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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.

Work first on $0<x<1$. Divide by $x-1$ to put the equation in leading-coefficient-one form. Its first-derivative coefficient is $-x/(x-1)$, so the integrating factor for the self-adjoint differential operator isThe resulting coefficients areIndeed $p^{\prime }(x)=x{e}^{-x}/(1-x{)}^2$, and multiplying the original equation by $-e^{-x}/(1-x{)}^2$ gives $(p{y}^{\prime }{)}^{\prime }+qy=F$. These transformed coefficients are singular at the right endpoint; the boundary value there is understood as the endpoint limit.

Substituting $y=x$ and $y=e^x$ in the homogeneous original equation gives zero in both cases. Their Wronskian iswhich is nonzero for $0<x<1$. Thus they are linearly independent on the solution interval.

To construct a Dirichlet Green function, choose homogeneous solutions adapted to the two endpoints:Their Wronskian is again $(x-1)e^x$, and $pW=-1$. Therefore the Green function for $L=(pd/dx{)}^{\prime }+q$ with zero endpoint values isFor fixed $s\in (0,1)$ it is continuous in $x$, vanishes at both endpoints, and satisfies the unit-flux jump conditionConsequently $L_{x}G(x,s)=\delta (x-s)$ in the sense of distributions. This is a singular-endpoint Green function with a double-zero right solution.

The endpoint data are nonhomogeneous, so use the homogeneous lifting $2x$ and writeAlthough $F(s)$ is singular at one, the integral exists for each interior $x$: the right solution satisfies $v(s)\sim e(1-s{)}^2/2$, cancelling the $(1-s{)}^{-2}$ forcing when $s>x$.

Splitting the integral at $s=x$ giveswhere elementary antiderivatives yieldFor the second equality, an antiderivative is $(1-e^{1-s})/(1-s)$, whose endpoint limit at one is $-1$. Simplifying the Green representation givesIt has the prescribed endpoint values. Direct substitution gives $(x-1)y^{\prime \prime }-x{y}^{\prime }+y=1$, and at the degenerate endpoint it also obeys $-y^{\prime }(1)+y(1)=1$. The only homogeneous solution satisfying both zero endpoint conditions is zero, so this solution is unique in the corresponding classical class.

The Lagrange identity for a Sturm-Liouville operator givesIn the intended class of functions whose values and first derivatives stay finite at the endpoints, the endpoint factor $a$ makes this entire boundary term vanish. Thus no separately prescribed endpoint values are needed to eliminate the boundary form. The same integration by parts argument then proves that the eigenvalues are real.

The admissible endpoint behaviour is part of this conclusion: it is the vanishing endpoint coefficient and the Sturm-Liouville boundary form, together with endpoint regularity, that replaces explicit boundary conditions. If singular functions are allowed, a vanishing coefficient alone need not suffice: $a(x)=x(1-x)$ and $v(x)=\log x$ have $a(x)v^{\prime }(x)\to 1$ as $x\downarrow 0$. A self-adjoint differential operator therefore still needs a specified domain of admissible functions.

The linearization at the zero solution is the self-adjoint differential operator $L=\mu -({\partial }_x^2+K{)}^2$, with domain satisfying both endpoint conditions. The Dirichlet boundary condition for $-{\partial }_x^2$ on $(-1,1)$ gives the orthogonal eigenfunctions $\cos ((n+\frac{1}{2})\pi x)$ and $\sin ((n+1)\pi x)$. Their second derivatives are multiples of themselves, so the additional condition ${\psi }_{xx}(\pm 1)=0$ is automatic. Completeness of the Dirichlet eigenfunctions also shows that these exhaust the spectrum of the polynomial operator $L$. Set $q_j=(j+1)\pi /2$; each eigenfunction has growth rate ${\sigma }_j=\mu -(K-q_j^2{)}^2$. Hence the stationary bifurcation thresholds areEquality of the first two thresholds requires $K$ to be the midpoint of ${\pi }^2/4$ and ${\pi }^2$, since these numbers are distinct. ThereforeAt this point the even and odd lowest eigenfunctions become neutral simultaneously.

