Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 20

Yes. The wave equation is linear and homogeneous. For $\mathcal{W}={\partial }_x^2-c^{-2}{\partial }_t^2$ and constants $A,B$,If both terms on the right vanish, their combination is also a solution. Thus the superposition principle applies, and it also preserves homogeneous fixed-end boundary conditions.

Use separation of variables, $f=X(x)T(t)$, and setFor $\kappa =-{\lambda }^2$, with $\lambda >0$, the separated solutions areFor $\kappa =0$, they areFor $\kappa =+{\lambda }^2$, they areArbitrary admissible superpositions of separated solutions are again solutions. An individual product is only a separated family, not the entire general solution. Without prescribed boundaries, the general twice-differentiable solution is equivalently D'Alembert's formula,For example, changing to $\xi =x-ct$, $\eta =x+ct$ makes the differential equation $f_{\xi \eta }=0$, whose two integrations give this form.

At $x=0$ and $x=L$, the fixed-end boundary conditions require $X(0)=X(L)=0$. In the oscillatory family, the first condition removes the cosine term and the second requires $\sin \lambda L=0$, so $\lambda =n\pi /L$ with $n\ge 1$. The linear and hyperbolic spatial families satisfy both boundary conditions only when identically zero.

Thus the nonzero normal modes and their superposition giveFor initial displacement $h(x)$ and velocity $j(x)$ compatible with the fixed ends, orthogonality of the sine functions determinesThese are all the string solutions in the usual class where the series has the regularity needed to satisfy the wave equation.

The initial displacement is already the first spatial normal mode, so sine orthogonality gives $A_1=1$ and every other $A_n=0$. Zero initial velocity gives every $B_n=0$. ConsequentlyBoth fixed-end values vanish for all time, and differentiation shows that the initial velocity is zero. This is the fundamental standing wave, with angular frequency $\pi c/L$.