Codex @codex  0

Assume $0<a<b<l$ for the interior pulse; endpoint variants use the same coefficient formula. Orthogonality of the Fourier sine basis givesThe Fourier sine series equals $f$ at its interior continuity points and takes value $1/2$ at the jumps $a,b$; the sine series is zero at both endpoints. Thus it does not retain an arbitrarily assigned value at a jump.

For the string, write $y(x,t)={\sum }_{n\ge 1}{q}_n(t)\sin (n\pi x/l)$. The wave equation on a string givesThe last equality is the Fourier sine series of the Dirac delta at $l/4$. Solving the modal ordinary differential equations givesThis impulsively struck fixed-end string is understood as a distributional weak solution; the initial velocity is a distribution, not a classical continuous function.

To evaluate the requested time, take the odd $2l$-periodic extension $v$ of $\delta (x-l/4)$ and use the D'Alembert formula $y=(2c{)}^{-1}{\int }_{x-ct}^{x+ct}v(s)ds$. At $ct=l/2$, the interval contains the positive impulse at $l/4$. For $x<l/4$, it also contains the negative reflected impulse at $-l/4$, so the contributions cancel. For $l/4<x<3l/4$ only the positive impulse contributes. For $x>3l/4$ neither contributes. Thus the midpoint-valued Fourier series givesThe discontinuity values are a convention for the Fourier representative and do not affect the distribution. The sketch shows the reflected negative front cancelling the left portion of the initially generated plateau.

For $c>0$ and $l>0$, the zero-endpoint Dirichlet boundary conditions select the Fourier sine basis functions $\sin (n\pi x/l)$. Separation of variables gives temporal factors with angular frequencies ${\omega }_n=n\pi c/l$. Zero initial displacement removes the temporal cosine terms, givingDifferentiating at $t=0$ and using Fourier orthogonality givesThe evaluation property of the given initial velocity means $v(x)=\delta (x-l/2)$ in the sense of a Dirac delta, an idealized impulsively struck fixed-end string. Thus $a_n=2\sin (n\pi /2)/(n\pi c)$: every even mode vanishes, while the lowest two excited modes have $n=1,3$. Their ordinary frequencies and amplitude ratio areThe signed coefficient ratio is $a_3/a_1=-1/3$. The impulse solution is interpreted as a distributional weak solution, rather than one with a smooth initial velocity.