# Solving Poisson's equation

Hello,

I have a problem seeing how a solution is reached??

The expression is poisson's equation for a positive charge which is debye shielded.

$$\nabla . \psi = (2 / \lambda^2) \psi$$

The solution of which is..

$$\psi = (Q / 4\pi\epsilon r )* exp(-2 \sqrt{r}/\lambda)$$

I was hoping I could express this as a Legendre polynomial but alas, this is only for Laplace's equation it seems.

How can I arrive at this solution?

Thanks,

Harry

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dextercioby
Homework Helper
Does $\psi$ depend upon $\theta$ and $\varphi$ ?

No $$\theta$$ or $$\varphi$$ dependence.

dextercioby
Homework Helper
Then the Poisson's eq becomes a second order ODE. Which can be solved.

Oh , yeah, I see that now...

Cheers!

Harry

What am I missing here...

The first equation expands out to...

V'' + (2/r)V' - (2/c)V = 0

Where I have replaced psi with V.

My standard repository of all mathematical knowledge (Mary L Boas. Mathematical Methods in the Physical Sciences) only covers ODEs which have constant coefficients.

I'm sure there must be a standard technique out there...

Well, first thing's first, you have the wrong equation. You have "divergence of scalar equals scalar".

As for your resulting ODE, expand $$\phi = \sum_j a_j r^j$$. That should get you a nice recursion relation for the coefficients, and then you're golden.

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As StatMechGuy said, the equation given isn't Poisson (Laplacian of a function equals something that can only depend on position and time)... and it's weird since a scalar function doesn't have a divergence (did you mean gradient?).

dextercioby
Homework Helper
First of all

$$\nabla^2 \psi(r)=\frac{2}{\lambda^2}\psi(r)$$

is not Poisson equation, but a Helmholtz equation.

It can be written

$$\frac{1}{r}\frac{d^2}{dr^2}\left[r\psi(r)\right]=\frac{2}{\lambda^2}\psi(r)$$

$$\frac{d^2}{dr^2}\left[r\psi(r)\right]=\frac{2}{\lambda^2}\left[r\psi(r)\right]$$

Can you solve it now ?

Ok I got it (said I.... after a little delay).

Thanks Dextercioby, StatMechGuy and Ahmes, I think thats the first time I've used a series solution in anger.

Oh and yes... I initially wrote the equation down incorrectly.. well done to those who spotted it (sorry).