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By considering three similar problems, show that, at the centre of the square, $T = \frac{H}{4}$.

The system I have is Laplace's equation in the plane: $$\frac{\partial ^2 T}{\partial x^2} + \frac{\partial ^2 T}{\partial y^2} = 0$$ for $(x,y) \in [0, L]^2$

With the boundary conditions: $ T(0, y) = T(L, y) = T(x, 0) = 0$, and $T(x, L) = H$ where $H$ is a constant.

Solving for $T(x,y)$ is routine, but I am struggling to figure out the solution to the following problem:

By considering three similar problems, show that, at the centre of the square, $T = \frac{H}{4}$.

I have tried solving for T on sections of the square using what I can deduce (e.g. $\frac{\partial T}{\partial x}(\frac{L}{2}, y) = 0$, by symmetry) but have then been unable to deduce that $T = \frac{H}{4}$ at the centre of the square.

I would be very grateful for any help.

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Consider "rotating your boundaries around", i.e.

  • $T$ is the solution to Laplace's equation with $T = H$ on the top side and $T = 0$ on other sides.

  • $T_1$ is the solution to Laplace's equation with $T_1 = H$ on the left side and $T_1 = 0$ on other sides.

  • $T_2$ is the solution to Laplace's equation with $T_2 = H$ on the bottom side and $T_2 = 0$ on other sides.

  • $T_3$ is the solution to Laplace's equation with $T_3 = H$ on the right side and $T_3 = 0$ on other sides.

Be aware that for any given choice of boundary conditions, the solution to Laplace's equation is unique. From this, can you argue that the solutions $T$, $T_1$, $T_2$ and $T_3$ are really just rotations of each other?

Therefore, can you see how the values of $T$, $T_1$, $T_2$ and $T_3$ at the centre of the square are related?

Finally, $T + T_1 + T_2 + T_3$ is a solution to Laplace's equation, obeying which boundary conditions? And what must this solution be? Can you finish off from here?

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  • $\begingroup$ Ah great! I was approaching this incorrectly - I was trying to solve for T on subsets of $[0, L]^2$ and for some reason hoping this would reveal additional information! I will keep this trick handy for similar situations - i.e. where the PDE and BCs are linear and solutions are unique, so that solutions to 'similar problems' can be 'added'. $\endgroup$ – John Don Apr 22 '17 at 10:33
  • $\begingroup$ Nicely explained too may I add! $\endgroup$ – John Don Apr 22 '17 at 10:39
  • $\begingroup$ sorry, I understood everything up to the point where 'And what must this solution be?' If the boundaries are all T=H for the square, does it imply that the whole square has T=H, if so why? $\endgroup$ – JustWandering Apr 16 '19 at 16:38
  • $\begingroup$ @JustWandering If $T = H$ on all boundaries, then it does imply that $T = H$. (One argue uses the maximum principle. This says that any solution to Laplace's equation takes its maximum value on the boundary. Now apply the maximum principle to $T$ and $-T$...) $\endgroup$ – Kenny Wong Apr 16 '19 at 20:23

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