Suppose that $\Psi (x,y)$ is a real valued function of the two real variables $x$ and $y$. Consider the following boundary value problem (BVP)

$$\begin{array}{c} \begin{array}{*{20}{l}} {{\nabla ^2}\Psi (x,y) = 0,}&{ - a \le x \le a}&{ - b \le y \le b}\\ {\Psi (a,y) = f(y)}&{}&{}\\ {\Psi ( - a,y) = f(y)}&{}&{}\\ {\Psi (x,b) = 0}&{}&{}\\ {\Psi (x, - b) = 0}&{}&{} \end{array}\\ \end{array}\tag{1}$$

where $f(y)$ is a function of $y$ satisfying

$$\left\{ \matrix{ f(y) = ( - y) \hfill \cr f(b) = f( - b) = 0 \hfill \cr} \right. \tag{2}$$

I want to prove the following theorem without directly obtaining $\Psi (x,y)$ from $(1)$.

Theorem. If $\Psi (x,y)$ satisfies the BVP in $(1)$ then $\Psi (x,y) = \Psi ( - x,y) = \Psi (x, - y) = \Psi ( - x, - y)$ over the rectangular domain $\left[ { - a,a} \right] \times \left[ { - b,b} \right]$.

I don't know how to start! Any hints and helps are appreciated. :)

  • $\begingroup$ I think you need to find another solution to the BVP, then use a uniqueness theorem to show that other solution is similar to previous. $\endgroup$
    – Michael
    Oct 26 '15 at 14:08
  • $\begingroup$ I don't think this works as we already know that $\Psi (x,y)$ exists and is unique due to the existence and uniqueness theorem for this problem. What we are looking for is a property of this unique solution! $\endgroup$ Oct 26 '15 at 14:12
  • $\begingroup$ Your response proves the second half of the result (the "then" part). But you still need to find another solution (or, a solution in a different form). $\endgroup$
    – Michael
    Oct 26 '15 at 14:15
  • $\begingroup$ I really don't get you! Please write what is in your mind formally as an answer so that I could understand it better. Many thanks. :) $\endgroup$ Oct 26 '15 at 14:18
  • $\begingroup$ Are you really saying you cannot find another solution? Have you tried? $\endgroup$
    – Michael
    Oct 26 '15 at 14:18

I just give this hint and you write this argument in formal language:

you have some defined Dirichlet's conditions. Think of the rectangle as being static in the $xy$ plane. Now invert the axes so that the same square is seen from this new $x'y'$ plane (think of it as taking the $x$ and $y$ axes and rotating them by a angle of $\pi$ to generate new axes $x'$ and $y'$). The conditions are invariant under this inversion, and the result follows.


I thought a little more and with the help of Michael and Vladimir Vargas I came up with something. The strategy is to first prove that $\Phi (x,y) = \Psi ( - x, - y)$ is also a solution of BVP $(1)$ in question above. Then I will apply the uniqueness theorem to conclude $\Phi (x,y) = \Psi (x,y)$.

So let's just do the first part. At the beginning, we may notice by using chain-rule that

$$\left\{ \begin{array}{l} \frac{{{\partial ^2}\Phi }}{{\partial {x^2}}}(x,y) = \frac{{{\partial ^2}\Psi }}{{\partial {x^2}}}( - x, - y)\\ \frac{{{\partial ^2}\Phi }}{{\partial {y^2}}}(x,y) = \frac{{{\partial ^2}\Psi }}{{\partial {y^2}}}( - x, - y) \end{array} \right.\tag{3}$$

Hence, we can write

$$\begin{array}{l} {\nabla ^2}\Phi (x,y) = \frac{{{\partial ^2}\Phi }}{{\partial {x^2}}}(x,y) + \frac{{{\partial ^2}\Phi }}{{\partial {y^2}}}(x,y)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{\partial ^2}\Psi }}{{\partial {x^2}}}( - x, - y) + \frac{{{\partial ^2}\Psi }}{{\partial {x^2}}}( - x, - y)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\nabla ^2}\Psi ( - x, - y) = 0 \end{array}\tag{4}$$

Also, the boundary conditions in $(1)$ become

$$\begin{array}{} {\Phi (a,y) = \Psi ( - a, - y) = f( - y) = f(y)}& \to &{\Phi (a,y) = f(y)} \\ {\Phi ( - a,y) = \Psi (a, - y) = f( - y) = f(y)}& \to &{\Phi ( - a,y) = f(y)}&{}&{}\\ {\Phi (x,b) = \Psi ( - x, - b) = 0}& \to &{\Phi (x,b) = 0}\\ {\Phi (x, - b) = \Psi ( - x,b) = 0}& \to &{\Phi (x, - b) = 0} \end{array}\tag{5}$$

it seems that we are done as we proved that $\Phi ( x, y)$ is a solution of BVP $(1)$ in question above. Other symmetries can be proved in a similar way. :)

  • $\begingroup$ Leaving the details, Is the strategy OK? :) $\endgroup$ Oct 26 '15 at 20:25
  • $\begingroup$ The strategy is good. $\endgroup$
    – Michael
    Oct 26 '15 at 20:26
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Michael
    Oct 26 '15 at 21:06
  • 1
    $\begingroup$ @H.R. It's good $\endgroup$ Oct 26 '15 at 21:32
  • $\begingroup$ @VladimirVargas: One more thing, In your hint, you didn't mention anything about the uniqueness theorem. Is there a way that we can conclude this without using uniqueness theorem? I don't think so! :) And how can I see your rotation in my solution? :) $\endgroup$ Oct 26 '15 at 22:05

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