Say I have a problem given by the 1D Laplace equation, $$ R (T(\alpha), \alpha) = \frac{d^2 T(x)}{dx^2} - \alpha(x) T (x) = 0, $$ with $x \in [0,1]$, Dirichlet boundary conditions on $x=0$ and $x=1$, and an objective function given by the following functional: $$ J = \int_0^1 \frac{1}{2} (T(x) - T_{def} (x))^2 dx. $$ I want to obtain the sensitivity of $J$ with respect to the spatially varying $\alpha(x)$ using the continuous adjoint. Introducing the adjoint variable $\psi(x)$ $$ \frac{\delta L}{\delta \alpha} = \frac{\delta J}{\delta \alpha} + \int_0^1 \psi (x) \frac{\delta R}{\delta \alpha} dx $$ Deriving the adjoint equation by integrating by parts twice gives (neglecting the boundary terms for now) $$ \frac{\delta L}{\delta \alpha} = \int_0^1 (T - T_{def}) \frac{\delta T}{\delta \alpha} dx + \int_0^1 \frac{d^2 \psi(x)}{dx^2} \frac{\delta T}{\delta \alpha} dx - \int_0^1 \psi \alpha \frac{\delta T}{\delta \alpha} dx - \int_0^1 \psi T dx $$

The adjoint equation is obtained by setting the sum of the first three terms to zero. $\psi(x)$ can then be used to obtain the gradient using the last term.

However, if I wanted to do the same for the following governing equation, $$ R (T(\alpha), \alpha) = \frac{d^2 T(x)}{dx^2} - \frac{d \alpha(x)}{dx} T (x) = 0, $$ using the same strategy for the second term, $$ \int_0^1 \psi \frac{\delta}{\delta \alpha} \left( \frac{d \alpha(x)}{dx} T(x) \right) dx = \int_0^1 \psi \frac{d}{d x} \left( \frac{\delta \alpha(x)}{\delta \alpha} \right) T(x) dx + \cdots = \int_0^1 \psi \frac{d}{d x} \left( 1 \right) T(x) dx + \cdots = 0 +\cdots, $$ the gradient resulting from the continuous adjoint will always be zero, while this is clearly not necessarily the case.

Where do I go wrong in my reasoning?


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