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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.

Consider $-y^{\prime \prime }-(n-1)y^{\prime }/x+x^{2}y=\lambda y$ for a positive integer $n$, with regular radial data at zero and square-integrable decay at infinity in the measure $x^{n-1}dx$. Its Sturm-Liouville form has $p=w=x^{n-1}$ and $q=x^{n+1}$. For the Gaussian trial $y=e^{a{x}^2}$, $a<0$, the Rayleigh quotient isThe Gamma function recurrence evaluates the ratio, and the minimum is $n$ at $a=-1/2$. It is the exact lowest eigenvalue: after the endpoint term $[x^n{y}^2{]}_0^{\infty }$ vanishes, integration by parts givesEquality gives the displayed positive Gaussian eigenfunction. For $n=1$, the radial regularity condition is the Neumann boundary condition at zero, not a vanishing value there. Changing the boundary condition changes the lowest eigenvalue and whether this trial is admissible.

With the weighted inner product ${\int }_{-1}^1\overline{f}gdx/\sqrt{1-x^2}$, the Chebyshev differential equation has the Sturm-Liouville form with coefficients $p=\sqrt{1-x^2}$, $w=(1-x^2{)}^{-1/2}$ and $q=0$. The substitution $x=\cos \theta$ converts its operator to $-d^2/d{\theta }^2$ on $(0,\pi )$ and its inner product to the ordinary one in $d\theta$. The natural endpoint conditions $p{y}^{\prime }\to 0$ become zero derivative conditions at $\theta =0,\pi$.

Consequently the eigenfunctions are the Chebyshev polynomials $T_n(x)=\cos (n\arccos x)$ with eigenvalues $n^2$, $n=0,1,2,\dots$. Their squared norms are $\pi$ for $n=0$ and $\pi /2$ otherwise. In particular $T_1=x$ and $T_3=4x^3-3x$ are orthogonal. The endpoints of the original differential expression are singular; specifying these transformed Neumann boundary conditions defines its self-adjoint realization. The Dirichlet eigenfunctions of the Chebyshev differential operator use a different endpoint condition and have sine rather than cosine profiles.

Differentiate the Legendre differential equation once and put $R_n=P_n^{\prime }$:Multiplying by $1-x^2$ gives its Sturm-Liouville form,Thus the weight is $w(x)=1-x^2$, not one. For $n\ge 1$, $R_n$ is a nonzero polynomial; $R_0=0$ is a trivial solution rather than a nonzero eigenfunction.

Multiply the equations for $n,m$ by $R_m,R_n$ respectively, subtract and integrate. The resulting boundary term is $[(1-x^2{)}^2(R_n^{\prime }{R}_m-R_m^{\prime }{R}_n){]}_{-1}^1=0$, since all these derivatives are finite at the endpoints. For distinct nonnegative indices, ${\lambda }_n-{\lambda }_m\ne 0$, giving the weighted orthogonality of Legendre polynomial derivatives:For $n=0$ the integral is zero simply because $R_0=0$.

Divide the differential equation by $\sqrt{1-x^2}$. Its Sturm-Liouville form isHere $p(x)=\sqrt{1-x^2}$ and weight $w(x)=(1-x^2{)}^{-1/2}$. Multiply the equations for $T_n,T_m$ by the other polynomial and subtract. Integration by parts gives a boundary term $p(T_m{T}_n^{\prime }-T_n{T}_m^{\prime })$, which vanishes at both endpoints because polynomial derivatives are finite and $p\to 0$. HenceWith the standard nonnegative degree indices $n,m$, this proves orthogonality for $n\ne m$. If signed integer indices are used literally, the condition is $|n|\ne |m|$: $T_{-n}=T_n$, so merely $n\ne m$ would not suffice.

Set $x=\cos \theta$ and $Y(\theta )=y(\cos \theta )$. The chain rule gives $(1-x^2)y^{\prime \prime }-x{y}^{\prime }=Y^{\prime \prime }$, so $Y^{\prime \prime }+n^{2}Y=0$. For $n>0$, $Y=A\cos (n\theta )+B\sin (n\theta )$. A polynomial $y$ has $Y^{\prime }(0)=0$, forcing $B=0$. At $n=0$, $Y=A+B\theta$ and the same endpoint condition gives $B=0$. Therefore every polynomial solution is proportional toThe cosine recurrence yields $T_{n+1}=2x{T}_n-T_{n-1}$, $T_0=1,T_1=x$, confirming that these functions are polynomials. In this normalization the Chebyshev polynomial weight identity isIt is simply cosine orthogonality after $x=\cos \theta$.

