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PARAMETRIC OPTIMIZATION
PROBLEM FOR ABSTRACT
PARABOLIC EQUATION
WITH VARIABLE OPERATOR
DOMAIN

Jan

J. SOKOLOWSKI

Rapport de Recherche N° 183

Août 1976

Institut de Recherche
d'Informatique
et d'Automatique

Domaine de Voluceau
Rocquencourt
B.P. 5 78150 - Le Chesnay
France
Tél.: 954 90 20
PARAMETRIC OPTIMIZATION PROBLEM FOR ABSTRACT
PARABOLIC EQUATION WITH VARIABLE OPERATOR DOMAIN

J. Sokolowski*

Résumé :

Dans cet article on considère un problème d'optimisation dans lequel le contrôle figure dans les coefficients d'une équation linéaire parabolique. On obtient une expression des conditions nécessaires d'optimalité en utilisant une formulation variationnelle où on a choisi $H^1$ comme espace pivot et non $L^2$, le domaine de l'opérateur dépendant du temps. On montre également la convergence d'un schéma d'approximation dans le cas d'un exemple simple.

Abstract :

In the paper, an optimization problem with control in coefficients of linear parabolic equation is considered.

Using a variational formulation of a parabolic equation with variable domain and $H^1$ as the pivot space, necessary conditions of optimality are obtained. Convergence of finite dimensional approximation for simple a model problem is proved.

<footnote>* Institut for Organization, Management and Control Sciences, Krajowej Rady Narodowej 55,
00-818 Warszawa, Poland.</footnote>
# 1. INTRODUCTION

In the paper an optimization problem for linear parabolic equation is considered. In the problem control appears in coefficients of the elliptic operator. Such problems appear in technology of solid state devices [7] where process of diffusion in semiconductor is described by nonlinear parabolic equation (state equation). We consider only linearized model (4.14) of the process, where the function $y = y(x,t)$ denotes the concentration of donors in semiconductor, and parameter (control) $u = u(t)$ denotes the temperature. Given set $U_{ad}$ of admissible controls, optimal control $\hat{u} = \hat{u}(t)$, $\hat{u} \in U_{ad}$ corresponds to minimale "difference" in $H^1$ norm between final concentration $y(\cdot, T)$ obtained as result of diffusion process and required concentration $z_d(\cdot)$. The paper is continuation of previous author's works [9], [10], [11], [12].

Using a variational formulation of the state equation with variable domain and $H^1$ instead of $L^2$ as the pivot space a form of necessary conditions of optimality much simpler than that in [12] is obtained. Such formulation seems have also some advantages for finite dimensional approximation of the problem.

In section 2 we recall some known results of existence, uniqueness and regularity of solutions for abstract linear parabolic equation. A variable domain variational formulation of the problem is introduced. Existence and uniqueness of the solution for the problem is proved. In section 3, necessary conditions of optimality are derived.

In section 4, convergence of a finite dimensional approximation of the problem is proved.
## 2. REGULARITY OF SOLUTION OF ABSTRACT PARABOLIC EQUATION

In the section we shortly recall some regularity results for the solutions of abstract parabolic equation.

Let there be given Hilbert spaces, $H$, $H_1$, $U$ with $H \subset H_1$, imbedding being continuous and dense, denote by $U_{ad}$ a bounded, convex and closed subset of $U$.

For given $u \in U_{ad}$, $t \in [0,T]$, $T > 0$, $T$ given, denote by $a_u(t;y,z)$ symmetric bilinear form

$$ (2.1) \qquad H \times H \ni (y,z) \to a_u(t;y,z) \in \mathbb{R} $$

We assume :

$$ (2.2) \quad \begin{array}{l} \text{i) } [0,T] \ni t \to a_u(t;y,z) \in C^1[0,T] \\ \phantom{\text{i) }} \forall u \in U_{ad}, \quad \forall y, z \in H \end{array} $$

ii) form (2.1) is continuous, that is

$$ (2.3) \qquad \begin{array}{l} |a_u(t;y,z)| \le M \|y\|_H^2 \|z\|_H, \\ \forall u \in U_{ad}, \quad \forall t \in [0,T], \quad \forall y, z \in H \end{array} $$

and coercive i.e.

$$ (2.4) \qquad \begin{array}{l} a_u(t;y,y) \ge \alpha \|y\|_H^2, \quad \alpha > 0, \quad \forall u \in U_{ad}, \quad \forall y \in H, \\ \phantom{a_u(t;y,y) \ge} \forall t \in [0,T] \end{array} $$

Let $t \in [0,T]$, $u \in U_{ad}$ be fixed for a moment. Denote by $V_t$ the domain in $H_1$ of unbounded linear operator $A_u(t)$ which is defined by bilinear form (2.1) in the following way :

$y \in V_t$ if linear mapping
$$
(2.5) \quad H \ni z \rightarrow a_u(t;y,z) \in \mathbb{R}
$$

is continuous in topology of the space H₁, and we have

$$
(2.6) \quad a_u(t;y,z) = (A_u(t)y,z)_{H_1}, \quad \forall y \in V_t, \forall z \in H
$$

Linear set V_t supplied with the norm :

$$
(2.7) \quad \| y \|_{V_t}^2 = \| y \|_{H}^{2} + \| A_u(t)y \|_{H_1}^{2}
$$

is the Hilbert space [6], and also [14]

$$
(2.8) \quad V_t \mathcal{G} H \mathcal{G} H_1
$$

with continuous and dense imbeddings. Symmetric, bilinear form a_u(t;.,.) cons-
titute the scalar product in H. We denote by <.,.>ₜ the extension of the scalar
product to the pair V_t, V'_t, where norm in V'_t is the dual norm, that is

$$
(2.9) \quad \begin{array}{@{}l@{\,}c@{\,}l@{\,}l@{}}
\| z \|_{V_t} & = & \sup_{y \in V_t} & |a_u(t;y,z)| \\
             & & \|y\|_{V_t} & \le 1
\end{array}
$$

