# Question about general solution to Poisson's Equation [closed]

I've never taken a course on solving partial differential equations before, but I was wondering if my understanding on how to solve these types of questions is correct. Let's say I have an equation of the form:

$$\nabla^2 \phi = A\phi + B \delta^3(\vec{r}).$$

Question: Is it correct to solve equations of this form in the following way?

First, solve the equation:

$\nabla^2 \phi = A*\phi$

for $\vec{r} \ne \vec{0}$

Second, solve the equation:

$\nabla^2 \phi = B *\delta(\vec{r})$

and then to solve for the arbitrary constants which arise just match the solutions at $\vec{r} = \vec{0}$.

An example: The potential due to a single electron (or ion) within a quasi-neutral plasma can be approximated by the differential equation

$\nabla^2 \phi = \frac{8\pi n_0 e^2}{kT}\phi - 4\pi e\delta(\vec{r})$

where $\phi = \phi(r)$ only (aka $\phi$ is rotationally symmetric)

Thus, solving the "first equation"

$\nabla^2 \phi = \frac{8\pi n_0 e^2}{kT}\phi$

$\frac{1}{r^2} \frac{d}{dr}(r^2 \frac{d\phi}{dr}) = \frac{8\pi n_0 e^2}{kT}\phi$

It turns out that (although I would not have intuitively thought of this):

$\frac{1}{r^2} \frac{d}{dr}(r^2 \frac{d\phi}{dr}) = \frac{1}{r}\frac{d^2}{dr^2}(r \phi)$

This simplifies our differential equation to:

$\frac{1}{r}\frac{d^2}{dr^2}(r \phi) = \frac{8\pi n_0 e^2}{kT}\phi$

$\frac{d^2}{dr^2}(r \phi) = \frac{8\pi n_0 e^2}{kT}r\phi$

thus,

$r \phi = \alpha * e^{-(\frac{8\pi n_0 e^2}{kT})^{1/2}r} + \beta * e^{(\frac{8\pi n_0 e^2}{kT})^{1/2}r}$,

Since $\phi$ must go to zero at infinity, $\beta = 0$ and we are left with:

$\phi(r) = \frac{\alpha}{r} * e^{-(\frac{8\pi n_0 e^2}{kT})^{1/2}r}$

for $r \ne 0$

Now solving the "second equation" for when $\vec{r} = \vec{0}$:

$\nabla^2 \phi = - 4\pi e\delta(\vec{r})$

$\nabla \cdot \nabla \phi = - 4\pi e\delta(\vec{r})$

$\int (\nabla \cdot \nabla \phi) dv = \int - 4\pi e\delta(\vec{r}) dv$

$\int (\nabla \phi) \cdot d\vec{A} = - 4\pi e$

$\nabla \phi 4\pi R^2 = -4 \pi e$

$\frac{d}{dr}(\phi) 4\pi R^2 = -4 \pi e$

$\frac{d}{dr}(\phi) = \frac{-e}{R^2}$

$\phi(r) = \frac{e}{r}$

So finally we match solutions:

The first solution must match the second solution as they both tend to $\vec{r} = \vec{0}$

In the limit of $\vec{r} \rightarrow \vec{0}$ the first solution goes to:

$$\lim_{\vec{r} \rightarrow \vec{0}} \frac{\alpha}{r} * e^{-(\frac{8\pi n_0 e^2}{kT})^{1/2}r} = \frac{\alpha}{r}$$

The second solution goes to:

$$\lim_{\vec{r} \rightarrow \vec{0}} \frac{e}{r} = \frac{e}{r}$$

Thus

$$\alpha = e$$

And the solution for all $\vec{r}$ is:

$$\phi(r) = \frac{e}{r} * e^{-(\frac{8\pi n_0 e^2}{kT})^{1/2}r}$$

## closed as off-topic by ACuriousMind♦Jun 2 '18 at 8:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Jun 2 '18 at 8:40