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We know that the 3D Poisson's equation $$-\Delta v(x)=\delta_{0}(x) $$ has the solution $$v(x)=\frac{1}{4\pi\left|x\right|}$$ because of the fundamental solution of Laplace equation. How can I interpret the $$-\Delta v(x)+Av(x)=\delta_{0}(x) $$ where A is a constant? Do I get the term $$C_{1}\exp({\sqrt{A}x})+C_{2}\exp({-\sqrt{A}x})$$ as product?

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how you arrived at $C_{1}\exp({\sqrt{A}x})+C_{2}\exp({-\sqrt{A}x})$? –  Manoel Jan 24 '13 at 14:41
    
Try to find radial solutions to the equation$-\Delta v+A v=0$ and see what happens. –  Tomás Jan 24 '13 at 14:55
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up vote 3 down vote accepted

The Helmholtz operator $H=(-\Delta+A)$ has for $A \geq 0$ the fundamental solution $$E(x)=\frac{\exp(-\sqrt{A} \vert x \vert)}{4 \pi \vert x \vert }$$ which can be obtained directly by applying the Fourier transformation onto $H\, E=\delta$ (in the sense of tempered distributions of course), dividing by the polynomial and inverting the Fourier transformation. You can also verify it directly - using $\Delta \frac{1}{\vert x \vert}=-4\pi \delta$ or Green's theorem - that this indeed must be true. For the operator $-\Delta-B,\, B \geq 0$, carrying out the Fourier transformation (shift the poles off the real line) by the residue theorem and making a choice of contour, one finds advanced/retarded solutions $$E_{\pm}(x)=\frac{\exp(\pm \mathcal{i}\sqrt{B} \vert x \vert)}{4 \pi \vert x \vert}.$$

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thank you very much @Dominik, for your nice answer. –  pcepkin Jan 25 '13 at 12:52
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