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An adjoint eigenfunction is a nonzero eigenfunction of the adjoint operator on its adjoint domain. For an inner product conjugate-linear in its first argument, $L^{\ast }g=\overline{\lambda }g$ implies $⟨g,(L-\lambda )f⟩=0$ for every $f$ in the domain of $L$. Consequently a necessary solvability condition for $(L-\lambda )f=h$ is $⟨g,h⟩=0$. If $L-\lambda I$ is a Fredholm operator, orthogonality to every element of its adjoint kernel is also sufficient. Boundary conditions and weights in a coupled system enter the adjoint domain and must be derived by integration by parts; copying the original eigenfunction without checking them can give a wrong projection.

For a self-adjoint Sturm-Liouville operator $L_{ϵ}=L_0+ϵL_1$ with common boundary conditions and an $L^2$-normalized lowest eigenfunction $U_0$ of $L_0$, evaluating the Rayleigh quotient on $U_0$ gives the displayed upper bound. More generally the lowest eigenvalue is the infimum of affine functions of $ϵ$, so it is concave in that parameter. The trial state must satisfy the boundary requirements of the variational domain; changing the operator without changing the trial function's normalization can still be legitimate when the weight is fixed.

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.

For a spatially periodic nonlinear pattern, a linear stability disturbance is written $e^{\lambda t+iq\cdot x}v(x)$ with $v$ periodic on the pattern cell. The conjugated operator $L(q)$ acts on that fixed cell and determines the sideband eigenvalues. Translation invariance supplies zero eigenfunctions at $q=0$, namely spatial derivatives of the base pattern. Their small-$q$ eigenvalues determine long-wavelength phase stability; other parts of the spectrum can give independent instabilities.

For real smooth coefficients with $p,w>0$, the Rayleigh quotient $\Lambda [y]=\int [p(y^{\prime }{)}^2+q{y}^2]dx/\int w{y}^{2}dx$ has the displayed first variation. Its stationary points solve the Sturm-Liouville equation on an admissible class whose boundary variations make $[p{y}^{\prime }\eta {]}_a^b=0$. Homogeneous Dirichlet boundary conditions do this by setting $\eta =0$ at the ends; homogeneous Neumann boundary conditions do it through $y^{\prime }=0$. More general self-adjoint boundary conditions can require a boundary contribution to the quadratic form.

For an eigenfunction, the weaker identity $[py{y}^{\prime }{]}_a^b=0$ suffices to give $\lambda =\Lambda [y]$ by integration by parts, but it does not by itself imply the condition on every variation. An explicit counterexample uses $p=q=w=1$ on $[0,1]$ and $y=\cos \pi x+\sin \pi x$. This solves $-y^{\prime \prime }+y=(1+{\pi }^2)y$ and has $[y{y}^{\prime }{]}_0^1=0$. Define a nearby familyIt keeps $y_{\epsilon }(0)=1$, $y_{\epsilon }^{\prime }(0)=\pi$, $y_{\epsilon }(1)=-1+\epsilon$, and $y_{\epsilon }^{\prime }(1)=\pi /(-1+\epsilon )$, so $[y_{\epsilon }{y}_{\epsilon }^{\prime }{]}_0^1=0$ for small $\epsilon$. Nevertheless, $\eta ={\partial }_{\epsilon }{y}_{\epsilon }{|}_0$ has $\eta (0)=0$, $\eta (1)=1$, and ${\int }_0^1{y}^{2}dx=1$, giving $\delta \Lambda =2[y^{\prime }\eta {]}_0^1=-2\pi \ne 0$. Boundary-energy cancellation alone therefore cannot replace a specified variational domain.

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 $f\in C([-1,1])$, 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.

A normal mode of a uniformly bending elastic filament fixed in position and slope at both ends satisfies $W^{\prime \prime \prime \prime }=k^{4}W$. With $\beta =kL>0$, the allowed wavenumbers solve $\cos \beta \cosh \beta =1$. The first root is $4.730040745$ and ${\beta }_n=(n+\frac{1}{2})\pi +O(e^{-(n+1/2)\pi })$ for $n\ge 1$. The apparent root at zero is not an eigenfunction: the zero-eigenvalue cubic polynomial satisfying all four clamped boundary conditions is identically zero. This spectrum differs from that of a filament clamped only at one end.

On a closed manifold with a Riemannian metric the positive Laplace-Beltrami operator is self-adjoint with compact resolvent. It has a complete smooth orthonormal eigenbasis, finite-dimensional eigenspaces, nonnegative eigenvalues tending to infinity, and polynomial eigenvalue-counting bounds. Elliptic regularity bounds derivatives of its eigenfunctions polynomially in their eigenvalues.

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.

The Coulomb radial factors satisfy $[A_l,A_l^{†}]=a^2(l+1)/r^2=m{a}^2(H_{l+1}-H_l)/{\hslash }^2$. Commuting $A_l$ through its factorized Hamiltonian gives the displayed intertwining relation. A nonzero transformed radial eigenfunction has the same energy in the next angular-momentum sector; this changes the radial sector, rather than being the usual magnetic-quantum-number raising operator within a fixed $l$ multiplet.

Distinct squared traction-free eigenfrequencies in a bounded lossless elastic body have eigenfunctions orthogonal in $L^2(\mathcal{D},\rho dV{)}^3$, by the Betti identity for elastodynamic fields. Normalize one field from each eigenspace. This is an orthonormal set in a separable Hilbert space, so it is countable: disjoint balls of radius less than $1/\sqrt{2}$ around its unit vectors must contain distinct points of a countable dense set. Positive and negative frequency partners add at most a factor of two. This argument proves countability without requiring a separate compact-resolvent proof.

