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I asked this question on MSE. However, I didn't get good answers there so I am seeking for it here. :)

Consider the following Laplace boundary value problem (BVP)

$$\matrix{ {{\nabla ^2}\Phi (x,y) = 0,} \hfill & { - a \le x \le a} \hfill & { - b \le y \le b} \hfill \cr {{{\partial \Phi } \over {\partial y}}(a,y) = f(y)} \hfill & {} \hfill & {} \hfill \cr {{{\partial \Phi } \over {\partial y}}( - a,y) = f(y)} \hfill & {} \hfill & {} \hfill \cr {{{\partial \Phi } \over {\partial x}}(x,b) = 0} \hfill & {} \hfill & {} \hfill \cr {{{\partial \Phi } \over {\partial x}}(x, - b) = 0} \hfill & {} \hfill & {} \hfill \cr } $$

where $f(-y)=-f(y)$. We have prescribed the tangential derivatives of $\Phi(x,y)$ on the boundary instead of the function value.

The more general form of the problem in 2D or 3D can be written as

$$\matrix{ {{\nabla ^2}\Phi ({\bf{x}}) = 0,} \hfill & {{\bf{x}} \in \Omega } \hfill \cr {\nabla \Phi ({\bf{x}}).{\bf{t}}({\bf{x}}) = f({\bf{x}})} \hfill & {{\bf{x}} \in \partial \Omega } \hfill \cr } $$

where $\Omega$ is the domain of interest and ${\partial \Omega }$ is its boundary. Also, ${{\bf{t}}({\bf{x}})}$ is the tangent unit vector to the boundary at point $\bf{x}$ on the boundary.


Questions

1) Is the solution to this BVP unique? If NO, what is the degree of non-uniqueness?

2) Is there a relation between the solution to this BVP and the one with Dirichlet boundary conditions, i.e., when we determine the function value on the boundary?


My Thought

I don't think that the solution is unique so I was thinking to relate this in some manners to the solution of the Laplace BVP with Dirichlet boundary conditions where the function value is prescribed over the boundary since this BVP has a unique solution.

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closed as off-topic by Michael Renardy, Joonas Ilmavirta, Stefan Kohl, Myshkin, Igor Khavkine Nov 23 '15 at 12:14

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ If you know the tangential derivatives of the boundary values of $\Phi$, then you know the boundary values of $\Phi$ up to a constant. In your case of a square you can easily solve the Dirichlet boundary values (up to a constant) explicitly in terms of $f$. And indeed, if you replace $\Phi$ with $\Phi+c$, you still have a solution, so solutions are not unique. (This is always the case for linear BVPs that only see derivatives of the function.) $\endgroup$ – Joonas Ilmavirta Nov 22 '15 at 20:09
  • $\begingroup$ @JoonasIlmavirta: I do agree with your claim about knowning boundary values of $\Phi$ up to a constant! But I cannot show this mathematically! Can you show it in the special or the general problem in the question? As an example when you integrate the boundary value in the special problem, four integration constants will appear! It is not just one constant! $\endgroup$ – H. R. Nov 23 '15 at 8:01
  • $\begingroup$ The four constants are tied together by requiring the boundary value to be continuous at the corners. I thought the question would be more appropriate at MSE than here (and not alone; there are 4/5 votes to close now) and I can give you a more complete answer there. I can't see the votes so I don't know if this will be migrated there after closure, but that sounds possible. $\endgroup$ – Joonas Ilmavirta Nov 23 '15 at 8:36
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    $\begingroup$ @JoonasIlmavirta: OK, I will delete it here, Take a look at here in MSE. Anyway, why it is not appropriate to be asked here? :) $\endgroup$ – H. R. Nov 23 '15 at 8:47
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    $\begingroup$ Oh, and you should have mentioned that the same question was already asked at MSE. $\endgroup$ – Joonas Ilmavirta Nov 23 '15 at 8:53