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I asked this question on Mathematics network but it didn't receive any answers. So I assume it is just beyond the classic things in PDEs and I decided to ask it here too.

Consider the following boundary value problem (BVP)

$$\matrix{ {{\Delta ^2}H = 0,} \hfill & {} \hfill & {{\rm{in}}\,} \hfill & \Omega \hfill \cr {\partial _y^2H = 0} \hfill & {{\partial _x}{\partial _y}H = 0} \hfill & {{\rm{on}}} \hfill & {{S_1}} \hfill \cr {\partial _y^2H = 0} \hfill & {{\partial _x}{\partial _y}H = 0} \hfill & {{\rm{on}}} \hfill & {{S_2}} \hfill \cr {\partial _x^2H = 0} \hfill & {{\partial _x}{\partial _y}H = 0} \hfill & {{\rm{on}}} \hfill & {{S_3}} \hfill \cr {\partial _x^2H = 0} \hfill & {{\partial _x}{\partial _y}H = 0} \hfill & {{\rm{on}}} \hfill & {{S_4}} \hfill \cr } \tag{1}$$

where

$$\eqalign{ & \Delta^2 = \partial_{x}^{4}+2\partial_{x}^{2}\partial_{y}^{2}+\partial_{y}^{4} \cr & \Omega = \left\{ {\left( {x,y} \right): - a < x < a, - b < y < b} \right\} \cr & {S_1} = \left\{ {\left( {x,y} \right):x = a, - b \le y \le b} \right\} \cr & {S_2} = \left\{ {\left( {x,y} \right):x = - a, - b \le y \le b} \right\} \cr & {S_3} = \left\{ {\left( {x,y} \right): - a \le x \le a,y = b} \right\} \cr & {S_4} = \left\{ {\left( {x,y} \right): - a \le x \le a,y = - b} \right\} \cr & \partial \Omega = \bigcup\limits_{i = 1}^4 {{S_i}} \cr} \tag{2}$$

The function $H:\mathbb{R^2} \to \mathbb{R}$ is considered to belong to $C^{\infty}(\mathbb{R}^2)$.

Then I want to show that

$$\begin{align} \partial_{x}^{2}H&=0 \qquad \text{in} \qquad \Omega\\ \partial_{y}^{2}H&=0 \qquad \text{in} \qquad \Omega\\ {\partial _x}{\partial _y}H&=0 \qquad \text{in} \qquad \Omega \end{align} \tag{3}$$

But I don't have any idea on how to proceed!

Any hints or help is appreciated. :)

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