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Answers for “狄利克雷、诺依曼和罗宾边界条件如何融入一般的施图姆-刘维尔问题边界形式?”

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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.

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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.

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A Sturm-Liouville problem is an eigenvalue problem of the form Ly=λyL y = \lambda y on an interval a<x<ba < x < b, together with homogeneous linear boundary conditions at the endpoints: 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. The coefficients at each endpoint cannot both be zero.

Conditions: The equation is posed on an interval a<x<ba < x < b.; Boundary conditions are homogeneous linear combinations of yy and y′y' at x=ax=a and x=bx=b.; At each endpoint, the pair of boundary coefficients is not both zero.