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Solution to Laplace's equation in parabolic coordinates

  1. Dec 29, 2015 #1
    I'm stuck on a seemingly simple 2D electrostatics problem. The problem is as follows:

    A parabolic interface ($$x(y)=cy^2$$) separates two regions of different conductivities, with a uniform electric field at infinity aligned with the x-axis.

    I write the Laplace operator in parabolic coordinates, where $$x=\frac{c(u^2-v^2)}{2}$$ and $$y=cuv$$ as

    $$\nabla^{2} \phi(u,v)=\frac{1}{c^2(u^2+v^2)}\left(\phi_{uu}+\phi_{vv}\right)=0$$

    which, using separation of variables, gives solutions of the form

    (1): $$\phi(u,v)=(A_{0}u+B_{0})(C_{0}v+D_{0})+\sum^{\infty}_{n=1}(A_{n}e^{knu}+B_{n}e^{-knu})(C_{n}\textrm{cos}(knv)+D_{n}\textrm{sin}(knv))$$.

    Now, at infinity (where the electric-field is uniform), the potential is $$\phi=-E_{0}x=-E_{0}\frac{c(u^2-v^2)}{2}$$.

    The problem I have is that none of the solutions in (1) admit this form for the potential... Please can someone help?
     
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  3. Dec 29, 2015 #2

    jambaugh

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    I believe that in general you would need to take the Fourier expansion of your boundary solution in terms of your series of general solutions.
     
  4. Dec 29, 2015 #3
    Thanks for the advice. I was hoping there'd be a simpler solution (all terms with n>1 vanish) like that for the case of an elliptical inclusion of a different conductivity.
    I stumbled across this: http://eqworld.ipmnet.ru/en/solutions/lpde/lpde301.pdf
    They list one of the particular solutions in Cartesian coordinates
    $$\phi(x,y)=A(x^2-y^2)+Bxy$$
    having the form I require, though I haven't figured out how this was obtained....
     
  5. Dec 30, 2015 #4

    vanhees71

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    Well, it doen't matter, how it was obtained! It's sufficient to know that it is a harmonic function. Now, because in the paraboloic coordinates
    $$\Delta \Phi(u,v) \propto (\partial_{u}^2+\partial_{v}^2) \Phi(u,v)$$
    also
    $$\phi(u,v)=A(u^2-v^2)+Bu v$$
    is a solution of Laplace's equation (i.e., a harmonic function).

    Note however that your parabolic coordinates seem not to be chosen cleverly enough, because it's not one of the coordinate lines of these. You have to shift the Cartesian coordinate system first such that the focus of the parabola lies in the origin since the coordinate lines ##u=\text{const}## or ##v=\text{const}## are confocal parabolae with the focus in the origin; see

    https://en.wikipedia.org/wiki/Parabolic_coordinates

    Then I also don't understand the formulation of the problem fully. What do you mean by "areas of different conductivity"? The electrostatic field within a conductor should vanish everywhere, because otherwise, you'd have non-zero currents in the conductor(s). So either you mean different non-conducting dielectrics or a magnetostatic (stationary) problem, i.e., all fields time-independent, but the current densities ##\vec{j}(\vec{x})## not necessarily vanishing. To know the proper definition of the problem is important since we need to know the boundary conditions on the field we aim to fulfill.
     
  6. Dec 30, 2015 #5
    Hi vanhees71, thanks for your reply. I can't figure out how shifting the parabola to the right (to get the focus coincident with the origin) in Cartesian coordinates before working in parabolic coordinates will help? Surely the potential field will just be scaled by a constant?

    The problem is an electrostatics problem; steady-state current (not zero current) and a discontinuity in $$\nabla\phi$$ (but with continuous normal flux) at the parabolic interface (where the material conductivity changes). Does that help?

    EDIT: Yes, it's a steady-state problem where the current density is everywhere finite
     
  7. Jan 1, 2016 #6

    vanhees71

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    It's not so simple, but I guess, one could use the method of conformal mapping in this 2D problem. First we define the new variables ##(u,v)## with help of an analytical function in the complex plane ##\eta=u+\mathrm{i} v \mapsto \zeta=x+\mathrm{i} y## such that your parabola becomes a coordinate line ##v=\text{const}##. One possibility is
    $$\zeta=c \eta^2+\frac{1}{4c}.$$
    Splitting in real and imaginary part gives something very similar to your parabolic coordinates:
    $$\vec{r}=\begin{pmatrix}x \\y \end{pmatrix}=\begin{pmatrix}c(u^2-v^2)+\frac{1}{4 c} \\ 2cuv \end{pmatrix}.$$
    Now for ##v=1/(2c)## you have ##y=u## and ##x=c u^2=c y^2##, which is your parabola.

