Let Ω be a bounded domain in \(\mathbb{R}^{d}\) (\(d \le3\)) with Lipschitz boundary ∂Ω. Let \(\mathcal{T}^{h}\) be a family of regular triangulation of Ω, such that \(\Omega=\bigcup_{T\in\mathcal{T}^{h}} T\). Denote \(h_{T}\) the diameter of the element T in \(\mathcal{T}^{h}\), and set \(h=\max_{T\in\mathcal{T}^{h}} h_{T}\). In this paper, we shall employ the usual notion for Lebesgue and Sobolev spaces; see Ref. [1] for details. Throughout, let C denote a strictly positive generic constant, not necessarily the same at each occurrence, but always independent of the mesh size h.
To give a detailed description of the model optimal control problem, let us define the state spaces
$$ Y=H_{0}^{1}(\Omega), \qquad \Sigma=L^{2}( \Omega)^{d}. $$
(2.1)
and the control space
$$ U=L^{2}(\Omega). $$
(2.2)
Furthermore, let \(U_{ad}\) be a closed convex subset of the control space U. In the following context, we will discuss some different cases on the choice of \(U_{ad}\).
In the following, we are ready to propose a simple but stable mixed numerical method. To this aim, we assume \(\mathcal{A}=(a_{i,j}(x))_{d\times d}\) is symmetric and positive definite, and there exist positive constants α and β such that
$$ \alpha|\mathcal{X}|^{2} \leq\mathcal{X}^{T}\mathcal{A} \mathcal{X} \leq \beta|\mathcal{X}|^{2}, \quad\forall\mathcal{X} \in \mathbb{R}^{d}. $$
Besides, suppose \(c=c(x)\) is bounded below and up by two constants \(c_{*}\) and \(c^{*}\) such that
$$ 0\le c_{*}\leq c\leq c^{*}. $$
We now first revisit the classical mixed formulation of problem (1.2)-(1.3): Find \((y,\sigma)\in L^{2}(\Omega) \times H(\operatorname{div};\Omega)\) such that for any \((v, \tau)\in L^{2}(\Omega)\times H(\operatorname{div};\Omega)\)
$$ \bigl(\mathcal{A}^{-1}\sigma, \tau \bigr)- (y, \operatorname{div} \tau)+(\operatorname{div} \sigma , v)+(cy, v)=(f+ \mathcal{B}u,v). $$
(2.3)
Then followed by integration by parts, and inspired by the stabilized finite element method [18–20], we propose a novel stabilized mixed weak formulation for (1.2)-(1.3): find \((y,\sigma)\in Y\times\Sigma\) such that
$$ \mathcal{A}_{\delta}(y,\sigma; v, \tau)=(f+ \mathcal{B}u, v),\quad \forall(v, \tau)\in Y\times\Sigma, $$
(2.4)
where the mixed bilinear formulation
$$\begin{aligned} \mathcal{A}_{\delta}(y,\sigma; v, \tau)={}& \bigl(\mathcal{A}^{-1}\sigma, \tau \bigr)+ (\nabla y, \tau)- ( \sigma, \nabla v)+(cy, v) \\ &{}-\sum_{T \in \mathcal{T}^{h}}\delta_{T} \bigl( \mathcal{A}^{-1}\sigma+\nabla y, \tau-\mathcal{A}\nabla v \bigr)_{T}. \end{aligned}$$
(2.5)
Here \(\delta_{T}\) is an elementwise mesh-free constant parameter, and \((\cdot,\cdot)_{T}\) denotes the inner product in \(L^{2}(T)^{d}\).
Define the corresponding stabilization norm
$$ \bigl|\!\bigl|\!\bigl|\{y,\sigma\}\bigr|\!\bigr|\!\bigr|_{\delta}^{2}= \bigl(\mathcal{A}^{-1}\sigma, \sigma \bigr) +\sum _{T \in \mathcal{T}^{h}}\delta_{T} (\mathcal{A}\nabla y, \nabla y)_{T}+(cy,y). $$
(2.6)
Then, we can easily derive the following coercive and bounded results.
