I have the Poisson Equation (with $\mathbf{x}\in\mathbb{R}^3$) with the following form: $$\nabla^2 \Phi (\mathbf{x}) = \delta(x)\delta(y)$$ I used 2 methods for the resolution of this PDE. I am frustated, I don't know why the first method doesn't work.
This method doesn't work: I know that the solution is:
$$ \Phi(\mathbf{x}) = -\frac{1}{4\pi}\int_{V} \frac{\delta(x')\delta(y')}{|\mathbf{x}-\mathbf{x'}|}\,dV'$$
Where $V$ is a cylinder with $z$-axis as the axis of the cylinder between $z'=\pm l$, $2l$ is an arbitrary length with the condition $|z|\ll l$. The Green's function of 3-D Laplacian $\nabla^2$ is $G(\mathbf{x},\mathbf{x'})=-\dfrac{1}{4\pi|\mathbf{x}-\mathbf{x'}|}$. So:
$$ \Phi(\mathbf{x}) = -\frac{1}{4\pi}\int_{-l}^{l} \left[(z'-z)^2 + \rho^2\right]^{-1/2}\,dz',\quad \rho=x^2+y^2$$
With $\rho$ the cylindrical radius. Using the substitution $z'-z=\rho\tan\phi$ and $dz'=\rho\sec^2\phi\,d\phi$:
$$ \Phi(\mathbf{x}) = -\frac{1}{4\pi}\int_{-\arctan(l+z/\rho)}^{\arctan(l-z/\rho)} \left[\rho^2\tan^2\phi + \rho^2\right]^{-1/2}\rho\sec^2\phi\,d\phi$$ With the identity $\tan^2\phi+1=\sec^2\phi$ : $$ \Phi(\mathbf{x}) = -\frac{1}{4\pi}\int_{-\arctan(l+z/\rho)}^{\arctan(l-z/\rho)} \sec\phi\,d\phi$$ $$ \Phi(\mathbf{x}) = -\frac{1}{4\pi}\Big[\ln|\tan\phi+\sec\phi|\Big]_{-\arctan(l+z/\rho)}^{\arctan(l-z/\rho)}$$
$$ \Phi(\mathbf{x}) = -\frac{1}{4\pi}\ln\left|\frac{ \frac{l-z}{\rho} + \sqrt{1+ \left(\frac{l-z}{\rho}\right)^2} }{ -\frac{l+z}{\rho} + \sqrt{1+ \left(\frac{l+z}{\rho}\right)^2} }\right|$$
$$ \Phi(\mathbf{x}) = -\frac{1}{4\pi}\ln\left|\frac{ 1-\frac{z}{l} + \sqrt{\frac{\rho^2}{l^2} + \left(1-\frac{z}{l}\right)^2 } }{ -1-\frac{z}{l} + \sqrt{\frac{\rho^2}{l^2} + \left(1+\frac{z}{l}\right)^2 } }\right|$$
If I send $l\to \infty$:
$$ \Phi(\mathbf{x}) = -\frac{1}{4\pi}\ln\left|\frac{2}{0}\right|=-\infty$$ This is not a satisfactory answer. Why does this not work?.
This method works: If $\mathbf{E}=\nabla\Phi$, then the equation is like Gauss law:
$$ \nabla\cdot \mathbf{E} = \delta(x)\delta(y) $$ $$ \int_{V} \nabla\cdot \mathbf{E} \,dV' = \int_{V} \delta(x')\delta(y') dV' $$ Where $V$ is a cylinder of lenght $L$, radius $\rho$ with $z$-axis as the axis of the cylinder. So: $$ \int_{V} \nabla\cdot \mathbf{E} \,dV' = L $$ With the Gauss theorem: $$ \oint_{\partial V} \mathbf{E} \cdot d\mathbf{A'} = L $$ With cylindrical symetry arguments we can say that $\mathbf{E}=E(\rho)\hat{\rho}$ doesn't depends of $z$, so: $$ E(\rho)2\pi\rho L= L $$ $$ E(\rho) = \frac{1}{2\pi\rho}$$ $$ \mathbf{E} = \frac{1}{2\pi\rho}\hat{\rho}$$ This implies: $$ \Phi(\rho) = \frac{1}{2\pi}\ln\left(\frac{\rho}{R}\right)$$
With $R$ as a constant. This is the Green's function for 2-D Laplacian $\nabla^2$ and it is correct.