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Answers for “为什么正则施图姆-刘维尔问题的特征函数可以用来构成正交归一基?”

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Understand why

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The eigenfunctions of a regular Sturm-Liouville problem are mutually orthogonal and total (meaning they form a complete basis). Because they are already orthogonal and complete, normalizing each eigenfunction to unit length upgrades the basis to an orthonormal basis (ONB) of eigenfunctions for the differential operator.

Conditions: The Sturm-Liouville problem is regular.; The eigenfunctions are mutually orthogonal and total.

Meet the concept

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Regular Sturm-Liouville problems have an infinite spectrum λn→∞\lambda_n \to \infty, and their eigenfunctions are orthogonal and total. Because they are orthogonal and total, they can form an orthonormal basis (ONB) after normalization.

Conditions: The Sturm-Liouville problem is regular (X=[a,b]X=[a,b] and p(x)≠0p(x) \neq 0 for all x∈Xx \in X).

Meet the concept

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A Sturm-Liouville problem is regular exactly when its domain is the finite closed interval X=[a,b]X=[a,b] and the coefficient pp does not vanish anywhere on XX (p(x)≠0p(x) \neq 0 for all x∈Xx \in X). A problem is singular when either the domain XX is infinite/unbounded or the coefficient pp vanishes at some point x0x_0 in XX.

Conditions: Regular: X=[a,b]X=[a,b] and p(x)≠0p(x) \neq 0 for all x∈Xx \in X.; Singular: XX is infinite, or p(x0)=0p(x_0)=0 for some x0∈Xx_0 \in X.; For unbounded XX, boundary conditions are interpreted using limits.