Theoretical Derivations
We want to solve the PDE-constrained shape optimization problem
\[
\min_{\Omega \subset \mathsf{D}} J(u)
:= \int_\Omega | \nabla u_\Omega - \nabla u_{ref}|^2 \, dx
\]
subject to \((\Omega,u_\Omega)\) satisfying
\[
\int_\Omega \nabla u_\Omega \cdot \nabla v + u_\Omega v \, dx
=
\int_\Omega f v \, dx
\quad \text{for all } v \in H_0^1(\Omega),
\]
where \(\Omega \subset \mathbb{R}^2\) and
\(u_d, f \in H^2(D)\) and \(\Omega\) belongs to
\[
\mathcal{A}(D) = \{\Omega \subset D: \Omega \;\text{is measurable and smooth} \}
\]
Exercise 1
Let \(\Omega\in \mathcal{A}(D)\) and assume \(f, u_{ref} \in C^2(D)\). We define \(\Theta :=C^1_c(D)^d\) and
\[
\mathcal{F}(\Theta) := \{I+X: \; X \in \Theta, \; I+X \text{ is bijective, } (I+X)^{-1}\in \Theta \},
\]
where \(I\) denotes the identity. We define the distance to the identity as follows:
\[
d(I,F):= ||F-I||_\Theta + ||F^{-1}-I||_\Theta, \quad \text{for all } F \in \mathcal{F}(\Theta).
\]
Show that the map \(\bar J: \; \mathcal{F}(\Theta) \to \mathbb{R}, \; F \mapsto J(F(\Omega))\) is continuous at the identity with respect to \(d\).
We want to show: For any sequence \(I_n \in \Theta\) holds: \(d(I, I_n)\xrightarrow{n \to \infty}0 \implies |J(\Omega)-J(I_n(\Omega))|\xrightarrow{n \to \infty} 0\). So we choose \(I_n\) with \(d(I, I_n)\xrightarrow{n \to \infty}0\), which means \(||I-I_n||_\Theta, \; ||I-I^{-1}||_\Theta \xrightarrow{n \to \infty}0\). Therefore we also know \(\Omega_n := I_n(\Omega) \xrightarrow{n \to \infty}\Omega.\)
We have
\[\begin{split}
\bar J (I) = J(\Omega)= \int_{\Omega} |\nabla u_\Omega - \nabla u_{ref}|^2 dx \\
\bar J (I_n) = J(\Omega_n)= \int_{\Omega_n} |\nabla u_{\Omega_n} - \nabla u_{ref}|^2 dx,
\end{split}\]
where \(u_\Omega\) solves
\begin{equation}
\int_{\Omega} \nabla u_\Omega \cdot \nabla \varphi + u_\Omega \varphi dx = \int_{\Omega} f \varphi dx \quad \forall \varphi \in H_0^1(\Omega)
\end{equation}
and \(u_{\Omega_n}\) solves
\[
\int_{\Omega_n} \nabla u_{\Omega_n} \cdot \nabla \varphi + u_{\Omega_n} \varphi dx = \int_{\Omega_n} f \varphi dx \quad \forall \varphi \in H_0^1(\Omega_n).
