# scattering

I have been writing a post on QTAT, though I haven’t updated it for a long time. Now I was focusing on a classic scattering problem.

Given far field pattern and determine the scatter media, i.e. if we are given the Helmholtz Equation:

$\Delta u+k^2 n(x) u =0$, with far field pattern $u_{\infty}$, then we can determine $n(x)$ uniquely by $u_{\infty}$.

The problem’s proof is not hard to understand, if not unique, then for there exists at least two solutions for $n(x)$, say $n_1$ and $n_2$, and $1-n_i$  is compactly supported. And the relative solution towards the equations are $u_1,u_2$. Then we shall try to build up a $L^2$ basis in the form of $u_1u_2$,

Since we have:

$\displaystyle\Delta u_1+k^2 n_1 u_1 =0$

$\displaystyle\Delta u_2+k^2 n_2 u_2 =0$

Set $\displaystyle v= u_1 - u_2$, then $\displaystyle v_{\infty}=0$.

$\displaystyle \Delta v + k^2 n_1 v = k^2 (n_1-n_2)u_2$, thus

$\displaystyle \int_{\Omega}u_1(\Delta v + k^2 n_1 v)=\int_{\Omega}k^2(n_1-n_2)u_2 u_1$. On the other hand,

$\displaystyle\int_{\Omega}v(\Delta u_1 + k^2 n_1 u_1)=0$, subtract one from the other, and use Green second identity,

$\displaystyle\int_{\partial\Omega}u_1v_{\nu}-v{u_1}_{\nu}=\int_{\Omega}k^2 (n_1-n_2)u_1u_2$, and use Rellich Lemma for Helmholtz equation(exterior), we have:

$\displaystyle\int_{\Omega}k^2 (n_1-n_2)u_1u_2=0$, thus if we can prove $u_1u_2$ can form a basis for $L^2(\Omega)$, then we have done.

It’s lucky we can expect the far field pattern is like some harmonic function $\exp(z\cdot x)$, where $z\cdot z =0$, and the multiplication of two harmonic functions are dense in $L^2$. Therefore, we just have to prove the solution of Helmholtz equation at a sufficient far-away place acts like an exponential function(complex). However, this result is not trivial.

The idea is if we have all incident waves in all directions, then we have almost all data of the scattering media(the wave must flow into the media, by dimension analysis, this should be true). Reconstruct the media is kind of wasting time. Newton’s method is just fine.