Second order differentiability of solution operator to nonlinear boundary value problem

Question

I am currently reading the paper [1]. The paper deals with the BVP: \begin{align*} \partial_t u - \Delta u + u(u-a) & = -qu \text{ in } ]0,T\mathclose[ \times \Omega \\[6pt] \partial_{n}u & = 0 \text{ in } ]0,T\mathclose[ \times \partial \Omega\\[6pt] u(0) & = u_0 \text{ in } \{0\} \times \Omega \end{align*}

where $u$ is the state, $q$ is the control and $u=S(q)$ the sol. operator.

To derive second-order optimality conditions, the authors calculate second-order derivatives $S''(q)[p,p]$, as the solution $v_{pp}$ to the BVP: \begin{align*} \partial_{t} v_{pp} - \Delta v_{pp} + (2u + q - a)\cdot v_{pp} & = -2v_p(v_p+p)\text{ in } \Omega_T \\[6pt] \partial_n v_{pp} & = 0 \text{ in } \Sigma_T \\[6pt] v_{pp}(0) & = 0 \text{ in } \Omega \end{align*} (Here $v_{p}=S'(q)(p)$)

And they reference [2]. Unfortunately, in [2], Thm. 5.16. the authors don't deal with mixed terms, like $-q\cdot u$. They consider only problems of the type: \begin{align*} \partial_t u - \Delta u +d(x,t,u) & = q \text{ in } \Omega_T\\[6pt] \partial_n u & = 0 \text{ in } \Sigma_T \\[6pt] u(0) & = u_0 \text{ in } \Omega \end{align*} That's why this is not directly applicable. Also the form of the BVP does not match the result of Thm. 5.16. So I am having the feeling that something else was used. I would be very grateful for any help or Ansatz to handle this.

Thanks in advance !

Reference

[1] Malte Braack, Martin F. Quaas, Benjamin Tews, Boris Vexler, "Optimization of fishing strategies in space and time as a non-convex optimal control problem", (English) Journal of Optimization Theory and Applications 178, No. 3, 950-972 (2018), DOI:10.1007/s10957-018-1304-7, MR3836262, Zbl 1409.49018.

[2] F. Troeltzsch. Optimale Steuerung partieller Differentialgleichungen. Vieweg-Teubner, 2. ed., 2009

Answer 1

This is a brief crash course in abstract optimal control theory, see e.g. [HPUU].

Write the original PDE as $e(u,q) = 0$ in appropriate function spaces. You have a solution/control-to-state operator $S \colon q \mapsto u$ such that $e(S(q),q) = 0$ for all $q$. Its derivative is given by (implicit function theorem) $$S'(q)p = - e_u(S(q),q)^{-1} e_q(S(q),q)p$$ where $e_u(\dots)^{-1}$ corresponds to the solution operator for the linearized equation whose differential operator corresponds to the left-hand side in the second equation in OP.

In order to determine $S''(q)[p,p]$, well, you do the unpleasant thing and differentiate the foregoing expression for $S'(q)$ once more. This should give you, with $v_p = S'(q)p$ and $y = (S(q),q)$, $$S''(q)[p,p] = -e_u(y)^{-1}\Bigl[e_{uu}(y)[v_p,v_p] + e_{uq}(y)[p,v_p] + e_{qu}(y)[v_p,p] + e_{qq}(y)[p,p]\Bigr].$$ Use that the derivative of the inverse $f(A) = A^{-1}$ of an operator is given by $f'(A)H = -A^{-1}HA^{-1}$.

Now, for the particular PDE, $e_{qq} = 0$, and $e_{uq}(y)[p,v_p] = e_{qu}(y)[v_p,p] = pv_p$ by symmetry of second derivatives and $e_{uu}(y)[v_p,v_p] = 2v_p^2$. Hence $$S''(q)[p,p] = e_u(y)^{-1}\bigl[-2v_p^2 - 2v_p p\bigr],$$ that is, $S''(q)[p,p]$ is given by the solution of the linearized state equation with right-hand side $$-2v_p^2 - 2v_p p = -2v_p(v_p + p),$$ as claimed.

Note that the initial value for the sensitivity equation which defines $S'(q)p$ becomes zero because there is no control in the initial value in the original problem. However, while the calculation above gives you the correct terms, it is clearly not extremely rigorous in the definition of $e$ and the calculation of its derivatives (because formally there should be a second component for the initial value in $e$, and well, there are no function spaces...), but it can be done.

[HPUU] Hinze, M.; Pinnau, R.; Ulbrich, M.; Ulbrich, S., Optimization with PDE constraints, Mathematical Modelling: Theory and Applications 23. Dordrecht: Springer xi, 270 p. (2009). ZBL1167.49001.