By (2.6) (2.8) we have :

$$
(2.10) \quad \langle y, z \rangle_t = a_u(t; y, z) , \quad \forall y \in V_t , \forall z \in H
$$

$$
(2.11) \quad \langle y, z \rangle_t = (A_u(t)y, z)_{H_1}, \quad \forall y \in V_t, \forall z \in H_1
$$

and also

$$
(2.12) \quad V_t \mathcal{G} H \mathcal{G} H_1 \mathcal{G} V'_t
$$

with continuous and dense imbeddings.
**Remark :**

Let us note, that symmetric, bilinear form on $V_t \times V_t$ :

$$ (2.13) \qquad b_u(t;y,z) = (A_u(t)y, A_u(t)z)_{H_1} + sa_u(t;y,z) $$

is continuous and coercive for every $s > 0$.

As usual, we denote by $W(0,T)$ the Hilbert space $[6]$ :

$$ (2.14) \qquad W(0,T) = \{y \in L^2(0,T,H) \mid \frac{dy}{dt} \in L^2(0,T,H')\} $$

with norm :

$$ (2.15) \qquad \| y \|^2_{W(0,T)} = \| y \|^2_{L^2(0,T,H)} + \| \frac{dy}{dt} \|^2_{L^2(0,T,H')} $$

It is well known [6], that there exists a constant $C$, such that

$$ (2.16) \qquad \| y \|_{C(0,T;H_1)} \le C \| y \|_{W(0,T)} , \quad \forall y \in W(0,T) $$

Let us assume, that there exists a Hilbert space $V$ such that :

(A1) :

$$ (2.17) \quad \text{i) } V \subset H \subset V', \quad H = [V,V']_{1/2} $$

$$ \text{i i) } V_t \subset V, \quad \forall u \in U_{ad}, \quad \forall t \in [0,T] $$

$$ C_1 \| y \|_{V_t} \le \| y \|_V \le C_2 \| y \|_{V_t}, \quad \forall y \in V_t $$

$C_1, C_2$ do not depend on $u \in U_{ad}$ and $t \in [0,T]$

$$ (2.18) \quad \text{iii) } V = V_t \oplus V_t^{\perp}, \quad \forall u \in U_{ad}, \forall t \in [0,T] $$

$P(\cdot)v$ is $V$-measurable for $v \in V$, where $P(t) : V \to V_t$ is the orthogonal projection.

From (A1) we can deduce that

$$ (2.19) \qquad V' = V_t' \oplus (V_t^{\perp})' $$
and also [3]

$$ (2.20) \qquad L^2(0,T;V) = L^2(0,T;V_t) \oplus L^2(0,T;V_t^{\perp}) $$

$$ (2.21) \qquad L^2(P,T;V') = L^2(0,T;V_t') \oplus L^2(0,T;(V_t^{\perp})') $$

where by $L^2(0,T;V_t)$ we denote simple integral [3] of Hilbert spaces $\{V_t\}_{0 \le t \le}$ with respect to Lebesgue measure $dt$, that is

$$ (2.22) \qquad L^2(0,T;V_t) \equiv \int dt V_t $$

**Lemma 1.**

Assume (A1). Let there be given an element $y$ such that

$$ (2.23) \qquad \left\{ \begin{array}{l} y \in L^2(0,T;V_t) \\ \frac{dy}{dt} \in L^2(0,T;V_t') \end{array} \right. $$

then there exists a constant $C$, such that

$$ (2.24) \qquad \| y \|_{C(0,T;H)} \le C (\| y \|_{L^2(0,T;V_t)}^2 + \| \frac{dy}{dt} \|_{L^2(0,T;V_t')}^2)^{\frac{1}{2}} $$

where constant $C$ does not depend on $y$.

**Proof :**

By (2.23) and (2.20) we have

$$ y \in L^2(0,T;V) $$

similarly by (2.24) (2.21)

$$ \frac{dy}{dt} \in L^2(0,T;V') $$

hence we may make use of a theorem given in [6] which implies that

$$ (2.25) \qquad y \in C(0,T; [V,V']_2) $$

with continuous imbedding, and by (2.17) we obtain (2.24).
**Definition :**

We denote by $W^1(0,T)$ the Hilbert space :

$$
(2.26) \quad W^1(0,T) = \{ y \in L^2(0,T;V_t) \mid \frac{dy}{dt} \in L^2(0,T;V_t^\prime) \}
$$

with the norm

$$
(2.27) \quad \| y \|_{W^1(0,T)}^2 = \| y \|_{L^2(0,T;V_t)}^2 + \| \frac{dy}{dt} \|_{L^2(0,T;V_t)}^2
$$

**Remark :**

By lemma 1, we have the following estimation a priori

$$
(2.28) \quad \| y \|_{C(0,T;H)} \le C \| y \|_{W^1(0,T)} , \quad \forall y \in W^1(0,T)
$$

Let us consider an abstract parabolic problem :

$$
(2.29) \quad \left\{
\begin{array}{ll}
\displaystyle \left( \frac{dy}{dt}(t), z \right)_{H'H} + a_u(t; y(t), z) = (f(t), z)_{H'H} & \forall z \in H \\[2ex]
\text{a.e. in } [0,T] \\[2ex]
\displaystyle y(0) = y_o
\end{array}
\right.
$$

where $y_o \in H_1$, $f \in L^2(0,T;H')$ are given elements. It can be shown that the
problem (2.29) has the unique solution $y_u \in W(0,T)$ and

$$
(2.30) \quad \| y_u \|_{W(0,T)} \le c ( \| y_o \|_H^2 + \| f \|_{L^2(0,T;H')}^{\frac{2}{1-2\alpha}} )^{\frac{1}{2}}
$$

