I'm trying to find to solve the following PDE:
$$\nabla ^2\phi(\vec{x}) +k^2 \phi(\vec{x})=0$$
In spherical coordinates and then use the Generalised Fourier Transform to extract 2 linearly independent solutions.
Using separation of variables, I have: $$\frac{d^2R}{dr^2} + \frac{2}{r}\frac{dR}{dr}-\frac{l(l+1)}{r^2}R + k^2R=0$$
Then using the transformations: $R=r^lX$ and $x=kr$, I end up with: $$\frac{d^2X}{dx^2} + \frac{2(1+l)}{x}\frac{dX}{dx} + X=0$$
At this point, I introduce the generalised fourier transform: $X(x) = \int_C e^{zx}\tilde{X}(z)dz$.
$$\int_C dz \tilde{X}(z)[2z(1+l) +(z^2+1)x]e^{zx}= 0$$
Now I need (apparently) to get rid of $x$ by stating that $x=\frac{d}{dz}e^{zx}$, therefore: $$\int_C dz \tilde{X}(z)[2z(1+l) +(z^2+1)\frac{d}{dz}]e^{zx}=0$$
Developing the integrals for the sum and using integration by parts:
$$[\tilde{X}(z)(z^2+1)e^{zx}] + \int_C dz e^{zx}[2z(1+l) - \frac{d}{dz}(z^2+1)]\tilde{X}(z) = 0$$
I need to make the integral term vanish:
$$\frac{d}{dz}\tilde{X}(z)(z^2+1)=2z(1+l)\tilde{X}(z)$$ $$\frac{1}{\tilde{X}(z)(z^2+1)}\frac{d}{dz}\tilde{X}(z)(z^2+1)=\frac{2z(1+l)}{z^2+1}$$ $$\tilde{X}(z) = (z^2+1)^l$$
This is where I am stuck. I need to determine the analytical structure of $\tilde{X}(z)$ and then determine what contour to use for the boundary term to be $0$. Then actually find $X(x)$.
From what I understand, the Riemann structure of $\tilde{X}(z)$ has zeros at $+i$ and $-i$ and when have a corkscrew type formation. Do we have one at each zeros ?
Then because of the $e^{zx}$ in the boundary term, our contour extends to $\Re (z)<0$ since $x>0$ as it is a radial component in this problem.
Other than that I don't have any other ideas. Any help?
Many thanks!
Why you don't use the general solution
$$X(x)\to x^{-l-\frac{1}{2}} \left(c_1 J_{l+\frac{1}{2}}(x)+c_2 Y_{l+\frac{1}{2}}(x)\right)$$
If you absolutely need an integral transformation, then use Laplace, see Laplace transform for nonlinear ODE.
$$\mathcal{L}_x\left[2 (l+1) X'(x)+x X''(x)+x X(x)\right](z)=2 l z \tilde{X}(z)-\left(z^2+1\right) \tilde{X}'(z)-((2 l+1) X(0))$$
Solution of this ODE is $\tilde{X}(z)\to -(2 l+1) X(0) z \left(z^2+1\right)^l \, _2F_1\left(\frac{1}{2},l+1;\frac{3}{2};-z^2\right)+c_1 \left(z^2+1\right)^l$
With the assumption $X(0)=0$ $$X(x)=\mathcal{L}_z^{-1}\left[\tilde{X}(z)\text{/.}\, X(0)\to 0\right](x)=\frac{\sqrt{\pi } c_1 2^{l+\frac{1}{2}} x^{-l-\frac{1}{2}} J_{-l-\frac{1}{2}}(x)}{\Gamma (-l)}$$
you get the BesselJ(x) function of the first kind which is one part of the general solution for $X(0)=0$.