The heat equation

Sine functions with these particular frequencies are exactly the shapes that keep zero temperature at both ends while decaying independently in time; since every reasonable initial shape can be written as a sum of such sine waves (a Fourier sine series), the same decomposition immediately produces the solution at all later times.

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.

Heat has nowhere to come from except the initial data and the boundary temperatures: if a hidden interior point were the single hottest spot at some later time, heat would have to be spontaneously created there, which the diffusion process forbids — diffusion only ever moves heat from hot to cold, it never manufactures a new peak in empty space.

Fix $\varepsilon>0$ and define $v(x,t) = u(x,t) - \varepsilon t$, so $v_t - \alpha v_{xx} = u_t - \varepsilon - \alpha u_{xx} = -\varepsilon < 0$ everywhere on the open rectangle. Suppose, for contradiction, that $v$ attains its maximum over the closed rectangle at an interior point $(x_0,t_0)$ with $0<x_0<L$ and $0<t_0\le T$ — i.e. off the parabolic boundary.

Because $x_0$ is an interior spatial maximum, the second-derivative test gives $v_{xx}(x_0,t_0)\le 0$. Because $t_0$ is where the maximum over $t\in[0,T]$ is reached (either an interior critical time, giving $v_t(x_0,t_0)=0$, or the endpoint $t_0=T$ approached from the left, giving $v_t(x_0,t_0)\ge 0$), in either case $v_t(x_0,t_0)\ge 0$.

Combining the two inequalities, $v_t(x_0,t_0) - \alpha v_{xx}(x_0,t_0) \ge 0 - \alpha\cdot 0 = 0$, which contradicts $v_t-\alpha v_{xx}=-\varepsilon<0$ at every point. Hence no such interior maximum exists, so $v$'s maximum lies on the parabolic boundary $\Gamma$ (the initial edge together with the two lateral edges), giving $u(x,t) \le \varepsilon t + \max_{\Gamma} u$ everywhere. Letting $\varepsilon \to 0^+$ proves the maximum principle for $u$ itself.