method of separation of variables for heat transfer with mixed conditions

Question

$u_t=c^2 u_{xx}, \quad \forall \quad 0<x<L \quad t>0$

$u_x(L,t)=0, \quad u(0,t)=0$

$u(x,0)=f(x)$

Can anyone please tell me how to apply the boundary conditions to this problem and arrive at a general solution.

Any help is much appreciated.

Answer 1

Since you specify "separation of variables", start by looking for solutions of the form u(x,t)= X(x)T(t). The differential equation becomes $XT'= c^2TX''$ which we can write as $\frac{T'}{4T}= c^2\frac{X''}{X}$. If we were to vary t while holding x constant, the right side would not change so the left side must be constant for varying t. Similarly, if we were to vary x while holding t constant, the left side would not change so the right side must be constant for varying x. The two sides must be equal to the same constant: we must have $\frac{T'}{T}= \lambda$ and $c^2\frac{X''}{X}= \lambda$ for some constant $\lambda$. The boundary conditions are $u(0, t)= X(0)T(t)= 0$ for all t so we must have X(0)= 0, and $u_x(L, t)= X'(L)T(t)= 0$ for all t so we must have X'(L)= 0.

That is, you want to solve $c^2X''= \lambda X$ with boundary conditions X(0)= 0, X'(L)= 0.

What kinds of solutions you get depends upon what $\lambda$ is. You should be able to recognize that, in order to get a non-trivial solution for X, $\lambda$ must be negative. Write $\lambda= -\omega^2$. The differential equation becomes $c^2X''= -\omega^2 X$. That has general solution $X(x)= A \cos\left(\frac{\omega}{c}x\right)+ B\sin\left(\frac{\omega}{c}x\right)$. The condition that $X(0)= 0$ give $A= 0$ so we must have $X(x)= B\sin\left(\frac{\omega}{c}x\right)$. Then $X'(x)= B\frac{\omega}{c}cos\left(\frac{\omega}{c}x\right)$ so that $X'(L)= B\frac{\omega}{c}cos\left(\frac{\omega}{c}\right)= 0$.

We cannot have B= 0 also- that would give the "trivial" solution for X so that u could not satisfy the initial condition. Instead we must have $cos\left(\frac{\omega}{c}L\right)= 0$. Cosine is 0 only for odd multiples of $\frac{\pi}{2}$ so we must have $\frac{\omega}{c}L= (2n+1)\frac{\pi}{2}$. We must have $\omega= (2n+1)\frac{c\pi}{2L}$ so $\lambda= -(2n+1)^2\frac{c^2\pi^2}{4L^2}$.

Now put that $\lambda$ into $T'= 4\lambda T$ and solve for T(t). The solution to the problem is the sum of all such X(x)T(t) summed over n.