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Use displacements $x_i$ from an equilibrium, so the two spring extensions are $x_2-x_1$ and $x_3-x_2$. The kinetic energy and elastic potential energy areThus $T={\dot{x}}^{T}M\dot{x}/2$ and $V=x^{T}Kx/2$, with the small-oscillation mass and stiffness matricesIf $x_i$ instead denote absolute positions, the appropriate natural spring lengths must first be subtracted from the differences in $V$. They do not affect the equations for displacements.

Euler-Lagrange equations give $M\ddot{x}+Kx=0$, so a normal mode $x=v{e}^{i\omega t}$ obeys the generalized eigenvalue problem for small oscillations $Kv={\omega }^{2}Mv$. Reflection symmetry separates antisymmetric shapes $(A,0,-A{)}^T$ from symmetric shapes $(A,B,A{)}^T$. In the antisymmetric sector the end equation is $kA=m{\omega }^{2}A$, giving ${\omega }^2=k/m$. In the symmetric sector the two independent equations areTwice the first plus the second gives $m{\omega }^2(2A+\mu B)=0$. If $\omega =0$, then $B=A$. Otherwise $B=-2A/\mu$, and substitution into the first equation gives ${\omega }^2=(k/m)(1+2/\mu )$. Thus the three independent shapes are:For $v_0$ both springs retain their lengths, so this is uniform translation. For $v_1$ the end masses move oppositely and their forces on the middle mass cancel. For $v_2$ the end masses move together, while the middle mass moves oppositely so the centre of mass remains fixed. Direct substitution verifies all three eigenpairs; for positive $m,k,\mu$ the three eigenvalues are distinct and exhaust the three degrees of freedom. These are the normal modes of three masses connected by two equal springs.

A useful normalization is the weighted inner product induced by $M$. Since the squared mass norms are $m(\mu +2)$, $2m$ and $m(2+4/\mu )$, respectively, takeIf ordinary Euclidean unit eigenvectors are intended instead, divide $v_0,v_1,v_2$ by $\sqrt{3},\sqrt{2},\sqrt{2+4/{\mu }^2}$, respectively. The frequencies and physical motions are independent of this normalization choice. The zero-frequency coordinate has general motion $A+Bt$ rather than an oscillation.

Substitute $T=e^{-\nu t}R(r)$ into the axisymmetric heat equation and cancel the exponential:Thus the radial ordinary differential equation and its Sturm-Liouville form areThe weight in this Sturm-Liouville problem is therefore $\boxed{w=r}$. The outer Dirichlet boundary condition is $R(1)=0$, and the physical solution is regular at the centre. The factor $r$ in the weighted inner product is also the radial part of the polar area element.

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.

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.

Two integrations by parts give the boundary form of the Sturm-Liouville operator:The interior terms cancel because $p,q$ are real. Thus the required endpoint condition isFor a regular interval, a standard self-adjoint domain is specified by one real homogeneous Robin boundary condition at each endpoint:where neither pair of real constants is identically zero. At each end, these conditions force the two boundary data vectors of $f,g$ to be proportional, making the displayed form zero. Dirichlet boundary conditions and Neumann boundary conditions are the cases $B=0$ and $A=0$. Coupled conditions also work, for example $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, whose two endpoint contributions cancel. Self-adjointness uses the maximal common domain with these conditions, not an arbitrary smaller symmetric restriction.

The weighted inner product must be nondegenerate. The usual interpretation is $w>0$ almost everywhere, allowing isolated endpoint zeros as in part (b), with the operator defined on its appropriate weighted domain. Merely allowing a weight that is zero on an entire interval would not by itself define the stated operator or a positive-definite inner product on ordinary functions.

Use the normalized eigenfunctions and their weighted inner product ${\int }_a^{b}w\overline{{\psi }_m}{\psi }_{n}dx={\delta }_{mn}$. Orthogonality givesFor complex trials, $F$ means the corresponding Hermitian quadratic form with $|{\psi }^{\prime }{|}^2,|\psi {|}^2$; it agrees with the displayed functional on real trials. If $a_0\ne 0$, division by $|a_0{|}^2$ yieldsSince ${\lambda }_n\ge {\lambda }_0$, this is at least ${\lambda }_0$. The Rayleigh-Ritz variational principle therefore gives an upper bound from any admissible trial, improved by minimizing over trial parameters or enlarging the trial space. Equality holds when the trial is entirely in the lowest eigenspace.

The displayed ratio using $b_n$ presupposes $a_0\ne 0$. If $a_0=0$, use the undivided quotient $\sum {\lambda }_n|a_n{|}^2/\sum |a_n{|}^2$ instead; the same upper-bound argument still applies. No ground-state coefficient can be divided out in that case.

Use the normalized eigenfunctions with the stated weighted inner product. Orthogonality gives $G[\hat{y}]={\sum }_n|c_n{|}^2$. The corrected Sturm-Liouville problem and the vanishing boundary term giveFor a complex trial the numerator means $\int (p|{\hat{y}}^{\prime }{|}^2+q|\hat{y}{|}^2)dx$, agreeing with the given formula for real trials. ThereforeIt is a convex combination of the eigenvalues, with equality only for a trial proportional to the lowest eigenfunction. The Rayleigh-Ritz variational principle consequently estimates ${\lambda }_0$ by choosing a nonzero admissible trial family and minimizing this quotient over its parameters:The result is an upper bound. A more flexible trial space can lower it, while equality requires that the exact lowest eigenfunction belong to the trial space. One need not know the expansion coefficients to calculate the quotient directly from the trial function.

