Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability / 5

The chain rule gives $\frac{d}{dx}\cos (\mu x)=-\mu \sin (\mu x)$. Dividing by the nonzero constant $-\mu$ gives the antiderivative

Apply integration by parts with $u=x$ and $dv=\sin (\mu x)dx$. Part (i) gives $v=-\cos (\mu x)/\mu$, so

For the Fourier sine series write $u_0(x)={\sum }_{n\ge 1}{b}_n\sin (n\pi x/L)$. Orthogonality of trigonometric functions givesIn the second half of the integral put $z=L-x$; then $u_0(L-z)=\rho z$ and $\sin [n\pi (L-z)/L]=(-1{)}^{n+1}\sin (n\pi z/L)$. HenceEvery even coefficient is zero. For odd $n$, part (a) or the supplied integral yieldsPutting $n=2m+1$ therefore givesThis is the Fourier sine series of a symmetric tent. The absolute values of its terms are uniformly bounded by a summable sequence, giving absolute convergence and uniform convergence; the piecewise smooth odd extension and the Dirichlet convergence theorem for Fourier series identify its value with the continuous $u_0$, including the zero endpoints and the peak.

The printed lower limit $m=1$ is an error. The original PDF visibly has this lower limit. It omits the nonzero fundamental coefficient $F_0=4L\rho /{\pi }^2$. In particular, the printed series equals $u_0(x)-F_0\sin (\pi x/L)$, rather than $u_0(x)$. At $x=L/2$ it gives $L\rho /2-4L\rho /{\pi }^2$ instead of $L\rho /2$. The corrected lower limit is displayed explicitly above.

Apply separation of variables to each mode of the damped wave equation $u_{tt}+2k^2{u}_t=c^2{u}_{xx}$, assuming $L,c>0$. The spatial ordinary differential equation is $X_n^{\prime \prime }+({\lambda }_n/c{)}^2{X}_n=0$. Its general solution is $C\sin ({\lambda }_{n}x/c)+D\cos ({\lambda }_{n}x/c)$. The boundary condition at $0$ gives $D=0$; a nonzero mode at $L$ requiresOnly odd $n=2m+1$ occur in the initial Fourier sine series. Put ${\omega }_m=(2m+1)\pi c/L$ and $d_m=\sqrt{k^4-{\omega }_m^2}$. The temporal equation $T_m^{\prime \prime }+2k^2{T}_m^{\prime }+{\omega }_m^2{T}_m=0$ has roots $-k^2\pm d_m$. DefineFor distinct roots, let the full amplitude be $q_m(t)=A_m{e}^{-{\alpha }_{m}t}+B_m{e}^{-{\beta }_{m}t}$. The initial displacement and zero initial velocity impose $A_m+B_m=F_m$ and ${\alpha }_m{A}_m+{\beta }_m{B}_m=0$. Solving givesThus the physically consistent superposition isFor ${\omega }_m<k^2$ there is overdamping, with two real decaying exponentials. For ${\omega }_m>k^2$ there is underdamping; set ${\nu }_m=\sqrt{{\omega }_m^2-k^4}$; the equivalent real expression isAt critical damping, the repeated-root case ${\omega }_m=k^2$, take the limit, giving $q_m(t)=F_m(1+k^{2}t)e^{-k^{2}t}$. These formulas all satisfy $q_m(0)=F_m$ and $q_m^{\prime }(0)=0$.

Three source qualifications matter. First, the sum must start at $m=0$, for the same reason as in (b). Second, the printed $X(x)=u_0(x)$ cannot hold for a single separated mode: the tent has $X^{\prime \prime }=0$ on each open linear segment but $X\ne 0$, contradicting $X^{\prime \prime }=-({\lambda }_n/c{)}^{2}X$ for positive ${\lambda }_n$, and its peak is not twice differentiable. Separation applies mode by mode, with the initial tent recovered by superposition. Third, no finite “large” $k$ makes every mode overdamped, because ${\omega }_m\to \infty$. The two-exponential expression permits complex conjugate roots, while a repeated root requires its displayed limiting form. The tent initial data are naturally interpreted through the convergent modal solution, rather than as a single smooth separated product.