Furthermore

**Lemma 2** [1]
Assume,

$$
(2.31) \quad y_o \in H
$$

$$
(2.32) \quad f \in L^2(0,T; H_1)
$$

then there exists the unique solution $y_u$ of the problem (2.29) such that

$$
(2.33) \quad y_u \in L^2(0,T;V_t)
$$
$$
(2.34) \quad \frac{dy_u}{dt} \in L^2(0,T; H_1)
$$

and

$$
(2.35) \quad \frac{dy_u}{dt} + A_u(t)y_u = f \quad \text{in } L^2(0,T; H_1)
$$

In the sequel we assume that for every $u \in U_{ad}$ the scalar product
$\langle \cdot, \cdot \rangle_t$ is differentiable with respect to $t$ on the interval $(0,T)$ in the
following sense :

there exists a bilinear form

$$
(2.36) \quad d_u(t, \dots) : H \times H \to \mathbb{R}, \quad t \in [0,T]
$$

which is continuous

$$
(2.37) \quad |d_u(t;y,z)| \le M_1 \|y\|_H \|z\|_H , \\
\quad \forall y,z \in H, \forall t \in [0,T], \forall u \in U_{ad}
$$

and measurable with respect to t, i.e. the mapping

$$
(2.38) \quad [0,T] \ni t \rightarrow d_u(t;y,z) \in \mathbb{R}
$$

$$
\text{is measurable } \forall u \in U_{\text{ad}}, \forall y, z \in H
$$

such that

$$
(2.39) \quad \int_0^s < \frac{dy}{dt}(t), z(t) >_t dt = a_u(s;y(s),z(s)) - a_u(0;y(0),z(0)) \\
\qquad - \int_0^s < y(t), \frac{dz}{dt}(t) >_t dt - \int_0^s d_u(t;y(t),z(t)) dt, \\
\qquad \forall y, z \in W^1(0,T), \forall s \in (0,T]
$$

Let us consider a linear parabolic problem with variable domain

$$
(2.40) \quad \left\langle \frac{dy}{dt}(t), z \right\rangle_t + b_u(t; y(t), z) = \left\langle f(t), z \right\rangle_t,
$$
$$
\forall z \in V_t, \quad \forall t \in [0, T]
$$

$$
(2.41) \quad y(0) = y_o
$$
where bilinear form $b_u(t;...)$ is defined by (2.13) with $s > M_1/\alpha$

**Remark :**

Under condition $s > M_1/\alpha$ we have

$$ (2.42) \qquad b_u(t;z,z) - d_u(t;z,z) \ge \alpha_1 \|z\|_{V_t}^2, \quad \forall z \in V_t $$

where $\alpha_1 = s - M_1/\alpha$

**Lemma 3 :**

Assume (2.31) (2.32) (2.42), then there exists the unique solution $y_u \in W^1(0,T)$ of the problem (2.40) (2.41)

**Proof :**

i) existence

Let us denote

$$ (2.43) \qquad \tilde{f} = e^{st} f $$

by lemma 2 there exists the unique function $\tilde{y} \in L^2(0,T;V_t)$ such that
$\frac{dy}{dt} \in L^2(0,T;H_1)$ which is the solution of the following problem.

$$ (2.44) \qquad \frac{d\tilde{y}}{dt}(t) + A_u(t)\tilde{y}(t) = \tilde{f}(t), \quad \text{in } H_1, \quad \text{a.e. in } [0,T] $$

$$ (2.45) \qquad \tilde{y}(0) = y_0 $$

Let $\varphi \in V_t$, hence $A_u(t)\varphi \in H_1$ and from (2.44) we have

$$ (2.46) \qquad \left( \frac{d\tilde{y}}{dt}(t), A_u(t)\varphi \right)_{H_1} + (A_u(t)\tilde{y}(t), A_u(t)\varphi)_{H_1} = (\tilde{f}, A_u(t)\varphi)_{H_1}, \\ \qquad \forall \varphi \in V_t, \text{ a.e. in } [0,T] $$

take

$$ (2.47) \qquad y = e^{-st} \tilde{y} $$

then from (2.46) we obtain
$$
\begin{align*}
(2.48) \quad & \left(\frac{dy}{dt}(t), A_u(t)\varphi\right)_{H_1} + \left(A_u(t)y, A_u(t)\varphi\right)_{H_1} + s(y, A_u(t)\varphi)_{H_1} \\
& = (f(t), A_u(t)\varphi)_{H_1}, \quad \forall \varphi \in V_t, \forall t \in [0,T]
\end{align*}
$$

$$
(2.49) \qquad y(0) = y_o
$$

then by (2.10) (2.11) we can rewrite (2.48) in the form :

$$
\begin{align*}
(2.50) \quad \begin{cases}
\langle \frac{dy}{dt}(t), \varphi \rangle_t + b_u(t; y(t), \varphi) = \langle f(t), \varphi \rangle_t, \\
y(0) = y_o
\end{cases} & \quad \forall \varphi \in V_t, \quad \forall t \in [0, T] \\
& \quad y = e^{-st} \tilde{y} \text{ is a solution of the problem (2.40) (2.41).}
\end{align*}
$$

hence $y = e^{-st} \tilde{y}$ is a solution of the problem (2.40) (2.41).

Remark :

Existence of a solution can be also proved using Galerkin approximation
of the problem (2.40) (2.41)

ii) uniqueness.

We have to prove that if we put y₀ = 0 and f = 0 then the only solution
of (2.40) (2.41) is equal to zero.

