(\lambda) + A \sin(\Omega t)] \sin[\omega(T-t)]dt = $$

$$ -d(\lambda) \int_{0}^{T} \sin[\omega(T-t)]dt + A \int_{0}^{T} \sin[\omega(T-t)] \sin(\Omega t)dt = $$

$$ d(\lambda) \int_{0}^{T} \sin[\omega(T-t)]d(T-t) + $$

$$ \frac{A}{2} \left( \int_{0}^{T} \cos[\omega(T-t) - \Omega t]dt - \int_{0}^{T} \cos[\omega(T-t) + \Omega t]dt \right) = $$

$$ d(\lambda) \left( -\frac{\cos[\omega(T-t)]}{\omega} \Big|_{0}^{T} \right) - \frac{A}{2} \left( \frac{\sin[\omega(T-t) - \Omega t]}{\omega + \Omega} \Big|_{0}^{T} \right) + \frac{A}{2} \left( \frac{\sin[\omega(T-t) + \Omega t]}{\omega - \Omega} \Big|_{0}^{T} \right) = $$

$$ d(\lambda) \left( \frac{\cos(\omega T)}{\omega} - \frac{1}{\omega} \right) - \frac{A}{2} \left( \frac{\sin(-\Omega T)}{\omega + \Omega} - \frac{\sin(\omega T)}{\omega + \Omega} \right) + \quad (2.5.6) $$

$$ + \frac{A}{2} \left( \frac{\sin(\Omega T)}{\omega - \Omega} - \frac{\sin(\omega T)}{\omega - \Omega} \right) = $$

$$ d(\lambda) \left( \frac{\cos(\omega T)}{\omega} - \frac{1}{\omega} \right) + $$

$$ \frac{A}{2} \frac{\sin(\Omega T) + \sin(\omega T)}{\omega + \Omega} + \frac{A}{2} \frac{\sin(\Omega T) - \sin(\omega T)}{\omega - \Omega} = $$

$$ d(\lambda) \left( \frac{\cos(\omega T)}{\omega} - \frac{1}{\omega} \right) + $$

$$ \frac{A}{2} \frac{[\sin(\Omega T) + \sin(\omega T)](\omega - \Omega) + [\sin(\Omega T) - \sin(\omega T)](\omega + \Omega)}{\omega^2 - \Omega^2} = $$

$$ d(\lambda) \left( \frac{\cos(\omega T)}{\omega} - \frac{1}{\omega} \right) + A \frac{\omega \sin(\Omega T) - \Omega \sin(\omega T)}{\omega^2 - \Omega^2} $$
$$\int_{0}^{T} g(\lambda, t) \sin[\omega(T-t)]dt = \int_{0}^{T} [-d(\lambda) + A \sin(\Omega t)] \sin[\omega(T-t)]dt =$$

$$-d(\lambda) \int_{0}^{T} \sin[\omega(T-t)]dt + A \int_{0}^{T} \sin[\omega(T-t)] \sin(\Omega t)dt =$$

$$d(\lambda) \int_{0}^{T} \sin[\omega(T-t)]d(T-t) +$$

$$\frac{A}{2} \left( \int_{0}^{T} \cos[\omega(T-t) - \Omega t]dt - \int_{0}^{T} \cos[\omega(T-t) + \Omega t]dt \right) =$$

$$d(\lambda) \left( -\frac{\cos[\omega(T-t)]}{\omega} \Big|_{0}^{T} \right) - \frac{A}{2} \left( \frac{\sin[\omega(T-t) - \Omega t]}{\omega + \Omega} \Big|_{0}^{T} \right) + \\ + \frac{A}{2} \left( \frac{\sin[\omega(T-t) + \Omega t]}{\omega - \Omega} \Big|_{0}^{T} \right) =$$

$$d(\lambda) \left( \frac{\cos(\omega T)}{\omega} - \frac{1}{\omega} \right) - \frac{A}{2} \left( \frac{\sin(-\Omega T)}{\omega + \Omega} - \frac{\sin(\omega T)}{\omega + \Omega} \right) + \quad (2.5.6)$$

$$+ \frac{A}{2} \left( \frac{\sin(\Omega T)}{\omega - \Omega} - \frac{\sin(\omega T)}{\omega - \Omega} \right) =$$

$$d(\lambda) \left( \frac{\cos(\omega T)}{\omega} - \frac{1}{\omega} \right) +$$

$$\frac{A}{2} \frac{\sin(\Omega T) + \sin(\omega T)}{\omega + \Omega} + \frac{A}{2} \frac{\sin(\Omega T) - \sin(\omega T)}{\omega - \Omega} =$$

$$d(\lambda) \left( \frac{\cos(\omega T)}{\omega} - \frac{1}{\omega} \right) +$$

$$\frac{A}{2} \frac{[\sin(\Omega T) + \sin(\omega T)](\omega - \Omega) + [\sin(\Omega T) - \sin(\omega T)](\omega + \Omega)}{\omega^2 - \Omega^2} =$$

$$d(\lambda) \left( \frac{\cos(\omega T)}{\omega} - \frac{1}{\omega} \right) + A \frac{\omega \sin(\Omega T) - \Omega \sin(\omega T)}{\omega^2 - \Omega^2}.$$
Using stationary phase method, from Eqs.(2.5.3)-(2.5.5) we obtain

$$
\begin{align}
\Psi(\bar{x}, \lambda, T) &= \frac{1}{\sqrt{2\pi\epsilon}|\cos\omega T|^{1/2}} \exp\left[-\frac{(y_{cr}(\bar{x}, \lambda, T) + \lambda)^2}{\epsilon}\right] \times \nonumber \\
&\quad \exp\left[\frac{i}{\hbar}\tilde{S}_{cl}(\bar{x}, y_{cr}, \lambda, T) + \frac{i\pi\gamma}{2} + \frac{i\pi\gamma'}{4}\right] + O(\hbar) = \nonumber \\
&= \frac{1}{\sqrt{2\pi\epsilon}|\cos\omega T|^{1/2}} \times \tag{2.5.7} \\
&\quad \times \exp\left[-\frac{1}{\epsilon}\left(\frac{\bar{x}}{\cos\omega T} - \frac{1}{m\omega\cos\omega T}\int_0^T g(\lambda, t)\sin\omega(T-t)dt + \lambda\right)^2\right] \times \nonumber \\
&\quad \exp\left[\frac{i}{\hbar}\tilde{S}_{cl}(\bar{x}, y_{cr}, \lambda, T) + \frac{i\pi\gamma}{2} + \frac{i\pi\gamma'}{4}\right] + O(\hbar). \nonumber
\end{align}
$$

Here

$$
\gamma' = \operatorname{sgn} \left( \frac{\partial^2 \tilde{S}_{cl}}{\partial y^2} \right). \qquad (2.5.8)
$$

From Eq.(2.5.7) we obtain

$$
|\Psi(\bar{x}, \lambda, T)|^2 = \frac{1}{\sqrt{2\pi\epsilon} |\cos \omega T|} \exp\left[-\frac{2[y_{cr}(\bar{x}, \lambda, T) + \lambda]^2}{\epsilon}\right] \quad (2.5.9)
$$

