System of non-linear partial differential eqs from electrostatics


by Madoro
Tags: differential, electrostatics, non-linear, partial
Madoro
Madoro is offline
#1
Dec12-12, 04:59 AM
P: 7
I have an electrostatics problem (shown here: http://www.physicsforums.com/showthread.php?t=654877) wich leads to the following system of differential equations:

[itex]\frac{\partial E_z}{\partial z}=\frac{\rho}{\epsilon_0}[/itex] (1)

[itex]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[/itex] (2)

Substituting eq. (1) into eq. (2):
[itex]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[/itex] (3)


Therefore I have a system of 2 equations (1 & 3) with 2 unknowns, the axial field [itex]E_z[/itex] and the charge density [itex]\rho(z,r)[/itex]. 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 [itex]z[/itex] to substitute in eq. (1), but I donīt get rid of [itex]E_z[/itex] and the eq. (3) becomes more complicated.

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

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.
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
haruspex
haruspex is offline
#2
Dec28-12, 07:15 PM
Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,162
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.
Madoro
Madoro is offline
#3
Jan4-13, 02:35 PM
P: 7
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 [itex]\rho(r,z)=R(r)Z(z)[/itex]
Then, the Poisson eq:
[itex]dE=\frac{RZ}{\epsilon_0}dz[/itex]
and the eq in [itex]\rho[/itex]:
[itex]AR'Z+BRZ'+CR^2Z^2=0[/itex]

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

Using #(2) into ρ eq: [itex]BZ'+CR_0Z^2=0[/itex]
where R(0)=R0

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

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

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.

haruspex
haruspex is offline
#4
Jan6-13, 07:15 PM
Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,162

System of non-linear partial differential eqs from electrostatics


Quote Quote by Madoro View Post
instead of separating Ez, which is supposed to only depend on z,
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?
Madoro
Madoro is offline
#5
Jan7-13, 04:43 AM
P: 7
oops, sorry about the confussion, subscripts denote components of the electrical field:
[itex]\vec{E}=E_r\vec{r}+E_z\vec{z}[/itex], neglecting the variation in the azimuthal direction.
The Poisson equation shown is the result of operating in cylindrical components:
[itex]\frac{1}{r}\frac{\partial (r E_r)}{\partial r}+\frac{\partial E_z}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}[/itex]
for a known [itex]E_r=\frac{U}{r Log(R_i/R_{tip})}[/itex] and unknowns [itex]E_z [/itex] and [itex]\rho(r,z)[/itex]

I hope the problem is better explained now.
haruspex
haruspex is offline
#6
Jan7-13, 04:00 PM
Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,162
Quote Quote by Madoro View Post
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:
[itex]\frac{\partial E_z}{\partial z}=\frac{\rho}{\epsilon_0}[/itex] (1)
[itex]\rho = \rho(r,z)[/itex]
Ez, which is supposed to only depend on z
Do you see my puzzlement?
Madoro
Madoro is offline
#7
Jan8-13, 04:39 AM
P: 7
Yes, now I see, since [itex]\rho = \rho(r,z)[/itex], [itex]\frac{\partial E_z}{\partial z}=f(r,z)[/itex].
Then I donīt know if it is a conceptual mistake from me, but I was thinking that since
[itex]\vec{E}=E_r\vec{r}+E_z\vec{z}[/itex],
each component was depending on each variable,
Er=E(r) and Ez=E(z),
and therefore the Poisson equation:
[itex]\nabla \vec{E}=\frac{\rho(r,z)}{\epsilon_0} [/itex]
would descompose in its form:
[itex]\frac{1}{r}\frac{\partial (r E_r(r))}{\partial r}+\frac{\partial E_z(z)}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}[/itex].
Since the term in Er is cancelled, the resulting equation would be:
[itex]\frac{\partial E_z(z)}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}[/itex].
But you are right and I donīt know why I conclude this "paradox"
Wouldnīt it be possible separate the effects on the r and z directions or maybe the correct expression is
[itex]\frac{\partial E_z(z)}{\partial z}=\frac{\rho(z)}{\epsilon_0}[/itex]?
haruspex
haruspex is offline
#8
Jan8-13, 02:48 PM
Homework
Sci Advisor
HW Helper
Thanks ∞
P: 9,162
Quote Quote by Madoro View Post
Then I donīt know if it is a conceptual mistake from me, but I was thinking that since
[itex]\vec{E}=E_r\vec{r}+E_z\vec{z}[/itex],
each component was depending on each variable,
Er=E(r) and Ez=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
Madoro is offline
#9
Jan9-13, 06:18 AM
P: 7
oh, I see my error, the field is actually discomposed as [itex]\vec{E}=E\vec{r}+E\vec{z}[/itex], where I have named [itex]E_r=E(r,z)\vec{r}[/itex] and [itex]E_z=E(r,z)\vec{z}[/itex], but I think I'm getting lost with the nomenclature, because then I only know Er at the discharge point, when z=0: [itex]E_r(r,0)=\frac{U}{rLog(R_i/R_{tip})}[/itex] 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.


Register to reply

Related Discussions
Linear Partial Differential Equations Science & Math Textbook Listings 5
partial differential of the flow of a system Calculus & Beyond Homework 0
first order non linear partial differential equations Calculus & Beyond Homework 1
First order linear partial differential equation Differential Equations 5
a Linear Partial Differential Eq. Calculus & Beyond Homework 0