# Solve the screened Poisson equation

1. Dec 29, 2013

### eoghan

1. The problem statement, all variables and given/known data
Solve the equation
$$\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=-\frac{q_T}{\epsilon_0}\delta(r)$$
substituting the $\delta$ representation
$$\delta(r)=\frac{1}{4\pi}\frac{q_T}{r}$$
and writing the laplacian in spherical coordinates. Use as your guess
$$\phi=\frac{1}{4\pi\epsilon_0}\frac{g(r)}{r}$$
and show
$$g(r)=\exp(-r/\lambda_D)$$

3. The attempt at a solution
I know how to solve this equation using the Green and Fourier formalism, but here I am asked to solve it "the easy way" (as said in "Introduction to plasma physics - Bellan").

All I can say is
$$\nabla^2\left(\phi-\frac{q_T}{4\pi\epsilon_0}\frac{1}{r}\right)-\frac{\phi}{\lambda^2}=0$$
Using the guess
$$\phi=\frac{1}{4\pi\epsilon_0}\frac{g(r)}{r}$$
I have
$$\frac{1}{r^2}\partial_r\left[r^2\partial_r\left(\frac{g(r)-q_T}{4\pi\epsilon_0r}\right)\right]-\frac{g}{4\pi\epsilon_0r\lambda^2}=0$$
$$\frac{1}{r^2}\partial_r\left[r^2\left(\frac{r\partial_r g-(g-q_T)}{r^2}\right)\right]-\frac{g}{r\lambda^2}=0$$
$$\frac{1}{r^2}(\partial_r g+r\partial_r^2g-\partial_r g)-\frac{g}{r\lambda^2}=0$$
And so the final differential equation for g(r) is
$$\frac{\partial_r^2g}{r}-\frac{g}{r\lambda^2}=0$$
which has the physical solution
$$g(r)=\exp(-r/\lambda)$$

This result is correct, but my doubt is: the $\delta$ term didn't play any role, I mean, if I solved the equation
$$\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=0$$
I would have obtained the same result.
I suppose that without the $\delta$ term there is some problem in r=0, but I don't fully understand what is the problem

2. Dec 29, 2013

### haruspex

How do you get that? Isn't $\nabla^2\left(\frac 1r\right) = 0$?
If so, don't we get $\nabla^2\left(\phi-\frac{q_T}{4\pi\epsilon_0}\frac{\lambda^2}{r}\right)-\frac{1}{\lambda^2}\left(\phi-\frac{q_T}{4\pi\epsilon_0}\frac{\lambda^2}{r}\right)=0$

3. Dec 29, 2013

### eoghan

No, the function 1/r is singular at r=0, so the laplacian is proportional to a delta, i.e., it is zero everywhere except at r=0.

4. Dec 29, 2013

### Dick

Right. All you can conclude from $\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=0$ is that $\phi=C \frac{e^{-\frac{r}{\lambda_D}}}{r}$ where C is some constant. If you don't have a $\delta$ term then C=0. The $\delta$ function source let's you fix C. The shortcut way is to consider a sphere around the origin of radius r and let r->0 applying the divergence theorem. Show you can ignore the correction from the exponential factor in that limit.

5. Dec 30, 2013

### eoghan

But the solution
$\phi=C \frac{e^{-\frac{r}{\lambda_D}}}{r}$
satisfies $\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=0$ for every value of the constant C, not only for C=0

6. Dec 30, 2013

### eoghan

I tried to do that:
$$\int_V\nabla\left(\nabla\phi\right)=\int_S-C\frac{\exp(-r/\lambda)}{\lambda r^2}(r+\lambda)=-C\frac{\exp(-r/\lambda)}{\lambda}(r+\lambda)4\pi\:\:\:\:\:\:\:(1)$$
But from $\nabla^2\phi-\frac{1}{\lambda^2}\phi$ I have
$$\int_V\nabla\left(\nabla\phi\right)=\int_V \frac{1}{\lambda^2}C\frac{\exp(-r/\lambda)}{r}=\int_0^r r^2dr\int_0^\pi \sin(\theta)d\theta\int_0^{2\pi}d\phi\,\left(-\frac{1}{\lambda^2}\right)C\frac{\exp(-r/\lambda)}{r}= \\ -\frac{4\pi}{\lambda}C\exp(-r/\lambda)(\lambda+r)-4\pi C \:\:\:\:\:\:\:(2)$$
For formula (1) and (2) to be the same the constant C must be 0.
The incorrect term $-4\pi C$ is indeed the value of (1) for $r\to 0$ and can be eliminated with a delta source.
Is this reasoning correct?

7. Dec 30, 2013

### Dick

That is EXACTLY my point. But you don't want to solve $\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=0$, that's only true for r>0. You want to solve $\nabla^2\phi-\frac{1}{\lambda^2_D}\phi=-\frac{q_T}{\epsilon_0}\delta(r)$. Integrate both sides over a sphere centered on the origin of radius r. The right side gives you $-\frac{q_T}{\epsilon_0}$ no matter what r is. The integral of $\frac{1}{\lambda^2_D}\phi$ will go to zero as r->0. That just leaves the integral of $\nabla^2\phi$. As you showed in part (1) of post 7 that consists of a constant part and another part that vanishes as r->0. Now just equate the parts that don't vanish as r->0. Solve for C.

8. Dec 30, 2013

### vanhees71

I'd use spherical coordinates + symmetry to begin with, i.e., setting $\phi(\vec{r})=\phi(r)$ with $r=|\vec{r}|$. Of course you can do that only for $r>0$, because the spherical coordinates are singular along the entire $z$ axis.

Then adopt the (two) free parameters of the general solution of the (ordinary!) differential equation for $\phi(r)$. One is $\phi \rightarrow 0$ for $r \rightarrow \infty$, which leaves you with the solution already found earlier in the tread.

Then use Dick's method to find the remaining constant $C$.

9. Dec 31, 2013

### eoghan

Thank you everybody for your help