 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/2/b/solution

Project the initial condition onto the orthogonal spatial eigenfunctions. Their squared norms in the ordinary inner product are

\int_{0}^{1} e^{2 x} d x = \frac{e ^{2} - 1}{2}, \int_{0}^{1} V_{n} (x)^{2} d x = \frac{n ^{2} π ^{2} + 1}{2} .

The latter follows from the integrals of

cos^{2} (nπ x)

and

sin^{2} (nπ x)

being

1/2

, with the sine-cosine cross term zero. Therefore

c_{0} = \frac{2}{e ^{2} - 1} \int_{0}^{1} f (x) e^{x} d x, c_{n} = \frac{2}{n ^{2} π ^{2} + 1} \int_{0}^{1} f (x) [nπ cos (nπ x) + sin (nπ x)] d x .

For square-integrable data the expansion converges to the data in the L2 norm at

t = 0

; for

t > 0

the decaying high modes are smoothed. Initial data with zero

e^{x}

projection have no growing term.

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