Solution

Multiplication by

(1 - x^{2})^{- 1/2}

puts the equation in Sturm-Liouville theory form:

\frac{d}{d x} (1 - x^{2} T_{n}^{'}) + \frac{n ^{2}}{1 - x ^{2}} T_{n} = 0 .

Multiply the equations for

T_{n}, T_{m}

T_{m}, T_{n}

, subtract and integrate. The boundary term vanishes because

1 - x^{2} = 0

at both endpoints and the polynomial derivatives are bounded. Therefore, for

n^{2} \neq = m^{2}

\int_{- 1}^{1} \frac{T _{n} ( x ) T _{m} ( x )}{1 - x ^{2}} d x = 0 .

This is the usual Chebyshev polynomial orthogonality.

Differentiating the original equation and writing

U_{n} = T_{n}^{'}

gives

(1 - x^{2}) U_{n}^{''} - 3 x U_{n}^{'} + (n^{2} - 1) U_{n} = 0 .

Its self-adjoint form is

\frac{d}{d x} ((1 - x^{2})^{3/2} U_{n}^{'}) + (n^{2} - 1) 1 - x^{2} U_{n} = 0 .

The same subtraction argument now yields, for

n^{2} \neq = m^{2}

\int_{- 1}^{1} U_{n} (x) U_{m} (x) 1 - x^{2} d x = 0 .

Solved by gpt-5.6-sol high.

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