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Multiply the eigenvalue equation of the Sturm-Liouville problem by $y$ and integrate. Integration by parts givesSince $y$ is a nonzero eigenfunction and $\omega >0$, the denominator is strictly positive. Thus the Rayleigh quotient isThe endpoint term must be retained for a general Robin boundary condition; it vanishes for homogeneous Dirichlet boundary conditions but need not vanish in the present general problem.

By part (c), write $\lambda ={\mu }^2$ with $\mu >0$. The equation is $y^{\prime \prime }+{\mu }^{2}y=0$, and the left Dirichlet boundary condition eliminates the cosine term. Thus $y=A\sin (\mu x)$, with $A\ne 0$. The right Robin boundary condition becomesPut $z=\mu \ell$ and $\kappa =k\ell >0$. A root cannot have $\cos z=0$, so this is equivalent to $\tan z=-z/\kappa$.

On each interval $((n-1/2)\pi ,n\pi )$, $n\ge 1$, the function $\tan z+z/\kappa$ rises from $-\infty$ to a positive value and has derivative ${\sec }^{2}z+1/\kappa >0$. The intermediate value theorem and strict monotonicity therefore give exactly one root $z_n$ there. On the intervening positive-tangent intervals both terms are positive, so there are no further positive roots. ConsequentlyThese intervals establish ${\lambda }_1<{\lambda }_2<\cdots$ and an infinite sequence of eigenvalues. The apparent algebraic root $z=0$ of the sine equation is excluded: it gives a zero sine function, and the direct zero-eigenvalue solution $y=Ax$ would require $A(1+k\ell )=0$.

For vector normalization to unit norm, the Dirichlet-Robin eigenvalues on a finite interval giveThe eigenfunctions can otherwise be scaled by any nonzero real constant.

When $r=p^{\prime }$, write the differential expression in divergence form,Two integrations by parts give the Lagrange identity for a Sturm-Liouville operator:The $q$ terms cancel because $q$ is real. Hence the exact boundary criterion, in the sense of self-adjointness defined in the question, isAt regular endpoints, all separated homogeneous conditions of the following type satisfy it:with real coefficients, not both zero at either endpoint. The allowed value-and-flux pairs at an endpoint are proportional, so their determinant $p(f{g}^{\prime }-f^{\prime }g)$ vanishes. This includes Dirichlet boundary conditions, Neumann boundary conditions, Robin boundary conditions and arbitrary mixtures of these at the two ends.

Coupled conditions can also work: periodic conditions $y(b)=y(a)$ and $p(b)y^{\prime }(b)=p(a)y^{\prime }(a)$ make the two endpoint expressions equal; simultaneous minus signs give the antiperiodic case. Thus requiring each endpoint term separately to vanish is sufficient, but not necessary.

More generally, for regular endpoints set $Y=(y(a),y(b){)}^{\mathsf{T}}$ and $N=(-p(a)y^{\prime }(a),p(b)y^{\prime }(b){)}^{\mathsf{T}}$. All maximal homogeneous self-adjoint boundary conditions can be written $AY+BN=0$ with real two-by-two matrices satisfying $\mathrm{rank}(AB)=2$ and $A{B}^{\mathsf{T}}=B{A}^{\mathsf{T}}$. If an endpoint is singular, the relevant limiting boundary form and admissible domain must instead be specified; the assumption $p>0$ only in the interior does not itself fix those endpoint limits.

Denote the unshifted differential expression by $L_{0}y=(p{y}^{\prime }{)}^{\prime }+qy$. The supplied Green's function satisfies $L_{0}G(x,\xi )=\delta (x-\xi )$, whereas the inhomogeneous equation is $L_{0}y=\lambda wy+wf$.

Multiply the latter equation by $G$ and the Green equation by $y$, subtract and integrate in $x$:The left-hand side is the boundary termIt vanishes because $G$ and $y$ satisfy the same homogeneous Robin boundary conditions. For example, where $\alpha \ne 0$, substituting $y^{\prime }=-\beta y/\alpha$ and $G_x=-\beta G/\alpha$ cancels the endpoint expression; where $\alpha =0$, both values vanish. The other endpoint works identically. HenceThe Dirac delta function contribution is valid for an interior evaluation point $\xi$; equivalently, $G$ is continuous there and has flux-derivative jump $[p{G}_x{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$.

This uses the existence of the unshifted Green's function stated in the question. Such a function is not automatic for every choice of $p,q$ and endpoint conditions: a nontrivial zero mode of $L_0$ would obstruct an ordinary inverse and would require a compatibility condition and a modified Green's function. Under the stated inverse assumption, the differential equation and the displayed integral equation are equivalent for functions in the boundary-condition domain.