Take $z = y(t)$ in (2.40) and integrate (2.40) on $(0,s)$, $s \in (0,T]$, hence

$$
(2.51) \quad \int_0^s \langle \frac{dy}{dt}(t), y(t) \rangle_t dt + \int_0^s b_u(t; y(t), y(t)) dt = 0
$$

using (2.39) gives

$$
\begin{align*}
(2.52) \quad 0 &= \frac{1}{2} a_u(s;y(s),y(s)) - \frac{1}{2} \int_0^s d_u(t;y(t),y(t))dt \\
&\quad + \int_0^s b_u(t;y(t),y(t))dt \ge \frac{\alpha_1}{2} \|y(s)\|_{H}^2 + \alpha_2 \|y\|_{L^2(0,s;V_t)}^2 \\
&\qquad \forall s \in (0,T]
\end{align*}
$$

hence $y = 0$.

<figure><img src="image_1.png" /></figure>
We define also so called adjoint state $p = p_u \in L^2(0,T;V_t)$ at point $u \in U_{ad}$ as the unique solution of the following adjoint state equation :

$$
(2.53) \quad \int_0^T \left( \left\langle \frac{d\phi}{dt}(t), p(t) \right\rangle_t + b_u(t; \phi(t), p(t)) \right) dt \\
\qquad = -(y_u(T) - z_d, \phi(T))_H, \quad \forall \phi \in W_o^1(0,T) \\
\text{where } W_o^1(0,T) = \{ \phi \in W^1(0,T) \mid \phi(0) = 0 \}
$$

**Lemma 4 :**

There exists the unique solution $p = p_u \in L^2(0,T;V_t)$ to the problem (2.53) for every $u \in U_{ad}$.

**Proof :**

Let us denote by $\Phi \subset L^2(0,T;V_t)$ the set $W_o^1(0,T)$ with the norm (it means, that
$\Phi = W_o^1(0,T)$ algebraically)

$$
\| \phi \|_{\Phi}^{2} = \| \phi \|_{L^{2}(0, T; V_{t})}^{2} + \| \phi(T) \|_{H}^{2}, \quad \forall \phi \in W_{o}^{1}(0, T)
$$

Left hand side of (2.53) is a bilinear form

$$
E(p, \phi) = \int_{0}^{T} \left( \left\langle p(t), \frac{d\phi}{dt}(t) \right\rangle_{t} + b_{u}(t; p(t), \phi(t)) \right) dt
$$

which maps $L^2(0,T;V_t) \times \Phi \to R$. is continuous in the first argument and $\Phi$-continuous,
that is

$$
\begin{align*}
E(\phi, \phi) &= a_u(T; \phi(T), \phi(T)) - \int_0^T d_u(t; \phi(t), \phi(t)) \, dt \\
&\quad + \int_0^T b_u(t; \phi(t), \phi(t)) \, dt \\
&\geq \alpha \| \phi(T) \|_H^2 + \alpha_1 \| \phi \|_{L^2(0,T;V_t)}^2 \\
&\qquad + \alpha, \alpha_1 > 0, \quad \forall \phi \in \Phi
\end{align*}
$$

Denote by $L(\cdot) \in \Phi'$ a linear, continuous form given by

$$
(2.54) \quad L(\phi) = -(y_u(T) - z_d, \phi(T))_H, \quad \forall \phi \in \Phi
$$
We can make use of the Lions projection theorem [3] (p. 169, theorem 3.1) which implies, that there exists an element $p \in L^2(0,T;V_t)$ such that

$$ (2.55) \qquad E(p, \phi) = J(\phi), \quad \forall \phi \in \Phi $$

To prove uniqueness of the solution let us observe [15], that for every $f \in L^2(0,T;H_1)$ there exists the unique solution $\phi \in W_0^1(0,T)$ of the problem

$$ E(p, \phi) = (f,p)_{L^2(0,T;H_1)}, \quad \forall p \in L^2(0,T;V_t) $$

hence $E(p,\phi) = 0$, $\forall \phi \in W_0^1(0,T)$ implies $(f,p)_{L^2(0,T;H_1)} = 0$, $\forall f \in L^2(0,T;H_1)$, whence $p = 0$.

■

### 3. NECESSARY CONDITIONS OF OPTIMALITY.

Let us consider the following parametric optimization problem :

(P) : minimize cost functional

$$ (3.1) \qquad J(u) = \frac{1}{2} \| y_u(T) - z_d \|_{H}^{2} + \frac{\varepsilon}{2} \| u \|_{U}^{2}, \quad \varepsilon > 0 $$

subject to (2.40) (2.41) and $u \in U_{ad}$.

In [10] are given sufficient conditions for existence of an optimal parameter $\hat{u} \in U_{ad}$. These conditions are obtained by proving continuity of cost functional $J(u)$ over the set $U_{ad}$ which is assumed to be compact, namely

**Theorem 1 :**

Assume that the set $U_{ad} \subset U$ is compact and the mapping

$$ (3.2) \qquad U \supseteq U_{ad} \ni u \to \tilde{A}_u(t) \in C^1(0,T; \mathcal{L}(V,H_1)) $$

is Lipschitzien, where $\tilde{A}_u(t)|_{V_t} = A_u(t)$, then there exists an optimal parameter $\hat{u} \in U_{ad}$ such that $J(\hat{u}) \le J(u)$, $\forall u \in U_{ad}$. Proof is given in [10].
**Remark :**

Operator $\tilde{A}_u(t)$ may be defined for example as follows :

$$ (3.3) \qquad \tilde{A}_u(t)y = \begin{cases} A_u(t)y, & \forall y \in V_t \\ 0, & \forall y \in V_t^{\perp} \end{cases} $$

