proper derivation of a functional for a time dependent parameter estimation problem

Question

Following my previous question and its answer, after some reading of the advised books, I'm still confused about how to get the derivative of the functional to find the best parameter of my reaction diffusion problem.

In short, I need to find the best (time independent, space dependent) parameters $D$ and $k$ which minimize the functional

$$ \mathcal{G}(u,D,k) = \int_{\Omega} (u(t=T_f)-u_f)^2 d\Omega + \int_{\Omega} (u(t=0)-u_i)^2 d\Omega + \frac{1}{2}\int_{\Omega} D^2 d\Omega + \frac{1}{2}\int_{\Omega} k^2 d\Omega $$

subject to $$ \partial_t{u} - \nabla \cdot (D\nabla{u}) - k u (1-u) = 0 $$ $$ u(t=0,x)=u_i \space \forall x$$ $$ u(t,x)=0 \space \forall x \in \Gamma $$ where $\Gamma$ is the border of $\Omega$

Using the adjoint method, I define the Lagrangian $\cal L$ as

$$ \mathcal{L} = \mathcal{G} - \int_t\int_\Omega{\lambda( \partial_t{u} - \nabla \cdot (D\nabla{u}) - k u (1-u)) d\Omega.dt }$$

So for exemple, in order to find the optimal equations according to the parameter $D$, my understanding is that I need to specify that $$ \frac{\mathcal{L}(u,D+\epsilon\tilde D,k) - \mathcal{L}(u,D,k)}{\epsilon}=0 \space\space \forall{\tilde D} $$

which gives me (considering $u$, $D$, $k$ and $\lambda$ as independent variables) $$ \int_\Omega{D \tilde D d\Omega} + \int_t\int_\Omega{\lambda \nabla \cdot (\tilde D \nabla u) d\Omega dt} = 0 \space \forall {\tilde D}$$

Using integration by parts, I get

$$ \int_\Omega{D \tilde D} d\Omega + \int_t\int_\Omega{\nabla (\lambda\tilde D\nabla u) d\Omega dt} - \int_t\int_\Omega{\tilde D\nabla u \nabla\lambda d\Omega dt }= 0 \space \forall {\tilde D}$$

and using the divergence theorem,

$$ \int_\Omega{D \tilde D} d\Omega + \int_t\oint_\Gamma{(\lambda\tilde D\nabla u) {\vec n} d\Gamma dt} - \int_t\int_\Omega{\tilde D \nabla u \nabla\lambda d\Omega dt }= 0 \space \forall {\tilde D}$$ the goal of these transformation being to factorize the expression with $\tilde D$ in order to get a relation between $D$, $k$, $\lambda$, $u$ and their derivatives.

However, with such an expression, while the $\tilde D$ is indeed factorized, the domains of each part of the equation are different (the first is time independent, the second is a surface integral and the third is volumetric), so I can't merge them together to get a unique equation like

$$ D + \lambda\nabla u - \nabla u\nabla\lambda =0 $$

I can eventually require each part to be 0 at the same time, but then

$$ \int_\Omega{D \tilde D} d\Omega= 0 \space \forall {\tilde D}$$

leads just to $D=0$ which is obviously wrong...

Any help would be very welcome...

Answer 1

You want to solve

$$ \arg \min_{D(x),k(x)} \mathcal{G}(u;D,k) := \frac{1}{2} \int_{\Omega} \left(u(t=T) - {\bar{u}}(T) \right)^2 d\Omega + \frac{1}{2}\int_{\Omega} D^2 d\Omega + \frac{1}{2}\int_{\Omega} k^2 d\Omega $$

subject to

\begin{align} \frac{\partial u}{\partial t} - \nabla \cdot (D \nabla{u}) - k u (1-u) &= 0 & \space \mathrm{on} \space \Omega \\ u(0,x) &= u_0 & \space \forall x \in \Omega \\ u(t,x) &= 0 & \space \forall x \in \Gamma := \partial \Omega. \end{align}

Let's define an inner product on $[0,T] \times \Omega$: $$ \left< v , w \right>_{[0,T] \times \Omega} := \int_{0}^{T} \int_\Omega v \space w \mathrm{d}\Omega \mathrm{d}t.$$

With this, the Lagrangian for the constrained problem reads $$ \mathcal{L}(u;\lambda;D,k) = \mathcal{G}(u;D,k) - \left< \lambda , \frac{\partial u}{\partial t} - \nabla \cdot (D \nabla{u}) - k u (1-u) \right>_{[0,T] \times \Omega} $$

We're now ready to write out the first order necessary conditions for a local minimum (the KKT system):

1. The primal equation ${\cal L}_\lambda (u; \lambda; D,k)(\delta \lambda) = 0 \space \forall \space \delta \lambda$. This is nothing other than the "forward" PDE constraint.

