Sturm-Liouville problem

A regular Sturm-Liouville eigenvalue problem has the formon a bounded interval with self-adjoint boundary conditions and positive weight $w$.

For $f\in C([-1,1])$, the equation $((1-x^2)y^{\prime }{)}^{\prime }=f$ has a bounded solution on $(-1,1)$ if and only if ${\int }_{-1}^{1}f=0$. All bounded solutions differ by a constant. To prove necessity, integrate the equation once. The quantity $(1-x^2)y^{\prime }$ has finite endpoint limits, and any nonzero endpoint limit would force logarithmic growth in $y$. Both limits must therefore vanish, yielding the zero-mean condition. For sufficiency set $F(x)={\int }_{-1}^{x}f(t)dt$ and $y(x)=C+{\int }_0^{x}F(s)/(1-s^2)ds$. Zero mean implies $|F(s)|\le \Vert f{\Vert }_{\infty }\min (1+s,1-s)$, so $y^{\prime }$ is bounded and $y$ extends continuously to the endpoints. The homogeneous solutions are a constant and $\frac{1}{2}\log ((1+x)/(1-x))$, proving uniqueness up to a constant. This is the Fredholm alternative for the zero eigenvalue of this singular Sturm-Liouville problem.

The eigenfunction expansion in Legendre polynomials is $y=C-{\sum }_{k\ge 1}{P}_k{\int }_{-1}^{1}f{P}_k/[k(k+1){\int }_{-1}^1{P}_k^2]$. It follows by projecting onto the eigenfunctions with eigenvalues $-k(k+1)$. Completeness follows because the Legendre polynomials span the polynomials, which are dense by the Weierstrass approximation theorem.

Continuity only in the open interval requires more care. Even if $f$ is integrable and has zero mean, the primitive of $F/(1-x^2)$ must still be bounded at both ends. For example, $F(x)=1/\log (e/(1-x^2))$ has zero endpoint limits and absolutely integrable derivative $f=F^{\prime }$, but $F/(1-x^2)$ has a divergent primitive. Thus zero mean alone is insufficient without endpoint control.

For $\omega >0$ on $x>0$, use the regular finite branch at zero, decay at infinity, and a finite quadratic-form energy. For trial functions with $[x{y}^2{]}_0^{\infty }=0$, integration by parts givesThe Rayleigh quotient is at least $\omega$, with equality exactly for a nonzero multiple of $e^{-\omega x}$. Direct substitution verifies that this is an eigenfunction with eigenvalue $\omega$, so the lower bound is attained. The zero coefficient $p(0)=0$ makes this a singular-endpoint Sturm-Liouville problem; regularity replaces a prescribed zero endpoint value. Continuous piecewise differentiable trials of finite form energy are allowed in the Rayleigh-Ritz method, even if their derivative has a jump away from the endpoint.

Suppose $p(x)=c(x-a)+O((x-a{)}^2)$ with $c>0$, and the other coefficients in $(p{y}^{\prime }{)}^{\prime }+(\lambda w-q)y=0$ extend continuously to $a$, with $w(a)>0$. A bounded eigenfunction has $(p{y}^{\prime }{)}^{\prime }=O(1)$, hence $p{y}^{\prime }=C+O(x-a)$. A nonzero $C$ would give $y^{\prime }\sim C/[c(x-a)]$ and an unbounded logarithmic singularity. Thus $C=0$, and the regular branch has bounded first derivative. The second local branch carries a logarithm. The title condition excludes that branch.

For two regular eigenfunctions, $p(\overline{u}{v}^{\prime }-\overline{u^{\prime }}v)\to 0$. This removes the singular-end contribution to the Lagrange identity for a Sturm-Liouville operator. Together with a separated self-adjoint condition at the other endpoint, it selects the regular self-adjoint differential operator realization. The precise operator domain uses the maximal differential-operator domain and the no-log endpoint condition; boundedness of classical eigenfunctions is its convenient local expression, not a replacement for specifying an operator domain.

For $\ell >0$ and $k>0$, the Sturm-Liouville problem with one Dirichlet boundary condition and one Robin boundary condition has positive eigenvalues. Indeed, integration by parts givesfor every nonzero eigenfunction. Let $z=\ell \sqrt{\lambda }$ and $\kappa =k\ell$. The eigenfunctions are proportional to $\sin (zx/\ell )$, withFor each $n\ge 1$, there is exactly one root $z_n$ in $((n-1/2)\pi ,n\pi )$: on that interval $\tan z+z/\kappa$ increases strictly from $-\infty$ to a positive value. There are no other positive roots. The apparent root $z=0$ gives only the zero sine function and is not an eigenvalue.

For fixed $k,\ell$, write $z_n=(n-1/2)\pi +{\epsilon }_n$. The root equation becomes $\kappa \cot {\epsilon }_n=(n-1/2)\pi +{\epsilon }_n$, henceAn optional unit normalization uses ${\int }_0^{\ell }{\sin }^2(\sqrt{{\lambda }_n}x)dx=\ell /2+k/[2(k^2+{\lambda }_n)]$. The positive sign of the Robin boundary term matters: reversing it can introduce nonpositive modes, as in nonpositive modes in a Dirichlet-Robin interval expansion.

For the radial Laplace operator $L=-r^{-2}(r^2{u}^{\prime }{)}^{\prime }$ on $0<r<R$, require a finite value at the origin and $u(R)=0$. The map $u\mapsto v=ru$ preserves the norm from $L^2((0,R),r^{2}dr)$ to $L^2(0,R)$ and sends $L$ to $-d^2/d{r}^2$. The regular endpoint condition becomes $v(0)=0$. The self-adjoint realization therefore corresponds to the ordinary Dirichlet boundary conditions for the one-dimensional second derivative. Its eigenvalues and eigenfunctions are displayed above, with $u_n(0)=n\pi /R$ understood by a limit. Their squared weighted norm is $R/2$. Completeness and orthogonality follow from the Fourier sine series on $(0,R)$.