■

In this section we derive necessary conditions of optimality for the problem (P). Let us note, that the bilinear form

$$ (3.4) \qquad E_u(\phi, p) = \int_0^T \left( \left( \frac{d\phi}{dt}(t), A_u(t)p(t) \right)_{H_1} + \left( A_u(t)\phi(t), A_u(t)p(t) \right)_{H_1} \right. \\ \left. \qquad + s a_u(t; \phi(t), p(t)) \right) dt $$

which is defined for $\phi, p \in L^2(0,T;V_t)$, $\frac{d\phi}{dt} \in L^2(0,T;H_1)$ may be extended, namely

$$ (3.5) \qquad \tilde{E}_u(d,p) = \int_0^T \left( \left( \frac{d\phi}{dt}, \tilde{A}_u(t)p(t) \right)_{H_1} + \left( \tilde{A}_u(t)\phi(t), \tilde{A}_u(t)p(t) \right)_{H_1} \right. \\ \left. \qquad + s a_u(t; \phi(t), p(t)) \right) dt, \quad \forall \phi, p \in L^2(0,T;V), \frac{d\phi}{dt} \in L^2(0,T;H_1) $$

and

$$ (3.6) \qquad \tilde{E}_u(\phi, p) = E_u(\phi, p), \quad \forall \phi, p \in L^2(0,T;V_t), \quad \frac{d\phi}{dt} \in L^2(0,T;H_1) $$

We will assume, that the mapping

$$ (3.7) \qquad V \supseteq U_{ad} \ni u \to \tilde{E}_u(\phi, p) \in R $$

is differentiable at the point $\hat{u} \in U_{ad}$ and we denote its differential by $\tilde{DE}_{\hat{u}}(\delta u; \phi, p)$, where the mapping $U \ni \delta u \to \tilde{DE}_{\hat{u}}(\delta u; \phi, p) \in R$ is linear. Then we obtain necessary conditions of optimality for the problem (P) in the following form :

**Theorem 2 :**

Assume that the mapping (3.7) is differentiable at $\hat{u} \in U_{ad}$ and $\hat{u}$ is a solution of (P), then
$$
(3.8) \quad \varepsilon(\hat{u}, u-\hat{u})_U + \tilde{D} \tilde{E}_{\hat{u}}(u-\hat{u}; \hat{y}, \hat{p}) \ge 0, \quad \forall u \in U_{ad}
$$

where $\hat{p} = p_{\hat{u}}$, $\hat{y} = y_{\hat{u}}$ denote adjoint state and state trajectories at the point
$\hat{u} \in U_{ad}$.

Proof :

Define Lagrangian :

$$
(3.9) \quad \begin{aligned}
\mathcal{L}(u,y,p) &= \frac{1}{2} \| y(T) - z_d \|_H^2 + \frac{\varepsilon}{2} \| u \|_U^2 + \tilde{E}_u(y,p) - \int_0^T \langle f(t), p(t) \rangle_t dt \\
&\quad u \in U_{ad} \subset U, \quad y \in W^1(0,T), \quad p \in L^2(0,T;V_t), \quad \frac{dy}{dt} \in L^2(0,T;H_1)
\end{aligned}
$$

Assume that $p = p_u$ is defined by (2.53), then

$$
(3.10) \quad \left( \frac{d}{dy} \mathcal{L}(u,y_u,p_u); \phi \right) = 0 \quad , \quad \forall \phi \in W_o^1(0,T)
$$

hence we can derive the gradient of the cost functional (3.1) in the following
form

$$
(3.11) \quad \left(\frac{dJ}{du}(\hat{u}); \delta u\right) = \left(\frac{\partial \mathcal{L}}{\partial u}(\hat{u}, y_{\hat{u}}, p_{\hat{u}}); \delta u\right) = \varepsilon(\hat{u}, \delta u) + \tilde{D} \tilde{E}_{\hat{u}}(\delta u; y_{\hat{u}}, p_{\hat{u}}), \quad \forall \delta u \in U
$$

Necessary conditions of optimality for the problem (P) has [5] the form

$$
(3.12) \quad \left(\frac{dJ}{du}(\hat{u}); u-\hat{u}\right) \ge 0, \quad \forall u \in U_{ad}
$$

hence using (3.12) and (3.11) we obtain (3.8).

<figure><img src="image_2.png" /></figure>

4. <u>APPROXIMATION.</u>

In the section we study approximation of the problem (P) in the case of a
simple model problem.

Denote

$$
I = [0.1], \quad J = [0,T], \quad T > 0
$$

and set
$$H = H^1(I), \quad H_1 = L^2(I)$$
$$
(4.1) \quad a_{u}(t;y,z) = a(u(t)) \int_{I} \frac{dy}{dx} \frac{dz}{dx} dx + g(u(t)) y(0) z(0) \\
u \in U_{ad} \subset H^{2}(0,T)
$$

where

$$
(4.2) \quad U_{ad} = \{ u \in H^2(0,T) \mid 0 \le u(t) \le 1, \left| \frac{d^2 u}{dt^2} \right| \le 1, \text{ a.e. in } [0,T] \}
$$

$$
(4.3) \quad a(\cdot), g(\cdot) \in C^2[0,T]
$$

$$
\left.
\begin{array}{l}
M \ge a(r) \ge \beta > 0 \\
M \ge g(r) \ge \beta > 0
\end{array}
\right\}
\quad \forall r \in [0,1], M, \beta \text{ are given constants}
\tag{4.4}
$$

It can be verified, that bilinear form (4.1) generates an operator $A_u(t)$
of the form

$$
(4.5) \quad A_u(t)y = -a(u(t)) \frac{d^2 y}{dx^2} , \quad \forall y \in H^2(I)
$$

unbounded in H¹(I) with the domain :

$$
(4.6) \quad V_t = \{y \in H^2(I) \mid -\frac{dy}{dx}(0) + g(u(t))y(0) = 0, \frac{dy}{dx}(1) = 0\}
$$

we set $V = H^2(I)$, it is obvious that

$$
(4.7) \quad A_u(t) \in \mathcal{L}(H^2(I); L^2(I)) \cap \mathcal{L}(V_t; L^2(I)) \\
\qquad \forall u \in U_{ad}, \forall t \in [0,T]
$$