Set $y=e^{-x/2}v$, the Dirichlet gauge transform for constant drift. Direct differentiation givesThe homogeneous endpoint conditions on $y$ become $v(0)=v(1)=0$. Thus the eigenvalue equation is $v^{\prime \prime }=(\lambda +1/4)v$. To exclude other branches without assuming the eigenvalue real, multiply by $\bar{v}$ and integrate by parts:for a nonzero eigenfunction. Therefore $\lambda +1/4=-k^2$ with $k>0$. The boundary conditions in $v=A\sin (kx)+B\cos (kx)$ give $B=0$ and $k=n\pi$, $n\ge 1$. HenceEquivalently, $L[y]=\lambda y$ is the Sturm-Liouville form $(e^x{y}^{\prime }{)}^{\prime }=\lambda e^{x}y$, with positive weight $e^x$. Its orthogonality relation isIt is the weighted inner product, not the unweighted one, that makes these eigenfunctions orthogonal.

Expand $x{e}^{-x/2}={\sum }_n{b}_n{y}_n$ with the coefficients from part a. Completeness and the nonzero eigenvalues give the Sturm-Liouville eigenfunction expansionThe coefficients are $O(n^{-3})$, so the function series converges uniformly and vanishes at both endpoints. Its first derivative also converges uniformly. Applying the transformed operator in the square-integrable sense gives the part-a expansion of $x$; equivalently $v^{\prime \prime }-v/4=x$ on the interior. Integrating this continuous right-hand side gives the classical twice-differentiable solution there. No nonzero homogeneous solution satisfies both endpoints, so this solution is unique. The mismatch of the forcing's sine series at a single endpoint does not affect the interior equation or these boundary conditions.

Multiply the Gegenbauer differential equation byIt becomes the Sturm-Liouville formFor two polynomial solutions of degrees $m\ne n$, multiply their equations crosswise, subtract, and integrate over $(-1,1)$. The boundary term vanishes because $\alpha >0$ and $(1-x^2{)}^{\alpha +1/2}\to 0$. Since the eigenvalues $n(n+2\alpha )$ are distinct,Thus $a=-1$, $b=1$, and the displayed function is the weight.

Every interior root is simple: a solution and its derivative cannot both vanish at an ordinary point of a second-order linear differential equation unless the solution is identically zero. Let $x_1,\dots ,x_k$ be all interior roots and put $q(x)={\prod }_i(x-x_i)$. If $k<n$, then $q$ has degree below $n$, so orthogonality givesBut $C_n^{\alpha }/q$ has no zero and constant sign on $(-1,1)$, whence the integrand $q(x{)}^2(C_n^{\alpha }(x)/q(x))w(x)$ has one strict sign except at finitely many points. Its integral cannot vanish, a contradiction. Therefore $k=n$: all $n$ roots are real, simple, and lie in $(-1,1)$.

Normalize the solution by its incoming amplitude at $+\infty$. The transmission coefficient and reflection coefficient areFor two solutions of the same equation in Sturm-Liouville form, the modified Wronskianis independent of $x$. Take ${\psi }_1={\varphi }_k$ and ${\psi }_2=\overline{{\varphi }_k}$. At $-\infty$ this gives $W=-2ik$, while at $+\infty$ it givesTherefore $|a{|}^2-|b{|}^2=1$, and division by $|a{|}^2$ yields conservation of scattering flux:

Write the operator in Sturm-Liouville form:The natural solution space is the Sobolev space $H_0^1(0,1)$. Multiplication by a test function $v\in H_0^1(0,1)$ and integration by parts gives the symmetric bilinear formIt is bounded andby the Poincaré inequality. Thus $L$ is positive definite and $a$ is coercive. The Lax-Milgram theorem gives a unique weak solution of the variational problem

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.

Multiply the differential equation by $-p(x)$ and compare the coefficients of $y^{\prime \prime }$ and $y^{\prime }$ with $-(p{y}^{\prime }{)}^{\prime }=-p{y}^{\prime \prime }-p^{\prime }{y}^{\prime }$. The required integrating factor satisfies $p^{\prime }=-bp$. Thus one may takeThen $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$, the desired Sturm-Liouville form, with the same homogeneous Dirichlet boundary conditions. A common positive constant multiplying $p,q,w$ is immaterial. For the usual regular Sturm-Liouville problem one also assumes real sufficiently regular coefficients and $d>0$, ensuring $p,w>0$.

