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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.

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.

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.

Use the normalization from part (ii), and writeThe Sturm-Liouville eigenfunction expansion coefficient is $c_k=N_k^{-1}{\int }_1^{e}ry{\hat{y}}_{k}dx$. Although ${\hat{y}}_k$ has zero endpoint values, $y$ generally does not. Retain the endpoint term in the Lagrange identity for a Sturm-Liouville operator:The positive sign comes from subtracting the contribution at $x=1$. The differential equation also gives $\hat{L}y=r{e}^x-(\beta -1)ry$. Substitution into the preceding identity yieldsHence, away from resonance in a differential equation,This is the effect of boundary data in a Sturm-Liouville eigenfunction expansion: omitting the endpoint contribution would discard the dependence on $\alpha$.

If $\beta =1+n^2{\pi }^2$ for a positive integer $n$, the Fredholm solvability condition isWhen it fails, there is no solution. When it holds, $c_n$ is arbitrary and the other coefficients retain the displayed formula, giving infinitely many solutions differing by a multiple of ${\hat{y}}_n$. Sufficiency follows by subtracting the smooth lift $h(x)=\alpha x(1-\log x)$, which has the required endpoint values and satisfies $Lh=0$, and applying the Fredholm alternative to the resulting homogeneous Dirichlet boundary conditions.

Finally, for $\alpha \ne 0$, the displayed Sturm-Liouville eigenfunction expansion represents the solution on $1<x<e$ and in the weighted inner product sense. Every term is zero at $x=1$, so that series cannot give the nonzero prescribed value by substitution at the endpoint. The boundary trace must be imposed separately. Equivalently, expand $y-h$ with homogeneous Dirichlet boundary conditions and add $h$ back afterwards.

The homogeneous solution is $A\cos x+B\sin x$. The left boundary condition removes $A$, and the right boundary condition gives $B\sin l=0$. Thus the kernel of a linear map associated with this boundary value problem isThe Fredholm alternative now gives the complete classification:For necessity at $l=m\pi$, multiply by the homogeneous solution $\sin x$ and use the Lagrange identity for a Sturm-Liouville operator to obtain ${\int }_0^{l}f\sin xdx=0$. For sufficiency, the particular solutionsatisfies $z(0)=0$ and $z^{\prime \prime }+z=f$. At the other endpoint,which is zero under the compatibility condition. All solutions are then $z+B\sin x$, with $B$ arbitrary. The assumed Lipschitz continuity of $f$ in particular makes $f$ continuous, so the displayed integral defines a classical solution. The length $l=0$ is excluded: it would not define a nonempty interval for this boundary value problem.

Take $l>0$ and use the inner product $⟨u,v⟩={\int }_0^l\overline{u}vdx$ on the Hilbert space $L^2(0,l)$. The differential operator has domainwhere $H^2$ is a Sobolev space. Twice applying integration by parts gives the Lagrange identity for a Sturm-Liouville operatorThe Dirichlet boundary conditions make this expression zero for $u,v\in D(\mathcal{L})$, and the formal adjoint is ${\mathcal{L}}^{\ast }=d^2/d{x}^2+1$. To obtain the domain of the adjoint operator, initially impose the boundary conditions only on $u$. The remaining expression is $-[\overline{u^{\prime }}v{]}_0^l$. Since $u^{\prime }(0)$ and $u^{\prime }(l)$ vary independently, it vanishes for all such $u$ exactly when $v(0)=v(l)=0$. The adjoint domain is therefore the original domain: this is full self-adjointness, including the boundary conditions.

Consequently the eigenvalues are real, eigenfunctions for distinct eigenvalues are orthogonal, and regular Sturm-Liouville theory supplies a complete orthonormal eigenbasis. Here these objects can be exhibited explicitly:Indeed $\mathcal{L}{\varphi }_n={\lambda }_n{\varphi }_n$, and the Fourier sine basis is complete in $L^2(0,l)$. There are no additional eigenvalues: for a nonzero eigenfunction $u$, integration by parts givesso $u^{\prime \prime }+(1-\lambda )u=0$ and the endpoint values force $\sqrt{1-\lambda }l=n\pi$. All these eigenvalues are simple. The eigenvectors are precisely the nonzero multiples of these eigenfunctions, regarded as vectors in the Hilbert space.

Write the operator in divergence form:The Legendre differential equation therefore givesThese are the eigenvalues of $\mathcal{L}$. With the common positive-sign convention for a Sturm-Liouville problem, use $-\mathcal{L}$ instead:This is a singular Sturm-Liouville problem, because $p$ vanishes at the endpoints. Bounded eigenfunctions select the nonlogarithmic branches and satisfy $p{P}_n^{\prime }\to 0$ there. The Lagrange identity for a Sturm-Liouville operator givesThe eigenvalues are distinct, proving the orthogonality conditionThe weight is one. No choice of normalization of the Legendre polynomials is needed for this conclusion.