Two integrations by parts give the boundary form of the Sturm-Liouville operator:The interior terms cancel because $p,q$ are real. Thus the required endpoint condition isFor a regular interval, a standard self-adjoint domain is specified by one real homogeneous Robin boundary condition at each endpoint:where neither pair of real constants is identically zero. At each end, these conditions force the two boundary data vectors of $f,g$ to be proportional, making the displayed form zero. Dirichlet boundary conditions and Neumann boundary conditions are the cases $B=0$ and $A=0$. Coupled conditions also work, for example $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, whose two endpoint contributions cancel. Self-adjointness uses the maximal common domain with these conditions, not an arbitrary smaller symmetric restriction.

The weighted inner product must be nondegenerate. The usual interpretation is $w>0$ almost everywhere, allowing isolated endpoint zeros as in part (b), with the operator defined on its appropriate weighted domain. Merely allowing a weight that is zero on an entire interval would not by itself define the stated operator or a positive-definite inner product on ordinary functions.

Choose homogeneous Dirichlet boundary conditions $\psi (a)=\psi (b)=0$, and take a nonzero real admissible trial $\psi$. Write the numerator as $F[\psi ]$ and denominator as $G[\psi ]>0$. For an endpoint-zero variation $\eta$, integration by parts givesThus the first variation of the Rayleigh quotient isBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiplying an eigenfunction equation by $\psi$ and integrating gives $F=\lambda G$, so its first variation vanishes. This establishes the equivalence for the chosen Sturm-Liouville problem. Homogeneous Neumann boundary conditions, or a mixed Dirichlet-Neumann pair, also cancel the boundary term; general Robin boundary conditions require adding the corresponding boundary contribution to the numerator.

Use separation of variables in the form $u=X(x)T(t)$, with $T^{\prime }/T=X^{\prime \prime }/X=-\lambda$. Then $T=e^{-\lambda t}$ and the spatial Sturm-Liouville problem isThese real separated Robin boundary conditions give a self-adjoint differential operator, so the eigenvalues are real. All three sign cases must be considered.

If $\lambda ={\mu }^2>0$, write $X=A\cos \mu x+B\sin \mu x$. The first condition gives $B=A/\mu$, while the second reduces toThus $\mu =n\pi$, and scaling the eigenfunction gives $V_n=n\pi \cos (n\pi x)+\sin (n\pi x)$.

If $\lambda =0$, then $X=A+Bx$. The first condition gives $B=A$, and the second gives $B=A+B$, so $A=B=0$; there is no zero eigenvalue.

If $\lambda =-{\kappa }^2<0$, write $X=A\cosh \kappa x+B\sinh \kappa x$. The first condition gives $B=A/\kappa$, and the second becomesHence $\kappa =1$, giving the single negative eigenvalue $-1$ and spatial eigenfunction $e^x$. Completeness of this regular Sturm-Liouville problem now givesThis is Robin diffusion with a growing mode. The growth is compatible with the stated conditions: they prescribe the same coordinate derivative at both ends, not identical dissipative outward-normal conditions.

The original PDF has a typographical omission: its displayed equation has $\lambda w(x)$ rather than $\lambda w(x)y(x)$ on the right. Taken literally, it is an inhomogeneous boundary value problem, not the stated Sturm-Liouville problem. The quotient and the later eigenfunction expansion identify the intended equation, which the following derivation produces explicitly.

Choose homogeneous Dirichlet boundary conditions $y(a)=y(b)=0$ and a nonzero real admissible $y$. Then $G[y]>0$ because $w>0$. For an endpoint-zero variation $\eta$, integration by parts givesThe first variation of the Rayleigh quotient is thusBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiply this corrected eigenvalue equation by $y$ and integrate. The same boundary cancellation gives $F[y]=\lambda G[y]$, so the first variation vanishes. This proves the requested equivalence for the intended equation. Homogeneous Neumann boundary conditions, or a Dirichlet condition at one end and a Neumann condition at the other, also eliminate the boundary term. General Robin boundary conditions would require the corresponding boundary contribution in the numerator.

The weight cancels the denominator in the Sturm-Liouville operator. Two integrations by parts, with real $p,q$, give the Lagrange identity for a Sturm-Liouville operatorThus the boundary form must vanish for every pair in the operator domain:Separated real Robin boundary conditions $A_{s}y(s)+B_{s}p(s)y^{\prime }(s)=0$ at $s=a,b$, with $(A_s,B_s)\ne (0,0)$, make each endpoint contribution vanish. This includes Dirichlet boundary conditions and Neumann boundary conditions. Coupled conditions can also work: periodic $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$ cancel the two endpoint contributions. For the regular problem, imposing a complete pair of such self-adjoint boundary conditions gives a self-adjoint differential operator; the displayed identity is the boundary-form criterion.

