System of non-linear partial differential eqs from electrostatics

In summary, the person is having an electrostatics problem and is looking for help to solve it. They have two equations and two unknowns. The first equation is E_z=ρ/ε0 and the second equation is R(r)=R0. The unknowns are E_r and \rho. They are considering two options, one is to solve by semi-implicit method and the other is to solve by separation of variables.
  • #1
Madoro
7
0
I have an electrostatics problem (shown here: https://www.physicsforums.com/showthread.php?t=654877) which 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 don´t 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.
 
Physics news on Phys.org
  • #2
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
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.
 
  • #4
Madoro said:
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?
 
  • #5
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.
 
  • #6
Madoro said:
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?
 
  • #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 don´t 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 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
[itex]\frac{\partial E_z(z)}{\partial z}=\frac{\rho(z)}{\epsilon_0}[/itex]?
 
  • #8
Madoro said:
Then I don´t 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.
 
  • #9
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.
 

1. What is a system of non-linear partial differential equations in electrostatics?

A system of non-linear partial differential equations in electrostatics is a set of equations that describe the behavior of electric fields and charges in a non-linear manner. These equations are used to model complex electrostatic systems, such as those involving multiple charges or varying electric fields.

2. How are these equations different from linear partial differential equations?

The main difference between linear and non-linear partial differential equations is that the latter involve terms that are not directly proportional to the variables being solved for. This means that the solutions to non-linear equations are often more complex and can exhibit unexpected behavior, compared to the more straightforward solutions of linear equations.

3. What is the importance of studying systems of non-linear partial differential equations in electrostatics?

Studying these equations allows us to better understand and predict the behavior of electric fields and charges in complex systems. This knowledge is crucial in various fields, including engineering, physics, and chemistry, where electrostatics plays a significant role in the behavior of materials and devices.

4. What are some real-world applications of these equations?

Systems of non-linear partial differential equations in electrostatics have numerous applications, such as in the design of electronic circuits, the study of plasma physics, and the modeling of biological systems. They are also used in the development of new technologies, such as semiconductors and medical imaging devices.

5. Are there any limitations or challenges in solving these equations?

Yes, there are several challenges in solving systems of non-linear partial differential equations in electrostatics. These equations often have complex solutions that can be difficult to obtain analytically, requiring numerical methods to approximate the solutions. Additionally, the computational cost of solving these equations can be high, making it challenging to model large-scale systems accurately.

Similar threads

  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
13
Views
2K
  • Differential Equations
Replies
1
Views
2K
  • Differential Equations
Replies
5
Views
2K
  • Differential Equations
Replies
2
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
  • Differential Equations
Replies
12
Views
3K
  • Differential Equations
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
879
Back
Top