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On $0<x<L$ with zero-temperature endpoints, let the initial temperature rise linearly from zero to height $H$ at $L/2$ and then fall linearly to zero. Integration by parts on the two halves gives Fourier sine series coefficients $8H\sin (n\pi /2)/({\pi }^2{n}^2)$. Under the heat equation $T_t=D{T}_{xx}$ with $D>0$, each coefficient decays by $e^{-D(n\pi /L{)}^{2}t}$, giving the displayed solution. Only odd spatial modes occur; absolute coefficient summability proves uniform convergence for $t\ge 0$ via the Weierstrass M-test. This spatial profile also occurs in a midpoint-plucked fixed-end string, where the time factors oscillate instead of decaying.

Separation of variables in the wave equation with fixed endpoints selects $H_n(x)=\sin (n\pi x/L)$ and angular frequencies ${\omega }_n=cn\pi /L$, $n\ge 1$. The time factors may be chosen as $F_n(t)=\cos {\omega }_{n}t$, $G_n(t)=\sin {\omega }_{n}t$. By Fourier sine series orthogonality, the initial data giveHence $u={\sum }_{n\ge 1}[a_n\cos {\omega }_{n}t+b_n\sin {\omega }_{n}t]\sin (n\pi x/L)$. The factor $1/{\omega }_n$ in $b_n$ is required because differentiating the time sine supplies ${\omega }_n$.

The given displacement is a symmetric triangular tent: it is zero at both endpoints and has height $k$ at $L/2$. Integrating each linear half gives $a_n=8k\sin (n\pi /2)/(n^2{\pi }^2)$. Release from rest gives $b_n=0$, so the midpoint-plucked fixed-end string isOnly odd modes occur, by symmetry about the midpoint. The initial slope corner makes this a finite-energy weak solution; it is a classical solution away from the propagating corner lines.