Proposition 2.1
(Coercivity and boundedness)
For
\(0<\delta _{0}\leq\delta_{T}\leq\delta_{1}<1\), we have
$$\begin{aligned} \mathcal{A}_{\delta}(y,\sigma; y,\sigma) \geq& c_{\delta} \bigl|\!\bigl|\!\bigl|\{y,\sigma \}\bigr|\!\bigr|\!\bigr|_{\delta}^{2},\quad \forall(y, \sigma)\in Y\times\Sigma. \end{aligned}$$
(2.7)
Moreover, for all
\((y, \sigma; v,\tau)\in(Y\times\Sigma)^{2}\)
$$\begin{aligned} \mathcal{A}_{\delta}(y,\sigma; v,\tau) \leq C_{\delta} \bigl|\!\bigl|\!\bigl|\{y,\sigma \}\bigr|\!\bigr|\!\bigr|_{\delta} \bigl|\!\bigl|\!\bigl|\{v,\tau \} \bigr|\!\bigr|\!\bigr|_{\delta}. \end{aligned}$$
(2.8)
Here
\(\delta_{i}\) (\(i=0,1\)) are two positive constants, \(c_{\delta }=1-\delta_{1}\)
and
\(C_{\delta}=\max\{4, 2+1/\delta_{0}\}\).
Proof
On the one hand, it follows from the definition that
$$\begin{aligned} &\mathcal{A}_{\delta}(y,\sigma; y, \sigma) \\ &\quad= \bigl(\mathcal{A}^{-1}\sigma, \sigma \bigr)+(cy, y) -\sum _{T \in \mathcal{T}^{h}}\delta_{T} \bigl( \mathcal{A}^{-1} \sigma, \sigma \bigr)_{T} +\sum _{T \in \mathcal{T}^{h}} \delta_{T} (\mathcal{A} \nabla y, \nabla y)_{T}. \end{aligned}$$
(2.9)
Hence, for \(\delta_{T}\leq\delta_{1}<1\), the bilinear form \(a(\diamond ,\diamond; \diamond, \diamond)\) is coercive on \(Y\times\Sigma\), and it satisfies (2.7).
On the other hand, we conclude from (2.5) that
$$\begin{aligned} \mathcal{A}_{\delta}(y,\sigma; v, \tau)={}& \bigl( \mathcal{A}^{-1} \sigma, \tau \bigr)+ (\nabla y, \tau)-(\sigma, \nabla v)+(cy,v) \\ &{} -\sum_{T \in \mathcal{T}^{h}}\delta_{T} \bigl( \mathcal{A}^{-1}\sigma, \tau \bigr)_{T} +\sum _{T \in \mathcal{T}^{h}}\delta_{T} ( \sigma, \nabla v)_{T} \\ &{} -\sum_{T \in \mathcal{T}^{h}}\delta_{T} (\nabla y, \tau)_{T} +\sum_{T \in \mathcal{T}^{h}}\delta_{T} ( \mathcal{A}\nabla y, \nabla v)_{T}. \end{aligned}$$
(2.10)
Note that
$$ \sum_{i} |a_{i}b_{i}|\leq \biggl(\sum_{i} a_{i}^{2} \biggr)^{\frac{1}{2}} \biggl(\sum_{i} b_{i}^{2} \biggr)^{\frac{1}{2}}. $$
Then for \(0<\delta_{0} \leq\delta_{T} \leq\delta_{1}<1\), we have
$$\begin{aligned} & \mathcal{A}_{\delta}(y,\sigma; v, \tau) \\ &\quad\leq \biggl[4 \bigl(\mathcal{A}^{-1}\sigma, \sigma \bigr)+ ( \mathcal{A} \nabla y, \nabla y) +2\sum_{T \in \mathcal{T}^{h}} \delta_{T} ( \mathcal{A}\nabla y, \nabla y)_{T}+(cy, y) \biggr]^{\frac{1}{2}} \\ &\qquad{}\times \biggl[4 \bigl(\mathcal{A}^{-1}\tau, \tau \bigr)+ ( \mathcal{A}\nabla v, \nabla v) +2\sum_{T \in \mathcal{T}^{h}} \delta_{T} ( \mathcal{A} \nabla v, \nabla v)_{T}+(cv, v) \biggr]^{\frac{1}{2}} \\ &\quad\leq \biggl[4 \bigl(\mathcal{A}^{-1}\sigma, \sigma \bigr) + \biggl(2+ \frac{1}{\delta_{0}} \biggr)\sum_{T \in \mathcal{T}^{h}}\delta _{T} ( \mathcal{A}\nabla y, \nabla y)_{T}+(cy, y) \biggr]^{\frac{1}{2}} \\ &\qquad{}\times \biggl[4 \bigl(\mathcal{A}^{-1}\tau, \tau \bigr) + \biggl(2+ \frac{1}{\delta_{0}} \biggr)\sum_{T \in \mathcal{T}^{h}}\delta _{T} (\mathcal{A} \nabla v, \nabla v)_{T}+(cv, v) \biggr]^{\frac{1}{2}} \\ &\quad\leq C_{\delta} \bigl|\!\bigl|\!\bigl|\{y,\sigma\}\bigr|\!\bigr|\!\bigr|_{\delta} \bigl|\!\bigl|\!\bigl|\{v, \tau \}\bigr|\!\bigr|\!\bigr|_{\delta}, \end{aligned}$$
(2.11)
which proves the boundedness result (2.8). □
Proposition 2.2
(Existence and uniqueness)
Assume the condition in Proposition
2.1
is valid. Furthermore, let
\(f\in L^{2}(\Omega)\). Then for given control
\(u\in L^{2}(\Omega)\), problem (2.4) admits a unique solution
\((y, \sigma)\in Y\times \Sigma\).
Proof
Proposition 2.1 implies that the mixed bilinear form \(\mathcal{A}_{\delta}(\diamond,\diamond; \diamond, \diamond)\) is coercive and bounded in a weighted norm (2.6). Then the Lax-Milgram lemma implies the existence and uniqueness of the solution pair \((y, \sigma)\in Y\times\Sigma\). □
Furthermore, suppose \(u=0\) in problem (2.4), we then have the following stability result with respect to the right-hand term f.
Proposition 2.3
(Stability)
Let
\(f\in L^{2}(\Omega)\). If the condition in Proposition
2.1
is valid, then we have
$$\begin{aligned} \bigl|\!\bigl|\!\bigl|\{y,\sigma\}\bigr|\!\bigr|\!\bigr|_{\delta} \leq C \|f\|_{L^{2}(\Omega)}, \end{aligned}$$
(2.12)
where the constant
C
depends on the Poincaré constant
\(C_{\Omega }\), and the reciprocal of
\(c_{\delta}\)
and
\(\delta_{0}\).
Proof
Let \((v, \tau)=(y, \sigma)\) in problem (2.4). Then it follows from Proposition 2.1, the Poincaré inequality in Lemma 4.3, and the definition of the stabilization norm that
$$ \begin{aligned} c_{\delta}\bigl|\!\bigl|\!\bigl|\{y,\sigma\}\bigr|\!\bigr|\!\bigr|_{\delta}^{2} &\leq\|f\|_{L^{2}(\Omega)}\|y\| _{L^{2}(\Omega)} \\ &\leq C_{\Omega} \|f\|_{L^{2}(\Omega)}\|\nabla y\|_{L^{2}(\Omega)} \\ &\leq\frac{C_{\Omega}}{\sqrt{\delta_{0}}} \|f\|_{L^{2}(\Omega)} \biggl(\sum _{T \in \mathcal{T}^{h}}\delta_{T} (\nabla y, \nabla y)_{T} \biggr)^{1/2}, \end{aligned} $$
which implies the conclusion. □
Denote
$$ \mathcal{J}(y,\sigma,u)= \frac{1}{2} \int_{\Omega} (y-y_{d})^{2} + \frac{1}{2} \int_{\Omega} (\sigma-\sigma_{d})^{2}+ \frac{\gamma}{2} \int _{\Omega} u^{2}, $$
where \(y=y(u)\) and \(\sigma=\sigma(u)\) are u-dependent.