With one orthonormal site orbital per cube vertex, common site energy $\alpha$, and coupling $\beta$ along every edge, the one-electron Hamiltonian is $\alpha I+\beta A$, where $A$ is the cube adjacency matrix. Its eigenvalues are $\alpha +3\beta ,\alpha +\beta ,\alpha -\beta ,\alpha -3\beta$ with multiplicities $1,3,3,1$.

One proof labels each vertex by three signs. The adjacency operator flips one sign at a time, so a product of $k$ coordinate signs is an eigenfunction of $A$ with eigenvalue $3-2k$. There are $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ such products. For $\beta <0$, eight electrons fill the lowest four spatial molecular orbitals, yielding total one-electron energy $8\alpha +12\beta$.

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 $a=9n^2{\pi }^2$.

Two distinguishable particles with equal one-particle spectra have degenerate ordered orbital products $|r,s⟩$ and $|s,r⟩$ for $r\ne s$. Suppose both individual position means vanish, as in a centred parity-symmetric well. The perturbation $\alpha x_1{x}_2$ then has zero diagonal matrix elements, but its restriction to this two-dimensional subspace isDegenerate perturbation theory gives normalized symmetric and antisymmetric eigenfunctions $(|r,s⟩\pm |s,r⟩)/\sqrt{2}$ with shifts $\pm \alpha |x_{rs}{|}^2$. Thus vanishing diagonal expectations in a chosen degenerate basis do not imply vanishing first-order energy shifts. Distinguishability permits both exchange symmetries; it does not forbid their energy splitting.

For an infinite square well of width $a$, the first two orbitals have $|x_{12}{|}^2=256a^2/(81{\pi }^4)$ by the position matrix elements in a symmetric infinite square well. If the particles are identical spin one-half fermions, the spatially symmetric and antisymmetric combinations must instead be paired with the singlet state and triplet states, respectively. A spin-independent perturbation leaves each allowed spin multiplicity intact.

A Dirichlet eigenfunction is an eigenfunction of a differential operator satisfying a homogeneous Dirichlet boundary condition. The term applies to more general boundary-value operators as well as the Dirichlet Laplacian.

The Chebyshev differential equation $-(1-x^2)y^{\prime \prime }+x{y}^{\prime }=\lambda y$ on $-1<x<1$, with zero endpoint values, has the Sturm-Liouville formThus $p=\sqrt{1-x^2}$ and the weight defining the weighted inner product is $w=(1-x^2{)}^{-1/2}$. The substitution $x=\cos \theta$ maps the problem to $-Y^{\prime \prime }=\lambda Y$ on $0<\theta <\pi$ with homogeneous Dirichlet boundary conditions. It also transforms $w(x)dx$ into $-d\theta$. Consequently the normalized eigenfunctions areTheir weighted inner product is the ordinary sine inner product on $[0,\pi ]$. Equivalently, $y_n=\sqrt{2/\pi }\sqrt{1-x^2}{U}_{n-1}(x)$, where $U_{n-1}$ is a Chebyshev polynomial of the second kind. These are not the cosine solutions $T_n$, which have different endpoint values. The original endpoints are singular because $p$ vanishes and $w$ diverges there, but the transformed Dirichlet problem specifies the self-adjoint realization. The eigenfunctions vanish at both endpoints and their quasi-derivatives $p{y}_n^{\prime }$ have finite limits, so the boundary form vanishes.

On $0<x<L_x$, $0<y<L_y$ with positive side lengths and zero Dirichlet boundary conditions, separation of variables for $-\Delta u=\lambda u$ giveswith the displayed eigenvalues. The one-dimensional endpoint conditions exclude constant and hyperbolic factors and force the positive integer sine frequencies. The product Fourier sine series basis is complete in the square-integrable functions on the rectangle. Consequently an eigenfunction with prescribed $\lambda$ is a linear combination of exactly those modes with ${\lambda }_{mn}=\lambda$; if none match, only the zero solution exists. Distinct pairs with the same sum give an eigenvalue with geometric multiplicity greater than one. On a square of side $\pi$, for instance, $\lambda =25$ has precisely the pairs $(3,4)$ and $(4,3)$, so its eigenspace has dimension two.

For $n_t=\mu \Delta n+\gamma n$ in a rectangular box with zero Dirichlet boundary conditions, separation of variables gives spatial eigenfunctions $\sin (p\pi x/L_x)\sin (q\pi y/L_y)\sin (r\pi z/L_z)$ and temporal rates $\gamma -\mu {\pi }^2(p^2/L_x^2+q^2/L_y^2+r^2/L_z^2)$. The largest rate is the displayed one. Every nonzero nonnegative initial density has positive projection onto this first eigenfunction, so its sign determines eventual exponential growth or decay; equality leaves a stationary first mode. At fixed volume, write $a_j=L_j^{-2}$: their product is fixed, and the arithmetic-geometric mean inequality minimizes their sum when all three are equal. Thus the cube maximizes the growth rate among rectangular boxes of fixed volume; making a box sufficiently thin suppresses growth.

A smooth uniformly elliptic differential expression with symmetric Dirichlet bilinear form has an unbounded self-adjoint operator realization $A_D$ on $L^2(U)$, whose domain is $H^2(U)\cap H_0^1(U)$ on a smooth bounded domain. If the form is strictly positive, the Lax-Milgram theorem and elliptic regularity make $A_D^{-1}$ a bounded map from $L^2$ to $H^2\cap H_0^1$. Its action on $L^2$ is compact by the Rellich-Kondrachov compactness theorem and self-adjoint by symmetry of the form. The spectral theorem for compact Hermitian operators therefore gives an orthonormal basis of smooth Dirichlet eigenfunctions with positive eigenvalues tending to infinity.