PDE-constrained inversion and adjoint error control — Examples and Counterexamples

Each of the four triangles has area $1\mathrm{/}4$. On each triangle, $\varphi$ is the affine barycentric coordinate of the center, so its gradient has magnitude $2$. The row-sum lumped mass contribution at the center is $\mathrm{\mid }K\mathrm{\mid }\mathrm{/}3=1\mathrm{/}12$ on each triangle, hence $m_h=4\mathrm{/}12=1\mathrm{/}3$. The stiffness is $k_h={\sum }_K{\int }_K\mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2=4\cdot 4\cdot (1\mathrm{/}4)=4$.

Dividing the semidiscrete equation by $m_h$ gives $\dot{U}=-(k_h\mathrm{/}{m}_h)U=-12U$. By the specified forward-Euler update, $U^1=U^0+\mathrm{\Delta }t{\dot{U}}^0=(1-12\mathrm{\Delta }t)U^0=-2U^0$. The scalar amplification has modulus at most one exactly when $-1\le 1-12\mathrm{\Delta }t\le 1$, which is equivalent to $0\le \mathrm{\Delta }t\le 1\mathrm{/}6$; $\mathrm{\Delta }t=1\mathrm{/}4$ violates it.

The model's admissibility predicate requires this explicit-step condition. Since the given step fails it, [F3] shows that the algorithm returns inadmissible before calling either state or adjoint solver. This conclusion concerns the declared discrete stability condition, not a stability theorem for the continuous heat equation.

At $q=0$, the unique state is $u=0$ because $R(0,u)=u=0$. Its state residual is $\mathrm{\mid }R(0,0)\mathrm{\mid }=0$, so the state solve is exact. Also $D_{u}R=1$ and $D_{q}R=-1$.

Here $J(q,u)=\frac{1}{2}(u-1{)}^2+\frac{1}{2}{q}^2$, so ${\mathrm{\nabla }}_{u}J(0,0)=-1$ and ${\mathrm{\nabla }}_{q}J(0,0)=0$. The exact adjoint equation is $p=-1$. By [F1], the exact reduced gradient is $0-(-1)(-1)=-1$. The same value follows by differentiating $\mathrm{\Phi }(q)=\frac{1}{2}(q-1{)}^2+\frac{1}{2}{q}^2$.

If an unchecked iterative adjoint returns $\tilde{p}=1$, its computed reduced gradient is $\tilde{g}={\mathrm{\nabla }}_{q}J(0,0)-(D_{q}R{)}^T\tilde{p}=0-(-1)(1)=1$, the opposite sign. Its exact adjoint-equation residual is $\mathrm{\mid }(D_{u}R{)}^T\tilde{p}-{\mathrm{\nabla }}_{u}J(0,0)\mathrm{\mid }=\mathrm{\mid }1-(-1)\mathrm{\mid }=2$.

With declared adjoint residual tolerance ${\tau }_p=1$, the residual in step 2.2 exceeds tolerance. By [F2], the algorithm rejects this vector as a solver result with status linear-solve-failed; it cannot receive converged. Step 2.2 nevertheless exhibits opposite gradient signs when the unchecked vector is substituted. Thus an exact state solve alone does not determine the gradient sign. Treating $\tilde{p}$ as a contract-valid solver result would violate the adjoint-residual requirement. The gradient-error certificate [F3] remains valid: take ${\eta }_u=0$, ${\eta }_p=2$, $P=M_q=L_{G,u}=1$, and $L_{J,u}=L_{R,u}=0$, which satisfy all the proposition's bounds and give ${\epsilon }_g=2=\mathrm{\mid }\tilde{g}-\mathrm{\nabla }\mathrm{\Phi }(0)\mathrm{\mid }$. This error bound permits the observed sign reversal.

The four triangles have area $1\mathrm{/}4$. On each triangle $\varphi$ is the barycentric coordinate of $O$. For $k=1,2,3$, integration on the reference triangle gives ${\int }_K{\varphi }^k=\mathrm{\mid }K\mathrm{\mid }2\mathrm{/}((k+1)(k+2))$; this follows by integrating $(1-x-y{)}^k$ over $0\le x\le 1$, $0\le y\le 1-x$. Hence ${\int }_{\mathrm{\Omega }}\varphi =1\mathrm{/}3$, ${\int }_{\mathrm{\Omega }}{\varphi }^2=1\mathrm{/}6$, and ${\int }_{\mathrm{\Omega }}{\varphi }^3=1\mathrm{/}10$. Each cell's base is a unit square edge and its altitude from $O$ is $1\mathrm{/}2$, so the affine barycentric coordinate has gradient magnitude $2$; therefore ${\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2=4$. By [F2], $w={\varphi }^2\in H_0^1(\mathrm{\Omega })$. Since $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }=1$, [F6] gives $\mathrm{\mid }{\int }_{\mathrm{\Omega }}q\mathrm{\mid }\le \mathrm{\parallel }q{\mathrm{\parallel }}_{L^2}$, and it also gives $\mathrm{\mid }a(w,z)\mathrm{\mid }\le \mathrm{\parallel }w{\mathrm{\parallel }}_V\mathrm{\parallel }z{\mathrm{\parallel }}_V$; hence $B$ is a bounded linear map into $V^{\ast }$. The same inequality gives $\mathrm{\mid }{\int }_{\mathrm{\Omega }}v\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }v{\mathrm{\parallel }}_V$, so the observation is bounded. The form has $a(v,v)=\mathrm{\parallel }v{\mathrm{\parallel }}_V^2$ and continuity constant $1$, so [F1] applies and gives a unique state.