Write the reaction term as $f(u;r)=u(1-u)(u-r)$ and assume $D>0$. At $r=1/2$, multiply the stationary travelling-wave reduction of a reaction-diffusion system $D{U}^{\prime \prime }+f(U;1/2)=0$ by $U^{\prime }$ and integrate once. Since $U^{\prime }=0$ at both asymptotic phases and ${\int }_0^{U}f(s;1/2)ds=-U^2(1-U{)}^2/4$, the integration constant is zero:The boundary orientation selects $U^{\prime }=-U(1-U)/\sqrt{2D}$. Separating variables gives the logistic front of a bistable cubic equation, with arbitrary center $q_0$:The arbitrary translation is expected from translation invariance.

To develop the requested perturbation theory, let $r=1/2+\epsilon a$ with signed $a=\pm 1$, introduce slow time $T=\epsilon t$, and put $z=(x-q(t))/\sqrt{D}$. Seekwhere $U_0(z)=(1+e^{z/\sqrt{2}}{)}^{-1}$. The explicit $T$ derivative of $\epsilon U_1$ is second order. Since ${\partial }_{r}f=-u(1-u)$, the first-order equation isDifferentiation of $U_0^{\prime \prime }+f(U_0;1/2)=0$ gives $L{U}_0^{\prime }=0$. The self-adjoint differential operator $L$ has this decaying translation mode, and $U_1\to 0$ at both ends because the phases remain exactly $1$ and $0$. Multiplying the first-order equation by $U_0^{\prime }$ and integrating by parts makes its left side zero. Therefore the translation-mode solvability for a bistable front requiresBoth integrals can be evaluated without a special-function formula. The decreasing profile takes $U$ from $1$ to $0$ and has $U_0^{\prime }=-U(1-U)/\sqrt{2}$, soandThe resulting speed isThis derives the velocity by the Fredholm solvability condition, rather than substituting a guessed speed.

With this value, the forcing actually cancels pointwise: $-c_1{U}_0^{\prime }+a{U}_0(1-U_0)=0$. Thus $L{U}_1=0$. The decaying kernel is only the translation direction: an independent solution is $U_0^{\prime }{\int }^z(U_0^{\prime }{)}^{-2}d\zeta$, which grows at at least one infinity. Fixing the center by $U_1(0)=0$ consequently gives $\boxed{U_1=0}$. In this particular bistable cubic reaction-diffusion equation the absence of shape deformation holds to all orders. Indeed $U_0^{\prime \prime }+f(U_0;r)=-(r-1/2)U_0(1-U_0)$, and the exact moving-frame equation has left side $-(\dot{q}/\sqrt{D})U_0^{\prime }$. Thereforeis the exact front for every $0<r<1$. For $r<1/2$, $c>0$ and the $u=1$ phase advances to the right; for $r>1/2$ the $u=0$ phase advances to the left. The small-bias speed is consistent with the prescribed $\dot{q}/\sqrt{D}=O(\epsilon )$ scaling.

The natural expectation is orbital stability, since a spatial translation of a travelling wave is another travelling wave. Thus convergence for ordinary perturbations, when it holds, is to a suitably translated front. It also matters which perturbations are allowed in the leading edge: the zero state ahead of this front is linearly unstable, and a sufficiently shallow positive tail can select a different speed. One should not assert asymptotic stability for arbitrary disturbances without a norm and a class of initial data. The requested essential spectrum calculation is a necessary part of such an analysis, not by itself a complete nonlinear stability proof.

In the moving coordinate $z=x-ct$, write $u(x,t)=w_c(z)+v(z,t)$. Expansion of the reaction term givesThe first equation to study is its linearization,and the associated eigenvalue equation $\mathcal{L}\varphi =\lambda \varphi$. Differentiating the front equation gives $\mathcal{L}{w}_c^{\prime }=0$, the neutral translation mode whenever it belongs to the chosen function space. Ahead of the front the limiting coefficient is $k$, so in unweighted L2 space the limiting dispersion relation is $\lambda =-{\xi }^2+ic\xi +k$: its rightmost real part is positive. An exponential weight penalizes precisely these disturbances ahead of the front.

