[Steps Shown] Consider the Sturm-Liouville problem -[ p(x)y^prime ]^prime +q(x)y=λ r(x)y , a_1y(0)+a_2y^prime (0)=0 , b_1y(1)+b_2y^prime (1)=0 , . Show


Question: Consider the Sturm-Liouville problem

\[\begin{aligned} & -{{\left[ p(x){{y}^{\prime }} \right]}^{\prime }}+q(x)y=\lambda r(x)y \\ & {{a}_{1}}y(0)+{{a}_{2}}{{y}^{\prime }}(0)=0 \\ & {{b}_{1}}y(1)+{{b}_{2}}{{y}^{\prime }}(1)=0 \\ \end{aligned}\]
  1. . Show that if \(\lambda\) is an eigenvalue and \(\phi\) the corresponding eigenfunction then
    \[\lambda \int_{0}^{1} r \phi^{2} d x=\int_{0}^{1}\left(p \phi^{\prime 2}+q \phi^{2}\right) d x+\frac{b_{1}}{b_{2}} p(1) \phi^{2}(1)-\frac{a_{1}}{a_{2}} p(0) \phi^{2}(0)\]
    provided that \(a_{2} b_{2} \neq 0 .\) How must this result be modified if \(a_{2}=0\) or \(b_{2}=0 ?\)
  2. . Show that if \(q(x) \geq 0\) and if \(b_{1} / b_{2}\) and \(-a_{1} / a_{2}\) are nonnegative, then the eigenvalue \(\lambda\) is nonegative.
  3. . Under the conditions of part(b), show that the eigenvalue \(\lambda\) is strictly positive unless \(q(x)=0\) for each \(x\) in \(0 \leq x \leq 1\) and also \(a_{1}=b_{1}=0\)

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