 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability-resit/5/c/ii/solution

Substitute the proposed Fourier sine series into the heat equation:

\sum_{n \geq 1} b_{n} T_{n}^{'} (t) sin (k_{n} x) = - D \sum_{n \geq 1} b_{n} k_{n}^{2} T_{n} (t) sin (k_{n} x) .

Projecting onto each sine using orthogonality gives

T_{n}^{'} = - D k_{n}^{2} T_{n}

for every mode with

b_{n} \neq = 0

T_{n} (0) = 1

, since

b_{n}

already describes the initial amplitude. Hence

T_{n} (t) = e^{- D k_{n}^{2} t}, Θ (x, t) = n = 1 \sum \infty b_{n} e^{- D (nπ / L)^{2} t} sin (\frac{nπ x}{L}) .

For a mode with

b_{n} = 0

, the value of

T_{n}

is immaterial and the same convention is convenient. The positive thermal diffusivity

D

damps every mode, with higher frequencies decaying faster. For square-integrable initial data the damping justifies termwise differentiation for

t > 0

and recovery of the initial data in L2 norm convergence as

t ↓ 0

The left panel illustrates the Fourier series in part (b). The right panel is a self-generated example with

D = L = 1

and

Θ_{0} (x) = sin (π x) + \frac{1}{2} sin (3 π x)

, showing the two normal modes decaying at rates

π^{2}

and

9 π^{2}

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