There is a small functional-analytic correction to the printed definition. Use the complete exponentially weighted L2 spaceThe additional requirement of unweighted L2 space membership in the printed set makes that set incomplete under its stated weighted norm. Indeed $v_N=1_{[-N,0]}$ belongs to the intersection and is a weighted Cauchy sequence, but its weighted limit $1_{(-\infty ,0]}$ does not belong to unweighted L2 space. The usual Hilbert space for this spectral argument is the completion $X_{\gamma }$. We use that completion explicitly, rather than treating the intersection as a complete space.

Multiplication $T_{\gamma }v=e^{\gamma z}v$ is a unitary map from $X_{\gamma }$ onto L2 space. Take the domain $T_{\gamma }^{-1}{H}^2(\mathbb{R})$ for $\mathcal{L}$. The transformed closed operator isThis follows from $T_{\gamma }{\partial }_z{T}_{\gamma }^{-1}={\partial }_z-\gamma$. The two limiting constant-coefficient dispersion relations, obtained by the Fourier transform, areIn particular the necessary strict leading-edge condition is ${\gamma }^2-c\gamma +k<0$. For $c>2\sqrt{k}$ this meansA particularly convenient choice isWith this choice there is no first derivative, and ${\mathcal{L}}_{\gamma }={\partial }_z^2+k-c^2/4-2w_c(z)$ is a self-adjoint differential operator on $H^2(\mathbb{R})$. This also avoids possible ambiguities between definitions of essential spectrum for nonnormal operators.

The relevant perturbation theorem is essential-spectrum invariance under relatively compact perturbations: if $A_0$ is a closed operator, $V:D(A_0)\to H$ is compact with respect to the graph norm, and $A_0+V$ is closed on the same domain, their Fredholm essential spectrum is the same. For self-adjoint operators this is the usual essential spectrum after removing isolated finite-multiplicity eigenvalues. To see why the theorem applies, regard $A_0-\lambda$ and $A_0+V-\lambda$ as bounded maps from the graph norm domain to $H$, and use compact perturbation invariance of Fredholm operators for each $\lambda$.

Define a step-potential comparison operatorBoth $A_0$ and ${\mathcal{L}}_{\gamma }$ are self-adjoint on $H^2(\mathbb{R})$. Their difference is multiplication byIt is relatively compact, not necessarily compact as a multiplication map on all of L2 space. To prove the needed compactness, take a sequence bounded in $H^2(\mathbb{R})$. On each bounded interval the Rellich-Kondrachov compactness theorem gives a subsequence converging in L2 space. A diagonal subsequence makes the products by $V$ converge locally. Outside a sufficiently large interval, $\Vert V{\Vert }_{\infty }$ is arbitrarily small and the sequence has a uniform L2 norm bound, so those tails are uniformly small. Thus the products converge globally in L2 space, proving graph-domain compactness.

Finally,Here is a direct verification. Its quadratic form is at most $(k-c^2/4)\Vert h{\Vert }_2^2$, so no spectrum lies above this endpoint. For every $\lambda \le k-c^2/4$, choose $\xi$ with $\lambda =k-c^2/4-{\xi }^2$. Normalized, increasingly long cutoffs of $e^{i\xi z}$, translated far to the right, have disjoint supports and satisfy $\Vert (A_0-\lambda )h_N{\Vert }_2\to 0$: differentiating the cutoff gives errors of orders $N^{-1}$ and $N^{-2}$. They converge weakly to zero. This is a singular sequence in the Weyl criterion, proving that every such $\lambda$ belongs to the essential spectrum. The relatively compact perturbation theorem therefore givesThis proves the weighted essential spectrum of a Fisher travelling front is strictly stable when $c>2\sqrt{k}$. In fact here the quadratic form gives more:Since $w_c\ge 0$, the full transformed spectrum lies below $-(c^2/4-k)$, and the evolution given by its linearization decays at least as $e^{-(c^2/4-k)t}$ in this weighted norm. This does not contradict translation invariance: the nonzero translation mode cannot lie in this weighted operator domain, since otherwise it would be a zero eigenfunction, contradicting the strictly negative quadratic-form bound. The weight restricts the permitted perturbations and excludes these neutral translations. Nor does a weighted L2 norm alone control the nonlinear remainder at $z\to -\infty$; a nonlinear theorem needs further boundedness or regularity assumptions.

