Reduced gradient and coercive Hessian for linear elliptic tracking control

For $h\in U$, let $w_h=u(q+h)-u(q)$. Subtracting the two state equations gives $a(w_h,v)=⟨Bh,v⟩$ for all $v$, so $w_h$ depends linearly and boundedly on $h$ by the state estimate in [F1]. Thus $Sh=J{w}_h$ is a bounded linear map from $U$ to $H$. The functional $h\mapsto ⟨Bh,p(q)⟩$ is bounded, so [F2] gives the unique $B^{\ast }p(q)\in U$ in the displayed definition.

Expanding the two squared norms in $\widehat{J}(q+h)$ yields $\widehat{J}(q+h)-\widehat{J}(q)=(Ju(q)-u_d,J{w}_h{)}_H+\alpha (q,h{)}_U+\frac{1}{2}\mathrm{\parallel }Sh{\mathrm{\parallel }}_H^2+\frac{\alpha }{2}\mathrm{\parallel }h{\mathrm{\parallel }}_U^2\mathrm{.}$ The adjoint equation with $v=w_h$ and the state variation equation with test $p(q)$ give $(Ju(q)-u_d,J{w}_h{)}_H=a(w_h,p(q))=⟨Bh,p(q)⟩=(B^{\ast }p(q),h{)}_U$. The last two terms are bounded by a constant times $\mathrm{\parallel }h{\mathrm{\parallel }}_U^2$, so the Fréchet derivative is $B^{\ast }p(q)+\alpha q$.

Since the state map is affine with linear part $S$, the same exact expansion at any $q$ shows the second derivative is $D^2\widehat{J}[h,k]=(Sh,Sk{)}_H+\alpha (h,k{)}_U$, represented by $S^{\ast }S+\alpha I$. Hence $((S^{\ast }S+\alpha I)h,h)=\mathrm{\parallel }Sh{\mathrm{\parallel }}_H^2+\alpha \mathrm{\parallel }h{\mathrm{\parallel }}_U^2\ge \alpha \mathrm{\parallel }h{\mathrm{\parallel }}_U^2$ and $\mathrm{\parallel }{S}^{\ast }S+\alpha I\mathrm{\parallel }\le \mathrm{\parallel }S{\mathrm{\parallel }}^2+\alpha$. The constant Hessian gives the stated Lipschitz-gradient bound; the exact quadratic expansion gives strong convexity.