Using now Laplace method, from Eq.(2.5.5) and Eq.(2.5.9) we obtain
Using stationary phase method, from Eqs.(2.5.3)-(2.5.5) we obtain

$$
\begin{align}
\Psi(\bar{x}, \lambda, T) &= \frac{1}{\sqrt{2\pi\epsilon}|\cos\omega T|^{1/2}} \exp\left[-\frac{(y_{cr}(\bar{x}, \lambda, T) + \lambda)^2}{\epsilon}\right] \times \nonumber \\
&\quad \exp\left[\frac{i}{\hbar}\tilde{S}_{cl}(\bar{x}, y_{cr}, \lambda, T) + \frac{i\pi\gamma}{2} + \frac{i\pi\gamma'}{4}\right] + O(\hbar) = \nonumber \\
&= \frac{1}{\sqrt{2\pi\epsilon}|\cos\omega T|^{1/2}} \times \tag{2.5.7} \\
&\quad \times \exp\left[-\frac{1}{\epsilon}\left(\frac{\bar{x}}{\cos\omega T} - \frac{1}{m\omega\cos\omega T} \int_0^T g(\lambda, t) \sin\omega(T-t)dt + \lambda\right)^2\right] \times \nonumber \\
&\quad \exp\left[\frac{i}{\hbar}\tilde{S}_{cl}(\bar{x}, y_{cr}, \lambda, T) + \frac{i\pi\gamma}{2} + \frac{i\pi\gamma'}{4}\right] + O(\hbar). \nonumber
\end{align}
$$

Here

$$
\gamma' = \operatorname{sgn} \left( \frac{\partial^2 \tilde{S}_{\text{cl}}}{\partial y^2} \right). \qquad (2.5.8)
$$

From Eq.(2.5.7) we obtain

$$
|\Psi(\bar{x}, \lambda, T)|^2 = \frac{1}{\sqrt{2\pi\epsilon} |\cos \omega T|} \exp\left[-\frac{2[y_{cr}(\bar{x}, \lambda, T) + \lambda]^2}{\epsilon}\right] \quad (2.5.9)
$$

Using now Laplace method, from Eq.(2.5.5) and Eq.(2.5.9) we obtain
$$ \int_{-\infty}^{\infty} d\bar{x}\bar{x} |\Psi(\bar{x}, \lambda, T)|^2 = $$

$$ \frac{1}{\sqrt{2\pi\epsilon}|\cos\omega T|} \int_{-\infty}^{\infty} d\bar{x}\bar{x} \exp\left(-\frac{2[v_{cr}(\bar{x}, \lambda, T) + \lambda]^2}{\epsilon}\right) = \qquad (2.5.10) $$

$$ \frac{1}{\sqrt{\epsilon} |\cos \omega T|} \times $$

$$ \times \int_{-\infty}^{\infty} d\bar{x}\bar{x} \exp\left\{-\frac{2}{\epsilon}\left[\left(\frac{\bar{x}}{\cos\omega T} - \frac{1}{m\omega\cos\omega T}\int_{0}^{T}g(\lambda,t)\sin\omega(T-t)dt + \lambda\right)^2\right]\right\}. $$

Therefore

$$ \bar{x}_{cr}(\lambda, T) = \frac{1}{m\bar{\omega}} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt - \lambda \cos \omega T, \qquad (2.5.11) $$

and master equation is:

$$ \bar{x}_{cr}(\lambda, T) = \frac{1}{m\bar{\omega}} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt - \lambda \cos \omega T = 0. \qquad (2.5.12) $$

Using Eq.(2.5.6) finally we obtain master equation of the form

$$ d(\lambda)\left(\frac{\cos(\omega T)}{\bar{\omega}} - \frac{1}{\bar{\omega}}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{\omega^2 - \Omega^2} - \lambda m\omega\cos(\omega T) = 0. \qquad (2.5.13) $$

Let us consider now quantum harmonic oscillator, i.e. $a = 0$ and therefore
$$ \int_{-\infty}^{\infty} d\bar{x}\bar{x} |\Psi(\bar{x}, \lambda, T)|^2 = $$

$$ \frac{1}{\sqrt{2\pi\epsilon}|\cos\omega T|} \int_{-\infty}^{\infty} d\bar{x}\bar{x} \exp\left(-\frac{2[v_{cr}(\bar{x}, \lambda, T) + \lambda]^2}{\epsilon}\right) = \qquad (2.5.10) $$

$$ \frac{1}{\sqrt{\epsilon}|\cos \omega T|} \times $$

$$ \times \int_{-\infty}^{\infty} d\bar{x}\bar{x} \exp\left\{-\frac{2}{\epsilon}\left[\left(\frac{\bar{x}}{\cos\omega T} - \frac{1}{m\omega\cos\omega T}\int_{0}^{T}g(\lambda,t)\sin\omega(T-t)dt + \lambda\right)^2\right]\right\}. $$

Therefore

$$ \bar{x}_{cr}(\lambda, T) = \frac{1}{m\bar{\omega}} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt - \lambda \cos \omega T, \qquad (2.5.11) $$

and master equation is:

$$ \bar{x}_{cr}(\lambda, T) = \frac{1}{m\bar{\omega}} \int_{0}^{T} g(\lambda, t) \sin \omega (T-t) dt - \lambda \cos \omega T = 0. \qquad (2.5.12) $$

Using Eq.(2.5.6) finally we obtain master equation of the form

$$ d(\lambda)\left(\frac{\cos(\omega T)}{\bar{\omega}} - \frac{1}{\bar{\omega}}\right) + A\frac{\omega \sin(\Omega T) - \Omega \sin(\omega T)}{\omega^2 - \Omega^2} - \lambda m\omega \cos(\omega T) = 0. \qquad (2.5.13) $$

Let us consider now quantum harmonic oscillator, i.e. $a = 0$ and therefore
$$d(\lambda) = m\omega^2\lambda - b, \tag{2.5.14}$$

$$\bar{\omega}(\lambda) = \omega.$$

Substitution Eq.(2.5.14) into Eq.(2.5.13) gives

$$(m\omega^2\lambda - b)\left(\frac{\cos(\omega t)}{\omega} - \frac{1}{\omega}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{\omega^2 - \Omega^2} - \lambda m\omega\cos(\omega T) = 0, \\ m\omega\lambda\cos(\omega t) - m\omega\lambda - b\left(\frac{\cos(\omega t)}{\omega} - \frac{1}{\omega}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{\omega^2 - \Omega^2} - \lambda m\omega\cos(\omega T) = 0, \tag{2.5.15}$$

$$-m\omega\lambda - b\left(\frac{\cos(\omega t)}{\omega} - \frac{1}{\omega}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{\omega^2 - \Omega^2} = 0.$$

Thus

$$\lambda = -\frac{b}{m\omega}\left(\frac{\cos(\omega t)}{\omega} - \frac{1}{\omega}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{m\omega(\omega^2 - \Omega^2)} \tag{2.5.16}$$

