Like this previous question, Solving the heat equation using Fourier series, I too am reading the same wikipedia article, http://en.wikipedia.org/wiki/Heat_equation#Solving_the_heat_equation_using_Fourier_series, and I have some specific questions.
1) $\frac{T'(t)}{\alpha T(t)}=\frac{X''(x)}{X(x)}$, why does this imply that both sides are equal to some constant value $-\lambda$?
2) Given $T(t)=Ae^{-\lambda\alpha t}$ and $X(x)=B\sin(\sqrt{\lambda}x)+C\cos(\sqrt{\lambda}x)$, $C=0$, how does one see that for some positive integer $n$; $\sqrt\lambda=n\frac{\pi}{ L}$?
3) Regarding the sum of solutions $u(x,t)=\Sigma_{n=1}^{\infty}D_n\sin(\frac{n\pi x}{L})e^{\frac{-n^2\pi^2\alpha t}{L^2}}$, where $D_n=\frac{2}{L}\int_0^L f(x)\sin(\frac{n\pi x}{L})\,dx $, where can I find a more thorough explanation for this? I am familiar with linear combinations of solutions also being solutions, but why is there suddenly a definite integral there?
Thanks!
It'll be clear to you if you solve it with your own hand.
Note that the left hand side is a variable depending only on $t$, while the right had side is one depending only on $x$. How can these two variables be the same. The only survival way is they both degenerate to the same constant.
Substitute $C=0$ into $X(x)$ and we see $X(x)=B\sin(\sqrt\lambda x)$. According to the boundary condition $X(0)=X(L)=0$, you can easily solve $\lambda$.
Combine the initial condition $u(x,0)=X(x)T(0)=f(x)$, using Fourier transformation, you can eventually get the solution.
So, why not write a solution by yourself.