    After some vector gymnastics you get
    $$\Delta \Phi=\frac{1}{4c^2(u^2+v^2)}(\partial_u^2 \Phi+\partial_v^2 \Phi).$$
    This means that
    $$\Delta \Phi=0 \; \Rightarrow \; \partial_u^2 \Phi+\partial_v^2 \Phi=0.$$
    Now you can solve the much simpler problem for the boundary being given by ##v=1/(2c)=\text{const}##.

    Now concerning your boundary conditions at infinity. The point is that the current density must fulfill the continuity equation (conservation of electric charge) everywhere. Now since the electric field should be homogeneous at infinity the obvious solution is ##\vec{j}=j \vec{e}_x## with ##j=\text{const}##.

    For the conductivity you have
    $$\sigma(u,v)=\begin{cases}
    \sigma_1 & \text{for} \quad v<1/(2c), \\
    \sigma_2 & \text{for} \quad v>1/(2c).
    \end{cases}$$

    So for ##x \rightarrow -\infty## you must have ##\vec{E}=j/\sigma_2 \vec{e}_x## and for ##x \rightarrow +\infty## it should be ##\vec{E}=j/\sigma_1 \vec{e}_x##. So our potential should be
    $$\Phi(u,v)=-j c (u^2-v^2) \left [\frac{1}{\sigma_1} \Theta(1/(2c)-v)+ \frac{1}{\sigma_2} \Theta(v-1/(2c)) \right ]. \qquad \text{WRONG!}$$
    I hope that's the solution you are after.

    Note: The above solution is WRONG, because it doesn't fulfill the boundary conditions on the surface, which are
    $$E_{u,<}=E_{u,>0}, \quad \sigma_1 E_{v,<}=\sigma_2 E(v,>).$$
     
    Last edited: Jan 2, 2016
  8. Jan 1, 2016 #7
    Thank you very much for this nice analysis - this has been bugging me for some time. The solution you obtained; is this field not discontinuous in the limit when approaching ##v\rightarrow1/(2c)## from either side?

    Before I considered the parabola, I considered the an elliptical inclusion of conductivity, say ##\sigma_{1}## surrounded by a different conductivity ##\sigma_{2}##. The solution to this system came nicely from separation of variables (in elliptical coordinates), and has an exponential decay in the radial direction which satisfies the BCs. I figured the solution to the parabolic interface would have a similar structure, as the parabola is a limiting case of an ellipse...but perhaps not!! Thank you again.
     
  9. Jan 2, 2016 #8

    vanhees71

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    I think my solution is wrong, because it doesn't fulfill the correct boundary conditions, i.e., ##E_{\parallel}## and ##j_{\perp}## should be continuous along the surface, i.e., and one should have ##E_{u<}(u,v_0)=E_{u>}(u,v_0)## and ##\sigma_1 E_{u<}=\sigma_2 E_{u>}##, where ##\vec{E}_{>,<}## is the electric field for ##v>v_0,v<v_0## respectively. I assumed the space left from the parabola ##v=v_0=1/(2c)## has conductivity ##\sigma_2## and right from it ##\sigma_1##. I've no solution for this problem yet, because the boundary conditions at infinity are not so easy to implement. The current for ##x \rightarrow \pm \infty## with ##y=\text{const}## should be the same ##j_{\infty} \vec{e}_x##, if I understood your boundary conditions right. This implies the ##\vec{E}## must be ##\vec{E}=j_{\infty}/\sigma_2 \vec{e}_x## for ##x \rightarrow -\infty## and ##\vec{E}=j_{\infty}/\sigma_1 \vec{e}_x## for ##x \rightarrow +\infty##.