Under assumptions (4.2) (4.3) we have $a(u(\cdot)) \in C^1[0,T]$, $\forall u \in U_{ad}$, hence

$$
(4.8) \quad A_u(\cdot) \in C^1(0,T; \mathcal{L}(H^2(I); L^2(I))) \quad \forall u \in U_{ad}
$$

It can be also verified that simple integral

$$
L^2(0,T; V_t) = \int_{\mathbb{R}} \oplus V_t dt
$$
constitutes a linear subspace of the space $L^2(0,T;H^2(I))$ of the form :

$$
(4.9) \quad \begin{aligned}
L^2(0,T;V_t) &= \{ y \in L^2(0,T; H^2(I)) \mid -\frac{\partial y}{\partial x}(0,t) + g(u(t))y(0,t) = 0, \\
&\qquad \frac{\partial y}{\partial x}(1,t) = 0, \quad \text{a.e. in } [0,T] \}
\end{aligned}
$$

By theorem given in [6] we have

$$
(4.10) \quad [H^2(I), L^2(I)]_1 = H^1(I)
$$

hence [6] if

$$
(4.11) \quad \left\{
\begin{array}{l}
y \in L^2(0,T;H^2(I)) \\
\\
\frac{dy}{dt} \in L^2(0,T;L^2(I))
\end{array}
\right.
$$

then

$$
(4.12) \quad \left\| y \right\|_{C(0,T;H^1(I))} \le C \left( \left\| y \right\|_{L^2(0,T;H^2(I))} + \left\| \frac{dy}{dt} \right\|_{L^2(0,T;L^2(I))} \right)
$$

Let there be given elements

$$
(4.13) \quad \left\{
\begin{array}{l}
f \in L^2(0,T;H^1(I)) \\
\\
y_o \in H^1(I)
\end{array}
\right.
$$

There exists (section 2) the unique function $y \in L^2(0,T;H^2(I)) \cap C(0,T;H^1(I))$ with $\frac{dy}{dt} \in L^2(0,T;L^2(I))$ such that :

$$
(4.14) \quad \left\{
\begin{array}{ll}
\displaystyle \frac{\partial y}{\partial x}(x,t) - a(u(t)) \frac{\partial^2 y}{\partial x^2}(x,t) = f(x,t), & x \in I, t \in J \\
\\
\displaystyle - \frac{\partial y}{\partial x}(0,t) + g(u(t))y(0,t) = 0, & t \in J \\
\\
\displaystyle \frac{\partial y}{\partial x}(1,t) = 0, & t \in J \\
\\
y(x,0) = y_o(x), & x \in I
\end{array}
\right.
$$
furthermore the following a priori estimation holds :

$$
(4.15) \quad \| y \|_{C(O,T;H^1(I))} \le C (\| y_o \|_{H^1(I)}^2 + \| f \|_{L^2(O,T;L^2(I))}^2)
$$

where constant C does not depend on u ∈ U_ad, y_o and f. Let there be given an
element z_d ∈ H^1(I), and a number ε > 0. Let us consider the following parame-
tric optimization problem :

(P) minimize cost functional :

$$
(4.16) \quad J(u) = \frac{1}{2} \| y_u(T) - z_d \|_{H^1(I)}^2 + \frac{\epsilon}{2} \| u \|_{H^2(O,T)}^2
$$

subject to state equation (4.14) and $u \in U_{ad}$.

We can apply results of previous section and deduce the existence of a solution
of (P) and to obtain necessary conditions of optimality.

Remark :

It can be shown, that in our case the bilinear form $d_u(t;...)$ takes on
the form :

$$
(4.17) \quad d_u(t;y,z) = \frac{da}{du} \frac{du}{dt} \int_I \frac{dy}{dx} \frac{dz}{dx} dx + \frac{dg}{du} \frac{du}{dt} y(0) z(0) , \\
\forall y,z \in H^1(I)
$$

In the sequel we consider approximation of the problem (P).

Let $\tau$ denote a parameter, $\tau = T/N_{\tau}$, $N_{\tau} = 1,2,...$ set

$$
(4.18) \quad U_{ad}^{\tau} = \{ u_{\tau} \in U_{ad} \mid \frac{d^2 u_{\tau}(t)}{dt^2} = u_{\tau}^{i}, \ t \in [i\tau, (i+1)\tau), \\
\phantom{(4.18) \quad U_{ad}^{\tau} = \{ u_{\tau} \in U_{ad} \mid \frac{d^2 u_{\tau}(t)}{dt^2} = } i = 0,1,\dots,N_{\tau}-1, \ u_{\tau}(0) = u_{\tau}^{N_{\tau}}, \frac{du_{\tau}(0)}{dt} = u_{\tau}^{N_{\tau}+1} \}
$$

The family {$U_{ad}^\tau$} constitute an internal, finite dimensional approximation of the set $U_{ad}$. $\tau > 0$
**Lemma 5 :**

For every element $\bar{u} \in U_{ad}$ there exists a sequence $u_{\tau} \in U_{ad}^t$ such that

$$ u_{\tau} \xrightarrow{\tau \to 0} \bar{u} \quad \text{strongly in } H^2(0,T) $$

**Lemma 6 :**

For every sequence $u_{\tau} \in U_{ad}^t$ there exists a subsequence of parameters $\tau' \to 0$ and an element $\bar{u} \in U_{ad}$ such that

$$ u_{\tau'} \to \bar{u} \quad \text{weakly in } H^2(0,T) $$

$$ u_{\tau'} \to \bar{u} \quad \text{strongly in } C^1[0,T] $$

we denote by $h$ a parameter which corresponds to discretization of the space variable $x$ and we assume that there is given a family of finite dimensional subspaces $V_h \subset H^2(I)$ such that

$$ (4.20) \qquad \bigcup_{h > 0} V_h = H^2(I) $$

For every $t \in [0,T]$ denote by $\eta_h(t) = V_h \cap V_t$ and by $\{\phi_j(x,t)\}$, $j = 0,1,\dots,N_h$ the basis in the space $\eta_h(t)$.

