# For a square plate heated to $T$ on only one edge, how can I show that the temperature in the centre is $T/4$?

I am struggling with the following problem:

One edge of the square plate with insulated faces is kept at uniform temperature $u_{0}$ and the other three edges are kept at temperature zero. Without solving a boundary value problem, but by superposition of solutions of like problems to obtain the trivial case in which all four edges are at temperature $U_{0}$, show why the steady temperature at the center of the given plate must be $U_{0}/4$.

What I tried:

Laplace equation of a PDE with four edges have the boundary conditions of the form

$$u(0,y)=g_{1}(y)$$ $$u(L,y)=g_{2}(y)$$ $$u(x,0)=f_{1}(x)$$ $$u(x,H)=f_{2}(x)$$

But all the boundary conditions here are non-homogenous, to get a homogeneous boundary conditions in order to solve the PDE we must split the solution into four parts and the add up the solutions of the four parts after solving each individually with the boundary conditions of each part being homogeneous (This can be done due to the linearity property). An example of the boundary conditions for one part is given below.

$$u(0,y)=0$$ $$u(L,y)=g_{2}(y)$$ $$u(x,0)=f_{1}(x)$$ $$u(x,H)=f_{2}(x)$$

Is my explnation correct and could it be improved upon?

Just looking at physics, if the initial problem has a solution and you rotate the plate of $\pi/2$ you obtain another solution with boundary conditions rotated of $\pi/2$. Perform the same procedure two other times and you end up with four solutions with corresponding four different boundary conditions rotated of $0$, $\pi/2$, $\pi$, $3/2 \:\pi$ respectively.

Each of these solutions, in view of the axial symmetry, attains the same value, say $u$, at the center of the plate. This $u$ is the unknown of the problem.

As the system is linear, if you sum all these solutions you obtain a solution with boundary conditions given by the sum of the four boundary conditions.

It is easy to see that the total boundary conditions is nothing but $U_0$ on every edge of the plate. A solution of this problem is trivially the constant one $u(x,y)=U_0$ and this is the only solution due to the uniqueness theorem. In particular, at the center the value is again $U_0$. It must coincide with $4u$.

Thus $u = U_0/4$.

• Actually I am not sure that the uniqueness theorem is valid here since the solutions we are considering are not continuous in the closed domain and the maximum principle does not hold... However it is evident that the idea of the exercise is the one I illustrated above, more looking at physics than at mathematics. Sep 8, 2016 at 19:31
• Have a look at the 'hard' solution, please.
– Gert
Sep 9, 2016 at 14:18
• @ValterMoretti You can fix the continuity problem by allowing a gradual continuous temperature change from $T$ to $0$ on the heated edge near the ends on a distance $dx$, and then look at the limit $dx\to 0$.
– JiK
Sep 12, 2016 at 19:40
• Or rather, make it $T$ to $T/2$ on the heated edge and $0$ to $T/2$ on the adjacent edges, so the sum of the boundary conditions of the four different cases is exactly $T$.
– JiK
Sep 12, 2016 at 19:42
• Well, yours is another problem with different boundary conditions...You are using some continuous dipendence of the solutions from boundary data, but also this should be proved in this case. Sep 12, 2016 at 20:17

Much as I love Valter's approach, I'm a little incredulous with regards to the $u_0/4$ value. So I decided to try and determine the value, 'the hard way'.

$$u_{xx}+u_{yy}=0$$ $$u(0,y)=0,u(L,y)=0$$ $$u(x,0)=0,u(x,L)=u_0$$ This yields: $$u(x,y)=\displaystyle \sum_{n=1}^{+\infty}A_n\sinh\Big(\frac{n\pi y}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$ And with the fourth boundary condition: $$u(x,L)=u_0=\displaystyle \sum_{n=1}^{+\infty}A_n\sinh(n\pi)\sin\Big(\frac{n\pi x}{L}\Big)$$ $$A_n\sinh(n\pi)=\frac{2}{L}\int_0^Lu_0\sin\Big(\frac{n\pi x}{L}\Big)dx$$ $$A_n=\frac{2u_0}{L\sinh(n\pi)}\int_0^L\sin\Big(\frac{n\pi x}{L}\Big)dx$$ $$A_n=\frac{2u_0}{n\pi\sinh(n\pi)}\big(1-(-1)^n\big)$$ So that: $$u(x,y)=\displaystyle \sum_{n=1}^{+\infty}\frac{2u_0}{n\pi\sinh(n\pi)}\big(1-(-1)^n\big)\sinh\Big(\frac{n\pi y}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$ $$u(x,y)=\frac{2u_0}{\pi}\displaystyle \sum_{n=1}^{+\infty}\frac{\big(1-(-1)^n\big)}{n\sinh(n\pi)}\sinh\Big(\frac{n\pi y}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$ At the centre of the plate $(L/2,L/2)$: $$u(L/2,L/2)=\frac{2u_0}{\pi}\displaystyle \sum_{n=1}^{+\infty}\frac{\big(1-(-1)^n\big)}{n\sinh(n\pi)}\sinh\Big(\frac{n\pi }{2}\Big)\sin\Big(\frac{n\pi }{2}\Big)$$ $$u(L/2,L/2)=\frac{2u_0}{\pi}\displaystyle \sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)\cosh ((k+1/2)\pi)}$$ where we have used $\sinh 2x = 2\sinh x \cosh x$, we have restricted the sum to odd integers $n= 2k+1$ since the terms corresponding to even $n$ vanish. Finally $\sin(\pi(k+1/2))= (-1)^k$.

From this site where $$1/\cosh x$$ is denoted by $$\mbox{Sech}[x]\:,$$ we have

$$\sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)\cosh ((k+1/2)\pi)}= \frac{\pi}{8}$$ so that $$u(L/2,L/2)=\frac{2u_0}{\pi} \frac{\pi}{8} = \frac{u_0}{4}\:.$$

(This answer is a joint work by Gert and V.Moretti)

Below are 3D and contour plots for $u(x,y)$ with $L=1$, $u_0=100$ and using the first five $(n=1,2,3,4,5)$ terms (the even terms are zero), using wolfram alpha's plotting tool:  Obviously more terms are needed to accurately represent the $u(x,1)=100$ condition.

• Its's ok, yours is just an approximate value, it makes sense :-) Sep 9, 2016 at 14:42
• with some elementary computations $u(L/2,L/2) = \frac{u_0}{\pi} \sum_{0}^{+\infty} \frac{1}{(2n+1)\cosh(n+1/2)}$...I looked through several tables of series but I could not find it unfortunately... Sep 9, 2016 at 15:19
• Ok. I'm going to stick that into a spreadsheet and see what gives. Ta.
– Gert
Sep 9, 2016 at 15:31
• I forgot a factor $\pi$ in the argument of $\cosh$!!! Sep 9, 2016 at 15:42
• Type it out again?
– Gert
Sep 9, 2016 at 15:45