Let $G(r,r') = \frac{\exp(-ik|r-r'|)} {|r-r'|} $, where $r$ and $r'$ are position vectors in a domain $D$ of $\mathbb R^3$ and $k$ is a positive real constant. Suppose that $h$ is a continuous real valued function defined in $D$.
Does $\int_D G(r,r') h(r') dr' = 0$ imply that $h = 0$ in $D$? Is there a well-known theorem that applies to this question and/or a proof that answers it?
Dear Dr. Wilson:
I think that the answer to your question is NO.
Let us call H(\vec{p}) to the 3D Fourier transform of h(r). It can be seen that the zero result of the integral only involves the values of H(\vec{p}) in an spherical shell domain of radius |\vec{p}|=k. Even if that zero means that H(\vec{p}) is zero in that domain, it does not constrain the whole H(\vec{p}), and hence h(r), to be zero.
The above conclusion can be reached using a decomposition of G(r,r') due to Weyl, which you can find in e.g. Eq. 3.4 in https://aip.scitation.org/doi/10.1063/1.1666629:
G(r,r') = \frac{i k}{2\pi} \int d shat \exp ( i k shat (r-r'))
Where shat is a unit vector and the integral is on a spherical shell (involving the polar and azimuthal angles).
Plugin such decomposition into the integral that gives zero the result and manipulating it a bit, yields the conclusion that I have written above.
Best regards
Ivan
Hi Ivan,
Thank you for your answer. Do you have a different reference I can use? The link provided above responds back with "The requested article is not currently available on this site.".
Thanks again for addressing this problem.
Greg
Hi Greg,
It is malfunctioning because of the extra ":" in the link. This should work
Article Multipole expansion and plane wave representations of the el...
Best regards
Ivan
No, it is not. The null space of this integral operator consists of all the functions of the type j_l(k'r')*Y_{lm}(\vec{r'}), with k' different from k. j is the spherical bessel function and Y_{lm} is a spherical function. To see this, you can look at the article in Phys. Rev. E vol. 65, p. 046609, (2002) , Eqs. (2.20) and (2.34).
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