Write the endpoint condition in its linear Robin boundary condition form $y^{\prime }(1)=y(1)/\alpha$. For a positive eigenvalue $\lambda =q^2$, the left Dirichlet boundary condition gives $y=C\sin (qx)$, and the right condition givesFor each positive root $q_n$, the squared norm isConsequently the sine coefficients in any complete expansion areWhen $\alpha >1$, these are all modes and the Sturm-Liouville eigenfunction expansion gives exactly the stated sine series, with the usual convergence interpretation. Positivity of all eigenvalues follows from integration by parts and the Cauchy-Schwarz inequality:for a nonzero admissible $y$, since $y(1)={\int }_0^1{y}^{\prime }dx$.

However, the original PDF states only $\alpha >0$, and with that range its sine-only completeness claim is false. The missing modes are as follows. At $\alpha =1$, $\lambda =0$ admits $y=x$, so the correct expansion isFor $0<\alpha <1$, a negative eigenvalue $\lambda =-{\kappa }^2$ admits $y=\sinh (\kappa x)$, withThere is exactly one positive root: $\tanh \kappa /\kappa$ decreases strictly from $1$ to zero. For example, its derivative has the sign of $\kappa {\mathrm{sech}}^2\kappa -\tanh \kappa <0$, because $\sinh \kappa \cosh \kappa >\kappa$. Thus the corrected expansion isThese are the nonpositive modes in a Dirichlet-Robin interval expansion. As an explicit counterexample to the printed claim, at $\alpha =1$ take $f(x)=x$: it satisfies both boundary conditions, and each positive root obeysbecause $\tan q_n=q_n$. Thus all $a_n$ vanish despite $f\ne 0$. Likewise $f=\sinh (\kappa x)$ is a counterexample for $\alpha <1$, by orthogonality of Sturm-Liouville eigenfunctions with distinct eigenvalues. The formulas above give the full solution without silently assuming away the omitted modes.

Put $s_n(x)=\sin (q_{n}x)$ and $s_m(x)=\sin (q_{m}x)$. Their differential equations are $-s_n^{\prime \prime }=q_n^2{s}_n$ and $-s_m^{\prime \prime }=q_m^2{s}_m$. Subtract the two cross-multiplied equations and use integration by parts:The lower boundary term is zero because both functions vanish there. At the upper endpoint, the Robin boundary condition gives $s_j^{\prime }(1)=s_j(1)/\alpha$, soSince distinct positive roots have $q_n^2\ne q_m^2$, we concludeThis orthogonality holds for every $\alpha >0$, even when the sine family requires the additional nonpositive mode for completeness as explained in part (c).

These Robin boundary conditions give the spatial Sturm-Liouville operator $-d^2/d{x}^2$ one negative eigenvalue, $-1$, with eigenfunction $e^x$. The other eigenvalues are $n^2{\pi }^2$, with eigenfunctions $V_n=n\pi \cos (n\pi x)+\sin (n\pi x)$, $n\ge 1$; there is no zero eigenvalue. Hence the diffusion equation has a growing contribution $c_0{e}^{t+x}$ in addition to decaying modes.

There is no contradiction with ordinary diffusive smoothing. The outward normal derivative is $-u_x$ at the left endpoint and $u_x$ at the right endpoint, so imposing the same coordinate derivative at both ends does not impose dissipative outward-normal data at both ends. The energy identity is $\frac{1}{2}{\partial }_t\int u^2=[u{u}_x{]}_0^1-\int u_x^2$, whose boundary term is $u(1{)}^2-u(0{)}^2$ here and can supply energy. Orthogonal projection of initial data $f$ gives $c_0=2{\int }_0^{1}f{e}^{x}dx/(e^2-1)$; it vanishes precisely when the growing mode is absent.

The reflected boundary conditions $u_x(0,t)=-u(0,t)$ and $u_x(1,t)=-u(1,t)$ have the same spectrum, because reflection $x\mapsto 1-x$ reverses the coordinate derivative. Here the growing eigenfunction is $e^{-x}$ and the decaying functions are $W_n=n\pi \cos (n\pi x)-\sin (n\pi x)$, givingFor square-integrable initial data $f$, orthogonal projection givesThe norms are $(1-e^{-2})/2$ and $(n^2{\pi }^2+1)/2$. The spatial Sturm-Liouville problem has no zero eigenvalue and exactly one negative eigenvalue for $-D^2$, namely $-1$. The energy boundary term now reads $u(0{)}^2-u(1{)}^2$, so the left endpoint can supply energy. The growing mode disappears exactly when ${\int }_0^1{e}^{-x}f(x)dx=0$.

