Existence of the solution by separation of variables

Seek a solution of the separated form $u(x,t)=X(x)T(t)$. Substituting into $u_t = \alpha u_{xx}$ gives $X(x)T'(t) = \alpha X''(x)T(t)$, and dividing by $\alpha X(x)T(t)$ separates the variables: $\dfrac{T'(t)}{\alpha T(t)} = \dfrac{X''(x)}{X(x)} = -\lambda$, where $\lambda$ must be a constant because the left side depends only on $t$ and the right side only on $x$.

The boundary conditions $u(0,t) = u(L,t) = 0$ force $X(0)=X(L)=0$. The eigenvalue problem $X'' + \lambda X = 0$ with these boundary conditions has nontrivial solutions only for $\lambda_n = (n\pi/L)^2$, $n=1,2,3,\dots$, with eigenfunctions $\sin\!\left(\frac{n\pi x}{L}\right)$; any other value of $\lambda$ forces $X \equiv 0$. Solving $T'(t) = -\alpha\lambda_n T(t)$ then gives $T_n(t) = e^{-\alpha (n\pi/L)^2 t}$.

Each product $u_n(x,t) = \sin\!\left(\frac{n\pi x}{L}\right)e^{-\alpha (n\pi/L)^2 t}$ solves the equation and boundary conditions. By linearity, so does any finite or (under mild convergence conditions) infinite sum $u(x,t) = \displaystyle\sum_{n=1}^{\infty} b_n \sin\!\left(\frac{n\pi x}{L}\right) e^{-\alpha (n\pi/L)^2 t}$. Setting $t=0$ and using the orthogonality relation $\int_0^L \sin(\frac{n\pi x}{L})\sin(\frac{m\pi x}{L})\,dx = 0$ for $n \ne m$ (and $=L/2$ for $n=m$) to project $f(x)$ onto each mode yields exactly the coefficients $b_n = \dfrac{2}{L}\displaystyle\int_0^L f(x)\sin\!\left(\frac{n\pi x}{L}\right)dx$, so the initial condition $u(x,0) = f(x)$ is matched termwise.