    For the case of a circle (or probably also for an ellipse), because there the region with a different ##\sigma## is a finite region in the ##xy## plane. I've checked it for the circle, where you can use of course modified polar coordinates by setting
    $$\zeta=x+\mathrm{i} y=r_0 \exp \eta=r_0 \exp u (\cos v+\mathrm{i} \sin v).$$
    Thus you have ##\rho=r_0 \exp u## and ##\varphi=v## in the usual polar coordinates. In the modified polar coordinates you have
    $$\Phi(u,v)=-j x+\Phi_2(u,v)=-\frac{j r_0}{\sigma_1} \exp u \cos v+\Phi_2(u,v).$$
    The boundary circle is ##u=0##, which is at ##\rho=r_0## and I assume that the conductivity is ##\sigma_1## for ##u>0## (i.e., ##\rho>r_0##) and ##\sigma_2## for ##u<0##, i.e., ##\rho<r_0##.

    Now one can make the ansatz
    $$\Phi_{2,<}(u,v)=\exp u (A_1 \cos v + A_2 \sin v), \quad \Phi_{2,>}=\exp(-u) (B_1 \cos v+B_2 \sin v),$$
    which vanishes for ##u \rightarrow \infty##, i.e., ##\rho \rightarrow \infty## as it should be to have the correct boundary condition at ##\rho \rightarrow \infty##, where we want to have the current ##\vec{j}_{\infty}=j \vec{e}_x## everywhere.

    Now you can take the gradients and solve for ##A_1##, ##A_2##, ##B_1##, ##B_2## using the boundary conditions
    $$E_{v,<}(u=0,v)=E_{v,>}(u=0,v), \quad \sigma_2 E_{u,<}(u=0,v)=\sigma_1 E_{u,>}(u=0,v),$$
    which gives
    $$A_1=B_1=\frac{j r_0}{\sigma_1} \, \frac{\sigma_2-\sigma_1}{\sigma_1+\sigma_2}, \quad A_2=B_2=0.$$
    This finally gives (written in terms of Cartesian coordinates again)
    $$\vec{E}_{>}=\frac{j}{\sigma_1} \vec{e}_x + \frac{j r_0^2(\sigma_2-\sigma_1)}{\sigma_1(\sigma_1+\sigma_2)(x^2+y^2)^2} [(x^2-y^2)\vec{e}_x+2x y \vec{e}_y]$$
    and
    $$\vec{E}_{<}=\frac{2j}{\sigma_1+\sigma_2}\vec{e}_x.$$
    Perhaps this example helps you to get an answer to your problem, which seems more tough to me, because the boundary extends to the infinity in the right half-plane!
     
    Last edited: Jan 4, 2016
  10. Jan 2, 2016 #9
    Hi Vanhees71, thanks again for looking into this. Yes I figured the previous solution was incorrect...as you said the boundary and interface conditions aren't satisfied. Perhaps the ansatz you proposed may lead somewhere - I will look at this next. If it doesn't lead somewhere I may just make a second-order approximation to the elliptic shaped solution in order to proceed as I have been working on this problem for an embarrassingly long time! :-(
     
  11. Jan 5, 2016 #10

    Andy Resnick

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  12. Jan 5, 2016 #11
    Hi Andy, thanks for your suggestion. However this is a problem which is independent of the z-dimension, so no special functions should be needed. I'll state the problem again and perhaps someone may have another idea: an electric field is applied at infinity ##\left.\mathbf{E}\right|_{x=\pm\infty}=E_0\mathbf{\hat{x}}## (aligned with the ##x##-axis) and the interface conditions on the parabola ##v=v_0## are (1); ##\phi(u,v=v^{+}_{0})=\phi_{v}(u,v=v^{-}_{0})## and (2); ##\sigma_{1}\phi_{v}(u,v=v^{+}_{0})=\sigma_{2}\phi(u,v=v^{-}_{0})##. These correspond to (1) continuity of potential and (2) continuity of flux. The subscript ##\phi_{v}## means ##\partial\phi/\partial v##. The problem is getting the solution to satisfy the boundary condition: ##\left.\mathbf{E}\right|_{x=\pm\infty}=-\left.\nabla\phi\right|_{x=\pm\infty}=E_0\mathbf{\hat{x}}##.... I don't think separation of variables in parabolic coordinates gives the correct form for the potential so a solution using conformal mapping to parabolic coordinates, as suggested by vanhees71, is probably the right approach, though I can't make it work!!
     
  13. Jan 5, 2016 #12
    PS - I tried to put this question on the physics stackexchange, but the over-vigilant moderators closed the thread for a reason I can't comprehend. This forum is much nicer! :)
     
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