At the minimal speed $c_{\ast }=2\sqrt{k}$, the best choice is $\gamma =\sqrt{k}$. Now ${\gamma }^2-c\gamma +k=(\gamma -\sqrt{k}{)}^2$, andThe essential spectrum has no positive part but touches zero; no positive exponential weight makes it strictly negative. Thus the critical front has no spectral gap and this argument gives no exponential decay rate there. Strict weighted spectral stability holds above the minimal speed; at the minimal speed it is marginal.

Let primes denote radial derivatives of the equilibrium and put $\Delta u=u^{\prime }+2u/r$. The linearized continuity, adiabatic pressure and Poisson equation areTake the time derivative of the Poisson equation and substitute continuity. Integration in radius givesCentre regularity excludes a changing point mass at the origin, so $C(t)=0$ andThe perturbed self-gravity must be retained; this calculation does not make the Cowling approximation.

The radial linear momentum equation is $\rho {\partial }_{t}u=-{\partial }_r\delta p-\delta \rho {\Phi }^{\prime }-\rho {\partial }_r\delta \Phi$. Differentiate it in time and use the preceding equations:The equilibrium has $p^{\prime }=-\rho {\Phi }^{\prime }$ and ${\Phi }^{\prime \prime }+2{\Phi }^{\prime }/r=4\pi G\rho$. The remaining terms simplify asConsequently the radial stellar oscillation equation isThe derivative acts on the full variable coefficient $\gamma (r)p(r)$; discarding ${\gamma }^{\prime }$ would change the result.

For $u=r\xi (r)e^{-i\omega t}$, $\Delta u=(3\xi +r{\xi }^{\prime })e^{-i\omega t}$. Substitute and multiply by $-r^3$. Combining the two terms proportional to ${\xi }^{\prime }$ into a total derivative gives the radial stellar pulsation equation,

Set $P=\gamma p{r}^4$, $W=\rho r^4$ and $Q=r^3[(4-3\gamma )p{]}^{\prime }$. The Sturm-Liouville operator isFor regular physical perturbations, $\xi$ is finite at the centre and $P{\xi }^{\prime }\to 0$ at the free surface. For a nonzero-frequency mode the fluid displacement is $u/(-i\omega )$, so vanishing Lagrangian pressure perturbation is equivalent to $\gamma p(3\xi +r{\xi }^{\prime })\to 0$. With $p(R_s)=0$ and finite $\xi$, this gives $P{\xi }^{\prime }\to 0$. Zero-frequency modes use the same displacement boundary condition directly. Assume bounded $\gamma$, positive $\rho ,p,\gamma$ in the interior and the usual finite-energy endpoint domain. Integration by parts givesThus this physical self-adjoint differential operator has real eigenvalues ${\omega }^2$. The endpoint domain is essential: the fact that $p$ vanishes does not by itself allow arbitrary singular trial functions. The regular free-surface realization of the Sturm-Liouville problem is the one used here.

The Rayleigh-Ritz variational principle gives the fundamental squared frequency as the infimum of the weighted stellar pulsation Rayleigh quotient,The infimum is over admissible finite-energy functions satisfying the physical endpoint conditions. If this quotient is nonnegative for every such function, all radial frequencies are real and there is no exponentially growing radial mode. A negative value for even one trial function proves a negative eigenvalue and an exponentially growing solution, because ${\omega }^2<0$. A zero lowest value is marginal and needs separate treatment of neutral motion.

