System of nonlinear partial differential eqs from electrostaticsby Madoro Tags: differential, electrostatics, nonlinear, partial 

#1
Dec1212, 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 semiimplicit 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. 



#2
Dec2812, 07:15 PM

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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"+CR^{2}(ZZ"+Z'^{2}) = 0 (where R' means differentiated wrt r, Z' wrt z) Integrating wrt z: AR'Z+BRZ'+CR^{2}ZZ' = 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
Jan413, 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 E_{z}, 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)=R_{0} 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 E_{z} 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 E_{z} should decrease as z increases (it moves away from origin). Thanks for helping. 



#4
Jan613, 07:15 PM

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System of nonlinear partial differential eqs from electrostaticsLooks like you meant ∂E/∂z = ρ/ε_{0}. 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? 



#5
Jan713, 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. 



#6
Jan713, 04:00 PM

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[itex]\frac{\partial E_z}{\partial z}=\frac{\rho}{\epsilon_0}[/itex] (1)Do you see my puzzlement? 



#7
Jan813, 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, E_{r}=E(r) and E_{z}=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 E_{r} 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]? 



#8
Jan813, 02:48 PM

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#9
Jan913, 06:18 AM

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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 E_{r} 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. 


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