\]
We want to show that \(u_n:=u_{\Omega_n} \circ I_n|_\Omega\) converges to \(u_\Omega\) with \(n \to \infty\):
Per definition is \(u_n\) a function in \(H_0^1(\Omega)\) for all \(n\) and from the chain rule follows:
\[
\nabla_{y} u_{\Omega_n}(y)=(\nabla_{x}I_n)^{-\top}\cdot \nabla_{x} u_n(x)
\]
for \(y:=I_n(x)\). Using this we get
\[\begin{split}
\int_{\Omega_n} |\nabla u_{\Omega_n} - \nabla u_{ref}|^2 dx =\\
\int_{\Omega}|\text{det }d I_n|\cdot [(\nabla I_n)^{-\top} \nabla u_n) \cdot (\nabla I_n)^{-\top} (\nabla \varphi) \circ I_n) + u_n \varphi \circ I_n] dy
\end{split}\]
for all \(\varphi \in H_0^1(\Omega_n)\). So the weak formulation for \(u_n\) becomes
\[\begin{split}
\int_{\Omega}\underset{\to 1 \text{ uniformly}}{\underbrace{|\text{det }d I_n|}}\cdot [\underset{\to I \text{ uniformly}}{\underbrace{(\nabla I_n)^{-\top}}} \nabla u_n \cdot \nabla \varphi + u_n \varphi] dy=\\
\int_{\Omega}\underset{\to 1 \text{ uniformly}}{\underbrace{|\text{det }d I_n|}}\cdot \underset{\to f \text{ uniformly}}{\underbrace{f \circ I_n}} \varphi dy \quad \forall \varphi \in H_0^1(\Omega)
\end{split}\]
So taking the limit we get
\[
\int_{\Omega_n} \nabla \text{lim}_n u_{n} \cdot \nabla \varphi + \text{lim}_n u_{n} \varphi dx = \int_{\Omega_n} f \varphi dx \quad \forall \varphi \in H_0^1(\Omega).
\]
and from (1) we get \(u_n \xrightarrow{n \to \infty} u_\Omega\) and \(\nabla u_{\Omega_n} \xrightarrow{n \to \infty} \nabla u_\Omega\). With that we see
\[\begin{split}
J(I_n(\Omega))=\int_{\Omega_n}|\nabla u_{\Omega_n}-\nabla u_{ref}|^2 dx = \int_{\Omega} |\text{det }d I_n| \cdot |(\nabla u_{\Omega_n}) \circ I_n - (\nabla u_{ref}) \circ I_n|^2 dx \\
\xrightarrow{n \to \infty} \int_{\Omega} |\nabla u_\Omega - \nabla u_{ref}|^2 dx = J(\Omega)
\end{split}\]
Exercise 2
Let \(F_t(x) := x + t X(x) \text{ for } t > 0 \text{ small},\) where \(X \in C_c^1(D)^d\).
We define
$\(
\xi(t) := \det(\nabla F_t), \qquad
\psi(t) := \nabla F_t^{-1} \nabla F_t^{-\top}.
\)$
Show that
(i) \(\quad \displaystyle \left. \partial_t \xi(t) \right|_{t=0} = \operatorname{div}(X)\)
(ii) \(\quad \displaystyle \left. \partial_t \psi(t) \right|_{t=0} = -(\nabla X + \nabla X^\top)\)
(i)
We first observe that \(\nabla F_t = I + t \nabla X.\)
Let \(A(t) := \nabla F_t\). Using the identity
$\(
\frac{d}{dt} \det A(t)
= \det A(t) \cdot \mathrm{tr}\!\left(A(t)^{-1} \frac{dA(t)}{dt}\right),
\)\(
we obtain
\)\(
\partial_t \xi(t)
= \det(\nabla F_t)\cdot
\mathrm{tr}\!\left((\nabla F_t)^{-1} \partial_t \nabla F_t\right).
\)$
Since \(\partial_t \nabla F_t = \nabla X\), we get
$\(
\partial_t \xi(0)
= \det(I)\, \mathrm{tr}(I \nabla X)
= \mathrm{tr}(\nabla X)
= \operatorname{div}(X).
\)$
(ii)
We use the identity
$\(
\frac{d}{dt} A^{-1}(t)
= -A^{-1}(t)\, \frac{dA(t)}{dt}\, A^{-1}(t).
\)$
Thus,
$\(
\partial_t \nabla F_t^{-1}
= -\nabla F_t^{-1} (\frac{d F_t(t)}{dt}) \nabla F_t^{-1} = -(I+t \nabla X)^{-1} \nabla X (I+ t \nabla X)^{-1}.