**Example.2.5.1.** (Linear Quantum harmonic oscillator, $a = 0$). **No quantum jumps.**

$$m = 1, \Omega = 0, \omega = 9, a = 0, b = 10, A = 0.$$
$$d(\lambda) = m\omega^2\lambda - b, \qquad (2.5.14)$$

$$\omega(\lambda) = \omega.$$

Substitution Eq.(2.5.14) into Eq.(2.5.13) gives

$$(m\omega^2\lambda - b)\left(\frac{\cos(\omega t)}{\omega} - \frac{1}{\omega}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{\omega^2 - \Omega^2} - \lambda m\omega\cos(\omega T) = 0, \\ m\omega\lambda\cos(\omega t) - m\omega\lambda - b\left(\frac{\cos(\omega t)}{\omega} - \frac{1}{\omega}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{\omega^2 - \Omega^2} - \lambda m\omega\cos(\omega T) = 0, \qquad (2.5.15)$$

$$-m\omega\lambda - b\left(\frac{\cos(\omega t)}{\omega} - \frac{1}{\omega}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{\omega^2 - \Omega^2} = 0.$$

Thus

$$\lambda = -\frac{b}{m\omega}\left(\frac{\cos(\omega t)}{\omega} - \frac{1}{\omega}\right) + A\frac{\omega\sin(\Omega T) - \Omega\sin(\omega T)}{m\omega(\omega^2 - \Omega^2)} \qquad (2.5.16)$$

**Example.2.5.1.** (Linear Quantum harmonic oscillator, $a = 0$). **No quantum jumps.**

$$m = 1, \Omega = 0, \omega = 9, a = 0, b = 10, A = 0.$$
<figure><img src="image_38.png" /></figure>

<figure><img src="image_39.png" /></figure>

Example.2.5.2.
<figure><img src="image_39.png" /></figure>

<figure><img src="image_40.png" /></figure>

Example.2.5.2.
$m = 1, \Omega = 0, \omega = 9, a = 3, b = 10, A = 0.$

<figure><img src="image_40.png" /></figure>

<figure><img src="image_41.png" /></figure>

$m = 1, \Omega = 0, \omega = 9, a = 3, b = 10, A = 0.$
$m = 1, \Omega = 0, \omega = 9, a = 3, b = 10, A = 0.$

<figure><img src="image_41.png" /></figure>

<figure><img src="image_42.png" /></figure>

$m = 1, \Omega = 0, \omega = 9, a = 3, b = 10, A = 0.$
<figure><img src="image_42.png" /></figure>

Example.2.5.3.

$m = 1, \Omega = 3, \omega = 9, a = 3, b = 10, A = 1.$
<figure><img src="image_43.png" /></figure>

Example.2.5.3.

$m = 1, \Omega = 3, \omega = 9, a = 3, b = 10, A = 1.$
<figure><img src="image_43.png" /></figure>

<figure><img src="image_44.png" /></figure>
<figure><img src="image_44.png" /></figure>

<figure><img src="image_45.png" /></figure>
<figure><img src="image_45.png" /></figure>

Example.2.5.4.

$m = 1, \Omega = 3, \omega = 15, a = 3, b = 10, A = 1.$

<figure><img src="image_46.png" /></figure>
<figure><img src="image_46.png" /></figure>

Example.2.5.4.

$m = 1, \Omega = 3, \omega = 15, a = 3, b = 10, A = 1.$

<figure><img src="image_47.png" /></figure>
<figure><img src="image_47.png" /></figure>

<figure><img src="image_48.png" /></figure>
<figure><img src="image_48.png" /></figure>

<figure><img src="image_49.png" /></figure>
<figure><img src="image_49.png" /></figure>

Example.2.5.5.

$m = 1, \Omega = 3, \omega = 10, a = 3, b = 10, A = 3.$

<figure><img src="image_50.png" /></figure>
<figure><img src="image_50.png" /></figure>

Example.2.5.5.

$m = 1, \Omega = 3, \omega = 10, a = 3, b = 10, A = 3.$

<figure><img src="image_51.png" /></figure>
<figure><img src="image_51.png" /></figure>

<figure><img src="image_52.png" /></figure>

Example.2.5.6.

$m = 1, \Omega = 12, \omega = 9, a = 1, b = 15, A = 2.$
<figure><img src="image_52.png" /></figure>

<figure><img src="image_53.png" /></figure>

Example.2.5.6.

$m = 1, \Omega = 12, \omega = 9, a = 1, b = 15, A = 2.$
<figure><img src="image_53.png" /></figure>

<figure><img src="image_54.png" /></figure>
<figure><img src="image_54.png" /></figure>

<figure><img src="image_55.png" /></figure>
<figure><img src="image_55.png" /></figure>

### III. Comparison with a perturbation theory.

#### III.1. Path-integrals calculation by using stationary-point approximation.

Let us consider classical dynamics of the form

$$ 
\begin{gathered}
\dot{x}(t,x_0) = F(x(t,x_0),t), x(0) = x_0, \\
F(x,t) = -V'(x) + A \sin(\Omega t), \\
\Omega = \frac{2\pi}{T}.
\end{gathered}
\qquad (3.1.1)
$$

Where $V(x)$ is an static potential. Let us consider now corresponding quantum dynamics. Note that the transition probability amplitude $K(x, t|x_0, 0) = K(x, t)$ is

governed by the Schrödinger type equation:
<figure><img src="image_56.png" /></figure>

### III. Comparison with a perturbation theory.

#### III.1. Path-integrals calculation by using stationary-point approximation.

Let us consider classical dynamics of the form

$$ 
\begin{aligned}
\dot{x}(t,x_0) &= F(x(t,x_0),t), x(0) = x_0, \\
F(x,t) &= -V'(x) + A \sin(\Omega t), \\
\Omega &= \frac{2\pi}{T}.
\end{aligned}
\qquad (3.1.1)
$$

Where $V(x)$ is an static potential. Let us consider now corresponding quantum dynamics. Note that the transition probability amplitude $K(x, t|x_0, 0) = K(x, t)$ is

governed by the Schrödinger type equation:
$$ -i\epsilon \frac{\partial}{\partial t} K(x,t) = \frac{\partial}{\partial x} \left\{ \left[ -F(x,t) + i\epsilon^2 \frac{\partial}{\partial x} \right] K(x,t) \right\}, \qquad (3.1.3) $$

$$ K(x, 0) = \delta(x - x_0). $$

Let us define the conditional average $\langle x^2(t) \rangle$ via formula $\langle x^2(t) \rangle \triangleq \int_{-\infty}^{\infty} x_f^2 K(x_f, t) dx_f$.

Hence for the conditional average $\langle x^2(t) \rangle$ we have the canonical path integral representation:

$$ \langle x^2(t) \rangle = \int_{-\infty}^{\infty} x_f^2 K(x_f, t) dx_f = $$

$$ \int_{-\infty}^{\infty} \int_{x(t_0)=x_0}^{x(t)=x_f} x_f^2 \exp\left[-\frac{i}{\epsilon} S[\dot{x}(t), x(t)]\right] dx_f [Dx(t)], \qquad (3.1.4) $$

$$ S[\dot{x}(t), x(t)] = \frac{1}{4} \int_{0}^{t} [\dot{x}(t') - F(x(t'), t')] dt'. $$

Note that for propagator $K_N(x_0, t_0|x_f, t_f)$ the time discretized path-integral representation is:

$$ K_N(x_f, t_f | x_0, t_0) = \underbrace{\int \dots \int}_{N-1} \exp \left\{ -\frac{iS_N(x_0, \dots, x_N)}{\epsilon} \right\} \frac{dx_1 \cdots dx_{N-1}}{(4\pi \epsilon \Delta t)^{N/2}}, \qquad (3.1.5) $$

with

$$ S_N(x_0, \dots, x_N) = \sum_{n=0}^{N-1} \frac{\Delta t}{4} \left[ \frac{x_{n+1} - x_n}{\Delta t} - F(x_n, t_n) \right]^2 = \qquad (3.1.6) $$

where the initial-$x_0$ and end-points $x_N$ are fixed by the prescribed $x_0$ and by the additional constraint $x_N = x_f$. From Eqs.(3.1.4)-(3.1.6) we obtain
$$ -i\epsilon \frac{\partial}{\partial t} K(x,t) = \frac{\partial}{\partial x} \left\{ \left[ -F(x,t) + i\epsilon^2 \frac{\partial}{\partial x} \right] K(x,t) \right\}, \qquad (3.1.3) $$

$$ K(x, 0) = \delta(x - x_0). $$

Let us define the conditional average $\langle x^2(t) \rangle$ via formula $\langle x^2(t) \rangle \triangleq \int_{-\infty}^{\infty} x_f^2 K(x_f, t) dx_f$.

