Laplacian and Fourier Transform in a differential equation I am dealing with the following equation
\begin{equation}
\lambda\nabla^2f(r)=f(r)
\end{equation}
where $f(r)$ is a function only of the radius $r$ in spherical coordinates and $\nabla^2$ is the Laplacian operator. Using that $f(r)$ doesn't depend on $\theta$ and $\phi$, the Laplacian of $f(r)$ can be then written as
\begin{equation}
\nabla^2f(r)=\dfrac{d^2f}{dr^2}+\dfrac{2}{r}\dfrac{df}{dr} \equiv f''(r)+\dfrac{2}{r}f'(r)
\end{equation}
Putting this in the first equation and giving it to Mathematica or WolframAlpha, it says that the solution is
\begin{equation}
f(r)=C_1\dfrac{\sqrt{\lambda}\;e^{r/\sqrt{\lambda}}}{r}+C_2 \dfrac{e^{-r/\sqrt{\lambda}}}{r}
\end{equation}
where $C_1$ and $C_2$ are arbitrary integration constants, fixed by the boundary conditions (not interested in them now). However, If I take the Fourier Transform on both sides,
\begin{equation}
f(r)=\int \dfrac{d^3k}{(2\pi)^3}\hat f(k)e^{ikr}
\end{equation}
the only information this gives is $-k^2=\lambda$, since $\hat f(k)$ appears in both side of the equation. How can the solution be derived via Fourier Transform?
 A: With your permission I will rewrite the initial equation in the form
$$\nabla^2f(r)=m^2f(r)$$
First, from the Laplace equation in spheric coordinates
$$\Delta f(r)=\frac{1}{r^2}\frac{d}{dr}\Bigl(r^2\frac{d}{dr}f(r)\Bigr)=\frac{1}{r^2}\frac{d}{dr}\Bigl(r^2\frac{d}{dr}(rf(r)\frac{1}{r})\Bigr)=\frac{1}{r^2}\frac{d}{dr}\Bigl(r\frac{d}{dr}(rf(r))-rf(r)\Bigr)=\frac{1}{r}\frac{d^2}{dr^2}(rf(r))=m^2f(r)\Rightarrow rf(r)=C_1e^{-mr}+C_2e^{mr}$$
$f(r)$ is defined for all $r$, except for $r=0$ and $r=\infty$, where the function has singularities.
In many cases a Green function $G(\vec r,\vec\rho)$ of the equation is required. In our case the equation for Green function (spherically symmetric over $r$) has a form
$$\Bigl(-\Delta_r+m^2\Bigr)G(\vec r,\vec\rho)=\delta(\vec r-\vec\rho)$$
$G(\vec r,\vec\rho)$ allows to find a general solution of the equation $\Bigl(-\Delta_r+m^2\Bigr)f(\vec r)=g(\vec r)$ with the right side $g(\vec r)$ in the form $f(\vec r)=\int_{R^3}G(\vec r, \vec\rho) g(\vec \rho)d^3\vec\rho$
At $\rho=0$ we get the initial equation $\Bigl(-\Delta_r+m^2\Bigr)f(\vec r)=\delta(\vec r)$ with the only difference: delta-function is in the RHS. But due to the fact that $\delta(\vec r)$ iz zero for all $r\neq0$ the solution of this equation will coincide with the solution of the initial equation for all $r\neq0$.
Let's find $G(\vec r, \vec\rho)$, and then we will put $\vec\rho=0$.
Applying FT to the both sides of the equation $\Bigl(-\Delta_r+m^2\Bigr)G(\vec r,\vec\rho)=\delta(\vec r-\vec\rho)$ we get $$\hat G(\vec k,\vec\rho)=\frac{e^{-i(\vec\rho,\vec k)}}{m^2+k^2}$$
Applying inverse FT
$$G(\vec r,\vec\rho)=\int \dfrac{d^3\vec k}{(2\pi)^3}\hat G(\vec k,\vec\rho)e^{i(\vec k,\vec r)}=\int \dfrac{d^3\vec k}{(2\pi)^3}\frac{e^{i(\vec r-\vec\rho,\vec k)}}{m^2+k^2}$$ $$=2\pi\int_0^{\infty}\int_0^{\pi} \dfrac{\sin\phi \,d\phi\, k^2dk}{(2\pi)^3}\frac{e^{i|\vec r-\vec\rho|k\cos\phi}}{(k+im)(k-im)}$$
$$=\frac{1}{(2\pi)^2}\int_0^{\infty} \int_0^{\pi}\frac{\sin\phi\,d\phi\, k^2dk}{(2\pi)^3}\frac{e^{i|\vec r-\vec\rho|k\cos\phi}}{(k+im)(k-im)}$$ $$=\frac{1}{2(2\pi)^2i|\vec r-\vec\rho|}\int_{-\infty}^{\infty}\frac{e^{i|\vec r-\vec\rho|k}-e^{-i|\vec r-\vec\rho|k}}{(k+im)(k-im)}k\,dk$$
To integrate we close the contour in the upper half plane for the first term and in the lower - for the second one (to get vanishing integrals along these half-circles at $R\to\infty$) and evaluate residuals at $k=\pm im$.
Finally we get
$$G(\vec r,\vec\rho)=\frac{2\pi i}{2(2\pi)^2i|\vec r-\vec\rho|}\frac{e^{-m|\vec r-\vec\rho}\,\, 2im}{2im}=\frac{e^{-m|\vec r-\vec\rho|}}{4\pi|\vec r-\vec\rho|}$$
At $\rho=0$ we have $$f(r)=\frac{e^{-mr}}{4\pi r}$$
It is not surprising that we got only finite (at $r\to\infty$) part of the solution: FT implies that smooth functions which we transform are finite.