Choose the homologous trial function $\xi =1$, which corresponds to radial velocity proportional to $r$. Its gradient contribution is zero, andThe denominator $\int \rho r^{4}dr$ is positive. HenceThis pressure-weighted radial instability criterion is sufficient, not necessary: another trial function can detect instability even if this one does not. For a constant stellar adiabatic exponent it recovers instability below $4/3$. At constant $\gamma =4/3$, the homologous mode is neutral. For constant $\gamma >4/3$, $Q=(4-3\gamma )r^3{p}^{\prime }>0$ in a normally stratified hydrostatic equilibrium, so the quotient is positive and the star is radially stable.

Boundary form. For real functions and $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }+qy$, integration by parts givesThe integral terms containing $p{y}_1^{\prime }{y}_2^{\prime }$ cancel, as do the $q$ terms. This is the Lagrange identity for a Sturm-Liouville operator. The boundary conditions must make this boundary form zero for every admissible pair. Several suitable choices are homogeneous Dirichlet boundary conditions at both endpoints; homogeneous Neumann boundary conditions at both endpoints; one of each; or separated homogeneous Robin boundary conditionswhere each fixed real pair $(\alpha ,\beta )$ is nonzero. At each endpoint the two boundary data vectors are then proportional, making their antisymmetric product zero. Another choice is periodic data $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, which cancel the two endpoint contributions. These domains give the required self-adjoint differential operator in the regular setting.

Eigenfunctions. Set $x=\cos \theta$, $0\le \theta \le \pi$, and $Y(\theta )=y(\cos \theta )$. The chain rule givesThe Chebyshev differential equation therefore becomes $Y^{\prime \prime }+n^{2}Y=0$, with $Y(0)=Y(\pi )=0$. For a positive integer $n$, the first condition excludes the cosine term and the second is satisfied by $\sin (n\theta )$. ThusThere is no nonzero $n=0$ eigenfunction: the transformed solution is linear in $\theta$ and vanishes at both endpoints only if identically zero. Negative $n$ merely repeats a positive mode with the opposite sign, so it is excluded from the indexing of an independent eigenbasis.

Weight and normalization. Multiplying the differential equation by $-1/\sqrt{1-x^2}$ puts it into Sturm-Liouville form:Hence $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. Under $x=\cos \theta$, the measure $w(x)dx$ becomes $-d\theta$, soThereforeThese endpoints are singular because $p$ vanishes and $w$ diverges there, but the weight is integrable. Moreover $p{y}_n^{\prime }=-N_{n}n\cos (n\theta )$ remains finite, so the vanishing values of $y_n$ make the boundary form zero. These are the Dirichlet eigenfunctions of the Chebyshev differential operator, not the first-kind Chebyshev polynomials $\cos (n\arccos x)$, which have nonzero endpoint values.

Mixed endpoint spectrum. For $y^{\prime \prime }+\lambda y=0$, integration against $y$ and integration by parts, using $y(0)=0,y^{\prime }(\pi /2)=0$, gives $\lambda \int y^2=\int (y^{\prime }{)}^2>0$ for every nonzero real eigenfunction. Thus $\lambda =k^2>0$ and the first condition makes $y=A\sin (kx)$. The second requires $\cos (k\pi /2)=0$, so $k=2n+1$, $n\ge 0$. Sincethe normalized eigenfunctions and eigenvalues areThe orthogonality also follows by applying integration by parts to two different eigenfunctions of the same self-adjoint differential operator.

Forced expansion. Let $k_n=2n+1$ and write $x=\sum f_n{u}_n$, $y=\sum c_n{u}_n$. The Fourier coefficients of the forcing arebecause the boundary cosine vanishes and $\sin (k_n\pi /2)=(-1{)}^n$. Applying $d^2/d{x}^2+\kappa$ to the eigenfunction expansion gives $(\kappa -k_n^2)c_n=f_n$. Since $\kappa$ is not an eigenvalue,The coefficients are $O(k_n^{-4})$, so the series and its first two differentiated series converge absolutely and uniformly on the interval. Each term satisfies the two endpoint conditions, and $y^{\prime \prime }+\kappa y$ equals the mixed-endpoint Fourier series of $x$. The forcing coefficients are $O(k_n^{-2})$, giving uniform convergence of that series as well. Uniqueness follows because the homogeneous problem has a nonzero solution only at one of the excluded eigenvalues.

