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

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.

For $N$ equal masses joined to fixed walls by equal springs, the stiffness matrix has diagonal $2k$ and adjacent entries $-k$. Its characteristic determinant obeys $D_N=d{D}_{N-1}-k^2{D}_{N-2}$, $d=2k-m{\omega }^2$, $D_0=1$, $D_1=d$. If ${\beta }_1+{\beta }_2=d$ and ${\beta }_1{\beta }_2=k^2$, then $D_N=({\beta }_1^{N+1}-{\beta }_2^{N+1})/({\beta }_1-{\beta }_2)$ for distinct roots. At a repeated root $\beta =\pm k$, the correct limit is $(N+1){\beta }^N$, which is nonzero. Thus the apparent ratio solution ${\beta }_1/{\beta }_2=1$ must be excluded. Writing $d=2k\cos q$ gives $D_N=k^N\sin [(N+1)q]/\sin q$, the Chebyshev polynomial of the second kind. There are exactly $N$ distinct positive normal frequencies $2\sqrt{k/m}\sin [j\pi /(2(N+1))]$, $1\le j\le N$.

With $x=\cos \theta$, the weight becomes ${\sin }^2\theta d\theta$. Sincethe sine orthogonality relations show that $P_n$ is a scalar multiple of the Chebyshev polynomial of the second kind $U_n$. Its roots areThe printed normalization $P_n(0)=1$ is possible only for even $n$; then $U_n(0)=(-1{)}^{n/2}$ andFor odd $n$, every such orthogonal polynomial is odd and vanishes at zero, so that normalization is inconsistent.

Let $Y(\theta )=y(\cos \theta )$ for $0\le \theta \le \pi$. The chain rule givesSubstitution cancels the first-derivative terms and leaves $-Y^{\prime \prime }=n^{2}Y$. The endpoint conditions become $Y(0)=Y(\pi )=0$. For $n\ne 0$, the first condition selects $Y=A\sin (n\theta )$, and the second requires $\sin (n\pi )=0$. Thus a nonzero solution requires an integer $n$. The case $n=0$ gives $Y=A+B\theta$ and both conditions force it to vanish. Negative integers give the same eigenvalues and eigenfunctions up to sign, so take $n=1,2,\dots$.

For vector normalization, use $w(x)dx=-d\theta$:Choosing a positive real normalization constant givesHere $U_{n-1}$ is the Chebyshev polynomial of the second kind. The factors $\sqrt{1-x^2}$ make both endpoint values zero. The derivatives need not stay finite there; instead $p{y}_n^{\prime }=-n\sqrt{2/\pi }\cos (n\theta )$ has finite endpoint limits, ensuring that the relevant boundary form vanishes.

For distinct nonzero recurrence roots, the determinant formula givesThe ratio one is excluded: at that repeated root, part (iii) gives $(N+1)(\pm k{)}^N\ne 0$. Thus the naive extra root corresponding to equal recurrence roots is spurious.

To count the physical solutions without ambiguity from exchanging ${\beta }_1$ and ${\beta }_2$, write ${\beta }_{1,2}=k{e}^{\pm iq}$ and $d=2k\cos q$ with $0<q<\pi$. ThenIts $N$ distinct zeros occur at $q=j\pi /(N+1)$, $j=1,\dots ,N$. They give exactly $N$ distinct positive normal frequencies, or equivalently $N$ distinct roots in ${\omega }^2$:If signed $\omega$ is counted, the characteristic equation has $2N$ roots $\pm {\omega }_j$. The sine quotient is the Chebyshev polynomial of the second kind representation of the fixed-end spring-chain determinant and its repeated-root limit.

Put $Y(\theta )=y(\cos \theta )$, with $0<\theta <\pi$. The chain rule givesConsequently the original differential expression becomes $-Y^{\prime \prime }-2\cot \theta Y^{\prime }$. If $Y=F/\sin \theta$, direct differentiation givesTaking $F=\sin n\theta$ therefore yieldsThe apparent endpoint singularities are removable. Expansion at $\theta =0,\pi$ givesThus both endpoint regularity conditions hold. This is the Chebyshev polynomial of the second kind $U_{n-1}(x)$.

The trigonometric addition formula givesDividing by $\sin \theta$ and writing $x=\cos \theta$ gives the recurrence relationThe first two members are $y_1=1$ and $y_2=2x$. If $y_n$ has degree $n-1$ and leading coefficient $2^{n-1}$, then $2x{y}_n$ has degree $n$ and leading coefficient $2^n$, whereas $y_{n-1}$ has lower degree and cannot cancel that term. Mathematical induction provesFor example the sequence begins $1,2x,4x^2-1,8x^3-4x$, agreeing with the Chebyshev polynomial of the second kind.