Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 1 / b

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- Solution b

Solution

Multiply the Sturm-Liouville form for

y_{n}

by

y_{m}

and subtract the equation for

y_{m}

multiplied by

y_{n}

. Integration gives

(n^{2} - m^{2}) \int_{- 1}^{1} w y_{n} y_{m} d x = [p (y_{n} y_{m}^{'} - y_{m} y_{n}^{'})]_{- 1}^{1} = 0.

The boundary expression is zero because

p \to 0

and the functions and first derivatives remain finite. Since

n, m

are distinct positive integers, this proves

\int_{- 1}^{1} y_{n} y_{m} 1 - x^{2} d x = 0.

For the derivatives, apply integration by parts once to the same Sturm-Liouville form:

\int_{- 1}^{1} p y_{n}^{'} y_{m}^{'} d x = [p y_{n}^{'} y_{m}]_{- 1}^{1} - \int_{- 1}^{1} (p y_{n}^{'})^{'} y_{m} d x = (n^{2} - 1) \int_{- 1}^{1} w y_{n} y_{m} d x = 0.

Thus the new weight is

w_{1} (x) = (1 - x^{2})^{3/2} .

This is derivative orthogonality when the Sturm-Liouville potential vanishes. For

n = 1

the derivative is identically zero, which is consistent with the identity.

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