Skip to content
START WITH A QUESTION

What would you like to understand?

Find an answer. See the moment it becomes clear. Follow the idea further.

← Concept directory

Answers for “正则施图姆-刘维尔问题和奇异施图姆-刘维尔问题有什么区别?”

4 keyword matches

Understanding your question. You can explore the search results below now.

Meet the concept

↗

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.

Find a method

↗

Dirichlet boundary conditions correspond to setting the derivative coefficients to zero (α2=β2=0\alpha_2 = \beta_2 = 0), leaving conditions on the values of yy at the endpoints. Neumann boundary conditions correspond to setting the function coefficients to zero (α1=β1=0\alpha_1 = \beta_1 = 0), leaving conditions on the derivatives.

Conditions: The general boundary conditions are α1y(a)+α2y′(a)=0\alpha_1 y(a) + \alpha_2 y'(a) = 0 and β1y(b)+β2y′(b)=0\beta_1 y(b) + \beta_2 y'(b) = 0.; Dirichlet: α2=β2=0\alpha_2 = \beta_2 = 0.; Neumann: α1=β1=0\alpha_1 = \beta_1 = 0.

Understand why

↗

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

↗

For real constants, the condition α12+α22≠0\alpha_1^2 + \alpha_2^2 \neq 0 is a compact way to state that α1\alpha_1 and α2\alpha_2 are not both zero. This ensures that the boundary condition at the endpoint x=ax=a is nontrivial.

Conditions: The constants α1,α2\alpha_1, \alpha_2 are real numbers.; The boundary condition is α1y(a)+α2y′(a)=0\alpha_1 y(a) + \alpha_2 y'(a) = 0.