Sturm-Liouville theory

The singular equation $2x{w}^{\prime \prime }-w^{\prime }+\lambda x^{2}w=0$ becomes $W^{\prime \prime }+(2\lambda /9)W=0$ under $z=x^{3/2}$, $w(x)=W(z)$. Its zero-endpoint finite-energy quotient transforms as $2{\int }_0^1{x}^{-1/2}(w^{\prime }{)}^{2}dx/{\int }_0^1{x}^{1/2}{w}^{2}dx=(9/2){\int }_0^1(W^{\prime }{)}^{2}dz/{\int }_0^1{W}^{2}dz$. The Poincaré inequality therefore proves the sharp minimum $9{\pi }^2/2$, attained by $w=\sin (\pi x^{3/2})$. This coordinate change turns an apparently singular weighted variational problem into the usual interval energy problem. The admissible eigenfunction is $O(x^{3/2})$, making the integration-by-parts boundary term $w{x}^{-1/2}{w}^{\prime }$ vanish and both energy integrals converge.

An eigenfunction solving a Sturm-Liouville problem, including its prescribed boundary conditions.

Suppose $u>0$ on an interval solves the Sturm-Liouville equation $-(p{u}^{\prime }{)}^{\prime }+qu=\lambda wu$. Writing $y=uv$ and expanding derivatives givesIf the endpoint term $[p{u}^{\prime }{y}^2/u]$ vanishes, integration yields the displayed identity. With $p>0$, it proves a lower bound $\lambda$ for the associated Rayleigh quotient, attained by $y=u$ when $u$ is admissible. For singular endpoints, establish the identity on compact subintervals and check the boundary limits; this is essential for the Bessel differential equation with $p=w=x$, $q=k^2/x$.

A second-order differential equation in the displayed divergence form. Under the usual regular spectral assumptions, $p>0$, $w>0$, the coefficients have suitable regularity, and separated self-adjoint boundary conditions specify a Sturm-Liouville problem. Multiplying by $y$ and using zero endpoint values gives the Rayleigh quotient $\lambda =\int (p{y}^{\prime 2}+q{y}^2)dx/\int w{y}^{2}dx$ for a nonzero real eigenfunction. Sign conventions for the potential and eigenvalue vary; they must be read from the stated equation.

For a real one-dimensional Sturm-Liouville problem with separated self-adjoint boundary conditions and discrete simple spectrum, the eigenfunction for the $n$th eigenvalue has exactly $n$ interior zeros, numbering from zero. For a confining real Schrödinger operator on the line, the same ordering makes the ground state nodeless and the first excited state have one node. If the potential is even, uniqueness and reflection symmetry imply that these states are respectively even and odd.

For $\mathcal{L}y=-y^{\prime \prime }-2a{y}^{\prime }$, setting $y=e^{-ax}v$ gives $\mathcal{L}y=e^{-ax}(-v^{\prime \prime }+a^{2}v)$. Dirichlet conditions are unchanged. The weighted self-adjoint form is $-e^{-2ax}(e^{2ax}{y}^{\prime }{)}^{\prime }$, with weight $e^{2ax}$. On an interval of length $L$, eigenvalues are $(n\pi /L{)}^2+a^2$ and eigenfunctions are $e^{-ax}\sin (n\pi x/L)$.

For a regular self-adjoint Sturm-Liouville problem with positive weight $w$ and self-adjoint boundary conditions, eigenfunctions with distinct eigenvalues are orthogonal in the weighted inner product $\int w\overline{u}v$. Integration by parts proves this by equating $\int \overline{u}Lv$ and $\int \overline{Lu}v$.

For the Sturm-Liouville problem $-(p{y}_n^{\prime }{)}^{\prime }={\lambda }_{n}w{y}_n$, assume that the boundary term $[p{y}_n^{\prime }{y}_m{]}_a^b$ vanishes. One integration by parts gives the displayed identity. Distinct eigenvalues make the right side zero by orthogonality of Sturm-Liouville eigenfunctions, so the derivatives are orthogonal with weight $p$, while the original functions are orthogonal with weight $w$. This conclusion uses the absence of the $qy$ term: for a general potential the extra integral $-\int q{y}_n{y}_m$ need not vanish. For the Chebyshev polynomial of the second kind of degree $n-1$, the two weights are $w=\sqrt{1-x^2}$ and $p=(1-x^2{)}^{3/2}$.

For $\alpha >0$, the Gegenbauer polynomial $C_n^{\alpha }$ is the degree-$n$ polynomial solution ofThese polynomials are orthogonal on $(-1,1)$ with weight $(1-x^2{)}^{\alpha -1/2}$.

The Gegenbauer differential equation is the displayed Sturm–Liouville eigenvalue equation defining the Gegenbauer polynomial. Multiplication by $(1-x^2{)}^{\alpha -1/2}$ puts it in self-adjoint form.

For $\alpha >0$, all $n$ roots of $C_n^{\alpha }$ are simple and lie in $(-1,1)$. If the product $q$ of its interior linear factors had degree below $n$, orthogonality would make $\int C_n^{\alpha }qw=0$, while $C_n^{\alpha }q=q^2(C_n^{\alpha }/q)$ has a constant strict sign away from finitely many points.

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.

