Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 73 / 1 / vi / Solution

For $0<s<1$, differentiation givesIts unique zero solves $e^{-(1-s)t/\tau }=s$. Therefore the mobile-volume peak of an exponentially forced porous current occurs atThe derivative is initially positive and is negative after this time. More explicitly, at the stationary point $V^{\prime \prime }=-Qs{e}^{-st/\tau }/\tau <0$. Thus it is the global positive maximum, withThe claim of a finite maximum requires nonzero capillary residual trapping. At $s=0$ the mobile volume rises monotonically towards $Q\tau$ and the peak time tends to infinity. As $s\to 1$, the limiting peak time is $\tau$ and the limiting maximum is $Q\tau /e$.