Hence for the conditional average $\langle x^2(t) \rangle$ we have the canonical path integral representation:

$$ \langle x^2(t) \rangle = \int_{-\infty}^{\infty} x_f^2 K(x_f, t) dx_f = $$

$$ \int_{-\infty}^{\infty} \int_{x(t_0)=x_0}^{x(t)=x_f} x_f^2 \exp\left[-\frac{i}{\epsilon} S[\dot{x}(t), x(t)]\right] dx_f [Dx(t)], \qquad (3.1.4) $$

$$ S[\dot{x}(t), x(t)] = \frac{1}{4} \int_{0}^{t} [\dot{x}(t') - F(x(t'), t')] dt'. $$

Note that for propagator $K_N(x_0, t_0|x_f, t_f)$ the time discretized path-integral representation is:

$$ K_N(x_f, t_f|x_0, t_0) = \underbrace{\int \dots \int}_{N-1} \exp \left\{ -\frac{iS_N(x_0, \dots, x_N)}{\epsilon} \right\} \frac{dx_1 \cdots dx_{N-1}}{(4\pi \epsilon \Delta t)^{N/2}}, \qquad (3.1.5) $$

with

$$ S_N(x_0, \dots, x_N) = \sum_{n=0}^{N-1} \frac{\Delta t}{4} \left[ \frac{x_{n+1} - x_n}{\Delta t} - F(x_n, t_n) \right]^2 = \qquad (3.1.6) $$

where the initial-$x_0$ and end-points $x_N$ are fixed by the prescribed $x_0$ and by the additional constraint $x_N = x_f$. From Eqs.(3.1.4)-(3.1.6) we obtain
$$
\begin{gather*}
|\langle x^2(t) \rangle| = \lim_{\Delta t \to 0} |\langle x_N^2(t) \rangle| = \\
\left| \int_{-\infty}^{\infty} x_N^2 K_N(x_0, t_0 | x_N, t) dx_N \right| = \tag{3.1.7}
\end{gather*}
$$

$$
\left| \underbrace{\int \dots \int}_{N} x_N^2 \exp \left\{ - \frac{i S_N(x_0, \dots, x_N)}{\epsilon} \right\} \frac{dx_1 \cdots dx_{N-1} dx_N}{(4 \pi \epsilon \Delta t)^{N/2}} \right| .
$$

Denoting an critical point of the discrete-time action S_N(x_0, ..., x_N) by **x**_k^# =
= (x_{0,k}, x_{1,k}^#, ..., x_{n,k}^#, ..., x_{N,k}^#) it follows that x^# satisfies the critical point conditions

$$
\frac{\partial S_N(\mathbf{x}_k^{\#})}{\partial x_{n,k}^{\#}} = 0, \qquad (3.1.8)
$$

for $n = 1, \dots, N$, supplemented by the prescribed boundary conditions for

$n = 0, n = N:$

$$
x_{0,k}^{\#} = x_0, x_{N,k}^{\#} = x_f \tag{3.1.9}
$$

From Eq.(3.1.7) in the limit $\epsilon \to 0$ we obtain
$$ \begin{gathered} |\langle x^2(t) \rangle| = \lim_{\Delta t \to 0} |\langle x_N^2(t) \rangle| = \\ \left| \int_{-\infty}^{\infty} x_N^2 K_N(x_0, t_0 | x_N, t) dx_N \right| = \tag{3.1.7} \\ \left| \underbrace{\int_N \dots \int_N}_{N} x_N^2 \exp \left\{ -\frac{i S_N(x_0, \dots, x_N)}{\epsilon} \right\} \frac{dx_1 \dots dx_{N-1} dx_N}{(4 \pi \epsilon \Delta t)^{N/2}} \right|. \end{gathered} $$

Denoting an critical point of the discrete-time action $S_N(x_0, \dots, x_N)$ by $\mathbf{x}_k^\# = (x_{0,k}, x_{1,k}^\#, \dots, x_{n,k}^\#, \dots, x_{N,k}^\#)$ it follows that $x^\#$ satisfies the critical point conditions

$$ \frac{\partial S_N(\mathbf{x}_k^\#)}{\partial x_{n,k}^\#} = 0, \qquad (3.1.8) $$

for $n = 1, \dots, N$, supplemented by the prescribed boundary conditions for
$n = 0, n = N$:

$$ x_{0,k}^{\#} = x_0, x_{N,k}^{\#} = x_f \qquad (3.1.9) $$

From Eq.(3.1.7) in the limit $\epsilon \to 0$ we obtain
$$ \langle x_N^2(t) \rangle = $$

$$ \int_{-\infty}^{\infty} x_N^2 K_N(x_0, t_0 | x_N, t) dx_N = $$

$$ \underbrace{\int \dots \int}_{N} x_N^2 \exp \left\{ - \frac{i S_N(x_0, \dots, x_N)}{\epsilon} \right\} \frac{dx_1 \cdots dx_{N-1} dx_N}{(4 \pi \epsilon \Delta t)^{N/2}} = \qquad (3.1.10) $$

$$ \mathbf{Z}_N(\mathbf{x}^{\#}) x_N^{\#2} \exp[-\frac{i}{\epsilon} S(\mathbf{x}^{\#})] + o(\epsilon), $$

where the prefactor $Z_N(\mathbf{x}^{\#})$ is given via $N$-dimensional Gaussian integral of the canonical form as

$$ \mathbf{Z}_N(\mathbf{x}^{\#}) = \underbrace{\int \dots \int}_{N} \exp \left\{ -\frac{i}{2\epsilon} \sum_{n,m=1}^{N} y_n \frac{\partial^2 S(\mathbf{x}^{\#})}{\partial x_n^{\#} \partial x_m^{\#}} y_m \right\} \frac{dy_1 \cdots dy_{N-1} dy_N}{(4\pi D \Delta t)^{N/2}}. \qquad (3.1.11) $$