The Sturm-Liouville operator satisfies the Lagrange identity for a Sturm-Liouville operatorIndeed the potential terms cancel, and differentiating the right side gives the remaining two derivative terms. Integrate over $[0,1]$. The common Dirichlet boundary conditions make the endpoint term zero, while $L(y_0;{\lambda }_0)=0$. ThereforeThis is the necessary solvability condition imposed by the nonzero homogeneous solution; its proof here does not rely on merely quoting the Fredholm alternative.

For the nonlinear problem, chooseThe orthogonality condition then givesWith $y=ϵy_0+{ϵ}^2{y}_1$, where the correction remains bounded as $ϵ\to 0$, the normalization givesDivide by ${ϵ}^3$ and use $\mu =O(1)$ to obtainThe positive quartic integral determines the leading direction of the nonlinear shift from the linear eigenvalue.

Multiply the two displayed equations by $e^{-x}$ and subtract the one with indices reversed. The product terms combine into the Lagrange identity for a Sturm-Liouville operator:Integrating from zero to infinity givesAll factors $L_j,L_j^{\prime }$ are polynomials. The endpoint term at zero vanishes because of the factor $x$, and the one at infinity vanishes because the exponential function dominates every polynomial. Therefore, if $m\ne n$,The integral is absolutely convergent. This is orthogonality in the weighted inner product with weight $e^{-x}$; the boundary cancellation is the self-adjoint mechanism behind the Laguerre differential operator on polynomials.

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.

For the real Sturm-Liouville operator $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }+qy$, two integrations by parts give the Lagrange identity for a Sturm-Liouville operator:Set $u=y_n,v=\tilde{y}$. Both vanish at the endpoints, so the boundary term is zero. Since $\mathcal{L}{y}_n={\lambda }_{n}w{y}_n$,ConsequentlyThis is self-adjointness with the homogeneous Dirichlet boundary conditions, and expresses the Fredholm solvability condition at the eigenvalue ${\lambda }_n$. In this unweighted integral $F_n$ already contains its factor of $w$; adding another weight would be incorrect.

Divide the equation by $\sqrt{1-x^2}$ to obtain its Sturm-Liouville form:Thus $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. For $L=-w^{-1}(pd/dx{)}^{\prime }$, the Lagrange identity for a Sturm-Liouville operator gives the endpoint form $[p({\bar{f}}^{\prime }g-\bar{f}{g}^{\prime }){]}_{-1}^1$. A natural self-adjoint choice isUnder $x=\cos \theta$, these become the ordinary zero-derivative conditions at $\theta =0,\pi$ for $-d^2/d{\theta }^2$. They therefore define the realization whose Neumann eigenfunctions of the Chebyshev differential operator are $T_n(x)=\cos (n\arccos x)$, with $n=0,1,\dots$.

Boundedness alone does not choose these endpoint conditions: $\sin (n\arccos x)$ is also bounded but generally has a nonzero endpoint value of $p{y}^{\prime }$. Requiring bounded $y^{\prime }$ is a sufficient stronger regularity condition for the displayed zero-flux limits. Other self-adjoint endpoint choices are possible; the boundary form and the operator domain must specify which is intended.

Define the weighted inner product $⟨u,v{⟩}_w={\int }_a^{b}w(x)\overline{u(x)}v(x)dx$. For the stated operator $L=-(pD{)}^{\prime }-q$, integration by parts gives the Lagrange identity for a Sturm-Liouville operatorThe endpoint terms vanish for Dirichlet boundary conditions. Taking $u=v$ proves each eigenvalue is real, since $w>0$ makes the squared weighted norm positive. For eigenfunctions $u_m,u_n$ of distinct eigenvalues, the same identity gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n{⟩}_w=0$, proving their orthogonality.

For the specific equation, divide by $x^2$ to obtain $-y^{\prime \prime }=\lambda x^{-2}y$, so its weight is $w=x^{-2}$, with $p=1,q=0$. Put $v=\ln x$, $\ell =\ln 2$ and $Y(v)=y(e^v)$. The logarithmic change of variable in a Sturm-Liouville problem gives $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$. Remove the first derivative with $Y=e^{v/2}Z$; the equation becomes $Z^{\prime \prime }+(\lambda -1/4)Z=0$, with $Z(0)=Z(\ell )=0$.