\)$
Evaluating at \(t = 0\) yields
$\(
\partial_t \nabla F_t^{-1}\big|_{t=0} = -\nabla X.
\)$
Now,
$$
\partial_t \psi(t)
= (\partial_t \nabla F_t^{-1}) \nabla F_t^{-\top}
At \(t = 0\) we have \(\nabla F_0^{-1} = I\) and
$\(
\partial_t \nabla F_t^{-\top}\big|_{t=0} = -(\nabla X)^\top.
\)$
Therefore,
$\(
\partial_t \psi(0)
= -\nabla X - (\nabla X)^\top.
\)$
Exercise 3
Let \(D \subset \mathbb{R}^d\) and \(F_t(x) := x + tX(x)\) for \(t > 0\) small, where \(X \in C_c^1(D)^d\). Let \(f^t := f \circ F_t\), where \(f \in C^2(D)\). Show that:
(i) \(\quad f_t \to f\) in \(L_2(D)\) for \(t \searrow 0\).
(ii) \(\quad \frac{f_t - f}{t} \to \nabla f \cdot X\) in \(L_2(D)\) for \(t \searrow 0\).
Argue by density properties that the previous convergences hold for \(f \in H^1(D)\).
(i)
First we observe that \((f(x + tX(x)) - f(x))|_{(\text{supp}X)^c}=0\). Using this we see
\[
\|f_t - f\|_{L_2(D)}^2 = \int_{\text{supp } X} (f(x + tX(x)) - f(x))^2 \, dx
\]
As \(t \to 0\), \(f(x + tX(x)) \to f(x)\) pointwise due to the continuity of \(f\). Because \(X\) has compact support, it follows from the Dominated Convergence Theorem:
\[
\lim_{t \to 0} \int_{\text{supp } X} |f(x + tX(x)) - f(x)|^2 dx = \int_{\text{supp } X} |f(x)-f(x)|^2 dx = 0
\]
(ii)
Using Taylor’s expansion for \(f(F_t(x))\):
\[
f(F_t(x)) = f(x) + t \nabla f(x) \cdot X(x) + \frac{t^2}{2} \sum_{|\alpha|=2} \frac{\partial^\alpha f}{\partial x^\alpha}(\xi) (X(x))^\alpha
\]
Rearranging for the difference quotient:
\[
\frac{f_t - f}{t} - \nabla f \cdot X = \frac{t}{2} \sum_{|\alpha|=2} \frac{\partial^\alpha f}{\partial x^\alpha}(\xi) (X(x))^\alpha
\]
Since \(f \in C^2\) and \(X\) is compactly supported, the second-order term is bounded. Thus:
\[
\lim_{t \to 0} \left\| \frac{f_t - f}{t} - \nabla f \cdot X \right\|_{L_2} = \lim_{t \to 0} \frac{t}{2} \left\| \sum_{|\alpha|=2} \frac{\partial^\alpha f}{\partial x^\alpha}(\xi) X^\alpha \right\|_{L_2} = 0
\]
(i) Extension to \(f \in H^1(D)\)
We want to show that \(f \mapsto f^t\) is bounded in \(L_2\). We observe that \(\text{det }(dF_t^{-1}) \xrightarrow{t \to 0}1\) uniformly. So we can choose \(t\) small enough such that \(det(dF_t^{-1}) \leq 2\). \(F_t\) is also a diffeomorphism for \(t\) small enough.
\[
||f^t||_{L_2}^2 = \int_{D}(f \circ F_t)^2 dx = \int_{F_t(D)} |\text{det } dF_t^{-1}| \cdot f^2 dy \leq 2 ||f||_{L_2}^2
\]
Choose a sequence \((f_n) \subset C^\infty(D)\) such that \(\|f_n - f\|_{H^1(D)} \to 0\).
For any \(\epsilon > 0\), we decompose the error:
\[
||f^t-f||_{L_2} \leq ||f^t-f^t_n||_{L_2} + ||f^t_n-f_n||_{L_2} + ||f_n-f||_{L_2}
\]
We just showed that the first term is \(< C \epsilon\).