The Gaussian integral in (3.1.11) is given via formula

$$ Z_N(\mathbf{x}^*) = \left[ 2 \Delta t \det \left( 2 \Delta t \frac{\partial^2 S(\mathbf{x}^*)}{\partial x_n^\# \partial x_m^\#} \right) \right]^{-\frac{1}{2}} \qquad (3.1.12) $$

$$ n = 1, \dots, N, m = 1, \dots, N, $$

Equation (3.1.12) we rewrite in the following form:

$$ Z_N(\mathbf{x}^*) = [2DQ_N^\#]^{-1/2}. \qquad (3.1.13) $$

Quantity $Q_N^\#$ in (3.1.13) can be calculated by using a second order linear recursion procedure given via formulae:
$$ \langle x_N^2(t) \rangle = $$

$$ \int_{-\infty}^{\infty} x_N^2 K_N(x_0, t_0 | x_N, t) dx_N = $$

$$ \underbrace{\int \dots \int}_{N} x_N^2 \exp \left\{ - \frac{i S_N(x_0, \dots, x_N)}{\epsilon} \right\} \frac{dx_1 \dots dx_{N-1} dx_N}{(4 \pi \epsilon \Delta t)^{N/2}} = \qquad (3.1.10) $$

$$ \mathbf{Z}_N(\mathbf{x}^{\#}) x_N^{\#2} \exp\left[-\frac{i}{\epsilon} S(\mathbf{x}^{\#})\right] + o(\epsilon), $$

where the prefactor $Z_N(\mathbf{x}^{\#})$ is given via $N$-dimensional Gaussian integral of the canonical form as

$$ \mathbf{Z}_N(\mathbf{x}^{\#}) = \underbrace{\int \dots \int}_{N} \exp \left\{ -\frac{i}{2\epsilon} \sum_{n,m=1}^{N} y_n \frac{\partial^2 S(\mathbf{x}^{\#})}{\partial x_n^{\#} \partial x_m^{\#}} y_m \right\} \frac{dy_1 \dots dy_{N-1} dy_N}{(4\pi D \Delta t)^{N/2}}. \qquad (3.1.11) $$

The Gaussian integral in (3.1.11) is given via formula

$$ Z_N(\mathbf{x}^*) = \left[ 2 \Delta t \det \left( 2 \Delta t \frac{\partial^2 S(\mathbf{x}^*)}{\partial x_n^\# \partial x_m^\#} \right) \right]^{-\frac{1}{2}} \qquad (3.1.12) $$

$$ n = 1, \dots, N, \quad m = 1, \dots, N, $$

Equation (3.1.12) we rewrite in the following form:

$$ Z_N(\mathbf{x}^*) = [2DQ_N^\#]^{-1/2}. \qquad (3.1.13) $$

Quantity $Q_N^\#$ in (3.1.13) can be calculated by using a second order linear recursion procedure given via formulae:
$$
\begin{align}
\frac{Q_{n+1}^* - 2Q_n^* - Q_{n-1}^*}{\Delta t^2} &= \frac{Q_{n+1}^* - 2Q_n^* - Q_{n-1}^*}{\Delta t^2} - \nonumber \\
&Q_n^* \left[ \frac{x_{n+1}^* - x_n^*}{\Delta t} - F(x_n^*, t_n) \right] F''(x_n^*, t_n) + \nonumber \\
&Q_n^* F'(x_n^*, t_n)^2 - Q_{n-1}^* F'(x_{n-1}^*, t_{n-1})^2, \tag{3.1.14}
\end{align}
$$

$Q_1^* = \Delta t,$

$$
\frac{Q_2^* - Q_1^*}{\Delta t} = 1 + o(\Delta t).
$$

As well known shall see later, we have to leave room for the possibility that even
for small noise-strengths $\epsilon \to 0$ more than one critical point of the action (3.1.6)
notably contributes to the path-integral expression (3.1.7). We label those various
critical points $x_k^\#$ by the discrete index $k$. Thus from Eq.(3.1.10) and Eq.(3.1.13) we
obtain

$$
\begin{align}
|\langle x_N^2(t_N, \omega) \rangle| &= \left| \int_{-\infty}^{\infty} x_N^2 K_N(x_0, t_0 | x_N, t) dx_N \right| \nonumber \\
&= \left| \sum_{k=0}^{M} \frac{(x_{N,k}^{\#})^2 \exp\left[ \frac{iS_N(x_k^{\#})}{\epsilon} \right]}{\sqrt{2Q_{N,k}^{\#}}} [1 + o(\epsilon)] \right|. \tag{3.1.15}
\end{align}
$$

The continuous-time limit of the action $S_N$ is

$$
S[\dot{x}(t), x(t)] = \frac{1}{4} \int_{t_0}^{t_f} [\dot{x}^2 - F(x, t)] dt \quad (3.1.16)
$$

The extremality conditions for the critical paths $x_k^\#(t)$ in the continuous-time limit are
obtained from (3.1.2) and (3.1.4) by letting $\Delta t \to 0$ is:

(I)
$$
\begin{align}
\frac{\underline{Q}_{n+1}^* - 2\underline{Q}_n^* - \underline{Q}_{n-1}^*}{\Delta t^2} &= \frac{\underline{Q}_{n+1}^* - 2\underline{Q}_n^* - \underline{Q}_{n-1}^*}{\Delta t^2} - \nonumber \\
&\quad Q_n^* \left[ \frac{x_{n+1}^* - x_n^*}{\Delta t} - F(x_n^*, t_n) \right] F''(x_n^*, t_n) + \nonumber \\
&\quad Q_n^* F'(x_n^*, t_n)^2 - \underline{Q}_{n-1}^* F'(x_{n-1}^*, t_{n-1})^2, \tag{3.1.14}
\end{align}
$$

$Q_1^* = \Delta t,$

$$
\frac{Q_2^* - Q_1^*}{\Delta t} = 1 + o(\Delta t).
$$

As well known shall see later, we have to leave room for the possibility that even
for small noise-strengths $\epsilon \to 0$ more than one critical point of the action (3.1.6)
notably contributes to the path-integral expression (3.1.7). We label those various
critical points $x_k^\#$ by the discrete index $k$. Thus from Eq.(3.1.10) and Eq.(3.1.13) we
obtain

$$
\begin{align}
|\langle x_N^2(t_N, \omega) \rangle| &= \left| \int_{-\infty}^{\infty} x_N^2 K_N(x_0, t_0 | x_N, t) dx_N \right| \nonumber \\
&= \left| \sum_{k=0}^{M} \frac{(x_{N,k}^{\#})^2 \exp\left[ \frac{iS_N(x_k^{\#})}{\epsilon} \right]}{\sqrt{2Q_{N,k}^{\#}}} [1 + o(\epsilon)] \right|. \tag{3.1.15}
\end{align}
$$

The continuous-time limit of the action $S_N$ is

$$
S[\dot{x}(t), x(t)] = \frac{1}{4} \int_{t_0}^{t_f} [\dot{x}^2 - F(x, t)] dt \quad (3.1.16)
$$

The extremality conditions for the critical paths $x_k^\#(t)$ in the continuous-time limit are
obtained from (3.1.2) and (3.1.4) by letting $\Delta t \to 0$ is:

(I)
$$ \dot{x}^{\#}(t, x_0) = F(x^{\#}(t, x_0), t), x^{\#}(0) = x_0. \qquad (3.1.17) $$

(II)

$$ \ddot{x}_k^{\#}(t, x_k^{\#}(t_f)) = \dot{F}(x_k^{\#}(t, x_k^{\#}(t_f)), t) + F(x_k^{\#}(t, x_k^{\#}(t_f)), t) F'(x_k^{\#}(t, x_k^{\#}(t_f)), t), $$ $$ x^{\#}(0, x_0, \tilde{x}_f) = x_0, x^{\#}(t_f, x_0, \tilde{x}_f) = \tilde{x}_f, \qquad (3.1.18) $$ $$ \left. \frac{dS[\dot{x}_k^{\#}(t, x_0, x_k^{\#}(t_f)), x_k^{\#}(t, x_0, x_k^{\#}(t_f))]}{d(x_k^{\#}(t_f))} \right|_{x_k^{\#}(t_f)=\tilde{x}_{f,k}} = 0. $$

Here we use the canonical definitions $F'(x,t) = \frac{\partial F(x,t)}{\partial x}$, $\dot{F}(x,t) = \frac{\partial F(x,t)}{\partial t}$.

