Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 2

For the linearly damped string, substitute $u(x,t)=X(x)T(t)$ and divide by $XT$:Thus separation of variables givesThe fixed-end Dirichlet boundary conditions require $X(0)=X(L)=0$. Nonzero solutions are $X_n(x)=\sin (n\pi x/L)$, with ${\alpha }_n=n\pi /L$, $n\ge 1$. Neither $\alpha =0$ nor a negative spatial eigenvalue produces a nonzero function obeying both endpoint conditions.

For each sine normal mode, the temporal characteristic equation is $s^2+2\lambda s+{\alpha }_n^2=0$, with roots $s=-\lambda \pm \sqrt{{\lambda }^2-{\alpha }_n^2}$. Since $\lambda L/\pi =M+1/2$, the first $M$ modes are overdamped and all higher modes are underdamped. No integer $n$ gives critical damping.

The zero initial displacement excludes the hyperbolic cosine and cosine terms. HencewhereTo impose the initial velocity, let $f_n=\frac{2}{L}{\int }_0^{L}f(x)\sin ({\alpha }_{n}x)dx$ be its Fourier sine series coefficients. Differentiating the expansion at $t=0$ gives $A_n{\Omega }_n=f_n$ for the overdamped modes and $A_n{\omega }_n=f_n$ for the underdamped modes. These determine all coefficients. Despite the hyperbolic sine, the first modes decay: ${\Omega }_n<\lambda$, so both temporal exponential rates are negative.

Here $\lambda =7/2$ and ${\alpha }_n=n$. LetTwo applications of integration by parts giveSolving these equations,Consequently the required amplitudes areThe stated initial velocity is nonzero at the fixed endpoints, whereas a classical solution differentiable there would require zero endpoint velocity. Thus it is understood on the open interval, or as an L2 space initial datum in the usual finite-energy solution. Its Fourier sine series takes value zero at the endpoints; this changes no initial energy or interior initial data.