2. The adjoint (or dual) equation:

   $${\cal L}_u (u; \lambda; D,k)(\delta u) = 0 \space \forall \space \delta u. $$

3. The gradient equations (the derivatives with respect to the control parameters $D$ and $k$):

\begin{align} {\cal L}_D (u; \lambda; D,k)(\delta D) &= 0 \space \forall \space \delta D \\ {\cal L}_k (u; \lambda; D,k)(\delta k) &= 0 \space \forall \space \delta k. \end{align}

Let's take the left-hand side of the adjoint equation:

\begin{align} {\cal L}_u (u; \lambda; D,k)(\delta u) &= \left( u(T,x) - {\bar{u}}(T,x) \right) \delta u(T,x) \\ & \underbrace{ - \int_{0}^{T} \int_\Omega \lambda \frac{\partial \delta u}{\partial t} \mathrm{d}\Omega \mathrm{d}t }_{\mathbf{A}} \space \underbrace{ + \int_{0}^{T} \int_\Omega \lambda \nabla \cdot (D \nabla u) \mathrm{d}\Omega \mathrm{d}t }_{\mathbf{B}} \space \underbrace{ + \int_{0}^{T} \int_\Omega \lambda k u (1-u) \mathrm{d}\Omega \mathrm{d}t }_{\mathbf{C}}. \end{align}

and look at each of the three integral terms in turn.

Integrating ($\mathbf{A}$) by parts, we get: $$ \mathbf{A} = - \int_\Omega \left. \left( \lambda \delta u \right) \right|_0^T \mathrm{d}\Omega + \int_{0}^{T} \int_\Omega \delta u \frac{\partial \lambda}{\partial t} \mathrm{d}\Omega \mathrm{d}t. $$

Similary for the second term, Green's second identity gives us $$ \require{cancel} \mathbf{B} = \int_{0}^{T} \int_\Gamma \lambda D (\nabla \delta u \cdot \vec{n}) \mathrm{d}\Gamma \mathrm{d}t - \cancelto{𝟶}{ \int_{0}^{T} \int_\Gamma \delta u D (\nabla \lambda \cdot \vec{n}) \mathrm{d}\Gamma \mathrm{d}t } + \int_{0}^{T} \int_\Omega \delta u \nabla \cdot (D \nabla \lambda) \mathrm{d}\Omega \mathrm{d}t, $$ where the second term vanishes because $\delta u(t,x) = 0$ on $\Gamma$.

Putting all the terms together, and requiring that all terms vanish for all $\delta u$, we arrive at the following adjoint problem:

\begin{align} \frac{\partial \lambda}{\partial t} + \nabla \cdot (D \nabla \lambda) + k(1-2u) & = 0 & \mathrm{on} \space \Omega \\ \lambda(T,x) &= u(T,x) - \bar{u}(T,x) & \forall x \in \Omega \\ \lambda(t,x) &= 0 & \forall x \in \Gamma, \space \forall t. \end{align}

Finally, on to the gradient equations. For $D$, this reads: $$ \int_\Omega D \delta D \mathrm{d}\Omega - \int_{0}^{T} \int_\Omega \lambda \nabla \cdot (\delta D \nabla u) \mathrm{d}\Omega \mathrm{d}t = 0, \space \forall \delta D. $$

Again, the integration by parts formula comes in handy here. After some algebra, this is what we're left with: $$ \int_\Omega \delta D D \mathrm{d}\Omega + \int_0^T \int_\Omega \delta D \left( \nabla u \cdot \nabla \lambda \right) = 0 \space \forall \space \delta D, $$

from which the equation for $D$ immediately follows:

$$ D(x) = \int_0^T \nabla u(x,t) \cdot \nabla \lambda(x,t)\mathrm{d}t. $$

The "correct" way to interpret this equation depends on the particular choice of function spaces for $u$, $\lambda$ and $D$. See this post for an example of why this is really important.

I'll let you derive a similar equation for $k$.