Odd fourth-power sum. Set $\kappa =0$. Direct integration of $y^{\prime \prime }=x$, with the same endpoint conditions, givesAt $x=\pi /2$, the eigenfunction expansion instead gives $y(\pi /2)=-(4/\pi ){\sum }_{n\ge 0}{k}_n^{-4}$. Equating the two values yieldsAbsolute and uniform convergence justify evaluation at the endpoint.

Let $y\ne 0$ be an eigenfunction, allowing complex values for the reality argument. The self-adjoint differential operator identity and integration by parts giveThe boundary term $[x(x-1)\overline{y}{y}^{\prime }{]}_1^R$ vanishes for the same reasons as in part ii. The denominator is real and positive; the numerator is real and strictly positive because $l\ge 1$ and $y\ne 0$. Hence the Rayleigh quotient isThe positivity uses a nondegenerate interval $R>1$.

The appropriate weighted inner product and Sturm-Liouville operator areFor two functions obeying the endpoint conditions, integration by parts gives the Lagrange identity for a Sturm-Liouville operatorAt $R$, both $u$ and $v$ vanish. At $1$, the factor $x(x-1)$ tends to zero while the functions and their first derivatives remain bounded. Thus the boundary form vanishes at both ends.

This gives the regular self-adjoint differential operator realization: its singular-end condition is ${\lim }_{x\downarrow 1}x(x-1)v^{\prime }=0$, together with $v(R)=0$. The bounded classical eigenfunctions satisfy exactly this no-log choice. More precisely the operator domain is the maximal differential-operator domain in the weighted space with these separated endpoint conditions; the preceding calculation verifies its self-adjoint boundary form, rather than merely the differential expression's formal symmetry.

For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The $q$ terms cancel. The resulting Lagrange identity for a Sturm-Liouville operator isIntegrating first over $[ϵ,R]$ and then taking the endpoint limits givesThe needed assumptions are that the improper integral exists and that the displayed boundary expression tends to zero at both endpoints. Bounded $y_m,y_n$ and their first derivatives are sufficient because $p\to 0$, but are stronger than necessary. Merely knowing that $p\to 0$ does not control arbitrarily growing eigenfunctions. With the vanishing boundary form and distinct eigenvalues, division by ${\lambda }_n-{\lambda }_m$ proves orthogonality of Sturm-Liouville eigenfunctions:For complex eigenfunctions in a real self-adjoint differential operator problem, use $\overline{y_m}$ instead, giving the corresponding weighted inner product.

Multiplying the particular equation by $e^{-x}$ puts it in Sturm-Liouville form:ConsequentlyIts polynomial eigenfunctions are the Laguerre polynomials. Products of these polynomials, with or without one derivative, grow only algebraically at infinity. The exponential in $p$ therefore makes the boundary form vanish there; its factor $x$ makes it vanish at zero. Their orthogonality is thus

For $\lambda =1$, substitution of $Ax+B$ gives $A+B=0$. For $\lambda =2$, substitution of $A{x}^2+Bx+C$ gives $4A+B=0$ and $B+2C=0$. Choosing $y_n(0)=1$, which fixes the otherwise arbitrary nonzero multiplicative constants, yieldsTheir product is $1-3x+\frac{5}{2}{x}^2-\frac{1}{2}{x}^3$. Using the Gamma function identity ${\int }_0^{\infty }{x}^k{e}^{-x}dx=k!$, also obtainable by repeated integration by parts, verifies the required orthogonality directly:

A weighted Sturm-Liouville operator has the formwhere $p,q,w$ are real, $p>0$ and $w>0$ in the interval's interior, and the coefficients have the regularity required by the derivatives. The corresponding eigenvalue equation is $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Its natural weighted inner product is $⟨f,g{⟩}_w=\int wf\bar{g}dx$. Suitable endpoint conditions must also annihilate $[p(f{\bar{g}}^{\prime }-f^{\prime }\bar{g})]$ for a self-adjoint differential operator; the displayed differential expression alone does not specify those conditions.