Solving the Poisson equation with spherically symmetric functions

[docnet]

Homework Statement

psb

Relevant Equations

psb

I tried to follow the method outlined in lectures, and ended up with an incorrect solution. My understanding of PDEs is a bit shaky so I thank anyone for constructive feedback or information.

The solution to the Poisson equation
\begin{equation}
-\Delta u(x)=\frac{q}{\pi a^3}e^{-\frac{2||x||}{a}}
\end{equation}
is given by
\begin{equation}
u(x)=\int_{R^3}\Phi(x-y)f(y)dy
\end{equation}
in $R^3$ and spherically symmetric $f(y)$ this is
\begin{equation}
u(x)=\frac{1}{2||x||}\int_0^\infty r\tilde{f}(r)\Big(||x||+r-\Big|||x||-r\Big|\Big)dr
\end{equation}
\begin{equation}
\frac{1}{2||x||}\int_0^\infty r\frac{q}{\pi a^3}e^{-\frac{2||x||}{a}}\Big(||x||+r-\Big|||x||-r\Big|\Big)dr
\end{equation}
\begin{equation}
\frac{1}{||x||}\frac{q}{\pi a^3}\int_0^\infty e^{-\frac{2r}{a}}r^2dr
\end{equation}
after integration by parts
\begin{equation}
u(x)=\frac{1}{||x||}\frac{q}{4\pi}
\end{equation}

 

[BvU]

My understanding of PDEs

Shouldn't be a problem in this case: no PDE involved !
First a question: it says

And what would that be ? My telepathic capabilities fail me.

You have (from e.g. here) $$ \Delta u(r) = {1\over r^2}{\partial \over \partial r} \Bigl ( r^2 {\partial u\over \partial r} \Bigr ) = - {q\over 4\pi a^3}\; e^{\displaystyle {-{2r\over a}}}$$ which is no longer a PDE, so we can replace all $\ \partial \$ by $\$ d $\$ !

But the $\ formula\ from\ the \ lecture\$ might offer a more comfortable path to the solution ?$\$

 

[docnet]

Shouldn't be a problem in this case: no PDE involved !
First a question: it says

View attachment 278868​

And what would that be ? My telepathic capabilities fail me.

You have (from e.g. here) $$ \Delta u(r) = {1\over r^2}{\partial \over \partial r} \Bigl ( r^2 {\partial u\over \partial r} \Bigr ) = - {q\over 4\pi a^3}\; e^{\displaystyle {-{2r\over a}}}$$ which is no longer a PDE, so we can replace all $\ \partial \$ by $\$ d $\$ !

But the $\ formula\ from\ the \ lecture\$ might offer a more comfortable path to the solution ?$\$

Thanks for the reply. The formula for the lecture goes like

$-\Delta u=f$ has a fundamental solution $\int_{R^n}\Phi(x-y)f(y)dy$

where $\Phi(x)=\frac{1}{n(n-2)w_n||x||^{n-2}}$ and $w_n=\frac{4\pi}{3}$ for $R^3$.

For spherically symmetric $f$, the formula becomes $$\frac{1}{2||x||}\int^\infty_0rf(r)\Big(||x||+r-\Big|||x||-r\Big|\Big)dr$$

One thing I am confused about is why the formula distinguishes between $||x||$ and $r$ since they are the same parameters?