 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-1/2/a/solution

θ (0, t) = θ (L, t) = 0

. The spatial modes are

sin (nπ x / L)

. Integrating the initial profile on its two straight segments gives

b_{n} = \frac{2}{L} [\int_{0}^{L /2} x sin \frac{nπ x}{L} d x + \int_{L /2}^{L} (L - x) sin \frac{nπ x}{L} d x] = \frac{4 L}{n ^{2} π ^{2}} sin \frac{nπ}{2} .

L /2

is therefore

θ_{D} (x, t) = \frac{4 L}{π ^{2}} n = 1 \sum \infty \frac{sin ( nπ /2 )}{n ^{2}} sin \frac{nπ x}{L} e^{- ν (nπ / L)^{2} t} .

Every nonzero mode decays, so

θ_{D} \to 0

t \to \infty

. There is no nonzero constant mode compatible with these endpoint temperatures.

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