Then for the given control set \(U_{ad}\), we reformulate the optimal control problem (1.1)-(1.3) as follows: (OCP)
$$ \mathcal{J} \bigl(y^{*},\sigma^{*},u^{*} \bigr)=\min_{u\in U_{ad}} \mathcal{J}(y,\sigma,u) $$
such that \((y, \sigma, u)\in Y\times\Sigma\times U\) and
$$ \mathcal{A}_{\delta}(y,\sigma; v, \tau)=(f+ \mathcal{B}u, v),\quad \forall(v, \tau)\in Y\times\Sigma. $$
It then follows from Ref. [3] that the optimal control problem (OCP) has a unique solution \((y^{*},\sigma^{*},u^{*})\in Y\times\Sigma\times U_{ad}\), and \((y^{*},\sigma ^{*},u^{*})\) is the solution of (OCP) if and only if there is a pair of adjoint state \((z^{*},\omega^{*})\in Y\times\Sigma\), such that \(( y^{*}, \sigma^{*}, z^{*},\omega^{*}, u^{*})\in (Y\times\Sigma)^{2}\times U_{ad}\) satisfies the following optimality system: (OS)
State equation:
$$ \mathcal{A}_{\delta} \bigl(y^{*}, \sigma^{*}; v, \tau \bigr)= \bigl(f+\mathcal{B}u^{*}, v \bigr),\quad \forall(v, \tau)\in Y \times \Sigma. $$
(2.13)
Adjoint state equation:
$$ \mathcal{A}_{\delta} \bigl( v, \tau; z^{*},\omega^{*} \bigr)=- \bigl(y^{*}-y_{d}, v \bigr)- \bigl(\sigma ^{*}- \sigma_{d}, \tau \bigr), \quad\forall(v, \tau)\in Y\times\Sigma. $$
(2.14)
Optimality condition:
$$ \bigl(\gamma u^{*}- \mathcal{B}^{*}z^{*}, u-u^{*} \bigr) \geq0, \quad\forall u\in U_{ad}. $$
(2.15)
Here \(\mathcal{B}^{*}\) is the adjoint operator of \(\mathcal{B}\), which satisfies
$$ (\mathcal{B} v, w)= \bigl(v, \mathcal{B}^{*}w \bigr), \quad\forall(v, w)\in L^{2}(\Omega)\times H_{0}^{1}(\Omega). $$
Remark 2.4
For the stabilization parameter \(\delta_{T}\) being chosen as a constant δ in the whole domain Ω, the adjoint states \(z^{*}\) and \(\omega^{*}\) in (2.14) satisfy the following strong forms:
$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\Delta z =-(y-y_{d})-\operatorname{div}(\sigma-\sigma_{d}), & \mbox{in } \Omega,\\ z =0, & \mbox{on } \partial\Omega, \end{array}\displaystyle \right . $$
(2.16)
and
$$ \omega= \nabla z-\frac{\sigma-\sigma_{d}}{1-\delta}, \quad\mbox{in } \Omega. $$
(2.17)
In the following, we introduce z and ω be the solutions of the dual problem such that
$$ \mathcal{A}_{\delta}( v, \tau; z, \omega)=-(g, v)-(q, \tau), \quad\forall(v, \tau)\in Y\times\Sigma. $$
(2.18)
Then, similar to the proof in Proposition 2.3, we have the following stability result.
