# System of non-linear partial differential eqs from electrostatics

1. Dec 12, 2012

I have an electrostatics problem (shown here: https://www.physicsforums.com/showthread.php?t=654877) wich leads to the following system of differential equations:

$\frac{\partial E_z}{\partial z}=\frac{\rho}{\epsilon_0}$ (1)

$Z_i E_r \frac{\partial \rho}{\partial r}+(u_g+ Z_i E_z) \frac{\partial \rho}{\partial z} + \rho Z_i \frac{\partial E_z}{\partial z}=0$ (2)

Substituting eq. (1) into eq. (2):
$Z_i E_r \frac{\partial \rho}{\partial r}+(u_g+ Z_i E_z) \frac{\partial \rho}{\partial z} + \frac{\rho^2 Z_i}{\epsilon_0}=0$ (3)

Therefore I have a system of 2 equations (1 & 3) with 2 unknowns, the axial field $E_z$ and the charge density $\rho(z,r)$. The rest of the variables are known so they can be supposed as constants.

I'm not sure on how to solve it, I'm considering two options:
- derivate eq. (3) with respect to $z$ to substitute in eq. (1), but I don´t get rid of $E_z$ and the eq. (3) becomes more complicated.

- Solve by semi-implicit method, considering that $z=du_z/dt$, but since is an equation in partial derivatives I'm not sure on how to manage the term in $r$

I'm totally stuck on this, I'm asking for a direction of solving it, not for a solution, so any help would be grateful.

2. Dec 28, 2012

### haruspex

How about separation of variables? Of course, being nonlinear it's not going to lead to a family of solutions which can be summed, but maybe towards one solution. I tried Ez = y(r,z) = R(r)Z(z). Substituting for ρ I get the form
AR'Z'+BRZ"+CR2(ZZ"+Z'2) = 0
(where R' means differentiated wrt r, Z' wrt z)
Integrating wrt z:
AR'Z+BRZ'+CR2ZZ' = f(r)
which should yield to a second round of integration wrt z easily enough.
Integrating wrt r gets hard unless you can figure out f(r) from boundary conditions.
That's all I can think of.

3. Jan 4, 2013

Hi Haruspex, first of all thank you very much for your help.
Your suggestion of separating variables gave me the idea of instead of separating Ez, which is supposed to only depend on z, separate $\rho(r,z)=R(r)Z(z)$
Then, the Poisson eq:
$dE=\frac{RZ}{\epsilon_0}dz$
and the eq in $\rho$:
$AR'Z+BRZ'+CR^2Z^2=0$

Considering the Neumann boundary conditions in the axial axis, when r=0:
$\frac{\partial E_r}{\partial r}=0 (1); \frac{\partial \rho}{\partial r}=0 (2)$
while in the walls, r=R: $E_z=0 (3)$

Using #(2) into ρ eq: $BZ'+CR_0Z^2=0$
where R(0)=R0

Integrating now wrt z: $\int B\frac{Z'}{Z^2} dz + \int CR_0 dz= 0$
$R_0Z = \rho(0,z) = \frac{B}{zC}$

Substituting and integrating for Ez in the axial line (r=0):
$E_z=\frac{\rho}{\epsilon_0}dz=\frac{B}{\epsilon_0C}Ln(z)$

What do you think of this kind of solution? I'm not very sure about having an electrical field which depends on a logarithm of distance, because is negative for small values of z and is contiunously growing, while Ez should decrease as z increases (it moves away from origin).

Thanks for helping.

4. Jan 6, 2013

### haruspex

Whoa, I'm confused. In the OP you wrote ∂Ez/∂z = ρ/ε0.
Looks like you meant ∂E/∂z = ρ/ε0.
Is that right? If so, please clarify the relationship between E, Ez and Er. Do the subscripts denote partial derivatives? Components? Something else?

5. Jan 7, 2013

oops, sorry about the confussion, subscripts denote components of the electrical field:
$\vec{E}=E_r\vec{r}+E_z\vec{z}$, neglecting the variation in the azimuthal direction.
The Poisson equation shown is the result of operating in cylindrical components:
$\frac{1}{r}\frac{\partial (r E_r)}{\partial r}+\frac{\partial E_z}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}$
for a known $E_r=\frac{U}{r Log(R_i/R_{tip})}$ and unknowns $E_z$ and $\rho(r,z)$

I hope the problem is better explained now.

6. Jan 7, 2013

### haruspex

It is, thanks, but I'm still stuck with an apparent contradiction. At different points in the thread you have written:
$\frac{\partial E_z}{\partial z}=\frac{\rho}{\epsilon_0}$ (1)
$\rho = \rho(r,z)$
Ez, which is supposed to only depend on z​
Do you see my puzzlement?

7. Jan 8, 2013

Yes, now I see, since $\rho = \rho(r,z)$, $\frac{\partial E_z}{\partial z}=f(r,z)$.
Then I don´t know if it is a conceptual mistake from me, but I was thinking that since
$\vec{E}=E_r\vec{r}+E_z\vec{z}$,
each component was depending on each variable,
Er=E(r) and Ez=E(z),
and therefore the Poisson equation:
$\nabla \vec{E}=\frac{\rho(r,z)}{\epsilon_0}$
would descompose in its form:
$\frac{1}{r}\frac{\partial (r E_r(r))}{\partial r}+\frac{\partial E_z(z)}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}$.
Since the term in Er is cancelled, the resulting equation would be:
$\frac{\partial E_z(z)}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}$.
But you are right and I don´t know why I conclude this "paradox"
Wouldn´t it be possible separate the effects on the r and z directions or maybe the correct expression is
$\frac{\partial E_z(z)}{\partial z}=\frac{\rho(z)}{\epsilon_0}$?

8. Jan 8, 2013

### haruspex

Yes, I'm afraid that's wrong. It's quite normal for a component in one direction to depend on location in another.

9. Jan 9, 2013

oh, I see my error, the field is actually discomposed as $\vec{E}=E\vec{r}+E\vec{z}$, where I have named $E_r=E(r,z)\vec{r}$ and $E_z=E(r,z)\vec{z}$, but I think I'm getting lost with the nomenclature, because then I only know Er at the discharge point, when z=0: $E_r(r,0)=\frac{U}{rLog(R_i/R_{tip})}$ and therefore I can only use it as an initial condition? It would change completely my approach to the problem...