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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 $p=(1-x^2{)}^{3/2}$.

Match the coefficient of $y^{\prime }$ with the derivative of the coefficient of $y^{\prime \prime }$. For Sturm-Liouville form this requiresA positive integrating factor is therefore $r(x)=x^{-3}$, with any positive constant multiple equally valid. Direct calculation givesThis uses the plus-divergence sign convention of the question. The usual minus-divergence Sturm-Liouville operator is its negative; no extra sign is inserted into the requested eigenvalue equation.

Because $r$ never vanishes on the interval, multiplying the original eigenvalue equation by $r$ leaves the requested weighted eigenvalue problem unchanged:The weight in the inner product is essential. The substitution $t=\log x$ givesThus orthogonality of Sturm-Liouville eigenfunctions holds for the weighted inner product with weight $x^{-3}$, and the squared norm is $N_k=1/2$.

Use the drift-removing substitution for a Dirichlet Sturm-Liouville problem, $y=e^{-x}v$. Direct differentiation gives $\mathcal{L}y=e^{-x}(-v^{\prime \prime }+v)$. The homogeneous boundary conditions become $v(0)=v(\pi )=0$, so the eigenvalues and eigenfunctions areThe self-adjoint weighted form isThus the weight and leading coefficient are both $e^{2x}$. The orthogonality of Sturm-Liouville eigenfunctions readsAfter multiplying the forcing equation by $e^x$, it becomes $-v^{\prime \prime }+v=x$. Expanding $x$ in the sine series of part (i) and dividing by $n^2+1$ givesEvery term satisfies the endpoint conditions. As a check, solving the transformed ODE directly gives $v=x-\pi \sinh x/\sinh \pi$, which has the same coefficients and forcing.

For the real coefficients of the stated self-adjoint Sturm-Liouville problem, define the weighted inner productSince $w>0$, a nonzero eigenfunction has positive squared norm. Let $Ly=-(p{y}^{\prime }{)}^{\prime }+qy$. Integration by parts and the Dirichlet boundary conditions giveTaking $u=v$ to be an eigenfunction shows that its eigenvalue is real, since $\lambda \int w|u{|}^2=\int (p|u^{\prime }{|}^2+q|u{|}^2)$ is real. For two eigenfunctions with eigenvalues $\lambda ,\mu$, the same identity givesHence distinct eigenvalues imply orthogonality in the weighted inner product. This is the orthogonality of Sturm-Liouville eigenfunctions. The calculation assumes the regularity needed for the stated self-adjoint boundary problem, so all boundary terms and integrals are defined.

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.

For real eigenfunctions $y_i,y_j$, multiply their Sturm-Liouville problems crosswise and subtract. Integration givesAt $R$, both eigenfunctions vanish, while at $1$ their bounded values and bounded first derivatives multiply a factor tending to zero. The right side is zero. Distinct eigenvalues therefore give the orthogonality of Sturm-Liouville eigenfunctionsFor complex eigenfunctions, replace $y_i$ by $\overline{y_i}$, in accordance with the weighted inner product.

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 real positive weight $w$, the weighted inner product isThe functions must have finite weighted norm; the conjugate is omitted for real functions. Multiplying the differential equation by $x{e}^{-x}$ givesThus the Sturm-Liouville form has $p=x{e}^{-x}$, $q=0$ and $w=e^{-x}$. With the positive-sign spectral convention defineThe existing Laguerre differential operator on polynomials uses the opposite operator $L=-\mathcal{A}$, with eigenvalues $-n$; keeping this sign convention explicit avoids confusing the two descriptions.

For admissible $f,g$, integration by parts givesThe boundary form must vanish for every pair in the self-adjoint domain. In the usual regular-at-zero Laguerre realization one imposes ${\lim }_{x\downarrow 0}x{e}^{-x}{y}^{\prime }=0$, excluding the logarithmic branch; at infinity one uses the weighted square-integrable domain with vanishing boundary form. A condition $y(\infty )=0$ would be inappropriate for polynomial eigenfunctions. For any pair of polynomials, the factor $x$ makes the boundary form zero at zero, while the exponential beats all polynomial growth at infinity. These functions also have finite weighted norms.

Self-adjointness yields $(n-m)⟨y_m,y_n{⟩}_w=0$, giving orthogonality of Sturm-Liouville eigenfunctions:For the specified polynomials, direct differentiation gives $\mathcal{A}u=u$ and $\mathcal{A}v=3v$. Hence their eigenvalues in the displayed convention are 1 and 3, respectively, or −1 and −3 for $L=-\mathcal{A}$.

