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

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.