Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 2

For separation of variables, take $u(x,t)=X(x)T(t)$ in the damped wave equation. Dividing by $c^{2}XT$, with $c>0$, gives $X^{\prime \prime }/X=(T^{\prime \prime }+\alpha T^{\prime })/(c^{2}T)=-k^2$. The separated ordinary differential equations are thereforeThe negative separation constant anticipates the positive spatial eigenvalues selected by the fixed endpoints.

An even function satisfying the spatial equation has $X=C\cos (kx)$. The Dirichlet boundary conditions require $\cos (kL)=0$, so the allowed wavenumbers areThere is no nonzero zero-wavenumber mode. With $T=e^{-\alpha t/2}S$, the time equation becomes $S^{\prime \prime }+{\omega }_n^{2}S=0$, where ${\omega }_n^2=c^2{k}_n^2-{\alpha }^2/4$. Thus superposition of the even normal modes givesThis uses completeness of the even Fourier series modes. Imaginary ${\omega }_n$ describe overdamping. At critical damping, ${\omega }_n=0$, the two displayed exponentials coincide: the complete time factor must instead be $e^{-\alpha t/2}(C_n+D_{n}t)$. This repeated-root exception is necessary for a genuinely general solution.

The angular frequency of an underdamped mode isFor ordinary damping, $\alpha \ge 0$, mode $n$ oscillates precisely when $\alpha <cn\pi /L$. In particular, all modes, including the fundamental, oscillate when $\alpha <c\pi /L$. There are some oscillatory modes for every finite $\alpha$, since the allowed $n$ are unbounded; no finite damping coefficient overdamps the entire ideal string spectrum.

If $\alpha <0$, the same mode has negative damping: oscillatory amplitudes grow as $e^{|\alpha |t/2}$. For modes with real characteristic roots, both roots are positive, since $|\alpha |/2>\sqrt{{\alpha }^2/4-c^2{k}_n^2}$. Hence those modes also grow. Negative damping makes the equilibrium unstable.

Let $d_n$ be the even Fourier coefficient of the initial displacement, with the normalization ${\int }_{-L}^L{\cos }^2(k_{n}x)dx=L$. Writing $s_n=\sin (n\pi /2)=(-1{)}^{(n-1)/2}$ givesZero initial velocity requires the real modal time factorOne consistent choice in the question's complex representation is thereforeFor real ${\omega }_n$, $B_n=\overline{A_n}$; the real-part notation otherwise makes the complex coefficients nonunique. For imaginary ${\omega }_n$, the same formula is interpreted using hyperbolic functions. Its limit at ${\omega }_n=0$ is $d_n{e}^{-\alpha t/2}(1+\alpha t/2)$.

For $l/L\to 0$ at fixed mode number,which can be substituted into $A_n,B_n$ above. The profile specified at $t=0$ is actually nonzero at $x=\pm L$. Thus for finite $l/L$ it cannot be the initial data of a solution continuous at the initial endpoint corners with fixed endpoints. The Fourier series reproduces it in the open interval, with zero endpoint values; it is an L2 space initial-data interpretation. The assumption $l\ll L$ makes this compatibility error exponentially small, rather than exactly zero.