The second term \(\to 0\) as \(t \to 0\) since \(f_n\) is smooth.
The third term is \(< \epsilon\) by the \(H^1\) convergence of \(f_n\).
Since \(\epsilon\) is arbitrary, the result holds for all \(f \in H^1(D)\).
(ii) Extension to \(f \in H^1(D)\)
We define \(T_t f = \frac{f^t - f}{t}\) and show it is bounded in \(H^1\). By the Fundamental Theorem of Calculus:
\[
f(x + tX(x)) - f(x) = \int_0^1 \nabla f(x + stX(x)) \cdot tX(x) ds
\]
Since \(X \in C^1_c(D)^d\) we know that there exists an upper bound \(M\) for \(X\). So we obtain
\[
\|T_t f\|_{L_2(D)} \leq M \int_0^1 \|\nabla f(x + stX(x))\|_{L_2(D)} ds
\]
Using the change of variables \(\Phi_s(x) = x + stX(x)\), which is a diffeomorphism for small \(t\). We choose \(t\) small enough such that \(|det(d\Phi_s^{-1})| \leq 2\). We can do that because \(|det(d\Phi_s^{-1})| = |det(I)| = 1\) for \(t=0\). We also observe that \(\Phi_s(D) \subseteq D\) holds.
\[
\int_D ||\nabla f(\Phi_s(x))||^2 dx = \int_{\Phi_s(D)} |det(d\Phi_s^{-1})| \cdot||\nabla f(y)||^2 dx \leq 2 \int_D ||\nabla f(y)||^2 dy
\]
This gives the uniform bound independent of \(t\):
$\(
\|T_t f\|_{L_2(D)} \leq \sqrt{2} M \|\nabla f\|_{L_2(D)}
\)$
Choose a sequence \((f_n) \subset C^\infty(D)\) such that \(\|f_n - f\|_{H^1(D)} \to 0\).
For any \(\epsilon > 0\), we decompose the error:
\[
\left\| \frac{f^t - f}{t} - \nabla f \cdot X \right\|_{L_2} \leq \left\| T_t(f - f_n) \right\|_{L_2} + \left\| T_t f_n - \nabla f_n \cdot X \right\|_{L_2} + \| (\nabla f_n - \nabla f) \cdot X \|_{L_2}
\]
We just showed that the first term is \(< C \epsilon\).
The second term \(\to 0\) as \(t \to 0\) since \(f_n\) is smooth.
The third term is \(< M \epsilon\) by the \(H^1\) convergence of \(f_n\).
Since \(\epsilon\) is arbitrary, the result holds for all \(f \in H^1(D)\).
Exercise 4
Use the direct method to compute the derivative
\[
DJ(\Omega)(X) = \lim _{t \downarrow 0} \frac{J(\Omega_t) - J(\Omega)}{t}
\]
Considering the perturbed state equation
\[
\int_{\Omega_t} \nabla u_{\Omega_t} \cdot \nabla \varphi
+ u_{\Omega_t}\,\varphi
=
\int_{\Omega_t} f \varphi
\qquad \forall \varphi \in H^1_0(\Omega_t),
\]
a change of variables with \(F_t(x) = y\) yields
\[
\int_\Omega A(t)\nabla u_t \cdot \nabla \varphi_t
+ \int_\Omega \xi(t)u_t \varphi_t
=
\int_\Omega \xi(t) f_t \varphi_t.
\]
Here,
\[\begin{split}
\begin{aligned}
A(t) &= \det(\partial F_t)\, \partial F_t^{-1}\partial F_t^{-\top}, \\
\xi(t) &= \det(\partial F_t), \\
u_t &= u_{\Omega_t} \circ F_t, \\
\varphi_t &= \varphi \circ F_t, \\
f_t &= f \circ F_t.