Hamiltonian counterpart of the Lagrangian dynamics given by (3.1.18) is

$$ H(x,p,t) = p\dot{x} - L = p^2 + pF(x,t), $$ $$ \dot{p}_k^{\#}(t) = -p_k^{\#}(t)F'(x_k^{\#}(t),t), \qquad (3.1.19) $$ $$ \dot{x}_k^{\#}(t) = 2p_k^{\#}(t) + F(x_k^{\#}(t),t). $$

From the last equation of the Eqs.(3.1.19) one obtain the momentum $p_k^{\#}(t)$

$$ p_k^{\#}(t) = \frac{1}{2}[\dot{x}_k^{\#}(t) - F(x_k^{\#}(t), t)]. \qquad (3.1.20) $$

From Eq.(3.1.16) and Eq.(3.1.20) one obtain:

$$ \Phi_k(x_f, t_f) = S[x_k^{\#}(t)] = \int_{t_0}^{t_f} p_k^{\#}(t)^2 dt. \qquad (3.1.20) $$

In the limit $\Delta t \to 0$ from the extremality conditions (3.1.17) for the critical paths $x^{\#}(t, x_0, \tilde{x}_f)$ and a second order linear recursion procedure given via formulae (3.1.10) we obtain the second order homogeneous linear differential equation
$$ \dot{x}^{\#}(t, x_0) = F(x^{\#}(t, x_0), t), x^{\#}(0) = x_0. \qquad (3.1.17) $$

(II)

$$ \ddot{x}_k^{\#}(t, x_k^{\#}(t_f)) = \dot{F}(x_k^{\#}(t, x_k^{\#}(t_f)), t) + F(x_k^{\#}(t, x_k^{\#}(t_f)), t) F'(x_k^{\#}(t, x_k^{\#}(t_f)), t), $$ $$ x^{\#}(0, x_0, \tilde{x}_f) = x_0, x^{\#}(t_f, x_0, \tilde{x}_f) = \tilde{x}_f, \qquad (3.1.18) $$ $$ \left. \frac{dS[\dot{x}_k^{\#}(t, x_0, x_k^{\#}(t_f)), x_k^{\#}(t, x_0, x_k^{\#}(t_f))]}{d(x_k^{\#}(t_f))} \right|_{x_k^{\#}(t_f)=\tilde{x}_{f,k}} = 0. $$

Here we use the canonical definitions $F'(x,t) = \frac{\partial F(x,t)}{\partial x}$, $\dot{F}(x,t) = \frac{\partial F(x,t)}{\partial t}$.

Hamiltonian counterpart of the Lagrangian dynamics given by (3.1.18) is

$$ H(x,p,t) = p\dot{x} - L = p^2 + pF(x,t), $$ $$ \dot{p}_k^{\#}(t) = -p_k^{\#}(t)F'(x_k^{\#}(t),t), \qquad (3.1.19) $$ $$ \dot{x}_k^{\#}(t) = 2p_k^{\#}(t) + F(x_k^{\#}(t),t). $$

From the last equation of the Eqs.(3.1.19) one obtain the momentum $p_k^{\#}(t)$

$$ p_k^{\#}(t) = \frac{1}{2}[\dot{x}_k^{\#}(t) - F(x_k^{\#}(t), t)]. \qquad (3.1.20) $$

From Eq.(3.1.16) and Eq.(3.1.20) one obtain:

$$ \Phi_k(x_f, t_f) = S[x_k^{\#}(t)] = \int_{t_0}^{t_f} p_k^{\#}(t)^2 dt. \qquad (3.1.20) $$

In the limit $\Delta t \to 0$ from the extremality conditions (3.1.17) for the critical paths $x^{\#}(t, x_0, \tilde{x}_f)$ and a second order linear recursion procedure given via formulae (3.1.10) we obtain the second order homogeneous linear differential equation
$$ \frac{1}{2} \ddot{Q}_k^\#(t, x_0, \tilde{x}_f) - \frac{d}{dt} [Q_k^\#(t, \tilde{x}_f) F'(x^\#(t, x_0, \tilde{x}_f), t)] + Q_k^\#(t) [x^\#(t, x_0, \tilde{x}_f)] \quad (3.1.1) $$

$$ Q_0^\#(t_0, x_0, \tilde{x}_f) = 0, \quad \dot{Q}_0^\#(t_0, x_0, \tilde{x}_f) = 1. $$

## IV. Wave function collapse and weak Colombeau solutions of the Colombeau-Schrödinger equation.

In quantum mechanics, wave function collapse is the phenomenon in which a wave function-initially in a superposition of several eigenstates-appears to reduce to a single eigenstate after interaction with a measuring apparatus.[57] It is the essence of measurement in quantum mechanics, and connects the wave function with classical observables like position and momentum. Collapse is one of two processes by which quantum systems evolve in time; the other is continuous evolution via the Schrödinger equation [58] However in this role, collapse is merely a black box for thermodynamically irreversible interaction with a classical environment.[59] Calculations of quantum decoherence predict apparent wave function collapse when a superposition forms between the quantum system's states and the environment's states. Significantly, the combined wave function of the system and environment continue to obey the Schrödinger equation [60].

When the Copenhagen interpretation was first expressed, Niels Bohr postulated wave function collapse to cut the quantum world from the classical.[60] This tactical move allowed quantum theory to develop without distractions from interpretational worries. Nevertheless it was debated, for if collapse were a fundamental physical phenomenon, rather than just the epiphenomenon of some other process, it would mean nature was fundamentally stochastic, i.e. nondeterministic, an undesirable property for a theory [61] This issue remained until quantum decoherence entered mainstream opinion after its reformulation in the 1980s.[62] Decoherence explains the perception of wave function collapse in terms of interacting large- and small-scale quantum systems, and is commonly taught at the graduate level (e.g. the Cohen-Tannoudji textbook).[63] The quantum filtering approach [64],[65] and the introduction of quantum causality non-demolition principle [66] allows for a classical-environment derivation of wave function collapse from the stochastic Schrödinger equation.
$$ \frac{1}{2} \ddot{Q}_k^\#(t, x_0, \tilde{x}_f) - \frac{d}{dt} [Q_k^\#(t, \tilde{x}_f) F'(x^\#(t, x_0, \tilde{x}_f), t)] + Q_k^\#(t) [x^\#(t, x_0, \tilde{x}_f)] \quad (3.1.1) $$

$$ Q_0^\#(t_0, x_0, \tilde{x}_f) = 0, \quad \dot{Q}_0^\#(t_0, x_0, \tilde{x}_f) = 1. $$

## IV. Wave function collapse and weak Colombeau solutions of the Colombeau-Schrödinger equation.

In quantum mechanics, wave function collapse is the phenomenon in which a wave function-initially in a superposition of several eigenstates-appears to reduce to a single eigenstate after interaction with a measuring apparatus.[57] It is the essence of measurement in quantum mechanics, and connects the wave function with classical observables like position and momentum. Collapse is one of two processes by which quantum systems evolve in time; the other is continuous evolution via the Schrödinger equation [58] However in this role, collapse is merely a black box for thermodynamically irreversible interaction with a classical environment.[59] Calculations of quantum decoherence predict apparent wave function collapse when a superposition forms between the quantum system's states and the environment's states. Significantly, the combined wave function of the system and environment continue to obey the Schrödinger equation [60].