For the real-coefficient differential operator $Mv=(p{v}^{\prime }{)}^{\prime }+qv$, integration by parts givesIf $p$ tends to zero at both endpoints and $u,v,u^{\prime },v^{\prime }$ remain bounded there, the boundary term vanishes without prescribing endpoint values. Thus regular endpoint behaviour can supply the domain condition that makes the expression symmetric. Endpoint degeneracy alone is insufficient for arbitrary singular functions: for $p=x(1-x)$ and $v=\log x$, one has $p{v}^{\prime }=1-x\to 1$ at zero. A self-adjoint differential operator requires a specified admissible domain, rather than only a formal differential expression.

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.

A differential operator is self-adjoint when integration by parts makes its inner products symmetric on its domain.

For real coefficient functions, the fourth-order scalar operatoris formally self-adjoint exactly when $q=p^{\prime }$. Indeed, $(p{D}^2{)}^{\ast }=p{D}^2+2p^{\prime }D+p^{\prime \prime }$ and $(qD{)}^{\ast }=-qD-q^{\prime }$. Under clamped boundary conditions, its quadratic form is

For $L=D^4+q(x)$ where $q$ is a real continuous function on $[a,b]$, integration by parts givesThe potential terms cancel, and differentiating the boundary expression verifies the identity. Under clamped boundary conditions $u=u^{\prime }=v=v^{\prime }=0$ at both endpoints, this form vanishes. Conversely, requiring its vanishing against every clamped test function forces the adjoint function and its first derivative to vanish at both endpoints, because the second and third test-function derivatives are independent endpoint traces. Thus, on a regular finite interval with the usual fourth-order Sobolev domain, the clamped realization is self-adjoint. The equality of domains is part of this conclusion; formal self-adjointness of the expression alone does not supply it.

A second-order scalar differential equation is in self-adjoint form when it can be writtenMultiplication by $y$ and integration by parts then gives an energy identity.

If $L(y_0;{\lambda }_0)=0$ and $y_0,y$ satisfy homogeneous separated boundary conditions, the Lagrange identity givesThus $L(y;{\lambda }_0)=f$ can be solved only if $f$ is orthogonal to the null mode $y_0$.

At a simple eigenvalue of a self-adjoint linear operator, an inhomogeneous equation is solvable only when its forcing is orthogonal to the null eigenfunction. In weakly nonlinear expansions, this condition determines the amplitude equation.

Suppose $u_t=D{u}_{xx}+f_0(u)+\epsilon h(u)$ has a balanced stationary front $U_0(z)$, where $z=(x-q)/\sqrt{D}$ and $U_0^{\prime \prime }+f_0(U_0)=0$. Take $\dot{q}/\sqrt{D}=\epsilon c_1+\cdots$ and $u=U_0+\epsilon U_1+\cdots$. At first order, $L{U}_1=-c_1{U}_0^{\prime }-h(U_0)$ with $L=d^2/d{z}^2+f_0^{\prime }(U_0)$. Differentiating the front equation proves $L{U}_0^{\prime }=0$. Integration by parts against the decaying translation mode givesThis is the Fredholm solvability condition determining the velocity, derived from the self-adjoint differential operator rather than assumed. An orthogonality condition or a fixed midpoint removes the freedom to add a multiple of $U_0^{\prime }$ to $U_1$. For the bistable cubic reaction-diffusion equation, $h=-a{U}_0(1-U_0)$ gives $c_1=-\sqrt{2}a$, and its first-order forcing vanishes pointwise, so the phase-fixed shape correction is zero.

At the Maxwell balance for a scalar reaction-diffusion front, let $q=3\alpha /4$, $\beta =\sqrt{3}\alpha /2$, and perturb the linear growth rate by $\epsilon K\delta (x+cT)$, $T=\epsilon t$. Projection onto the front's translation mode gives $X_T=-K{\mathrm{sech}}^2(\beta Z/2)$, where $Z=X+cT+{\beta }^{-1}\log q$. Hence $Z_T=c-K{\mathrm{sech}}^2(\beta Z/2)$. Finite locked positions exist for $0<c/K\le 1$. For a strict inequality, the root with $KZ<0$ is stable and the other unstable; they meet in a saddle-node bifurcation at $c=K$. This is a leading-order solvability condition, not an exact finite-perturbation solution.

For $L(y;\lambda )=y^{m+1}$ near a normalized simple null mode $y_0$, put $y=\epsilon y_0+O({\epsilon }^2)$ and $\lambda -{\lambda }_0={\epsilon }^m\mu$. Orthogonality to $y_0$ yields

Suppose a real simple leading eigenfunction is $y_0=C\varphi$ and a first-order correction satisfies $L_{0}v=-\nu w{y}_0-y_0^p$, where $p\ge 2$ is an integer, $L_0$ is self-adjoint with null space spanned by $\varphi$, the boundary conditions are homogeneous and independent of the parameter, and $w>0$. The Fredholm solvability condition givesIndeed, orthogonality of the forcing to $\varphi$ gives $\nu C\int w{\varphi }^2+C^p\int {\varphi }^{p+1}=0$, and $C\ne 0$ permits division. Thus the numerical eigenvalue correction depends on the chosen leading amplitude; fixing an eigenfunction normalization is part of defining the nonlinear branch. This calculation assumes a perturbative branch and does not by itself establish that branch's existence.