\end{aligned}
\end{split}\]
Using the identity
\[
\nabla v \circ F_t = \partial F_t^{-\top}\nabla (v \circ F_t),
\]
and replacing the test function (space) accordingly, we obtain
\[
\int_\Omega A(t)\nabla u_t \cdot \nabla \psi
+ \int_\Omega \xi(t)u_t \psi
=
\int_\Omega \xi(t) f_t \psi
\qquad \forall \psi \in H^1_0(\Omega).
\]
Differentiating with respect to \(t\) and evaluating at \(t=0\) yields
\[
\int_\Omega A'(0)\nabla u \cdot \nabla \psi
+ \int_\Omega \nabla \dot{u} \cdot \nabla \psi
+ \int_\Omega \operatorname{div}(X) u \psi
+ \int_\Omega \dot{u}\,\psi
=
\int_\Omega \operatorname{div}(X)f\psi
+ \int_\Omega \nabla f \cdot X\,\psi.
\]
With
\[
A'(0) = \operatorname{div}(X)\,I - \partial X - \partial X^\top,
\]
we obtain
\[\begin{split}
\begin{aligned}
\int_\Omega \nabla \dot{u} \cdot \nabla \psi
+ \int_\Omega \dot{u}\,\psi
&=
- \int_\Omega \operatorname{div}(X) u \psi \\
&\quad
- \int_\Omega
\bigl(\operatorname{div}(X)I - \partial X - \partial X^\top\bigr)
\nabla u \cdot \nabla \psi \\
&\quad
+ \int_\Omega (\nabla f \cdot X)\,\psi \\
&\quad + \int_\Omega \operatorname{div}(X)f\psi.
\end{aligned}
\tag{2}
\end{split}\]
Here, \(\dot{u}\) denotes the material derivative of \(u\).
Following the approach of Lemma 4.217 (adapted to our setting), we obtain
$$
\dot{u} \in H^1_0(\Omega).
$$
Shape derivative of cost functional
We differentiate \(t \mapsto J(\Omega_t)\):
\[\begin{split}
\begin{aligned}
DJ(\Omega)(X)
&= \left.\frac{d}{dt} J(\Omega_t)\right|_{t=0} \\
&= \left.\frac{d}{dt} \int_\Omega
\det(\partial F_t)\,|\nabla u_t - \nabla u_{\mathrm{ref}}|^2 \right|_{t=0} \\
&= \int_\Omega \operatorname{div}(X)|\nabla u - \nabla u_{\mathrm{ref}}|^2
+ 2 \int_\Omega \nabla \dot{u} \cdot (\nabla u - \nabla u_{\mathrm{ref}}) \\
&\quad
- 2 \int_\Omega (\nabla u - \nabla u_{\mathrm{ref}}) \nabla^2 u_{\mathrm{ref}} \cdot X.
\end{aligned}
\tag{3}
\end{split}\]
The adjoint equation
To eliminate the explicit occurrence of the material derivative \(\dot{u}\),
we introduce the Lagrangian
\[
\mathcal{L} :
\mathbb{R} \times H^1_0(\Omega) \times H^1_0(\Omega)
\rightarrow \mathbb{R},
\]
defined by
\[
\mathcal{L}(t,u,p)
=
\int_\Omega A(t)\nabla u \cdot \nabla p
+ \int_\Omega \xi(t) u p
- \int_\Omega \xi(t)f_t p
+ \int_\Omega (A(t)(\nabla u - \nabla u_{\mathrm{ref},t})) \cdot (\nabla u - \nabla u_{\mathrm{ref},t}).
\]
The formal Lagrangian approach (see Tröltzsch) suggests that the adjoint
\(p\) satisfies
\[
\partial_u \mathcal{L}(0,u,p)(\psi) = 0
\qquad \forall \psi \in H^1_0(\Omega).
