 Solution

ID: 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 gives

u (x, t) = \frac{G}{ν n ^{2} π ^{2}} (1 - e^{- ν n^{2} π^{2} t}) sin (nπ x) .

Integrating the sine across the channel gives

\int_{0}^{1} sin (nπ x) d x = [1 - (- 1)^{n}] / (nπ)

. Thus

Q (t) = \frac{G [ 1 - ( - 1 ) ^{n} ]}{ν n ^{3} π ^{3}} (1 - e^{- ν n^{2} π^{2} t}) .

For 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.

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