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Consider a rectangular area, defined by the region $x=0,x=a,y=0,y=b$. Now, there is a potential $\phi(x,y)$ defined in this region, which satisfies, $\nabla^2 \phi=0$, and the following boundary conditions: $\phi(0,y)=\phi(x,0)=\phi(a,y)=0$ and $\phi(x,b)=Ax/a$.

My Argument What I am really confused about, is whether a boundary value problem is solvable, if it boundary values themselves contradict each other. For e.g: At $(a,b)$ in the above problem, the potential is both A and 0. As such, even if I consider an infinite series (of sines and sinhs), it does not converge to a single value near the corner $(a,b)$ as the $\phi(x,y)$ itself is not single valued 'everywhere'.

Can someone help me out with this conceptual difficulty ? Is there a formal solution to such 'contradicting boundary' value problems?

Lelouch
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1 Answers1

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Sorry for my english !

You'll have a discontinuity in your potential function. But the discontinuity can be handled by the serie of function.

For example, if you decompose a square signal in Fourier series, you'll have Gibbs oscillations localised around the point of discountinuity but it is not always a real problem for computations.

In some boundary values problem, you have a localised Dirac with a Fourier representation and it works !

Hope it can help.

Vincent Fraticelli
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  • A discontinuity is slightly different than being multivalued at a point, is it not ? If I am not very wrong, Fourier expansions and Gibbs oscillations handle discontinuities like you said, but if the function is multivalued at a point, the expansion cannot possibly converge at that point right ? – Lelouch Feb 12 '19 at 08:52
  • If there is a discontinuity in $t0$, the limit when $t$ in less than $t0$ and $t$ greater than $t0$ is different and so the function has no definite value in $t0$. You could say that it is multivalued in $t0$. You could have a boundary value problem with $\Phi $ equal to zero if $x <0$ and $\Phi $ equal to 1 of $x>0$. The Fourier series converge toward the half value, but not uniformly. – Vincent Fraticelli Feb 12 '19 at 09:00