The Dirichlet spectrum with inverse-square weight is thereforeIndeed ${\int }_1^2{x}^{-2}{y}_m{y}_{n}dx=(2/\ell ){\int }_0^{\ell }\sin (m\pi v/\ell )\sin (n\pi v/\ell )dv={\delta }_{mn}$. Values $\lambda \le 1/4$ give a linear or hyperbolic transformed solution and no nonzero solution vanishing at both ends.

Multiply the original equation by $p(x)=\exp ({\int }^{x}u(s)ds)>0$. Since $p^{\prime }=up$, it becomes $(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0$ with $q=pv$, $r=pw>0$. This is Sturm-Liouville form. For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$, subtract the equation for $y_m$ multiplied by $y_n$, and integrate. The Lagrange identity for a Sturm-Liouville operator givesEach separated homogeneous boundary condition makes the corresponding determinant zero, provided its pair of coefficients is not both zero. For distinct eigenvalues, this proves the orthogonality of Sturm-Liouville eigenfunctions.

For the constant drift, the drift-removing substitution for a Dirichlet Sturm-Liouville problem is $y=e^{-\alpha x}v$. It reduces the equation to $v^{\prime \prime }+\lambda v=0$ with $v(0)=v(\pi )=0$. Nonzero solutions occur exactly atThe zero and negative values give only the zero solution under both endpoint conditions. The self-adjoint expression is $(e^{2\alpha x}{y}^{\prime }{)}^{\prime }+{\alpha }^2{e}^{2\alpha x}y+\lambda e^{2\alpha x}y=0$. Hence the weight is $r=e^{2\alpha x}$, and for the displayed normalization,

Multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The common $h$ terms cancel, and integration by parts gives the Lagrange identity for a Sturm-Liouville operator:At each endpoint the two pairs $(y_n^{\prime },y_n)$ and $(y_m^{\prime },y_m)$ satisfy the same nontrivial homogeneous linear equation. They therefore lie on the same one-dimensional line, and their determinant vanishes. The boundary expression is zero. Since the eigenvalues are distinct, division gives the required orthogonality of Sturm-Liouville eigenfunctions:For complex-valued eigenfunctions of a real self-adjoint problem one may choose real eigenfunctions, or use the conjugated version as a weighted inner product.

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:

For the weighted inner product, the factor $w$ cancels the $w^{-1}$ in the Sturm-Liouville operator. Assuming the coefficients are real, two applications of integration by parts giveThis is the boundary form in the Lagrange identity for a Sturm-Liouville operator. The potential terms cancel; if $q$ were complex, an additional integral of $(q-q^{\ast })f^{\ast }g$ would remain, so suitable real coefficients are part of formal self-adjointness. The endpoint restrictions in part (a) eliminate the displayed form for every pair of functions in the domain.

Use the ordinary real inner product $⟨u,v⟩={\int }_a^{b}uvdx$ on the domain of functions obeying the two separated boundary conditions. Integration by parts gives the Lagrange identity for a Sturm-Liouville operatorAt $a$, the two boundary-value vectors $(u(a),u^{\prime }(a))$ and $(v(a),v^{\prime }(a))$ satisfy the same homogeneous linear relation. If $A_2=0$, both function values vanish; if $A_2\ne 0$, their derivatives are the same constant multiple $-A_1/A_2$ of their values. Either way $v{u}^{\prime }-u{v}^{\prime }=0$. The same argument applies at $b$. HenceThis proves symmetry, and the regular separated Sturm-Liouville problem has the same adjoint domain, so $L$ is self-adjoint. To see the domain point, require the boundary form to vanish against every admissible $u$: the independently variable endpoint data force an adjoint-domain function to obey those same endpoint relations.

As usual, each endpoint pair $(A_1,A_2)$ and $(B_1,B_2)$ must be nonzero, and the coefficients must have the regularity needed for these operations. An identically zero pair would impose no condition at that endpoint and would not give the claimed self-adjoint operator. The related operator ${\omega }^{-1}L$ is self-adjoint in the weighted inner product ${\int }_a^b\omega uvdx$.

Seek a positive weight such that the differential operator becomes $-w^{-1}(p{y}^{\prime }{)}^{\prime }$. Matching the second derivative gives $p=w(1-x^2)$, while matching the first derivative requires $p^{\prime }=-xw$. ConsequentlyChoosing the multiplicative constant to be one givesThis is Sturm-Liouville form with zero potential term. Multiply the equations for $y_n,y_m$ by $y_m,y_n$, subtract, and integrate. The Lagrange identity for a Sturm-Liouville operator givesThe boundary expression vanishes because $p\to 0$ and the functions and their derivatives are finite. Since $n,m\ge 0$ are distinct, their eigenvalues $n^2,m^2$ are distinct, provingThe endpoint singularities of the weight are integrable, so these weighted inner products exist.