Answer to Question #188624 in Differential Equations for chad

Question

A rectangular steel sheet is bounded by the axis x = 0, y = 0, x = a and y = b. The temperature along the edge x = 0 are kept at 100°C and other edges are at 0°C. Let u(x,y) denote the temperature satisfying the equation [15] + d^2u/dx^2 + d^2u/dy^2 = 0 Find the steady state temperature u(x, y), by assuming the solution to be of the form u(x, y) = (AePX + Be-px)(C cospy + D sin py).

Answer 1

Solution

Solution u(x, y) = (Ae^px + Be^-px)(C cospy + D sin py) satisfy the equation ∂²u/∂x² + ∂²u/∂y² = 0 because ∂²u/∂x² = p²u, ∂²u/∂y² = -p²u.

Satisfying boundary conditions at y = 0 and y = b we’ll get C = 0, D sin pb = 0. From the last equality p = πn/b for any natural n and the solution is

$u(x,y)\ =\ \sum_{n=1}^{\infty}{(A_ne^{\pi nx/b}\ +\ B_ne^{-\pi nx/b})}sin\left(\frac{\pi n}{b}y\right)$

Expanding function f(y) on interval (0,b) into the series by sin(πny/b)  we’ll get f_n = 2[1-(-1)ⁿ]/ πn

Satisfying boundary conditions at x = 0 and x = a we’ll get

A_n + B_n = 200[1-(-1)ⁿ]/ πn, A_ne^πna/b + B_ne^-πna/b= 0   =>   B_n =  - A_ne^2πna/b , A_n (1- e^2πna/b ) = 200[1-(-1)ⁿ]/ πn =>   A_n = 200[1-(-1)ⁿ]/ [πn (1- e^2πna/b )], B_n =  - 200[1-(-1)ⁿ]e^2πna/b / [πn (1- e^2πna/b )].

Therefore solution is

$u(x,y)\ =\ \frac{200}{\pi}\sum_{n=1}^{\infty}\frac{e^{\pi nx/b}\ -\ e^{2\pi na/b}e^{-\pi nx/b}}{1-e^{2\pi na/b}}\frac{1-{(-1)}^n}{n}sin\left(\frac{\pi n}{b}y\right)\ =\$

$=\frac{400}{\pi}\sum_{m=0}^{\infty}\frac{sinh(\pi(2m+1)(a-x)/b)}{sinh(\pi(2m+1)a/b)}\frac{1}{(2m+1)}sin\left(\frac{\pi(2m+1)}{b}y\right)$

Answer

$u(x,y)=\frac{400}{\pi}\sum_{m=0}^{\infty}\frac{sinh(\pi(2m+1)(a-x)/b)}{sinh(\pi(2m+1)a/b)}\frac{1}{(2m+1)}sin\left(\frac{\pi(2m+1)}{b}y\right)$