Evaluating Solution to Laplace's Equation at Centre of Square

Question

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.

Answer 1

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?