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For $\ell >0$, the logarithmic change of variable in a Sturm-Liouville problem sends this equation to $-Y^{\prime \prime }-\alpha Y=1$ on $(0,\ell )$. The normalized eigenfunctions are $\sqrt{2/\ell }\sin (n\pi t/\ell )$, with eigenvalues ${\lambda }_n=(n\pi /\ell {)}^2-\alpha$. If every ${\lambda }_n$ is nonzero, the Sturm-Liouville eigenfunction expansion givesIndeed the constant function has sine coefficients $2[1-(-1{)}^n]/(n\pi )$; division by each eigenvalue solves the equation in the operator domain. The series is uniformly convergent, with coefficients $O(n^{-3})$, while applying the differential operator is justified in the weighted square-integrable sense. At a zero eigenvalue the solvability condition at a Sturm-Liouville eigenvalue must instead be checked: an odd resonant index makes the constant forcing incompatible, whereas an even resonant index permits solutions with an arbitrary multiple of that zero mode.

Define the weighted inner product $⟨u,v{⟩}_w={\int }_a^{b}w(x)\overline{u(x)}v(x)dx$. For the stated operator $L=-(pD{)}^{\prime }-q$, integration by parts gives the Lagrange identity for a Sturm-Liouville operatorThe endpoint terms vanish for Dirichlet boundary conditions. Taking $u=v$ proves each eigenvalue is real, since $w>0$ makes the squared weighted norm positive. For eigenfunctions $u_m,u_n$ of distinct eigenvalues, the same identity gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n{⟩}_w=0$, proving their orthogonality.

For the specific equation, divide by $x^2$ to obtain $-y^{\prime \prime }=\lambda x^{-2}y$, so its weight is $w=x^{-2}$, with $p=1,q=0$. Put $v=\ln x$, $\ell =\ln 2$ and $Y(v)=y(e^v)$. The logarithmic change of variable in a Sturm-Liouville problem gives $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$. Remove the first derivative with $Y=e^{v/2}Z$; the equation becomes $Z^{\prime \prime }+(\lambda -1/4)Z=0$, with $Z(0)=Z(\ell )=0$.

The Dirichlet spectrum with inverse-square weight is thereforeIndeed ${\int }_1^2{x}^{-2}{y}_m{y}_{n}dx=(2/\ell ){\int }_0^{\ell }\sin (m\pi v/\ell )\sin (n\pi v/\ell )dv={\delta }_{mn}$. Values $\lambda \le 1/4$ give a linear or hyperbolic transformed solution and no nonzero solution vanishing at both ends.

Multiplying by $y$ and integrating with the zero endpoint values first gives $\lambda {\int }_1^b{y}^{2}dx/x={\int }_1^{b}x(y^{\prime }{)}^{2}dx>0$ for a nonzero real eigenfunction; hence $\lambda >0$. Using the printed change of variables $\xi =\lambda \ln x$, write $y(x)=Y(\xi )$. Then $(x{y}^{\prime }{)}^{\prime }={\lambda }^2{Y}_{\xi \xi }/x$, so $Y_{\xi \xi }+Y/\lambda =0$. The first endpoint gives $Y=A\sin (\xi /\sqrt{\lambda })$, and the other gives $\sin (\sqrt{\lambda }\ln b)=0$. ThusThis is the logarithmic change of variable in a Sturm-Liouville problem. The measure $dx/x=d(\ln x)$ gives squared weighted norm $A_n^2\ln b/2$, so an orthonormal choice is $A_n=\sqrt{2/\ln b}$.

The useful coordinate for this scale-dependent differential operator is $t=\ln x$. With $Y(t)=y(e^t)$, the chain rule givesThe interval becomes $0\le t\le 1$, and the endpoint values become $Y(0)=Y(1)=0$. This logarithmic change of variable in a Sturm-Liouville problem reduces the calculation to a constant-coefficient Dirichlet problem. Also $dx/x=dt$, so the transformation preserves the appropriate weighted inner product.

Use a logarithmic change of variable in a Sturm-Liouville problem: set $t=\log x$ and $Y(t)=y(e^t)$. The chain rule givesThe endpoint conditions become $Y(0)=Y(1)=0$. Multiplying $-Y^{\prime \prime }=(\lambda +1)Y$ by $Y^{\ast }$ and integrating shows that $\lambda +1>0$ for a nonzero solution. Put $\mu =\sqrt{\lambda +1}$. The first endpoint selects $Y=A\sin (\mu t)$ and the second requires $\mu =n\pi$, $n\ge 1$. HenceThese eigenfunctions are orthonormal in the specified weighted inner product, because $w(x)dx=dx/x=dt$ and