Proposition 2.5
(Stability)
Let
\(g\in L^{2}(\Omega)\)
and
\(q\in L^{2}(\Omega)^{d}\)
in problem (2.18). Assume the condition in Proposition
2.1
is valid. Then we have
$$\begin{aligned} \bigl|\!\bigl|\!\bigl|\{z,\omega\}\bigr|\!\bigr|\!\bigr|_{\delta} \le C \bigl(\|g\|_{L^{2}(\Omega)}+\|q\| _{L^{2}(\Omega)^{d}} \bigr), \end{aligned}$$
(2.19)
where the constant
C
depends on the Poincaré constant
\(C_{\Omega }\), and the reciprocal of
\(c_{\delta}\)
and
\(\delta_{0}\).
Remark 2.6
Let \(f, u^{*}\in L^{2}(\Omega)\). If the condition in Proposition 2.1 is valid, then from Proposition 2.3 we have
$$\begin{aligned} \bigl|\!\bigl|\!\bigl|\bigl\{ y^{*},\sigma^{*} \bigr\} \bigr|\!\bigr|\!\bigr|_{\delta} \le C \bigl(\|f \|_{L^{2}(\Omega)}+\bigl\| u^{*}\bigr\| _{L^{2}(\Omega)} \bigr). \end{aligned}$$
(2.20)
Furthermore, let \(y_{d}\in L^{2}(\Omega)\) and \(\sigma_{d}\in L^{2}(\Omega)^{d}\). Then from Proposition 2.5 and the above conclusion we have
$$\begin{aligned} \bigl|\!\bigl|\!\bigl|\bigl\{ z^{*},\omega^{*} \bigr\} \bigr|\!\bigr|\!\bigr|_{\delta} &\le C \bigl( \bigl\| y^{*}-y_{d}\bigr\| _{L^{2}(\Omega )}+\bigl\| \sigma^{*}-\sigma_{d} \bigr\| _{L^{2}(\Omega)^{d}} \bigr) \\ &\le C \bigl(\|f\|_{L^{2}(\Omega)}+\bigl\| u^{*}\bigr\| _{L^{2}(\Omega)}+\|y_{d}\| _{L^{2}(\Omega)}+\|\sigma_{d}\|_{L^{2}(\Omega)^{d}} \bigr). \end{aligned}$$
(2.21)
In the end of this section, we pay special attention on the solution of the variational inequality (2.15). It depends heavily on the structure of the convex set \(U_{ad}\). For some cases (see, e.g. [3, 4]), we have the following explicit results.
Case I. \(U_{ad}=U\)
Then the solution is
$$ u^{*}(x)= \frac{1}{\gamma} \mathcal{B}^{*} z^{*}(x). $$
(2.22)
Case II. \(U_{ad}=\{u\in U: u\geq0, \mbox{a.e. in } \Omega\} \)
Then the solution is
$$ u^{*}(x)= \max \biggl\{ 0, \frac{1}{\gamma} \mathcal{B}^{*} z^{*}(x) \biggr\} . $$
(2.23)
Case III. \(U_{ad}=\{u\in U: a\leq u\leq b, \mbox{a.e. in } \Omega\} \) where the bounds \(a, b\in\mathbb{R}\) fulfill \(a< b\).
Then the solution is
$$ u^{*}(x)=\max \biggl\{ a, \min \biggl\{ b, \frac{1}{\gamma} \mathcal{B}^{*} z^{*}(x) \biggr\} \biggr\} . $$
(2.24)
Case IV. \(U_{ad}=\{u\in U: \int_{\Omega} u\geq0\} \)
Then the solution is
$$ u^{*}(x)=\max \biggl\{ 0, -\frac{1}{\gamma}\overline{\mathcal{B}^{*} z^{*}(x)} \biggr\} +\frac{1}{\gamma} \mathcal{B}^{*} z^{*}(x), $$
(2.25)
where \(\overline{\mathcal{B}^{*} z^{*}(x)}=\frac{\int_{\Omega}\mathcal{B}^{*} z^{*}(x)}{\int_{\Omega} 1}\).