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A Sturm-Liouville problem is regular exactly when its domain is the finite closed interval $X=[a,b]$ and the coefficient $p$ does not vanish anywhere on $X$ ($p(x) \neq 0$ for all $x \in X$). A problem is singular when either the domain $X$ is infinite/unbounded or the coefficient $p$ vanishes at some point $x_0$ in $X$.

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

A Sturm-Liouville problem is regular exactly when its domain is the finite closed interval $X=[a,b]$ and the coefficient $p$ does not vanish anywhere on $X$ ($p(x) \neq 0$ for all $x \in X$). A problem is singular when either the domain $X$ is infinite/unbounded or the coefficient $p$ vanishes at some point $x_0$ in $X$.

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

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

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

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

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