Sturm-Liouville form

A scalar second-order differential operator is in Sturm–Liouville form when it is written aspossibly together with a positive weight in an eigenvalue problem.

The weighted operator $L=W^{-1}[-(P\xi \prime )\prime +Q\xi ]$, with real coefficients and $P,W>0$ in the interior, defines a Sturm-Liouville problem. Its realization as a self-adjoint differential operator requires a domain whose endpoint conditions annihilate the boundary form $[P(\eta {\prime }^{\ast }\xi -{\eta }^{\ast }\xi \prime )]$. Its Rayleigh quotient uses the weight $W$ in the denominator.

For $0<x<1$ and zero endpoint values, write $L=w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]$ with $p=x^{-2}$, $w=x^2$ and $q=-a{x}^2$. The change of variable $z=x^3/3$ has $dz=x^{2}dx$ and sends $y^{\prime }=x^2{Y}_z$, $y^{\prime \prime }=2x{Y}_z+x^4{Y}_{zz}$. Thus it is an isometry from the weighted inner product to the ordinary integral on $0<z<1/3$, and the apparent singular operator becomes $-Y_{zz}-aY$. Its Dirichlet eigenvalues are $9n^2{\pi }^2-a$ and its normalized eigenfunctions are $\sqrt{6}\sin (n\pi x^3)$, $n\ge 1$. The singular endpoint is handled through this transformed regular domain; a zero eigenvalue occurs exactly when $a=9n^2{\pi }^2$.

For continuous $p>0$ and bounded real $q$, the form $a(u,v)=\int (p{u}^{\prime }{v}^{\prime }+quv)$ on the zero-trace Sobolev space defines the weak operator $L=-(p{u}^{\prime }{)}^{\prime }+qu$. Its literal $L^2$ operator domain is the displayed flux domain. If a weak solution has right-hand side $f\in L^2$, then $(p{u}^{\prime }{)}^{\prime }=qu-f\in L^2$, so it belongs to that domain. This avoids assuming that a merely differentiable coefficient has bounded derivative. With $p\in C^1$ bounded away from zero, the domain reduces to $H^2\cap H_0^1$.

For $L(v)=(p{v}^{\prime }{)}^{\prime }+av$, direct differentiation gives $uL(v)-vL(u)=[p(u{v}^{\prime }-v{u}^{\prime }){]}^{\prime }$. The potential terms cancel. Integrating with common homogeneous Dirichlet boundary conditions eliminates the endpoint term. If $L(u)=0$ and $L(v)=f$, this gives $\int uf=0$, a necessary solvability condition for the inhomogeneous equation.