Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 2

For the established time-periodic response, take a complex separated solution $v=X(x)e^{i\omega t}$ and set $u=\mathrm{Re}v$. The diffusion equation gives $X^{\prime \prime }=(i\omega /\nu )X$. For $\omega >0$, define the penetration depth $\delta =\sqrt{2\nu /\omega }$; the spatial roots are $\pm (1+i)/\delta$. Decay at infinity selectsIts real part is the solution of Stokes' second problem:The velocity amplitude decays exponentially, while the phase lags by $x/\delta$. Direct substitution verifies the diffusion equation, and the two boundary conditions follow immediately.

These data select a unique periodic response, but do not specify an arbitrary initial transient. With no initial velocity supplied, one may add any suitable solution of the homogeneous-boundary diffusion problem. The displayed solution is the usual established oscillatory flow after transients have decayed.

Separate variables as $u=X(x)T(t)$. The zero values at both plates give the Fourier sine basisExpand the initial profile using orthogonality. With $s=x/L$, its coefficients areConsequentlyOnly odd modes occur because the initial profile is symmetric about the midpoint. The coefficients are absolutely summable, so dominated convergence gives $\boxed{u(x,t)\to 0}$ uniformly as $t\to \infty$. The slowest decay rate is $\nu {\pi }^2/L^2$, associated with $j=1$.