To find the quadratic, put $h=a{x}^2+bx+c$. The equation $x{h}^{\prime \prime }+(1-x)h^{\prime }+2h=0$ reduces to $(4a+b)x+(b+2c)=0$. Thus $b=-4a$, $c=2a$. Since $h(1)=-a=1$,These are scalar multiples of the Laguerre polynomials of degrees one, three and two.

For $n=0$, the first integral is $x{e}^{-x}{y}^{\prime }=C$, so all solutions with the prescribed interior normalization areTwo independent choices areThey both equal one at $x_0$, and their Wronskian is $e^x/x\ne 0$. For sufficiently large $x$, the last unit interval contributes at least $e^{x-1}/x$, proving that $y_2\to \infty$. These are the zero-eigenvalue solutions of the Laguerre differential equation, with the second expressible through the exponential integral. It also has a logarithmic singularity at zero and is not an admissible polynomial or weighted square-integrable eigenfunction. An interior value alone does not enforce the endpoint conditions of the spectral problem.

Use the real eigenfunctions, or insert a complex conjugate on the first factor for complex functions. Multiply $-(p{y}_n^{\prime }{)}^{\prime }=n^{2}w{y}_n$ by $y_m$, do the same with $m,n$ interchanged, and subtract. Integration by parts givesThe boundary term vanishes because both functions and derivatives are bounded while $p\to 0$. For distinct squared indices, orthogonality of Sturm-Liouville eigenfunctions therefore yieldsA second integration by parts, using the equation for $y_n$, givesThus derivative orthogonality when the Sturm-Liouville potential vanishes givesThe index qualification matters. The source initially allows integer indices, so $n\ne m$ alone does not imply distinct eigenvalues: $m=-n$ gives the same $n^2$. Taking the conventional nonnegative labels $n,m\ge 0$ makes the stated conclusion valid. For the polynomial solutions below, $y_{-n}=y_n$, so opposite indices are an explicit counterexample to an unqualified orthogonality claim.

Let $L{y}_n={\lambda }_n{y}_n$. By self-adjointness,Taking $m=n$ first shows that each eigenvalue is real: $⟨y_n,y_n{⟩}_w>0$ for a nonzero eigenfunction, since $w>0$. ThereforeFor distinct eigenvalues this proves orthogonality of Sturm-Liouville eigenfunctions:The weight is essential: this need not be orthogonality under the unweighted integral.

On its specified domain of admissible functions, a self-adjoint operator equals its adjoint, including the adjoint domain. Its integration identity on the interval isFor a differential operator, the boundary conditions must make the boundary form from integration by parts vanish. Formal symmetry of the differential expression alone does not specify a self-adjoint boundary-value problem.

Apply this identity to two eigenfunctions. Since the weight is real,First take $m=n$: positivity of ${\int }_a^{b}w|y_n{|}^{2}dx$ shows ${\lambda }_n={\lambda }_n^{\ast }$, so the eigenvalues are real. For distinct eigenvalues, subtraction therefore givesThis is orthogonality of Sturm-Liouville eigenfunctions in the weighted inner product, rather than necessarily in the unweighted inner product.

Multiplication by a positive weight puts the differential equation into Sturm-Liouville form if $(xw{)}^{\prime }=(1-x)w$. Thus $x{w}^{\prime }=-xw$, and the constant factor can be chosen so thatThis fixes the otherwise arbitrary normalization of the weight consistently with the requested unit norms.

Write the polynomial as $p_n(x)={\sum }_{j=0}^n{c}_j{x}^j$. Comparing coefficients givesConsequently $c_j=(-1{)}^j(\genfrac{}{}{0ex}{}{n}{j})/j!$, identifying the solutions as Laguerre polynomials. In particular,Repeated integration by parts gives the moments ${\int }_0^{\infty }{e}^{-x}{x}^{j}dx=j!$. ThereforeFor later use, integration by parts in the weighted eigenvalue equations also givesThe endpoint expression vanishes because of the factor $x$ at zero and exponential decay at infinity. Hence distinct Laguerre polynomials are orthogonal, an instance of orthogonality of Sturm-Liouville eigenfunctions.

