Finite difference nodal temperatures in a square plate

Finite-difference method

The region is covered by a grid of nodes spaced $\Delta x$ and $\Delta y$. The Laplace equation $\dfrac{\partial^2T}{\partial x^2}+\dfrac{\partial^2T}{\partial y^2}=0$ is replaced using central differences:

$$\frac{\partial^2T}{\partial x^2}\approx\frac{T_{m+1,n}-2T_{m,n}+T_{m-1,n}}{\Delta x^2},\qquad \frac{\partial^2T}{\partial y^2}\approx\frac{T_{m,n+1}-2T_{m,n}+T_{m,n-1}}{\Delta y^2}$$

For $\Delta x = \Delta y$ this gives the interior-node equation

$$T_{m,n} = \frac{1}{4}\left(T_{m+1,n}+T_{m-1,n}+T_{m,n+1}+T_{m,n-1}\right)$$

so each node is the average of its four neighbours. Writing it for every unknown node gives a set of linear equations, solved by matrix inversion or iteration (Gauss-Seidel).

Nodal equations

   500   500   500   500
 100 [ 1 ]-[ 2 ] 100
      |      |
 100 [ 3 ]-[ 4 ] 100
   100   100   100   100

Node 1 neighbours: top 500, left 100, node 2, node 3.

$$\begin{aligned} 4T_1 &= 500 + 100 + T_2 + T_3\\ 4T_2 &= 500 + 100 + T_1 + T_4\\ 4T_3 &= 100 + 100 + T_1 + T_4\\ 4T_4 &= 100 + 100 + T_2 + T_3 \end{aligned}$$

Solution

The conditions are symmetric about the vertical centre line, so $T_1 = T_2$ and $T_3 = T_4$:

$$\begin{aligned} 3T_1 - T_3 &= 600\\ 3T_3 - T_1 &= 200 \end{aligned}$$

From the second, $T_1 = 3T_3 - 200$. Substitute in the first: $9T_3 - 600 - T_3 = 600$, so $T_3 = 150$ °C and $T_1 = 250$ °C.

Check in node 1 equation: $(500+100+250+150)/4 = 250$. Correct.

Answer: $T_1 = T_2 = 250\ ^\circ$C; $T_3 = T_4 = 150\ ^\circ$C.