 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/2/b/solution

The transformed endpoint values are

u (0, t) = 0

and

u (π, t) = 3

u = 3 x / π + w

. The lifted profile is linear, so

w_{t} = w_{xx}

w (x, 0) = \frac{s i n 2 x}{π} .

w = X (x) T (t)

. Then

X^{''} / X = T^{'} / T = - λ

. The endpoint conditions give nonzero spatial modes only for

λ = n^{2}

n = 1, 2, \dots

, with

X_{n} = sin n x

and

T_{n} = e^{- n^{2} t}

. Hence

w (x, t) = \sum_{n = 1}^{\infty} b_{n} e^{- n^{2} t} sin n x, b_{n} = \frac{2}{π} \int_{0}^{π} \frac{s i n 2 x}{π} sin n x d x = \frac{δ _{n 2}}{π},

by orthogonality of the sine modes. Only the

n = 2

mode is present. Undoing both substitutions gives

v (x, t) = \frac{( t + 1 ) ^{2}}{π} (3 x + e^{- 4 t} sin 2 x) .

t = 0

this has the prescribed initial profile, and at

x = π

it is

3 (t + 1)^{2}

. Differentiating also verifies the original evolution equation directly.

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