Answer to Question #214930 in Differential Equations for Vimal

Question

ut =50uxx

u(0,t) = u(π,t) = 0, t > 0

u(x,0) = { x , 0 < x < π / 2

{ 4 , π / 2 < x < π

Answer 1

Solution

Let u(x,t) = X(x)T(t)

Substitution into equation gives X(x)T’(t) = 50 X’’(x)T(t) => T’(t)/[50 T(t)] = X’’(x)/ X(x)

Variables t and x are independent. So T’(t)/[50 T(t)] = X’’(x)/ X(x) = -k²  

NOTE: variant k² at the right side is impossible because equation X’’(x) - k² X(x) = 0 with the given boundary conditions have only trivial solution.

Therefore we’ll get two ordinary differential equations

T’(t) + k² 50 T(t)] = 0

 X’’(x) + k² X(x)  = 0

From this equations T(t) = Aexp(-k²50t), X(x) = Bsin(kx) + Ccos(kx) (A, B, C – arbitrary constants)

From boundary conditions u(0,t) = u(π,t) = 0, t > 0 => X(0) = X(π) = 0 and C = 0 X(x) = Bsin(k π) = 0 => k = ±n (n = 1,2, . . .).

Solution is

$u(x,t)=\sum_{n=1}^{\infty}A_ne^{-50n^2t}sin(nx)$

From initial condition

$u(x,0)=\sum_{n=1}^{\infty}A_nsin(nx)$

Multiplying by sin(mx) and integrating with bounds 0, π

$A_n=\frac{2}{\pi}\int_{0}^{\pi}{u(x,0)}sin(nx)dx=\frac{2}{\pi}\int_{0}^{\pi/2}xsin(nx)dx+\frac{8}{\pi}\int_{\pi/2}^{\pi}{sin(nx)}dx=\frac{2}{\pi n^2}\left(sin(nx)-xncos(nx)\right)|_0^{\pi/2}-\frac{8}{\pi n}cos(nx)|_{\pi/2}^\pi$

$A_n=\frac{2}{\pi n^2}\left[sin(\pi n/2)-\frac{\pi n}{2}cos(\pi n/2)\right]-\frac{8}{\pi n}\left[cos(\pi n)-cos(\pi n/2)\right]$