\]
Computing the derivative yields
\[
\partial_u \mathcal{L}(0,u,p)(\psi)
=
\int_\Omega \nabla \psi \cdot \nabla p
+ \int_\Omega \psi p
+ 2 \int_\Omega (\nabla u - \nabla u_{\mathrm{ref}}) \cdot \nabla\psi.
\]
Adjoint problem
We define the adjoint solution \(p \in H^1_0(\Omega)\) by
\[
\int_\Omega \nabla \psi \cdot \nabla p
+ \int_\Omega \psi p
=
-2 \int_\Omega (\nabla u - \nabla u_{\mathrm{ref}}) \cdot \nabla\psi
\qquad \forall \psi \in H^1_0(\Omega).
\tag{4}
\]
Choosing \(\psi = \dot{u}\) in (4) gives
\[
\int_\Omega \nabla \dot{u} \cdot \nabla p
+ \int_\Omega \dot{u} p
=
-2 \int_\Omega (\nabla u - \nabla u_{\mathrm{ref}}) \cdot \nabla\dot u.
\]
On the other hand, choosing \(\psi = p\) in (2) yields
\[\begin{split}
\begin{aligned}
\int_\Omega \nabla \dot{u} \cdot \nabla p
+ \int_\Omega \dot{u}\,p
&=
- \int_\Omega \operatorname{div}(X) u p \\
&\quad
- \int_\Omega
\bigl(\operatorname{div}(X)I - \partial X - \partial X^\top\bigr)
\nabla u \cdot \nabla p \\
&\quad
+ \int_\Omega (\nabla f \cdot X)\,p \\
&\quad + \int_\Omega \operatorname{div}(X)fp.
\end{aligned}
\end{split}\]
Combining both identities yields
\[\begin{split}
\begin{aligned}
2 \int_\Omega (\nabla u - \nabla u_{\mathrm{ref}}) \cdot \nabla\dot u
&=
\int_\Omega \operatorname{div}(X) u p \\
&\quad
+ \int_\Omega
\bigl(\operatorname{div}(X)I - \partial X - \partial X^\top\bigr)
\nabla u \cdot \nabla p \\
&\quad
- \int_\Omega (\nabla f \cdot X)\,p \\
& \quad
- \int_\Omega \operatorname{div}(X)f p
\end{aligned}
\tag{5}
\end{split}\]
Final expression for the shape derivative
Inserting (5) into (3) yields
\[\begin{split}
\begin{aligned}
DJ(\Omega)(X)
&=
\int_\Omega \operatorname{div}(X)|\nabla u - \nabla u_{\mathrm{ref}}|^2
- 2 \int_\Omega (\nabla u - \nabla u_{\mathrm{ref}}) \nabla^2 u_{\mathrm{ref}} \cdot X\\
&\quad
+ \int_\Omega \operatorname{div}(X) u p \\
&\quad
+ \int_\Omega
\bigl(\operatorname{div}(X)I - \partial X - \partial X^\top\bigr)
\nabla u \cdot \nabla p \\
&\quad
- \int_\Omega (\nabla f \cdot X)\,p\\
& \quad
- \int_\Omega \operatorname{div}(X)fp .
\end{aligned}
\end{split}\]
with \(u = u_0\).
Exercise 5.
Use the average adjoint method to compute the derivate
\[
DJ(\Omega)(X) = \lim _{t \downarrow 0} \frac{J(\Omega_t) - J(\Omega)}{t}
\]
Considering the parametrised Lagrangian
\[
\mathcal{L}(t,u,p)
=
\int_\Omega A(t)\nabla u \cdot \nabla p
+ \int_\Omega \xi(t) u p
- \int_\Omega \xi(t)f_t p
+ \int_\Omega (A(t)(\nabla u - \nabla u_{\mathrm{ref},t})) \cdot (\nabla u - \nabla u_{\mathrm{ref},t}).
