Eigen values of BVP $(x^3y’)’+\lambda xy=0, y(1)=y(e)=0$ .

Question

How to find eigen values of the Sturm Liouville problem $$(x^3y’)’+\lambda xy=0, y(1)=y(e)=0?$$ I only know that it’s Cauchy Euler equation as $$x^3y’’+3x^2y’+\lambda xy=0$$ which after dividing by $x$, we get $$x^2y’’+3xy’+\lambda y=0$$ with corresponding Auxiliary equation as $$m^2+2m+\lambda =0$$ having roots as $$-1\pm\sqrt{1-\lambda}$$ How to find eigen values ? I am stuck as I know that eigen values sequence always tends to infinty. Thank you .

Answer 1

The solution associated with the roots $m_{\pm}(\lambda)=-1\pm\sqrt{1-\lambda}$ has the form$^{(*)}$ $$ y_{\lambda}(x)=Ax^{m_+}+Bx^{m_-}. \tag{1} $$ Applying the boundary conditions, we obtain the system of linear equations$^{(\dagger)}$ $$ \begin{cases} A+B=0, \\ Ae^{m_+}+Be^{m_-}=0, \end{cases} \tag{2} $$ which has non-trivial solutions iff its determinant is equal to zero, i.e. $$ e^{m_-(\lambda)}-e^{m_+(\lambda)}=0. \tag{3} $$ Solve Eq. $(3)$ to find the eigenvalues. (Suggestion: consider separately the cases $\lambda<1$, $\lambda=1$, and $\lambda>1$.)

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$^{(*)}$Except if $\lambda=1$; in this case, the solution is $$ y_1(x)=\frac{A+B\ln x}{x}. \tag{$1'$} $$

$^{(\dagger)}$In the case $\lambda=1$, the system $(2)$ must be replaced with $$ \begin{cases} y_1(1)=0, \\ y_1(e)=0. \end{cases} \tag{$2'$} $$

Answer 2

As the OP mentioned we have to solve equation $x^2y’’+3xy’+\lambda y=0$. The general solution is $$y(x)=C_1(x^{\sqrt{1-\lambda }-1}-x^{-\sqrt{1-\lambda }-1})$$ The boundary conditions only are satisfied if $$e^{-\sqrt{1-\lambda }-1}=e^{\sqrt{1-\lambda }-1}$$ We can solve this equation for $\lambda\ge 1$: $$e^{-1-i \sqrt{\lambda -1}}-e^{-1+i \sqrt{\lambda -1}}=-\frac{2 i \sin \left(\sqrt[4]{(\lambda -1)^2}\right)}{e}=0$$ $$\implies \sqrt{\lambda-1}=k \pi$$ Hence eigenvalue $$\lambda = 1 + k^2 \pi^2,\quad k\in \mathbb{Z}\land k\geq 0$$

Answer 3

This problem is a regular Sturm-Liouville problem on the interval $[1,e]$ because the highest order coefficient of the equation does not vanish on $[1,e]$. The standard form of the equation is $$ -x^2 y''(x)-3xy(x) = \lambda y(x). $$

Because of this, you know that the eigenvalues form an ordered set of values $$ \lambda_1 < \lambda_2 < \lambda_3 < \cdots, $$ and $\lambda_n\rightarrow\infty$ as $n\rightarrow\infty$. In order to solve for the eigenvalues, it suffices to find the unique solution $\psi_{\lambda}$ of $$ -x^2y_{\lambda}''(x)-3xy_{\lambda}(x)=\lambda y_{\lambda}(x), \\ \psi_{\lambda}(1)=0, \psi_{\lambda}'(1)=1, $$ and then find all $\lambda$ for which $\psi_{\lambda}(e)=0$. These values of $\lambda$ are the eigenvalues and, as mentioned, they form a strictly ordered set tending to $+\infty$. Moreover, the corresponding eigenfunctions $\{y_{\lambda_j}\}_{j=1}^{\infty}$ form a complete orthogonal basis of $L^2[1,e]$.