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I am curious about the Fréchet derivative of a misfit regularization term, defined as follows: Let

$$ \phi(\mathbf{m}) = \frac{1}{2}||\nabla^2(\mathbf{m} - \mathbf{m}^{\textrm{ref}})||_2^2 $$ be said regularization term, where $\mathbf{m}\in\mathbb{R}^n = \mathbf{m}(\mathbf{x})$, $\mathbf{x}\in\mathbb{R}^n$, $\mathbf{m}^{\textrm{ref}}\in\mathbb{R}^n$ is constant, and

$$ \nabla^2 f(\mathbf{x}) \triangleq \sum_{i = 1}^n\frac{\partial^2 f}{\partial x_i^2}, $$ where $x_i$ denotes the $i$-th entry of the $\mathbf{x}$ $n$-th dimensional vector.

How can I compute the Fréchet derivative with respect to $\mathbf{m}$, which we can denote as $\nabla_m \phi(\mathbf{m})$?

I have researched the concept and the properties of Fréchet derivatives as generalized directional derivatives over Banach spaces. On preliminary efforts, I have been tempted to take the Laplacian operator as the bounded operator, such that, following (Wikipedia), I conclude that $$ \nabla_m \phi(\mathbf{m}) = \nabla^2(\mathbf{m} - \mathbf{m}^{\textrm{ref}}). $$ However, works such as (Grayver, 2013) when introducing a discrete analog of $\mathbf{L}$ of $\nabla^2$ they compute $\nabla_m$---neglecting any relation to a Fréchet derivative---obtaining:

$$ \nabla_m \phi = \mathbf{L}^T\mathbf{L}(\mathbf{m} - \mathbf{m}^{\textrm{ref}}), $$ which baffles me, since $\mathbf{L}^T\mathbf{L}$ is not a discrete analog for $\nabla^2$.

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Generally, one replaces $\mathbf m$ with $\mathbf m+\delta$ and then extracts the term that is linear in $\delta$. In this case, $$\frac{1}{2}\|\nabla^2(\mathbf{m} +\delta - \mathbf{m}^{\textrm{ref}})\|_2^2 = \frac12\langle \nabla^2(\mathbf{m} +\delta - \mathbf{m}^{\textrm{ref}}), \nabla^2(\mathbf{m} +\delta - \mathbf{m}^{\textrm{ref}})\rangle $$ so the linear term is $$\langle \nabla^2 \delta, \nabla^2 (\mathbf{m} - \mathbf{m}^{\textrm{ref}}) \rangle $$ Under reasonable boundary conditions (such as vanishing at infinity), the Laplacian is self-adjoint, which allows us to move it to the other side, getting $$\langle \delta, \nabla^4 (\mathbf{m} - \mathbf{m}^{\textrm{ref}}) \rangle$$ So, the Frechét derivative is the biLaplacian of $(\mathbf{m} - \mathbf{m}^{\textrm{ref}})$, or, more precisely, the linear functional indiced by it on $L^2$.