Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 20

In a steady state, $\nu u_s^{\prime \prime }=-G\sin (n\pi x)$. Since the second derivative of the sine is $-(n\pi {)}^2$ times itself,The zero boundary values at $x=0,1$ force $A=B=0$. Therefore

The steady profile has no time derivative. Substituting $u=\tilde{u}+u_s$ and using its steady equation cancels the forcing, givingThe initial condition isThe resulting heat equation is independent of $G$, as are its boundary conditions. Its initial amplitude still depends on $G$; removing a source from the evolution equation does not remove it from the whole problem.

Separate $\tilde{u}=f(t)g(x)$. On a nonzero product region,A decaying temporal factor requires $\lambda >0$, with $f=C{e}^{-\nu \lambda t}$. The zero spatial boundary values give $g=A\sin (\sqrt{\lambda }x)$ and $\sin \sqrt{\lambda }=0$, hence $\lambda =j^2{\pi }^2$ with $j=1,2,\dots$. Zero or negative separation parameters give no nonzero spatial mode with both boundary values zero.

The initial profile is already exactly the $j=n$ sine eigenfunction. Its coefficient therefore determines a single separated solution:The exponential tends to zero because $\nu >0$. More generally sine orthogonality would determine a sum of such modes, but all other coefficients vanish here. This is the transient of the heat equation with a single stationary sine forcing.

Adding the steady and transient parts givesIntegrating the sine across the channel gives ${\int }_0^1\sin (n\pi x)dx=[1-(-1{)}^n]/(n\pi )$. ThusFor even $n$ the signed flow rate is zero, because positive and negative velocity lobes cancel. For odd $n$ the bracket is two. Both expressions vanish initially and approach their steady values exponentially; $G=0$ gives the trivial zero solution throughout.