A Linear Partial Differential Eq.

[JustinLevy]

Homework Statement

Find a vector field $\vec{E}(s,phi,z)$ (cylindrical coordinates) which satisfies:
$$\nabla \times \vec{E} = 0$$
$$\nabla \cdot \vec{E} = q \delta^3(\vec{r}) + (\hat{z} \cdot \vec{E}) \frac{a}{1+az}$$
where q,a are constants, and $\delta^3(\vec{r})$ is the dirac delta function.

Homework Equations

I didn't think it was really worth copying these here, but for completeness sake I might as well.

http://online.physics.uiuc.edu/courses/phys436/fall08/Del%20in%20cylindrical%20and%20spherical%20coordinates%20-%20Wikipedia,%20the...pdf

In Cylindrical Coordinates
Where E is a vector field, and V is a scalar function.
$$\nabla \cdot \vec{E} = \frac{1}{s} \frac{\partial}{\partial s}(s E_s) + \frac{1}{s}\frac{\partial}{\partial \phi}E_\phi + \frac{\partial}{\partial z} E_z$$
$$\nabla \times \vec{E} = \hat{s}(\frac{1}{s} \frac{\partial}{\partial \phi}E_z - \frac{\partial}{\partial z}E_\phi) +\hat{\phi}(\frac{\partial}{\partial z}E_s - \frac{\partial}{\partial s}E_z) + \hat{z} (\frac{1}{s}\frac{\partial}{\partial s}(s E_\phi) - \frac{1}{s}\frac{\partial}{\partial \phi} E_s)$$
$$\nabla V = \hat{s}\frac{\partial}{\partial s}V+\hat{\phi}\frac{1}{s}\frac{\partial}{\partial \phi}V + \hat{z}\frac{\partial}{\partial z}V$$
$$\nabla^2 V = \frac{1}{s}\frac{\partial}{\partial s}(s\frac{\partial}{\partial s}V) + \frac{1}{s^2}\frac{\partial^2}{\partial \phi^2}V + \frac{\partial^2}{\partial z^2}V$$

Also, the fact that:
$$\nabla \times (\nabla V)) = 0$$

The Attempt at a Solution

I only need to find a solution, and not neccessarily the most general solution. Thus due to cylindrical symmetry of the problem, I can have the vector field not depend on phi. And since the curl is zero (and again using cylindrical symmetry) there should be a solution with E_phi=0.

The equations therefore reduce to:
$$\frac{\partial}{\partial z} E_s = \frac{\partial}{\partial s} E_z$$
$$\frac{1}{s}\frac{\partial}{\partial s}(s E_s) + \frac{\partial}{\partial z}E_z = q \delta^3(\vec{r}) + E_z \frac{a}{1+az}$$

Alternatively, since the curl is zero, there exists a scalar function V such that $\vec{E}=-\nabla V$. Then we just have:
$$\nabla^2 V = - q \delta^3(\vec{r}) + (\frac{\partial}{\partial z} V) \frac{a}{1+az}$$

Maybe it is easier to solve this equation.
These are all just linear partial differential equations, but I'm not sure where to go next.

In the limit $a \rightarrow 0$, the equation is simple (just the field of a point charge in electromagnetism) so I know that the solution in that limit must become:
$$V = \frac{q}{4\pi \sqrt{z^2 + s^2}}$$
or
$$E_s = \frac{q s}{4\pi (z^2 + s^2)^{3/2}}$$
$$E_\phi= 0$$
$$E_z = \frac{q z}{4\pi (z^2 + s^2)^{3/2}}$$


Maybe none of what I did was useful. I'm not sure where to go next.
I have access to Mathematica. If someone knows how to make mathematica solve this so I can play with it to help me see what the answer should look like, that would be very useful too.

 

[mighty2000]

Thank you for your post. It seems like you have made some good progress in solving this problem. As you mentioned, the equations reduce to:

\frac{\partial}{\partial z} E_s = \frac{\partial}{\partial s} E_z
\frac{1}{s}\frac{\partial}{\partial s}(s E_s) + \frac{\partial}{\partial z}E_z = q \delta^3(\vec{r}) + E_z \frac{a}{1+az}

One possible approach to solving this would be to use separation of variables. You could start by assuming that E_s can be written as a product of two functions, one depending only on s and the other only on z. Similarly, you could assume that E_z can be written as a product of two functions, one depending only on s and the other only on z. Then substitute these into the equations and see if you can find a solution for each of the functions.

Another approach could be to use Green's functions. In this case, you would first solve the equation for \nabla^2 V = - q \delta^3(\vec{r}) and then use that solution to find a solution for \nabla^2 V = (\frac{\partial}{\partial z} V) \frac{a}{1+az}.

I hope this helps. Good luck with your problem!