Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 16 / b

For a classical solution take $F,G$ twice differentiable, so that the second derivatives exist. Direct use of the chain rule giveswith the two profile arguments as specified. Thus $4y_{tt}=25y_{xx}$.

To prove generality, introduce characteristic coordinates $\xi =2x+5t$, $\eta =2x-5t$, whose Jacobian determinant is $-20\ne 0$. For $y=U(\xi ,\eta )$ the differential operator isThe equation becomes $U_{\xi \eta }=0$. Integrating once in each variable yieldsThis is the countertravelling decomposition of the one-dimensional wave equation, with wave speed $5/2$. Once differentiability alone is not sufficient for a classical second-order equation; the displayed argument uses the required second derivatives.

Choose $F(s)=G(s)=\frac{1}{2}\sin s$. The two travelling profiles then add by the sine addition identity:This is a standing wave. At $x=0$ and $x=\pi$ it vanishes for all $t$. At $t=0$ its displacement is $\sin 2x$, and its time derivative $-5\sin 2x\sin 5t$ is zero. Also $y_{tt}=-25y$ and $y_{xx}=-4y$, directly checking the original equation.