I want to solve $u_{xx} + u_{yy}+u_{zz} = \delta(x,y,z)$ with $u \to 0$ as $x,y,z \to \infty$, given $u(x,y,0)=0$ and $u_z(x,y,0) = f(x,y)$ using Fourier transform.
Then if I do Fourier transform respect to $x,y,z$, I will get $-(a^2 +b^2 +c^2) \hat{u} = \mathcal{F}(\delta(x,y,z)) = 1$
so $u = -\frac{1}{8 \pi^3} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{a^2+b^2+c^2} e^{i(ax + by+cz)} \mathrm{d}a \,\mathrm{d}b\, \mathrm{d}c$, but then Mathematica shows this integral will diverge, is there anything I did wrong? Thank you for any help.
I just realized that in order to use the IC, I may only fourier transform on $x$ and $y$ not $z$ ? so I will get $\hat{u}_{zz} - \delta(z) = (a^2 + b^2) \hat{u}$ . Is that right? Then how can I solve it since it involves delta function, thank you.
Then for this ODE, I apply the Laplace transform to it and get $s^2 \bar{u} - s \bar{u}(a,b,0) - \bar{u}_{z} (a,b,0) - (a^2 + b^2)\bar{u} = 1$ Then I can get the solution of $\bar{u} = \frac{1+\mathcal{F} (f(x,y))/s}{s^2 + a^2 + b^2}$, Is this right? Then I need to do a inverse laplace and then inverse Fourier to get my answer. I believe after inverse laplace I get $\hat{u} = \frac{\sin(\sqrt{a^2+b^2} z)}{\sqrt{a^2+b^2}} + \int_0^z \sin(\sqrt{a^2+b^2}) v \frac{\mathcal{F} (f(x,y))}{\sqrt{a^2+b^2}}dv$.
I am not sure what I did is right, so thank you for help me checking and if it is right, how can I do inverse Fourier to this $\hat{u}$ Thank you.
The only solution to Poisson's equation that is $0$ at the origin and approaches $0$ at infinity is the trivial solution. We proceed to derive the Green (or Green's) function for Poisson's equation in $\mathbb{R}^3$ subject to the condition at infinity.
Define the $3$-D Fourier transform of $u(x,y,z)$ as
$$\begin{align} U(a,b,z)&=\mathscr{F}\{u\}(a,b,z)\\\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty \int_{-\infty}^\infty u(x,y,z)e^{-i(ax+by+cz)}\,dx\,dy \end{align}$$
Applying this to the PDE $u_{xx}+u_{yy}+u_{zz}=\delta(x)\delta(y)\delta(z)$, we obtain the ODE
$$U(a,b,z)=\text{PV}\left(-\frac1{a^2+b^2+c^2}\right)\tag1$$
where the distribution $U(a,b,c)=\text{PV}\left(-\frac1{a^2+b^2+c^2}\right)$ is the Principal Value distribution.
Using spherical coordinates, we find that the inverse Fourier Transform of $U(a,b,c)$ is given by
$$\begin{align} u(R)&=\frac1{8\pi^3}\lim_{L\to\infty}\int_0^L\int_0^{2\pi} \int_0^\pi \frac{-e^{irR\cos(\theta)}}{r^2}\,r^2\,\sin(\theta)\,d\theta \,d\phi\,dr\\\\ &=\frac{-1}{4\pi^2}\lim_{L\to\infty}\int_0^L \int_0^\pi e^{irR\cos(\theta)}\,\sin(\theta)\,d\theta \,dr\\\\ &=-\frac{1}{4\pi^2}\lim_{L\to\infty}\int_0^L \frac{e^{irR}-e^{-irR}}{irR}\,dr\\\\ &=-\frac{1}{2\pi^2R}\int_0^\infty \frac{\sin(rR)}{r}\,dr\\\\ &=-\frac1{4\pi R} \end{align}$$
as expected!