Answer to Question #128652 in Differential Equations for Duaa

Question

Solve the boundary value problem:
∂u/∂t = β(∂^2u/∂x^2), 0 < x < L, t > 0
u(0, t) = 0, u(L, t) = 0, t > 0
u(x, 0) = f(x) 0 < x < L

Answer 1

$u=X(x)*T(t)\\ \dfrac{T'}{\beta T}=\dfrac{X'}{X}=const=-\lambda\\ X''+\lambda X=0\\ X(x)=C_1\cos(\sqrt{\lambda}x)+C_2\sin(\sqrt{\lambda}x)\\ X(0)=C_1*1+C_2*0=C_1=0=>X(x)=\sin(\sqrt{\lambda}x)\\ X(L)=\sin(\sqrt{\lambda}L)=0\\ \sqrt{\lambda}L=\pi *k=>\lambda=(\dfrac{\pi *k}{L})^2\\ X(x)=\sin(\sqrt{\lambda}x),\\ T(t)=C*e^{-\beta\lambda t}, \lambda=(\dfrac{\pi *k}{L})^2=>\\ We \space know \space that \space:u=X(x)*T(t), \space so\\ u(x,t)=\underset{k=0}{\sum}C*e^{-\beta\lambda t}*\sin(\sqrt{\lambda}x),\\ where\\ C=\dfrac{2}{L}\overset{L}{\underset{0}{\int}}f(x)*\sin(\sqrt{\lambda}x)dx,is \space coefficient \space of \space a \space Fourier \space series\\ \lambda=(\dfrac{\pi *k}{L})^2$