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What is the difference between a regular and a singular Sturm-Liouville problem?

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. For unbounded domains, boundary conditions are understood in terms of limits.

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.

Reasoning, step by step

  1. Check the domain XX: if it is a finite closed interval [a,b][a,b], it satisfies the first regularity condition.
  2. Check the coefficient pp: if p(x)≠0p(x) \neq 0 for all x∈Xx \in X, it satisfies the second regularity condition.
  3. If both conditions hold, the problem is regular.
  4. If the domain is infinite or pp vanishes somewhere in the domain, the problem is singular.

Example

The board states: The SLP is regular iff X=[a,b]X=[a,b] and p(x)≠0p(x) \neq 0 for x∈Xx \in X. The SLP is singular iff XX is infinite or p(x0)=0p(x_0)=0 for some x0∈Xx_0 \in X.

Common misconceptions

  • Treating p(x)≠0p(x) \neq 0 alone as sufficient for a regular Sturm-Liouville problem, ignoring the finite interval requirement.
  • Assuming that singular problems cannot have boundary conditions; for unbounded domains, they are understood in terms of limits.

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