I'm trying to follow the derivation of the adjoint Poisson equation provided in this video, but I'm getting tripped up on some of the steps that are skipped. Here is my derivation in full. Questions at the end.

Consider the minimization problem

\begin{align} \min_a J(u, a, \vec{x}) = \int_{\Omega} c u(a, \vec{x}) \ dV \end{align}

over the domain $\Omega$ bounded by $\partial \Omega$ subject to the PDE constraint (Poisson's equation) given by

\begin{align} \nabla \cdot (a(\vec{x}) \nabla u(\vec{x})) - f(\vec{x}) = 0 \end{align}

where $a(\vec{x})$ is the design variable and $f(\vec{x})$ is prescribed.

To approach this problem, we are looking to find the sensitivity of $J$ with respect to $a$ or $\frac{\delta J}{\delta a}$. To get this quantity, use the method of Lagrange multipliers to add the constraint to the objective function

\begin{align} J = \int_{\Omega} c u \ dV + \int_{\Omega}\lambda \left[\nabla \cdot (a \nabla u) - f \right] dV \end{align}

Note that I'm dropping the function arguments to be more concise. Take the variation of $J$

\begin{align} \delta J = \int_{\Omega} c (\delta u) \ dV + \int_{\Omega}\lambda\left[\nabla \cdot (\delta a \nabla u) + \nabla \cdot (a \nabla \delta u)\right] dV \end{align}

Now integrate the following term by parts twice \begin{align} \int_{\Omega} \lambda \nabla \cdot (a \nabla \delta u) &= \int_{\partial \Omega} \lambda a (\nabla \delta u) \cdot d\vec{S} - \int_{\Omega} (a \nabla \delta u) \cdot \nabla \lambda dV \\ &= \int_{\partial \Omega} \lambda a (\nabla \delta u) \cdot d\vec{S} - \int_{\partial \Omega} \delta u (a \nabla \lambda) \cdot d\vec{S} + \int_{\Omega} \delta u (\nabla \cdot (a \nabla \lambda )) dV \end{align}

Recombining this term with the expression for $\delta J$ we get

\begin{align} \delta J = \int_{\Omega} \left(c + \nabla \cdot (a \nabla \lambda ) \right)\delta u \ dV + \int_{\Omega}\lambda\nabla \cdot (\delta a \nabla u) dV + \int_{\partial \Omega} \lambda a (\nabla \delta u) \cdot d\vec{S} - \int_{\partial \Omega} \delta u (a \nabla \lambda) \cdot d\vec{S} \end{align}

Then since we are looking for a relationship between $\delta J$ and $\delta a$ we can choose $\lambda$ such that the first term goes to zero. Since $\delta u$ is not, in general, $0$ then we must have \begin{align} c + \nabla \cdot (a \nabla \lambda) = 0 \end{align} which is our adjoint PDE (also Poisson's equation) which can be solved for $\lambda$.

Now integrate the second term by parts and we get

\begin{align} \delta J = - \int_{\Omega} \delta a \nabla u \cdot \nabla \lambda dV + \int_{\partial \Omega} \lambda \delta a \nabla u \cdot d\vec{S} + \int_{\partial \Omega} \lambda a (\nabla \delta u) \cdot d\vec{S} - \int_{\partial \Omega} \delta u (a \nabla \lambda) \cdot d\vec{S} \end{align}

But here is where my understanding breaks down. In the video, it is declared that

\begin{align} \frac{\delta J}{\delta a} = \nabla \lambda \cdot \nabla u \end{align}

So I have the following questions:

  1. What happens to the boundary terms? Can we assume that all the variations are $0$ at the boundaries? Are their gradients also $0$? How do the boundary conditions of the forward problem factor in here?
  2. How do we pull $\delta a$ out of the integral? and what happened to the minus sign?
  3. What are the boundary conditions of the adjoint PDE?

Any tips or resources would be greatly appreciated!

  • $\begingroup$ Note that u=0 on the boundary. $\endgroup$ – Brian Borchers Jan 29 '18 at 4:25

I was watching the same video and had the same questions. He explicitly says that we have $\lambda = 0$ on the boundaries in order to get rid of the first surface integral. (We are free to chose the boundary conditions for $\lambda$)

Since we apply a Dirichlet boundary condition to $u$, the variation $\delta u$ is zero on the boundaries since $u$ is fixed. Which removes the other surface integral too.

The inegral relation below is like a global view, it gives you the change in the objective functional $J$ when you perturb the function $\delta a$. Note that $\delta a$ is a function of x.

$\delta J = -\int_{\Omega} \delta a \nabla \phi \nabla u dx$

It makes intuitively sense to go from the integral relation to the differential relation below. Since the integral change is like a large sum of local changes, the differential relation is like a local view. If you change $\delta a(x)$ locally, it tells how this affects $J$ locally.

$\frac{\partial J}{\partial a} = -\nabla \phi \nabla u$

(Ans yes, he forgot the minus in his derivation)


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