Multiplication by $x^2$ puts the eigenvalue problem into Sturm-Liouville form, $-(x^2{y}^{\prime }{)}^{\prime }=\lambda x^{2}y$, with weight $x^2$. If $u,v$ correspond to distinct eigenvalues $\lambda ,\mu$, subtract their multiplied equations and integrate to getThe upper boundary term vanishes by the Dirichlet boundary condition. For a bounded solution near zero, $x^2{y}^{\prime }$ has a limit because $(x^2{y}^{\prime }{)}^{\prime }=-\lambda x^{2}y$ is integrable. A nonzero limit would force a $1/x$ divergence in $y$, so that limit is zero; integration then gives $y^{\prime }=O(x)$. The lower boundary term therefore also vanishes, proving the weighted orthogonality.

Set $z=xy$. For $x>0$, $y^{\prime \prime }+2y^{\prime }/x=z^{\prime \prime }/x$, and finiteness of $y(0)$ gives $z(0)=0$. Thus $z^{\prime \prime }+\lambda z=0$ with zero values at zero and $\pi$. The radial Dirichlet spectrum in three dimensions givesThe zero and negative eigenvalues yield only the zero solution under these boundary conditions. Also ${\int }_0^{\pi }{x}^2{y}_n^{2}dx=\pi /2$.

For the forced equation, the same substitution gives $z^{\prime \prime }-\beta z=x$. The Fourier sine series of $x$ on $(0,\pi )$ has coefficientsExpanding $z=\sum c_n\sin nx$ gives $-(n^2+\beta )c_n=b_n$, henceThe value at zero is defined by a limit. In fact $|\sin nx/x|\le n$, so uniform convergence holds even there, by comparison with $2\sum (n^2+\beta {)}^{-1}$. An independent check comes from solving the constant-coefficient equation directly:It satisfies both endpoint conditions and the forcing for $x>0$, and its finite origin limit agrees with the series. For $\beta >0$ the homogeneous boundary value problem has no nonzero solution, so the solution is unique.

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.

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,

Use the weighted inner product $⟨f,g{⟩}_w={\int }_a^{b}w(x)\overline{f(x)}g(x)dx$ with positive weight $w$. For a self-adjoint operator $L$ and nonzero eigenfunction $Ly=\lambda y$, self-adjointness gives $\lambda ⟨y,y{⟩}_w=⟨y,Ly{⟩}_w=⟨Ly,y{⟩}_w=\overline{\lambda }⟨y,y{⟩}_w$. The positive norm therefore forces $\lambda$ to be real. For two eigenfunctions of distinct eigenvalues, the same calculation gives $({\lambda }_2-{\lambda }_1)⟨y_1,y_2{⟩}_w=0$, proving their orthogonality.

The Chebyshev differential equation takes the Sturm-Liouville formIts operator is $Ly=-(1-x^2)y^{\prime \prime }+x{y}^{\prime }$, so $Lx=x$ as required. For $y=x^3+Bx$, direct substitution gives $Ly=9x^3+(B-6)x$. Comparing coefficients yieldsExplicitly, using $x=\cos \theta$,The endpoints are singular because $p$ vanishes and $w$ diverges there. The polynomial solutions use the natural condition $p{y}^{\prime }\to 0$ at each end; under $x=\cos \theta$ these are the Neumann eigenfunctions of the Chebyshev differential operator, proportional to $\cos (n\theta )$. They differ from the zero-value endpoint realization of the same differential expression.

Multiplying by $x^2$ makes the first two terms a total derivative:The resulting Sturm-Liouville form isEquivalently, an integrating multiplier $\mu$ must satisfy $(\mu (1-1/x){)}^{\prime }=\mu (2/x-1/x^2)$, which reduces to ${\mu }^{\prime }/\mu =2/x$ and gives $\mu =x^2$ up to a constant. The coefficient $p$ is positive for $1<x<R$ and vanishes linearly at the left endpoint; the weight $w$ is positive throughout. The boundedness condition removes the logarithmic local branch, giving the no-log boundary condition at a linearly vanishing Sturm-Liouville endpoint.

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.

Multiply the differential equation by $(1-x^2{)}^{-1/2}$. Sincethe Sturm-Liouville form isThe weighted inner product is consequently $⟨u,v{⟩}_w={\int }_{-1}^1{u}^{\ast }vdx/\sqrt{1-x^2}$. These endpoints are singular: $p$ vanishes and $w$ diverges, though $w$ is integrable. The required boundedness of $y,y^{\prime }$ makes the boundary factors $p{y}^{\prime }$ vanish, which is the fact needed in the integration identities below.

Divide the eigenvalue equation by $x$ and collect the first two derivative terms into a single derivative:Thus one choice of Sturm-Liouville form isIndeed $w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]=-x^2{y}^{\prime \prime }-x{y}^{\prime }-y$. Both $p$ and $w$ are positive and regular throughout this interval. The zero endpoint values make the boundary form vanish.