Fix one self-adjoint choice of separated homogeneous Dirichlet boundary conditions or Neumann boundary conditions at each endpoint. The stated product condition then makes the boundary term $[awy{y}^{\prime }{]}_{\alpha }^{\beta }$ vanish. Multiplication of $Ly=\lambda wy$ by $y$ and integration by parts giveHence define the quadratic functionalsand $\Lambda [y]=F[y]/G[y]$ for nonzero admissible $y$. The first expression for $F$ is linear in the operator $L$, and $G$ is linear in the weight $w$.

To prove the minimum statement, expand an admissible function in a complete weighted orthonormal system of eigenfunctions of the Sturm-Liouville problem, $y={\sum }_n{c}_n{y}_n$, with ordered eigenvalues ${\lambda }_0\le {\lambda }_1\le \cdots$. Self-adjointness and orthogonality of Sturm-Liouville eigenfunctions giveEquality is attained by a lowest eigenfunction, soThis is the Rayleigh-Ritz variational principle. The endpoint product condition is shorthand for fixed choices of boundary type: allowing each trial function to choose different endpoint types would not define one fixed self-adjoint eigenvalue problem.

Let $L\varphi =\lambda \varphi$ with $\varphi \ne 0$ in the self-adjoint domain. By the boundary identity from part (i),Since its squared weighted norm is positive, $\boxed{\lambda \in \mathbb{R}}$.

For $L{\varphi }_n={\lambda }_n{\varphi }_n$ and $L{\varphi }_m={\lambda }_m{\varphi }_m$, the same identity givesThus ${\lambda }_n\ne {\lambda }_m$ impliesThis is orthogonality of Sturm-Liouville eigenfunctions. Degenerate eigenspaces can be given an orthonormal basis, although vectors in a degenerate eigenspace are not automatically orthogonal.

First, setting $f=g=y_n$ in the self-adjoint operator identity shows that ${\lambda }_n⟨y_n,y_n{⟩}_w$ is real, so every eigenvalue is real. For two eigenfunctions,Distinct eigenvalues therefore implyThis is orthogonality of Sturm-Liouville eigenfunctions.

Use the standard regular Sturm-Liouville problem hypotheses: positive weight $w$, real coefficients, and real self-adjoint boundary conditions. The positive weight and boundary-domain assumptions are implicit in the requested real orthonormal basis; real differential coefficients alone would not guarantee them for arbitrary complex boundary conditions.

Multiply the eigenfunction equation by $\overline{y_n}$ and integrate. By self-adjointness, the number ${\int }_a^b\overline{y_n}\mathcal{L}{y}_{n}dx$ is real, whereasThe second integral is strictly positive, so $\boxed{{\lambda }_n\in \mathbb{R}}$. Because the coefficients and boundary conditions are real, the real and imaginary parts of any complex eigenfunction separately obey the same equation and conditions. At least one is nonzero, so a real eigenfunction can be chosen. More generally, each eigenspace has a real basis, which can be orthonormalized using the weighted inner product.

The orthogonality of Sturm-Liouville eigenfunctions, with normalization chosen to one, isFor the real choice, omit the complex conjugate.

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.

Multiplication by $x^2$ gives the Sturm-Liouville formso the weight is $w(x)=x^2$. Use the weighted inner product $⟨f,g{⟩}_w={\int }_0^{\pi /2}{x}^2\overline{f}gdx$ and the Sturm-Liouville operator $L=-x^{-2}(x^{2}d/dx{)}^{\prime }$. Integration by parts gives its boundary form:The upper endpoint term vanishes by the Dirichlet boundary conditions, and the lower one vanishes for the regular solutions. More precisely, the radial Dirichlet spectrum in three dimensions identifies $L$ under the unitary transformation $v=xu$ with $-v^{\prime \prime }$ on $(0,\pi /2)$ with zero endpoint values; this specifies the self-adjoint realization at the singular endpoint as well as proving the formal identity.

For eigenfunctions $u_a,u_b$ with different eigenvalues, the same identity gives $({\lambda }_b-{\lambda }_a)⟨u_a,u_b{⟩}_w=0$. Taking real eigenfunctions proves the required orthogonality of Sturm-Liouville eigenfunctions.

Multiply the equation for $y_n$ by $y_m$ and that for $y_m$ by $y_n$, then subtract. Integrating the resulting identity yieldsAll endpoint terms vanish because both functions satisfy the zero Dirichlet boundary conditions. Since the multipliers differ,This is the weighted orthogonality of Sturm-Liouville eigenfunctions.