For $y^{\prime \prime }+\lambda x^{-2}y=0$ on $0<a<x<b$ with zero endpoint values, write $t=\log (x/a)$ and $y=\sqrt{x}w(t)$. The Euler-Cauchy equation becomes $w^{\prime \prime }+(\lambda -1/4)w=0$. Thus $y_n=\sqrt{x}\sin [n\pi \log (x/a)/\log (b/a)]$ and the exact spectrum is displayed above. Leading WKB quantization omits the constant $1/4$, while producing the same leading eigenfunction shape after quantization.

A Dirichlet eigenvalue is an eigenvalue for an eigenfunction constrained to vanish at the endpoints of its interval or on the boundary of its domain.

For a bounded domain $\Omega$, the first Dirichlet eigenvalue of $-\Delta$ has the variational characterization

For a regular self-adjoint Sturm-Liouville problem, eigenfunctions are orthogonal in the weighted inner productand the coefficient of $y_n$ is $⟨f,y_n{⟩}_w/⟨y_n,y_n{⟩}_w$.

Let $Ay=(p{y}^{\prime }{)}^{\prime }+qy$ have real eigenfunctions $u_n$ obeying $A{u}_n={\lambda }_{n}w{u}_n$ and homogeneous Dirichlet boundary conditions, with $N_n={\int }_a^{b}w{u}_n^2>0$. Suppose $Ay+\mu wy=f$, with possibly nonzero endpoint values. The Lagrange identity for a Sturm-Liouville operator givesDivision by $({\lambda }_n+\mu )N_n$ gives the displayed coefficient whenever the denominator is nonzero. At resonance in a differential equation, equality of the forcing projection and the boundary term is the Fredholm solvability condition. The Sturm-Liouville eigenfunction expansion represents the interior solution in the weighted inner product sense; its zero endpoint terms do not reproduce nonzero boundary traces by pointwise substitution. A smooth lift of the endpoint data reduces the calculation to homogeneous Dirichlet boundary conditions.

For $-y^{\prime \prime }=\lambda y$, positive eigenvalues have eigenfunctions $\sin (q_{n}x)$, where $\tan q_n=\alpha q_n$ and $q_n>0$. Their squared norms are $N_n=1/2-\sin (2q_n)/(4q_n)$. A zero eigenvalue occurs precisely at $\alpha =1$, with eigenfunction $x$. A negative eigenvalue occurs precisely at $0<\alpha <1$: it is $-{\kappa }^2$ with eigenfunction $\sinh (\kappa x)$, where $\tanh \kappa =\alpha \kappa$. There is exactly one such $\kappa$, since $\tanh \kappa /\kappa$ decreases strictly from $1$ to $0$. The integration by parts identityand the Cauchy-Schwarz inequality show positivity for $\alpha >1$ and nonnegativity at equality. The complete expansion of a square-integrable function therefore includes the extra term $3x{\int }_0^{1}tf(t)dt$ when $\alpha =1$, or $\sinh (\kappa x){\int }_0^{1}f(t)\sinh (\kappa t)dt/[\sinh (2\kappa )/(4\kappa )-1/2]$ when $\alpha <1$, in addition to ${\sum }_n[{\int }_0^{1}f(t)\sin (q_{n}t)dt/N_n]\sin (q_{n}x)$. Omitting that extra mode makes the system incomplete. In particular, $f=x$ at $\alpha =1$ is orthogonal to every positive sine mode and is an immediate counterexample to a sine-only expansion.

Write $\mathcal{A}y=-(p{y}^{\prime }{)}^{\prime }+qy$ and $\mathcal{A}{y}_n={\lambda }_{n}r{y}_n$, where $r$ is the positive weight. Normalize the real eigenfunctions using $\int r{y}_m{y}_n={\delta }_{mn}$. For $\mu$ outside the spectrum, applying $\mathcal{A}-\mu r$ to the displayed Green function expansion gives $r(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$ by weighted completeness. This is the kernel with respect to ordinary source integration $d\xi$; using the weighted measure would change its normalization.

Let $L=w^{-1}[-(d/dx)(pd/dx)+q]$ be the regular self-adjoint operator of a Sturm-Liouville problem, with eigenfunctions orthonormal for the weighted inner product $\int w\overline{u}v$. For $\mu$ outside its spectrum, the displayed kernel satisfies $(L_x-\mu )G_{\mu }=\delta (x-\xi )$ with an ordinary, unweighted Dirac delta. Indeed weighted completeness gives $w(\xi ){\sum }_n{y}_n(x)y_n(\xi {)}^{\ast }=\delta (x-\xi )$. Thus $u(x)=\int G_{\mu }(x,\xi )f(\xi )d\xi$ solves $(L-\mu )u=f$. A kernel integrated against $w(\xi )d\xi$ instead omits the prefactor. For real coefficients, the weighted reciprocity relation is $G_{\mu }(x,\xi )/w(\xi )=G_{\mu }(\xi ,x)/w(x)$ when the spectral parameter is real and outside the spectrum. A zero eigenvalue prevents inversion at $\mu =0$ without a forcing compatibility condition or a projected inverse.

For $L=d^2/d{x}^2$ on $(0,\pi )$ with zero endpoints, the zero-parameter Green function is $-x_{<}(\pi -x_{>})/\pi$. Its eigenfunctions are $\sqrt{2/\pi }\sin (nx)$ with eigenvalues $-n^2$. At $x=\xi =\pi /2$, the kernel is $-\pi /4$ and its expansion contains precisely the odd reciprocal squares, giving the displayed identity.