An unchecked under-solved adjoint can reverse the reduced gradient

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.