Adjoint

The adjoint solver is nothing but the transpose of the solver. Suppose we have a equation as

$P(\alpha )u-f=0$

where $P(\alpha )$ is a differential operator depending on $\alpha$. Then it is easy to invert it to solve our solution $u=u(\alpha ,f)$.

However, what if we want something else, like a functional of our interest $J(u)$. And usually this is a functional which we want to minimize, then common optimization methods will need the gradient.

$\frac{\partial J(u)}{\partial \alpha }=\frac{\partial J(u)}{\partial u}\frac{\partial u}{\partial \alpha }$

where $\frac{\partial u}{\partial \alpha }$ can be deducted from the explicit solution (if we have), otherwise, we are getting another equation:

$\frac{\partial P(\alpha )u}{\partial \alpha }=P\frac{\partial u}{\partial \alpha }+\frac{\partial P}{\partial \alpha }u=0$

which means

$\frac{\partial J}{\partial \alpha }=-\frac{\partial J}{\partial u}{P}^{-1}\frac{\partial P}{\partial \alpha }u$

transpose it,

${\frac{\partial J}{\partial \alpha }}^{\ast }=-u^{\ast }\frac{\partial P}{\partial \alpha }\underline{P^{-\ast }{\frac{\partial J}{\partial u}}^{\ast }}$

We do this is because $\frac{\partial J}{\partial u}$ is vector, can be easily solved through adjoint solver $P^{-\ast }$.

Another problem is the derivative of $P$, i.e. $\frac{\partial P}{\partial \alpha }$. If $P$ is simply a $n\times n$ matrix, and there is $m$ dofs of $\alpha$, then $\frac{\partial P}{\partial \alpha }$ will be a $m\times 1$ vector of matrix of $n\times n$. However, it is not a problem with numerical.

• solve the forward problem and get $u$, with $O(n^3)$
• solve the adjoint problem and get $\lambda =\underline{P^{-\ast }{\frac{\partial J}{\partial u}}^{\ast }}$, with $O(n^3)$
• calculate the matrix multiplication. Roughly $O(n^{2}m)$.

It will be much more faster if $P$ is sparse.