3
$\begingroup$

I'm working on an inverse problem where the solution is the values of a 2D scalar field at the vertices of a 2D triangular mesh, such that the field can be defined continuously inside the mesh via barycentric interpolation.

The problem is however quite sensitive to noise, and without any kind of regularisation to impose spatial smoothness on the solution we obtain a very non-smooth field.

Can anyone suggest methods for imposing the required smoothness of the field?

Importantly, whatever cost function is used to promote smooth solutions, I need to be able to find the vector-derivative of its value with respect to the field values on each vertex, as gradient-based optimisation is being used to solve the problem.

In case it's important - the meshes are mostly made up of equilateral triangles, but not exclusively.

$\endgroup$
1
  • 1
    $\begingroup$ An easy cost function is the $L^2$ norm of the gradient. For barycentric interpolation this becomes simply an area-weighted sum over the triangles of the mesh. $\endgroup$
    – user3883
    Commented Feb 20, 2019 at 11:03

1 Answer 1

3
$\begingroup$

In microwave imaging, a great chunk of literature is devoted to regularization and its effect on the solution process and inversion results. One of the common methods for microwave imaging is the Contrast Source Inversion (CSI) method, which is essentially a gradient-based optimization.

In CSI, one would formulate the inverse problem as the optimization of the functional $\mathcal F(\chi)$: $$ \mathcal F_n(\chi) = F_{1,n}(\chi)+\ldots+ F_{m,n}(\chi) \tag{1} $$ In (1), $F_{i,n}$ is the $i$th cost function (in particular for CSI, $m=2$) for the $n$th iteration. Now, there are several ways to perform a regularization on (1).

  • additive regularization with a penalty function $F^R(\chi)$ and weighting parameter $\gamma$:

$$ \mathcal F_n(\chi) = F_{1,n}(\chi)+\ldots+ F_{m,n}(\chi)+\gamma^2F^R_n(\chi) \tag{2} $$

  • multiplicative regularization: $$ \mathcal F_n(\chi) = \big(F_{1,n}(\chi)+\ldots+ F_{m,n}(\chi)\big)F^R_n(\chi) \tag{3} $$

The problem with additive regularization is that looking for good $\gamma$ usually requires a lot of numerical experimentation and, possibly, prior information about the desired reconstruction of $\chi$.

Multiplicative regularization stems from the idea that $\gamma$ should depend on the $\mathcal F_{n-1}(\chi)$, the functional value at the previous iteration.

I suggest looking into

In this free-access paper, the authors present a multiplicative regularizer that fits into the gradient-optimization framework.

Now, regarding the usual suspects from the regularizers' family:

$$ F^R_n(\chi) = \frac{\int\limits_S ds(\vec{\rho}) \sqrt{\left|\nabla \chi(\vec{\rho})\right|^2+\delta_n^2}}{\int\limits_S ds(\vec{\rho}) \sqrt{\left|\nabla \chi_n(\vec{\rho})\right|^2+\delta_n^2}} \tag{4} $$

  • $L^2$-norm of the total variation - prefers smooth profile (that's probably what you are looking for), but you will lose the distinct object edges. Although, one can employ some edge-preserving algorithm, - but that's a topic for another discussion.

$$ F^R_n(\chi) = \frac{\int\limits_S ds(\vec{\rho}) \left(\left|\nabla \chi(\vec{\rho})\right|^2+\delta_n^2\right)}{\int\limits_S ds(\vec{\rho}) \left(\left|\nabla \chi_n(\vec{\rho})\right|^2+\delta_n^2\right)} \tag{5} $$

In (4) and (5), $\rho$ is the 2-D position vector (wrt to which we perform spatial differentiation and integration), $S$ is your 2-D domain, and $\delta_n$ can be defined, as follows:

$$ \delta_n^2 = h^2\sum\limits_{j}F_{j,n}(\chi) $$ By choosing $j$ one can control, which cost functions from (1) participate in regularization. If you have just one functional ($m=1$), you have a simpler formula. $h$ is the reciprocal of your mesh element size. Now, choosing non-zero $\delta_n$ helps to restore differentiability of regularizers $F_n^R$ in (4) and (5).

In this answer, I simplified notation from the cited paper to only one inverse function $\chi$ (CSI uses two different ones) and 2-D. The paper (and especially references in it) should help in finding regularizers that work better for your particular inverse problem or are just simpler versions of them.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.