System of non-linear partial differential eqs from electrostatics

[Madoro]

I have an electrostatics problem (shown here: https://www.physicsforums.com/showthread.php?t=654877) which 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.

Thanks in advance.

 

[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 E_z = 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.

 

[Madoro]

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 E_z, 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 E_z 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 E_z should decrease as z increases (it moves away from origin).

Thanks for helping.

 

[haruspex]

instead of separating E_z, which is supposed to only depend on z,

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

 

[Madoro]

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.

 

[haruspex]

I hope the problem is better explained now.

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)
$\rho = \rho(r,z)$
E_z, which is supposed to only depend on z​

Do you see my puzzlement?

 

[Madoro]

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,
E_r=E(r) and E_z=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 E_r 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
$\frac{\partial E_z(z)}{\partial z}=\frac{\rho(z)}{\epsilon_0}$?

 

[haruspex]

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,
E_r=E(r) and E_z=E(z),

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

 

[Madoro]

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 E_r 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...


Sorry for the long discussion, but is a key step in my work, and thank you very much for the help and ideas.