**Example :**

As $V_h$ we can use the space of Hermite cubic polynomials, that is the basis of the space $V_h$ has the following form :

$$ (4.21) \qquad \omega_o(x) = \begin{cases} 2h^{-3} x^3 - 3h^{-2} x^2 + 1, & x \in [0,h] \\ 0, & x \notin [0,h] \end{cases} $$

where $h = 1/(N+1)$. $N = 0,1,\dots$

$$ (4.22) \qquad \omega_i(x) = \begin{cases} -\omega_o(x+h-ih) , & x \in [(i-1h, ih] \\ \omega_o(x-ih) , & x \in [ih, (i+1)h] \\ 0 , & x \notin [(i-1)h, (i+1)h] \end{cases} $$
$$
(4.23) \quad \omega_{N+1}(x) \equiv \begin{cases} -\omega_o(x-Nh) & , \ x \in [Nh, 1] \\ 0 & , \ x \notin [Nh, 1] \end{cases}
$$

$$
(4.24) \quad \omega_o^1(x) \equiv \begin{cases} h^{-2}x(h-x)^2 & , \ x \in [0,h] \\ 0 & , \ x \notin [0,h] \end{cases}
$$

$$
(4.25) \quad \omega_i^1(x) \equiv \begin{cases} -\omega_o^1(x+h-ih) & , \ x \in [(i-1)h, ih] \\ \omega_o^1(x-ih) & , \ x \in [ih, (i+1)h] \\ 0 & , \ x \notin [(i-1)h, (i+1)h] \end{cases} \\
i=1, \dots, N
$$

$$
(4.26) \quad \omega_{N+1}^{1}(x) \equiv \begin{cases} - \omega_o(x-Nh) & , \ x \in [Nh, 1] \\ 0 & , \ x \notin [Nh, 1] \end{cases}
$$

Hence, we can derive the basis in $\mathcal{M}_h(t)$ of the form :

$$
(4.27) \quad \varphi_o = -g(u(t)) \omega_o^1(x) + \omega_o(x)
$$

$$
(4.28) \quad \varphi_j = \omega_j, \quad j = 1, \dots, N+1
$$

$$
(4.29) \quad \varphi_{i+N+1} = \omega_i^1, \quad i = 1, \dots, N
$$

and $N_h = \dim \mathcal{M}_h(t) = 2N+1$

Remark :

Only $\varphi_o = \varphi_o(x,t)$ depends on time variable $t$ in the basis of the space
$\mathcal{M}_h(t)$.

For given parameter $\tau > 0$, we denote by $y_h = y_h(x,t)$ the function

$$
(4.30) \quad y_h : [0,T] \to \mathcal{M}(.)
$$

which is determined as the unique solution of the problem :
$$
\begin{align*}
(4.31) \quad & a(u_{\tau}(t)) \int_I \frac{\partial^2 y_h}{\partial x \partial t} \frac{\partial \varphi_j}{\partial x} \, dx + g(u_{\tau}(t)) \frac{\partial y_u}{\partial t}(0,t) \varphi_j(0,t) \\
& + a^2(u_{\tau}(t)) \int_I \frac{\partial^2 y_h}{\partial x^2} \frac{\partial^2 \varphi_i}{\partial x^2} \, dx + s a(u_{\tau}(t)) \int_I \frac{\partial y_u}{\partial x} \frac{\partial \varphi_j}{\partial x} \, dx \\
& + s g(u_{\tau}(t)) y_h(0,t) \varphi_j(0,t) \\
& = a(u_{\tau}(t)) \int_I \frac{\partial f}{\partial x} \frac{\partial \varphi_i}{\partial x} \, dx + g(u_{\tau}(t)) f(0,t) \varphi_j(0,t), \\
& \qquad \forall \varphi_j \in \mathcal{M}_h(t), \quad \forall t \in [0,T]
\end{align*}
$$

$$
(4.32) \quad y_h^0(x,0) = y_h^0(x)
$$

where

$$
(4.33) \quad y_h^0 = \sum_{i=1}^{N_h} y_{oi} \varphi_i(0,x)
$$

is a $H^1(I)$ projection of $y_o \in H^1(I)$ on $\mathcal{M}_h(0)$. Hence $\mathcal{M}_h(t)$ is finite dimensional subspace, then

$$
(4.34) \quad y_h^0(x,t) = \sum_{i=1}^{N_h} y_{hi}(t) \varphi_j^0(x,t)
$$

$$
(4.35) \quad y_h^0(x,0) = y_h^0
$$

and (4.31) (4.32) yields $N_h$-dimensional system of ordinary differential equations
of the form :

$$
(4.36) \quad \left\{
\begin{array}{l}
\underline{w}(t) \quad \displaystyle\frac{dy_h}{dt}(t) + \underline{a}(t) \quad \vec{y}_h(t) = \vec{f}(t) \\
\\
\vec{y}_h(0) = \vec{y}_o
\end{array}
\right.
$$

where

$$
(4.37) \quad \underline{w}(t) = [m_{ij}(t)], \quad m_{ij}(t) = a_{u\tau}(t; \varphi_i, \varphi_j) \\
m_{ij}(\cdot) \in H^2(0,T)
$$
(4.38) $$ \underline{a}(t) = [a_{ij}(t)] $$
$$ a_{ij}(t) = b_{u\tau}(t; \varphi_i, \varphi_j) + a_{u\tau}(t; \frac{\partial \varphi_i}{\partial t}, \varphi_j), $$
$$ a_{ij}(.) \in H^1(0,T) $$

