Question

Use the method of separation of variables to determine the general solution to the one dimensional heat equation ∂²u/∂x²=1/α(∂u/∂t),0≤ x≤ l,t≥0 subject to the boundary conditions u(0,t)=u((l,t)=0 and the initial condition u(x,0)=f(x). Determine the particular solution when f(x)=x²

Answer 1

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

$X(x)T'(t)=1/\alpha\cdot T(t)X''(x)$

$\alpha T'(t)/T(t)=X''(x)/X(x)=-\lambda$

$T'(t)=-\lambda T(t)/\alpha$

$X''(x)+\lambda X(x)=0$

$X(0)=0,X(l)=0$

$\lambda>0$

$X(x)=c_1cos(\sqrt{\lambda}x)+c_2sin(\sqrt{\lambda}x)$

$0=X(0)=c_1$

$0=X(l)=c_2sin(l\sqrt{\lambda})$

$sin(l\sqrt{\lambda})=0\implies l\sqrt{\lambda}=n\pi,\ n=1,2,3,...$

$\lambda_n=(\frac{n\pi}{l})^2,\ X_n(x)=sin(\frac{n\pi x}{l})$

$\lambda=0$

$X(x)=c_1+c_2x$

$0=X(0)=c_1,\ 0=X(l)=c_2l\implies c_2=0$

So, in this case the only solution is the trivial solution and so λ=0 is not an eigenvalue for this boundary value problem.

$\lambda<0$

$X(x)=c_1cosh(\sqrt{-\lambda}x)+c_2sinh(\sqrt{-\lambda}x)$

$0=X(0)=c_1$

$0=X(l)=c_2sinh(l\sqrt{-\lambda})$

$\lambda_n=(\frac{n\pi}{l})^2,\ X_n(x)=sin(\frac{n\pi x}{l})$

for the time differential equation:

$T(t)=ce^{-k\lambda_nt}=ce^{-k(\frac{n\pi}{l})^2t}$

the general solution:

$u_n(x,t)=B_nsin(\frac{n\pi x}{l})e^{-k(\frac{n\pi}{l})^2t}$

where

$B_n=\frac{2}{l}\int^l_0 f(x)sin(\frac{n\pi x}{l})dx=\frac{2}{l}\int^l_0 x^2sin(\frac{n\pi x}{l})dx=$

$=\frac{2}{n^3\pi^3}(2\pi lnxsin(n\pi x/l)+(2l^2-n^2\pi^2x^2)cos(n\pi x/l))|^l_0=$

$=\frac{2}{n^3\pi^3}((2l^2-n^2\pi^2l^2)cos(n\pi )-2l^2)$

for odd n:

$u_n(x,t)=-\frac{2l^2}{n^3\pi^3}((4+n^2\pi^2)sin(\frac{n\pi x}{l})e^{-k(\frac{n\pi}{l})^2t}$

for even n:

$u_n(x,t)=-\frac{2l^2}{n\pi}$