Given $u_{tt} - u_{xx} = 0$ where $0 < x <3$ with boundary conditions $u(0,t) = u(3,t) = 0$
To solve it, separate the variables $$u(x,t) = T(t)X(x)$$ and get $$\frac{T''}{T} = \frac{X''}{X} = - k$$ let $k = \lambda^2$, we have $$X'' + \lambda ^2 X = 0$$ The general solution can be written in terms of sine and cosine. $$X(x) = C cos \lambda x + D sin \lambda x$$ Now use the boundary conditions to determine constants $C$ and $D$ $$X(0) = C = 0$$ $$X(3) = 0 \cdot cos 3 \lambda + D sin 3 \lambda = 0$$ $$X(3) = D sin(3\lambda) = 0$$ To avoid getting the trivial solution, we let $D \neq 0$. So it left $$sin(3\lambda) = 0$$ Solve for $3\lambda$ $$3\lambda = n \pi, n = 3,6,9,..$$ $$ \lambda = \lambda_n = \frac{n \pi}{3}, n = 3,6,9,..$$ I'm stuck here because it seems doesn't make sense to me. In general when $0 < x < L$, we would have $$\lambda_n = \frac{n \pi}{L}, n = 1,2,3...$$ Now in this case, $n$ has to be multiple of 3 otherwise $sin(3 \lambda) \neq 0$. However, if $n$ begin at 3, can we still use principle of of linear superposition, such that the solution of PDE is a linear combination of all product of $T_n(t)X_n(x)$ over all eigenvalues? I'm not sure what I'm missing here, this problem just seems straight to me, any suggestion would be really appreciated!!