(4.39) $$ \vec{f}(t) = \operatorname{col}(f_0(t), \dots, f_{N_h}(t)) $$
$$ f_i(t) = a_{u\tau}(t; f(t), \varphi_i) $$

(4.40) $$ \vec{y}_o = \operatorname{col}(y_{oi}) $$

We can discretize (4.36) with respect to time variable, for example by using Crank - Nicolson scheme with time step $\Delta t$, then we obtain the system of algebraic equations of the form :

$$ \frac{1}{\Delta t} \underline{w}(t_{k+\frac{1}{2}})(\vec{y}^{k+1} - \vec{y}^k) + \underline{a}(t_{k+\frac{1}{2}})\vec{y}^{k+\frac{1}{2}} = \vec{f}(t_{k+\frac{1}{2}}) $$
$$ k = 0,1,\dots,T/\Delta t $$

$$ \vec{y}^o = \operatorname{col}(y_{oi}) $$

where

$$ t_{k+\frac{1}{2}} = (k + \frac{1}{2})\Delta t $$

$$ t_k = k\Delta t $$

$$ \vec{y}^k = \operatorname{col}(y_o^k, \dots, y_{N_k}^k) $$

$$ \vec{y}^{k+\frac{1}{2}} = (\vec{y}^{k+1} + \vec{y}^k)/2 $$

$$ y_h^k = \sum_{i=0}^{N_h} y_i^k \varphi_j(t_k, x) \in \mathcal{M}_h(t_k) $$

It can be shown that there exists the unique solution of the system (4.41) furthermore

### Lemma 7.

Assume that for $\tau \to 0$ we have
$u_{\tau} \to \bar{u}$ strongly in $C^1[0,T]$

then, for $h$, $\Delta t \to 0$

(4.42) $$y_h^{N_1} \to y_{\bar{u}}(T)$$ strongly in $H^1(I)$

where $N_1 = T/\Delta t$

The lemma can be proved in similar way as in general theory of approximation of linear parabolic problems [13].

For given vector parameter $\xi = (h, \Delta t, \tau)$ define finite dimensional optimization problem :

(P_ξ) : minimize cost functional

(4.43) $$J(u_{\tau}) = \frac{1}{2} \| y_h^{N_1} - z_d^{h} \|_{H^1(I)}^2 + \frac{\epsilon}{2} \| u \|_{H^2(0,T)}^2$$

subject to (4.41) and $u_{\tau} \in U_{ad}^{T}$ $z_d^h$ denotes $H^1(I)$ projection of $z_d$ onto $\eta_h^T(T)$.

**Lemma 8 :**

There exists a solution $\hat{u}_{\xi} \in U_{ad}^{T}$ of the problem (P$_{ξ}$).

**Theorem 3 :**

Let $\xi \to 0$, that is $|h|^2 + |\Delta t|^2 + |\tau|^2 \to 0$, then for some subsequence $\xi' \to 0$

(4.44) $$\hat{u}_{\xi} \to \hat{u}$$ weakly in $H^2(0,T)$

(4.45) $$\hat{u}_{\xi} \to \hat{u}$$ strongly in $C^1[0,T]$

where $\hat{u}$ is a solution of (P), and

$J(\hat{u}_{\xi}) \to J(\hat{u})$.

**Proof of lemma :**

Problem (P_ξ) corresponds to minimization of a continuous function
$J(u_{\tau}) = J(u_{\tau}^{0}, \dots, u_{\tau}^{N_{\tau}+1})$ of $N_{\tau} + 1$ variables over a compact subset of $R^{N_{\tau}+2}$,
hence there exists a solution of this problem.

### Proof of theorem.

Since the set $U_{ad}$ is weakly compact there exists a subsequence of
$\{\hat{u}_{\xi}\} \subset U_{ad}$ (also denoted by $\{\hat{u}_{\xi}\}$) and an element $\bar{u} \in U_{ad}$ such that

$$ (4.47) \qquad \hat{u}_{\xi} \to \bar{u} \quad \text{weakly in } H^{2}(0,T) $$

From (4.47) by imbedding theorem [6] we obtain

$$ (4.48) \qquad \hat{u}_{\xi} \to \bar{u} \quad \text{strongly in } C^{1}[0,T] $$

hence by lemma 7

$$ (4.49) \qquad y_{k}^{N} \to \bar{y}_{\bar{u}}(T) \quad \text{strongly in } H^{1}(I) $$

whence

$$ (4.50) \qquad \frac{1}{2} \| y_h^N - z_d^k \|_{H^1(I)}^2 \to \frac{1}{2} \| \bar{y}_{\bar{u}}(T) - z_d \|_{H^1(I)}^2 $$

By (4.47), (4.50) we have

$$ (4.51) \qquad \underline{\lim} J(\hat{u}_{\xi}) \geq J(\bar{u}) $$

Let there be given a sequence $u_{\xi} \in U_{ad}^{T}$ such that

$$ (4.52) \qquad u_{\xi} \to \hat{u} \quad \text{strongly in } H^{2}(0,T) $$

such sequence exists by lemma 5. By lemma 7 and (4.52) we have

$$ (4.53) \qquad \lim_{\xi \to \bar{u}} J(u_{\xi}) = J(\bar{u}) $$

It is obvious, that $J(u_{\xi}) \geq J(\hat{u})$, $\forall \xi > 0$ hence

$$ (4.54) \qquad J(\hat{u}) = \lim_{\xi \to \bar{u}} J(u_{\xi}) \geq \underline{\lim_{\xi \to \bar{u}}} J(\hat{u}_{\xi}) \geq \underline{\lim_{\xi \to \bar{u}}} J(\bar{u}_{\xi}) = J(\bar{u}) $$

which implies, that $\bar{u} \in U_{ad}$ is a solution of the problem (P).
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