\]
we define the average adjoint \(p^t \in H^1_0(\Omega)\) as the solution to
\[
\int_0^1 \partial _u \mathcal{L}(t, s u_t + (1-s)u_0, p^t)(v) ds = 0 \quad \forall v \in H^1_0(\Omega)
\]
Similar to Lemma 3.63 we get
\[
\mathcal{L}(t, u_t, 0) - \mathcal{L}(0, u_0, 0) = \mathcal{L}(t, u_0, p^t) - \mathcal{L}(0, u_0, p^t)
\]
Therefore we have
\[
DJ(\Omega)(X) = \lim _ {t \downarrow 0} \frac{J(\Omega_t) - J(\Omega)}{t} = \lim _ {t \downarrow 0} \frac{\mathcal{L}(t, y^t, 0)- \mathcal{L}(0, y^0, 0)} =\lim _{t \downarrow 0} \frac{\mathcal{L}(t, u_0, p^t) - \mathcal{L}(0, u_0, p^t)}{t}
\]
Since
\[\begin{split}
\begin{aligned}
\frac{\mathcal{L}(t, u_0, p^t) - \mathcal{L}(0, u_0, p^t)}{t} &= \int _\Omega \frac{A(t) - \operatorname{I}}{t} \nabla u_0 \cdot \nabla p^t \\
&+ \int _\Omega \frac{\xi(t)- 1}{t} u_0 p^t \\
&- \int _\Omega \frac{\xi(t)f_t-1 \cdot f_0}{t} p^t \\
&+ \int_\Omega \frac{(A(t)(\nabla u_0 - \nabla u_{\mathrm{ref},t})) \cdot (\nabla u_0 - \nabla u_{\mathrm{ref},t}) - (\nabla u_0 - u_{ref}) \cdot (\nabla u_0 - u_{ref})}{t}
\end{aligned}
\end{split}\]
Therefore
\[ \begin{align}\begin{aligned}\begin{split}
\begin{aligned}
DJ(\Omega)(X) &= \int _\Omega A'(0) \nabla u_0 \cdot \nabla p ^0 \\
& + \int _\Omega \xi'(0) u_0 p^0 \\
& - \int _\Omega \partial _t f_t\bigg |_{t=0} \xi(0) p ^0\\
& - \int _\Omega \xi'(0) f p^0 \\
& + \int _\Omega \frac{d}{dt}(A(t)(\nabla u_0 - \nabla u_{\mathrm{ref},t})) \cdot (\nabla u_0 - \nabla u_{\mathrm{ref},t}) \bigg |_{t=0}\end{split}\\
\end{aligned}
\end{aligned}\end{align} \]
with exercise 2 we get
\[\begin{split}
\begin{aligned}
DJ(\Omega)(X)
&=
\int_\Omega (\operatorname{div}(X))|\nabla u - \nabla u_{\mathrm{ref}}|^2
- 2 \int_\Omega (\nabla u - \nabla u_{\mathrm{ref}}) \nabla^2 u_{\mathrm{ref}} \cdot X\\
&\quad
+ \int_\Omega \operatorname{div}(X) u p \\
&\quad
+ \int_\Omega
\bigl(\operatorname{div}(X)I - \partial X - \partial X^\top\bigr)
\nabla u \cdot \nabla p \\
&\quad
- \int_\Omega (\nabla f \cdot X)\,p\\
& \quad
- \int_\Omega \operatorname{div}(X)fp .
\end{aligned}
\end{split}\]
with \(p^0 = p, u_0 = u\).
Exercise 9b.
find X by solving
\[
\min _{X \in H^1(\Omega)^2} \frac{1}{p} \int _D |\nabla X | ^p dx - DJ(\Omega) (X)
\]
This p -Laplacian can be rewritten by the formulation
\[
\int _D |\nabla(X)| ^{p-2} \nabla X :\nabla Y dx - DJ(\Omega)(Y) = 0 \qquad \forall Y \in H^1_0(D)^2
\]
which is a nonlinear equation in X and requires therefore something like Newton’s method.