When the Copenhagen interpretation was first expressed, Niels Bohr postulated wave function collapse to cut the quantum world from the classical.[60] This tactical move allowed quantum theory to develop without distractions from interpretational worries. Nevertheless it was debated, for if collapse were a fundamental physical phenomenon, rather than just the epiphenomenon of some other process, it would mean nature was fundamentally stochastic, i.e. nondeterministic, an undesirable property for a theory [61] This issue remained until quantum decoherence entered mainstream opinion after its reformulation in the 1980s.[62] Decoherence explains the perception of wave function collapse in terms of interacting large- and small-scale quantum systems, and is commonly taught at the graduate level (e.g. the Cohen-Tannoudji textbook).[63] The quantum filtering approach [64],[65] and the introduction of quantum causality non-demolition principle [66] allows for a classical-environment derivation of wave function collapse from the stochastic Schrödinger equation.
IV. 1. Weak Colombeau solutions of the
Colombeau-Schrödinger equation.

Let us consider Colombeau-Schrödinger equation:

$$
(ih\epsilon \partial \Psi_{\epsilon}(\mathbf{x},t) / \partial t)_{\epsilon} = (\hat{\mathbf{H}}_{\epsilon} \Psi_{\epsilon}(\mathbf{x},t)), (\Psi_{\epsilon}(\mathbf{x},t_0))_{\epsilon} = (\Psi_{\epsilon}(\mathbf{x}))_{\epsilon}, \qquad (4.1.1)
$$

$$
\mathbf{x} \in \mathbb{R}^d, \mathbf{x} = (x_1, x_2, \dots, x_d), (h_\epsilon)_\epsilon \approx 0,
$$

$$
\text{where } [(\Psi_{\epsilon}^{\#}(\mathbf{x}))_{\epsilon}] \in G(\mathbb{R}^{d}).
$$

**Definition 4.1.1.** Assume that: (i) $[(\Psi_{\epsilon}^{\#}(\mathbf{x}))_{\epsilon}] \in G(\mathbb{R}^d)$, (ii) $[(\Psi_{\epsilon}^{\#}(\mathbf{x},t))_{\epsilon}] \in G(\mathbb{R}^d \times \mathbb{R}_+)$.

IV. 2. Gamow theory of the alpha decay via tunneling. Weak Colombeau solutions of the Colombeau-Schrödinger equation corresponding to alpha decay via tunneling.

By 1928, George Gamow had solved the theory of the alpha decay via tunneling [].
The alpha particle is trapped in a potential well by the nucleus. Classically, it is
forbidden to escape, but according to the (then) newly discovered principles of
quantum mechanics, it has a tiny (but non-zero) probability of "tunneling" through
the barrier and appearing on the other side to escape the nucleus. Gamow solved
a model potential for the nucleus and derived, from first principles, a relationship
between the half-life of the decay, and the energy of the emission.
The particle has total energy *E* and is incident on the barrier from the left.
IV. 1. Weak Colombeau solutions of the
Colombeau-Schrödinger equation.

Let us consider Colombeau-Schrödinger equation:

$$
(ih\epsilon \partial \Psi_{\epsilon}(\mathbf{x},t) / \partial t)_{\epsilon} = (\hat{H}_{\epsilon} \Psi_{\epsilon}(\mathbf{x},t))_{\epsilon}, (\Psi_{\epsilon}(\mathbf{x},t_0))_{\epsilon} = (\Psi_{\epsilon}(\mathbf{x}))_{\epsilon}, \quad (4.1.1)
$$

$$
\mathbf{x} \in \mathbb{R}^d, \mathbf{x} = (x_1, x_2, \dots, x_d), (h_\epsilon)_\epsilon \approx 0,
$$

$$
\text{where } [(\Psi_{\epsilon}^{\#}(\mathbf{x}))_{\epsilon}] \in G(\mathbb{R}^{d}).
$$

**Definition 4.1.1.** Assume that: (i) $[(\Psi_{\epsilon}^{\#}(\mathbf{x}))_{\epsilon}] \in G(\mathbb{R}^d)$, (ii) $[(\Psi_{\epsilon}^{\#}(\mathbf{x},t))_{\epsilon}] \in G(\mathbb{R}^d \times \mathbb{R}_+)$.

IV. 2. Gamow theory of the alpha decay via tunneling. Weak Colombeau solutions of the Colombeau-Schrödinger equation corresponding to alpha decay via tunneling.

By 1928, George Gamow had solved the theory of the alpha decay via tunneling [].
The alpha particle is trapped in a potential well by the nucleus. Classically, it is
forbidden to escape, but according to the (then) newly discovered principles of
quantum mechanics, it has a tiny (but non-zero) probability of "tunneling" through
the barrier and appearing on the other side to escape the nucleus. Gamow solved
a model potential for the nucleus and derived, from first principles, a relationship
between the half-life of the decay, and the energy of the emission.

The particle has total energy *E* and is incident on the barrier from the left.
<figure><img src="image_56.png" /><figcaption>Pic.4.2.1. The particle has total energy *E* and is incident on the barrier *V(x)* from the left.</figcaption></figure>

The Schrodinger equation in each of regions I = {x|x < 0}, II = {x|0 ≤ x ≤ l} and III = {x|x > l} takes the following form

$$ \left( \frac{\partial^2 \Psi_\epsilon(x)}{\partial x^2} + \frac{2m}{\hbar_\epsilon^2} [E - U(x)] \Psi_\epsilon(x) \right)_\epsilon = 0, \quad (4.2.1) $$

where

$$ U(x) = \begin{cases} 0, & \text{при } x < 0 \\ U_0, & \text{при } 0 \le x \le l \\ 0, & \text{при } x > l \end{cases} \tag{4.2.2} $$

The solutions are

$$ \begin{align} (\Psi_{\epsilon,I}(x))_{\epsilon} &= (C_{\epsilon,+} \exp(ik_{\epsilon}x))_{\epsilon} + (C_{\epsilon,-} \exp(-ik_{\epsilon}x))_{\epsilon} \nonumber \\ (\Psi_{\epsilon,II}(x))_{\epsilon} &= (B_{\epsilon,+} \exp(k'_{\epsilon}x))_{\epsilon} + (B_{\epsilon,-} \exp(-k'_{\epsilon}x))_{\epsilon} \tag{93} \end{align} $$

$$ (\Psi_{\epsilon,III}(x))_{\epsilon} = (A_{\epsilon} \exp(ik_{\epsilon}x))_{\epsilon}, $$

where
<figure><img src="image_57.png" /><figcaption>Pic.4.2.1. The particle has total energy *E* and is incident on the barrier *V(x)* from the left.</figcaption></figure>