The exact state is $u(\gamma )=\gamma w$, since substituting it gives $a(u(\gamma ),z)=\gamma a(w,z)$ for every $z\in V$; uniqueness from [F1] identifies this as the state map. Therefore $F(\gamma )=\gamma {\int }_{\mathrm{\Omega }}{\varphi }^2=\gamma \mathrm{/}6$.

Write the Galerkin state as $u_h(\gamma )=c\gamma \varphi$. Testing its weak equation with $\varphi$ gives $ca(\varphi ,\varphi )=a({\varphi }^2,\varphi )$. Here $a(\varphi ,\varphi )=\int ({\varphi }^2+\mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2)=1\mathrm{/}6+4=25\mathrm{/}6$, while $a({\varphi }^2,\varphi )=\int ({\varphi }^3+2\varphi \mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2)=1\mathrm{/}10+2\cdot 4\cdot (1\mathrm{/}3)=83\mathrm{/}30$. Thus $c=(83\mathrm{/}30)\mathrm{/}(25\mathrm{/}6)=83\mathrm{/}125$, and $F_h(\gamma )=c\gamma {\int }_{\mathrm{\Omega }}\varphi =83\gamma \mathrm{/}375$. The scalar Galerkin stiffness $25\mathrm{/}6$ is positive, so this state solve is unique and its exact algebraic residual is zero.

The exact objective is $\mathrm{\Phi }(\gamma )=\frac{1}{2}(\gamma \mathrm{/}6-1\mathrm{/}6{)}^2+\frac{1}{2}{\gamma }^2$. Its derivative is $(1\mathrm{/}36+1)\gamma -1\mathrm{/}36$, so its unique minimizer is ${\gamma }_{\ast }=1\mathrm{/}37\in [0,1]$. The discrete objective is ${\mathrm{\Phi }}_h(\gamma )=\frac{1}{2}(83\gamma \mathrm{/}375-1\mathrm{/}6{)}^2+\frac{1}{2}{\gamma }^2$. Its derivative is $((83\mathrm{/}375{)}^2+1)\gamma -(83\mathrm{/}375)(1\mathrm{/}6)$. Since $(83\mathrm{/}375)(1\mathrm{/}6)=83\mathrm{/}2250$ and $(83\mathrm{/}375{)}^2+1=147514\mathrm{/}140625$, its unique zero is ${\gamma }_h=(83\mathrm{/}2250)(140625\mathrm{/}147514)=10375\mathrm{/}295028\in [0,1]$. Both second derivatives are positive, so these stationary points are the unique minimizers on the interval. Moreover $10375\cdot 37=383875>295028$, hence ${\gamma }_h>1\mathrm{/}37={\gamma }_{\ast }$.

For every $\gamma \in [0,1]$, $\mathrm{\mid }F(\gamma )-F_h(\gamma )\mathrm{\mid }=\mathrm{\mid }\gamma \mathrm{\mid }\mathrm{\mid }1\mathrm{/}6-83\mathrm{/}375\mathrm{\mid }\le 41\mathrm{/}750$, with equality at $\gamma =1$. Thus the forward mesh discrepancy has the explicit bound $41\mathrm{/}750$. The scalar Galerkin state matrix is $25\mathrm{/}6>0$, and the scalar adjoint matrix is its transpose, so both equations can be solved exactly with zero algebraic residuals. By [F4], [F5] and [F7], the exact discrete reduced gradient at ${\gamma }_h$ is ${\mathrm{\Phi }}_h^{\mathrm{\prime }}({\gamma }_h)=0$. With the algorithm inputs specified in the Example, ${\gamma }_h$ is interior and $P_{[0,1]}({\gamma }_h-{\mathrm{\Phi }}_h^{\mathrm{\prime }}({\gamma }_h))={\gamma }_h$, so the projected-gradient residual is zero. The state and adjoint solves use the two solve calls, while the projection uses one of the two evaluation calls, leaving the evaluation budget unexhausted at step 4; hence the algorithm returns converged before its discrepancy check. This status certifies only discrete stationarity while the continuous model has the different minimizer proved in step 3.1; the mesh bound remains separate.