Boundary condition for electrostatics problem - found issue?

  • #1
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Main Question or Discussion Point

Hey everyone

Just a picture of my configuration.

YlCEu.png


The assumption here is $$\epsilon_a,\epsilon_b,\epsilon_c$$ are different from one another. Really the interest of this problem is to find the scalar potential $$\phi$$, such that $$\nabla^2 \phi = 0$$.

So now my question, about jump conditions,
Surface at $$y=0$$ has tangent $$\vec{E}$$ continous, thus
\begin{align}
-\hat{x} \cdot \nabla \phi_a = -\hat{x} \cdot \nabla \phi_b \\
-\hat{x} \cdot \nabla \phi_a = -\hat{x} \cdot \nabla \phi_c \\
\end{align}

However if we look at $$x=0$$ then normal $$\vec{D}$$ is continous thus

\begin{align}
-\epsilon_b \hat{x} \cdot \nabla \phi_b = - \epsilon_c \hat{x} \cdot \nabla \phi_c \\
\end{align}

From our relation above this implies that $$\epsilon_b=\epsilon_c$$, which we made no such assumption. So this looks like a contradiction to me.

Can someone tell me where I have gone wrong?

Thank you!
 
Last edited:

Answers and Replies

  • #2
33,804
9,516
Mathematically, there is also the solution ##\hat{x} \cdot \nabla \phi_a = 0##.

There might a smooth solution that has a violation of the boundary conditions in an arbitrarily small region where the three materials cross, and the actual fields then will depend on the non-exact material structure there.
 
  • #3
5
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Mathematically, there is also the solution ##\hat{x} \cdot \nabla \phi_a = 0##.

There might a smooth solution that has a violation of the boundary conditions in an arbitrarily small region where the three materials cross, and the actual fields then will depend on the non-exact material structure there.
Hey mfb,

Thanks for the response. I guess I'm in the game of solving these problems by numerical methods. I suppose my worry right now is if I implement this with $$\epsilon_b \neq \epsilon_c$$, then there is an underlying inconsistency in the system. You know how I can get around this?
 
  • #4
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9,516
You can check if an iterative solution converges to something stable.
 
  • #5
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You can check if an iterative solution converges to something stable.
Hi mfb,

I do not believe it would converge, or at the very best converge slowly, the underlying assumption would be that the matrix is well conditioned. Lets assume we did some sort of finite differencing and obtained an $$Ax=b$$ system. This inconsistency in the equations, will cause $$A^{-1}$$ not to exist analytically. Thus numerically, $$A$$ will be ill-conditioned, so iterative linear methods will converge slowly in order to find a $$x$$ such that $$Ax=b$$. Do you think GMRES, would be the best hope to find something reasonable?
 
Last edited:
  • #6
33,804
9,516
I would put it in a program and see what happens.
 

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