The Schrodinger equation in each of regions I = {x|x < 0}, II = {x|0 ≤ x ≤ l} and III = {x|x > l} takes the following form

$$ \left( \frac{\partial^2 \Psi_\epsilon(x)}{\partial x^2} + \frac{2m}{\hbar_\epsilon^2} [E - U(x)] \Psi_\epsilon(x) \right)_\epsilon = 0, \quad (4.2.1) $$

where

$$ U(x) = \begin{cases} 0, & \text{при } x < 0 \\ U_0, & \text{при } 0 \le x \le l \\ 0, & \text{при } x > l \end{cases} \qquad (4.2.2) $$

The solutions are

$$ \begin{align} (\Psi_{\epsilon,I}(x))_{\epsilon} &= (C_{\epsilon,+} \exp(ik_{\epsilon}x))_{\epsilon} + (C_{\epsilon,-} \exp(-ik_{\epsilon}x))_{\epsilon} \nonumber \\ (\Psi_{\epsilon,II}(x))_{\epsilon} &= (B_{\epsilon,+} \exp(k'_{\epsilon}x))_{\epsilon} + (B_{\epsilon,-} \exp(-k'_{\epsilon}x))_{\epsilon} \tag{.93} \end{align} $$

$$ (\Psi_{\epsilon,III}(x))_{\epsilon} = (A_{\epsilon} \exp(ik_{\epsilon}x))_{\epsilon}, $$

where
$$ (k_{\varepsilon})_{\varepsilon} = \frac{2\pi}{(\hbar_{\varepsilon})_{\varepsilon}} \sqrt{2mE}, \qquad (VI.94) $$
$$ (k'_{\varepsilon})_{\varepsilon} = \frac{2\pi}{(\hbar_{\varepsilon})_{\varepsilon}} \sqrt{2m(U_0 - E)}. $$

### IV. 3. Quantum trajectories induced from a strong Colombeau solutions of the Colombeau-Schrödinger equation.

Let us consider Colombeau-Schrödinger equation:

$$ (i\hbar\varepsilon\partial\Psi_{\varepsilon}(\mathbf{x},t)/\partial t)_{\varepsilon} = (\hat{\mathbf{H}}_{\varepsilon}\Psi_{\varepsilon}(\mathbf{x},t),)_{\varepsilon}, $$
$$ (\Psi_{\varepsilon}(\mathbf{x},t_0))_{\varepsilon} = (\Psi_{\varepsilon}(\mathbf{x}))_{\varepsilon} \in G(\mathbb{R}^d), \qquad (4.3.1) $$
$$ \mathbf{x} \in \mathbb{R}^d, \mathbf{x} = (x_1, x_2, \dots, x_d), (\hbar\varepsilon)_{\varepsilon} \approx 0, $$

where initial data $[(\Psi_{\varepsilon}(\mathbf{x}))_{\varepsilon}]$ defined by formula
$$ (k_{\varepsilon})_{\varepsilon} = \frac{2\pi}{(\hbar_{\varepsilon})_{\varepsilon}} \sqrt{2mE}, \qquad (VI.94) $$
$$ (k'_{\varepsilon})_{\varepsilon} = \frac{2\pi}{(\hbar_{\varepsilon})_{\varepsilon}} \sqrt{2m(U_0 - E)}. $$

### IV. 3. Quantum trajectories induced from a strong Colombeau solutions of the Colombeau-Schrödinger equation.

Let us consider Colombeau-Schrödinger equation:

$$ (ih\varepsilon\partial\Psi_{\varepsilon}(\mathbf{x},t)/\partial t)_{\varepsilon} = (\hat{\mathbf{H}}_{\varepsilon}\Psi_{\varepsilon}(\mathbf{x},t),)_{\varepsilon}, $$
$$ (\Psi_{\varepsilon}(\mathbf{x},t_0))_{\varepsilon} = (\Psi_{\varepsilon}(\mathbf{x}))_{\varepsilon} \in G(\mathbb{R}^d), \qquad (4.3.1) $$
$$ \mathbf{x} \in \mathbb{R}^d, \mathbf{x} = (x_1, x_2, \dots, x_d), (\hbar_{\varepsilon})_{\varepsilon} \approx 0, $$

where initial data $[(\Psi_{\varepsilon}(\mathbf{x}))_{\varepsilon}]$ defined by formula
$$ (\Psi_{\varepsilon}(\mathbf{x}))_{\varepsilon} = (\Psi_{\varepsilon}(\mathbf{x}, \mathbf{x}_0))_{\varepsilon} = \\ = \left( \frac{\mu(\mathbf{x}_0)}{\sqrt[4]{2\pi\eta_{\varepsilon}}} \Psi_{\varepsilon}(\mathbf{x}) \exp\left[ \frac{\langle \mathbf{x} - \mathbf{x}_0, \mathbf{x} - \mathbf{x}_0 \rangle}{2\eta_{\varepsilon}} \right] \right)_{\varepsilon} = \\ = \left( \frac{1}{(\sqrt[4]{2\pi\eta_{\varepsilon}})^d} \exp\left[ \frac{\langle \mathbf{x} - \mathbf{x}_0, \mathbf{x} - \mathbf{x}_0 \rangle}{2\eta_{\varepsilon}} \right] \exp\left[ \frac{i}{\hbar_{\varepsilon}} S(\mathbf{x}) \right] \right)_{\varepsilon}, \quad (4.3.2) $$

$$ (\eta_{\varepsilon})_{\varepsilon} \approx 0 \text{ and } \left(\frac{\hbar_{\varepsilon}}{\eta_{\varepsilon}}\right)_{\varepsilon} \approx 0, $$

**Definition 4.3.1.** Quantum trajectory $\{(x_{\varepsilon,i}(t, \mathbf{x}_0; \hbar))\}_{\varepsilon}^{d}\}_{i=1}^{d}$ with initial data $\{(x_{\varepsilon,i}(0, \mathbf{x}_0; \hbar))\}_{\varepsilon}^{d}\}_{i=1}^{d} = \mathbf{x}_0$ is defined by following formula

$$ (x_{\varepsilon,i}(t, \mathbf{x}_0; \hbar))_{\varepsilon} = \left( \int_{\mathbb{R}^d} x_i |\Psi_{\varepsilon}(\mathbf{x}, t, \mathbf{x}_0; \hbar)|^2 d^d\mathbf{x} \right)_{\varepsilon}. \quad (4.3.3) $$

## V. Schrödinger's cat paradox resolution using GRW collapse model.

In this subsection an possible solution of the Schrödinger's cat paradox is considered. We pointed out that: the collapsed state of the cat always shows definite and predictable measurement outcomes even if Schrödinger's cat consists of a superposition: $|\text{cat}\rangle = c_1|\text{live cat}\rangle + c_2|\text{death cat}\rangle$.

Contrary to van Kampen's [61] and some others' opinions, "looking" at the outcome changes nothing, beyond informing the observer of what has already happened.

We remain: there are widespread claims that Schrödinger's cat is not in a definite alive or dead state but is, instead, in a superposition of the two. van Kampen, for