Finite element pde constrained optimization — Examples

The form is the full $H^1$ inner product, so it is bounded and coercive with constants one, and $\mathrm{\mid }⟨Bq,v⟩\mathrm{\mid }\le \mathrm{\parallel }q{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$. Since $f=0$, $-u_{\ast }^{\mathrm{\prime }\mathrm{\prime }}+u_{\ast }=q_{\ast }$, and $u_{\ast }(0)=u_{\ast }(1)=0$, integration by parts first for compactly supported smooth tests and then by $H_0^1$ density gives $a(u_{\ast },v)=(q_{\ast },v{)}_{L^2}$ for every $v\in H_0^1(0,1)$; uniqueness in [F1] identifies the state. Since $-p_{\ast }^{\mathrm{\prime }\mathrm{\prime }}+p_{\ast }=-\frac{1}{2}({\pi }^2+1)\sin (\pi x)=u_{\ast }-u_d$ and $p_{\ast }$ has zero boundary values, the same argument verifies the adjoint equation.

The load identity gives $(B^{\ast }{p}_{\ast },r{)}_U=⟨Br,p_{\ast }⟩=(p_{\ast },r{)}_U$ for every $r\in U$, hence $B^{\ast }{p}_{\ast }=p_{\ast }$. Since $0\le q_{\ast }\le 1\mathrm{/}2$, it is admissible, and $B^{\ast }{p}_{\ast }+\alpha q_{\ast }=p_{\ast }+q_{\ast }=0$. Thus its variational inequality holds with equality for every admissible comparison control, and [F2] makes $q_{\ast }$ the unique minimizer.

By [F2], $v$ has an absolutely continuous representative. By [F3], the hypotheses of [F4] hold. Applying [F4] to $F(x)=x$ and $G(x)=v(x)$ gives $v(1)={\int }_0^{1}v(x)dx+{\int }_0^{1}x{v}^{\mathrm{\prime }}(x)dx$. Cauchy--Schwarz [F5] yields $\mathrm{\mid }v(1)\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_2+\mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_2\le \sqrt{2}\mathrm{\parallel }v{\mathrm{\parallel }}_V$. Applying [F4] to $F=u^{\mathrm{\prime }}$ and $G=v$ for a smooth state gives ${\int }_0^1(u^{\mathrm{\prime }}{v}^{\mathrm{\prime }}+uv)dx={\int }_0^{1}fvdx+gv(1)$. Thus the boundary control operator $B_{\mathrm{\partial }}g:v\mapsto gv(1)$ is bounded on $V$.

Let $v_n(x)=\max \{1-n(1-x),0\}$. Then $v_n\in H^1(0,1)$, $v_n(1)=1$, and $\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2=(3n{)}^{-1\mathrm{/}2}$. If $g\mathrm{\ne }0$ and $gv(1)={\int }_0^{1}qv$ for every $v\in V$ with some $q\in L^2(0,1)$, Cauchy--Schwarz [F5] gives $\mathrm{\mid }g\mathrm{\mid }=\mathrm{\mid }{\int }_0^{1}q{v}_n\mathrm{\mid }\le \mathrm{\parallel }q{\mathrm{\parallel }}_2(3n{)}^{-1\mathrm{/}2}\to 0$, a contradiction. For $g=0$, $q=0$ represents the same zero functional. The boundary input space is nevertheless the scalar $L^2(\{1\})$ space, while the distributed input space in [F1] is $L^2(0,1)$; their operators and control sets must be specified separately.

If $x<1\mathrm{/}2$, then $s(x)<0$, so $s(x)q(x)\ge s(x)$ for every $q(x)\in [0,1]$, with equality only when $q(x)=1$. If $x>1\mathrm{/}2$, then $s(x)>0$, so $s(x)q(x)\ge 0$, with equality only when $q(x)=0$.

Integrating the pointwise inequalities gives $\mathcal{L}(q)\ge \mathcal{L}(q_{\ast })$. The gap $s(q-q_{\ast })$ is nonnegative almost everywhere and integrable; if equality holds, [F3] makes it zero almost everywhere. Since $s(x)\mathrm{\ne }0$ away from the single point $x=1\mathrm{/}2$, equality forces the stated endpoint values almost everywhere, so the minimizer is unique as an $L^2$ equivalence class. It is bang-bang by construction. Because $\alpha =0$, the strict-convexity and unique-control conclusions that use $\alpha >0$ are not invoked.

The exact box projection formula from [F2] concerns clipping a point to the interval; this example instead minimizes a linear integrand and reaches the interval endpoints according to the sign of $s$. Thus this is a separate limiting/example branch, not a consequence of substituting $\alpha =0$ into a formula that divides by $\alpha$.