Solving Partial Differential Equation with separation of variables.

Question

I have been given this question as an assignment

So far I have made a start solving one of the possible solutions and have made it this far.

My question may seem simple but am I right so far? If so how do I solve for the constants D and E?

Answer 1

You don't have to care about the $T(t)$ part of the solution. The boundary conditions are true for all $t$, so you necessarily have

\begin{array}{rrr} u(0,t) = X(0)T(t) = 0 &\implies & X(0)=0 \\ u_x(1,t) = X'(1)T(t) = 0 &\implies & X'(1) = 0 \end{array}

Since the only way for a function of $t$ to be zero for all $t$ is if it's zero everywhere, $T(t)\equiv 0$, which we don't want.

Now go through the 3 possible options for $X(x)$ and check whether there exists a non-zero pair $(A,B)$ that can satisfy both conditions.

Example: The second option is $X(x) = Ax+B$ which gives

\begin{align} X(0) = B &= 0 \\ X'(1) = A &= 0 \end{align}

Clearly, the only solution is $A=B=0$. Therefore, this can't be a solution.

Note: This solution need not be unique. One of the remaining options will leave you with a free constant. This is fine as long as the solution isn't identically zero.

Spoilers:

1)

$X(0) = A + B = 0 \implies B = -A \implies X(x) = A(e^{kx}-e^{-kx})$. Then, $X'(1) = kA(e^k + e^{-k}) = 0$. This only works if $k=0$ or $A=0$, both of which result in $X(x)\equiv 0$. Hence, this can't be a solution.

3)

$X(0)=A=0 \implies X(x) = B\sin(kx)$. Then, $X'(1) = kB\cos(k) = 0$. This is valid for $B\ne 0$ if you select $k$ such that $\cos(k)=0$, or $k=(2n+1)\frac{\pi}{2}$