Show that the singular Sturm-Liouville boundary value problem consisting of the differential equation

$x y^{\prime\prime}+y^{\prime}+\lambda x y=0$

with boundary conditions that both y and y′ remain bounded as x approaches 0 from the right and that $\beta_{1}y(1)+\beta_{2}y^{\prime}(1)=0$ is self-adjoint.

Question

Show that the singular Sturm-Liouville boundary value problem consisting of the differential equation

[latex]x y^{\prime\prime}+y^{\prime}+\lambda x y=0[/latex]

with boundary conditions that both y and y′ remain bounded as x approaches 0 from the right and that [latex]\beta_{1}y(1)+\beta_{2}y^{\prime}(1)=0[/latex] is self-adjoint.

Accepted Answer

From the differential equation, we see that p(x) = x. If both u and [latex] u[/latex] and their first derivatives are bounded as [latex]x o0^{+}[/latex], it is clear that equation (19)

[latex]\lim_{\epsilon o0^{+}}p(\epsilon)(u^{\prime}(\epsilon) u(\epsilon)-u(\epsilon) u^{\prime}(\epsilon))=0[/latex] (19)

will hold. Hence this singular boundary value problem, with any boundary condition of the form [latex]\beta_{1}y(1)+\beta_{2}y^{\prime}(1)=0[/latex] at x = 1, is self-adjoint.

Question 11.4.1: Show that the singular Sturm-Liouville boundary value proble......

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