Writing $u=e^{-\beta x/2}v$ transforms $u_t=u_{xx}+\beta u_x$ into $v_t=v_{xx}-{\beta }^{2}v/4$. A prescribed coordinate derivative $u_x=g$ at an endpoint becomes $v_x-(\beta /2)v=e^{\beta x/2}g$ there. Thus a Neumann boundary condition becomes a Robin boundary condition. The signs are coordinate-derivative signs at both endpoints; outward normal derivatives require a sign change at the left endpoint.

For a monic second-order operator, choose a homogeneous solution $u$ satisfying the left Robin boundary condition and a solution $v$ satisfying the right decay condition. With $W=u{v}^{\prime }-u^{\prime }v\ne 0$, the displayed kernel is continuous and its first derivative has jump one. It therefore satisfies $LG=\delta$ and the two boundary conditions. The operator's first derivative term generally makes $W$ nonconstant and the resulting Green function nonsymmetric in the unweighted variables.

The effective boundary coefficient is $\tilde{\Gamma }=-(2k-ic)b^{♯}(-k)/(a\mathcal{D})$. All its ingredients come from initial scattering and the Robin boundary condition. Genuine zeros produce pole data; common zeros in numerator and denominator can instead be removable. It is incorrect to assert absence of boundary poles for every real Robin parameter.

For the elastic filament operator $K=A{\partial }_x^4$, two integration by parts operations give boundary form $A[\overline{u}{v}^{\prime \prime \prime }-\overline{u^{\prime }}{v}^{\prime \prime }+\overline{u^{\prime \prime }}{v}^{\prime }-\overline{u^{\prime \prime \prime }}v]$. Four elementary homogeneous choices at each endpoint are $h=h^{\prime }=0$ (clamped), $h=h^{\prime \prime }=0$ (hinged), $h^{\prime \prime }=h^{\prime \prime \prime }=0$ (free), and $h^{\prime }=h^{\prime \prime \prime }=0$ (slope-constrained and transverse-force-free). Each defines a self-adjoint operator on a finite interval when imposed at both ends. More general real Robin boundary conditions also give self-adjointness; the four choices are not an exhaustive classification of all boundary subspaces.

Let ${\lambda }_n$ be the transverse eigenvalues of a Sturm-Liouville problem for the two Robin boundary conditions. For a harmonic strip solution decaying in the longitudinal direction, projection of the prescribed end Neumann boundary condition onto every nonpositive eigenvalue must vanish. Positive modes give $q_n(x)=-g_n{e}^{-\sqrt{{\lambda }_n}x}/\sqrt{{\lambda }_n}$. Zero modes are affine and negative modes are oscillatory, so neither can contribute to a decaying solution. The condition also establishes uniqueness within the decay class.

A stratified layer between unstratified regions can trap internal gravity waves, because the exterior disturbance is an evanescent wave. Its discrete normal modes obey Robin boundary conditions representing the exterior layers. Periodic boundary forcing at a trapped normal mode frequency produces resonance in the ideal undamped model.

For a decaying harmonic function on a semi-infinite strip with equal outward Robin boundary conditions and reflection-symmetric end data, the top and bottom value traces agree. The global relation and its reflected spectral companion express their transforms through $D(k)$. With $H(k)=\frac{1}{2}{\int }_0^{\ell }{e}^{ky}g(y)dy$, the relation is $\psi (-ik)+\psi (ik)=2H(k)/D(k)$. Real zeros of $D$ require compatible data; unrestricted negative Robin parameters do not guarantee a decaying solution.

For $Lu=-{\partial }_j(a^{ij}{\partial }_{i}u)+b^i{\partial }_{i}u+cu$ with smooth coefficients on a bounded smooth domain, the weak formulation of $(L+\lambda )u=f$ and conormal Robin boundary condition $a^{ij}{\partial }_{i}u{\nu }_j+\beta u=0$ isThe bilinear form is bounded by the Sobolev trace theorem. If $a^{ij}$ satisfy the positive uniformly elliptic operator inequality, the multiplicative trace inequality on a bounded smooth domain and the Young inequality show that the form is coercive when $\lambda$ is sufficiently large, even if $\beta$ and $c$ have negative parts. The Lax-Milgram theorem then gives a unique weak solution for every $f\in L^2(U)$.