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

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- Solution ii

Solution

Separate variables as

u = X (x) T (t)

. The zero values at both plates give the Fourier sine basis

X_{j} (x) = sin \frac{jπ x}{L}, T_{j} (t) = e^{- ν j^{2} π^{2} t / L^{2}}, j \geq 1.

Expand the initial profile using orthogonality. With

s = x / L

, its coefficients are

A_{j} = \frac{2}{L} \int_{0}^{L} U_{0} (\frac{x}{L} - \frac{x ^{2}}{L ^{2}}) sin \frac{jπ x}{L} d x = 2 U_{0} \int_{0}^{1} s (1 - s) sin (jπ s) d s = \frac{4 U _{0} [ 1 - ( - 1 ) ^{j} ]}{j ^{3} π ^{3}} .

Consequently

u (x, t) = \frac{8 U _{0}}{π ^{3}} j \geq 1 j m o dd \sum \frac{sin ( jπ x / L )}{j ^{3}} e^{- ν j^{2} π^{2} t / L^{2}} .

Only odd modes occur because the initial profile is symmetric about the midpoint. The coefficients are absolutely summable, so dominated convergence gives

u (x, t) \to 0

uniformly as

t \to \infty

. The slowest decay rate is

ν π^{2} / L^{2}

, associated with

j = 1

.

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