Use a logarithmic change of variable in a Sturm-Liouville problem: set $t=\log x$ and $Y(t)=y(e^t)$. The chain rule givesThe endpoint conditions become $Y(0)=Y(1)=0$. Multiplying $-Y^{\prime \prime }=(\lambda +1)Y$ by $Y^{\ast }$ and integrating shows that $\lambda +1>0$ for a nonzero solution. Put $\mu =\sqrt{\lambda +1}$. The first endpoint selects $Y=A\sin (\mu t)$ and the second requires $\mu =n\pi$, $n\ge 1$. HenceThese eigenfunctions are orthonormal in the specified weighted inner product, because $w(x)dx=dx/x=dt$ and

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.

Apply the identity from (c) to the two real eigenfunctions. The equations of their Sturm-Liouville problem giveSince the eigenvalues differ,This is orthogonality in the weighted inner product.

A fixed circular elastic membrane of radius $a$ has radial normal modes satisfying the displayed Sturm-Liouville problem, with weight $r$, $R(a)=0$ and regularity at the origin. The regular solutions are Bessel functions of the first kind $J_0(\sqrt{\lambda }r)$. Integration by parts gives the positive Rayleigh quotient $\lambda ={\int }_0^{a}r|R^{\prime }{|}^{2}dr/{\int }_0^{a}r|R{|}^{2}dr$. Distinct modes are orthogonal in the weighted inner product ${\int }_0^{a}r\overline{R}Sdr$.

For noninteger real $\gamma$, expand a smooth forcing as $F={\sum }_{n\ge 0}{F}_n{T}_n$ in Chebyshev polynomials. Because $(1-x^2)T_n^{\prime \prime }-x{T}_n^{\prime }=-n^2{T}_n$, division of each coefficient by ${\gamma }^2-n^2$ gives the regular solution. The coefficients use the weighted inner product with weight $(1-x^2{)}^{-1/2}$. Choosing the same endpoint domain as the regular eigenfunctions of the Sturm-Liouville problem is essential: finite values alone also allow homogeneous solutions $\cos (\gamma \arccos x)$ and $\sin (\gamma \arccos x)$, which generally have divergent endpoint derivatives. Setting $x=\cos \theta$ converts the regular endpoint condition into the Neumann boundary conditions $Z_{\theta }(0)=Z_{\theta }(\pi )=0$. The homogeneous equation then has no nonzero solution when $\gamma$ is noninteger, proving uniqueness in this domain.

For a uniform string with tension $T>0$, linear mass density $\rho >0$, left end fixed, and a freely transverse-moving point mass $M\ge 0$ at $x=L$, the wave equation and endpoint law areThe last equation is Newton's second law: the string pulls the mass back toward the interior, giving the minus sign. With wave speed $c=\sqrt{T/\rho }$, a nonzero normal mode has spatial part $X(x)=\sin (\omega x/c)$ and satisfiesFor $M>0$, its dimensionless roots $q=\omega L/c$ occur once in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$, because $\cot q$ decreases from infinity to zero while $Mq/(\rho L)$ increases. For $M=0$, $q=(n+1/2)\pi$: the right end has a homogeneous Neumann boundary condition. There is no nonzero zero-frequency mode, since a linear spatial profile vanishing at the left end and with zero right slope must vanish everywhere.

Multiplying the spatial equation by $X$ and using integration by parts givesFor two different modes, the same calculation gives the weighted inner product orthogonality $\rho {\int }_0^L{X}_m{X}_{n}dx+M{X}_m(L)X_n(L)=0$. The endpoint mass therefore contributes to both inertia and mode orthogonality.

Let $A$ and $M$ be real symmetric positive-definite matrices. The principal square root of a positive semidefinite matrix construction gives an invertible matrix for each. SetThen $Ax=b$ is equivalent to $By=c$. This is a matrix congruence, and $B$ is a symmetric positive-definite matrix: $B^T=B$, and for every nonzero vector $v$,since $S^{-1}v\ne 0$. This short proof establishes both requirements for applying the Conjugate gradient method to the transformed linear system.

The left transformed matrix has the same eigenvalues becauseIts correct geometry uses the weighted inner product $⟨u,v{⟩}_M=u^{T}Mv$: $⟨u,M^{-1}Av{⟩}_M=u^{T}Av=⟨M^{-1}Au,v{⟩}_M$. Thus it is symmetric in this inner product, even when it is not a symmetric matrix in ordinary coordinates. The spectral condition number governing the transformed symmetric problem is ${\kappa }_2(B)$, not generally ${\kappa }_2(M^{-1}A)$.

The Preconditioned conjugate gradient method realizes this transformation using auxiliary solves $M{z}_k=r_k$, without constructing $S$ or its matrix inverse. With fixed $M$, initialize $r_0=b-A{x}_0$ and $p_0=z_0$; while $r_k\ne 0$, usethen solve $M{z}_{k+1}=r_{k+1}$ and setStop at a zero linear-system residual before forming another ratio. Positive definiteness makes the denominators positive before termination.^[ref]

An invertible matrix alone is insufficient: with $A=I$ and $M=\mathrm{diag}(-1,1)$, the left transformed matrix has eigenvalues $-1,1$, and $M$ has no real symmetric positive-definite matrix square root. Even a valid symmetric positive-definite matrix can worsen conditioning if its scaling is unsuitable, as the next worked example shows.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.

A symmetric positive-definite matrix $M$ defines an inner product on real coordinate vectors byFor diagonal $M$, positive diagonal entries weight the squared coordinate errors. This is the finite-coordinate version of a weighted inner product; the positive-definite matrix example appears in Francis Narcowich, Inner Products and Norms, standard inner products, example 8.

For $M=\mathrm{diag}(4,1)$, the squared norm of $(a,b{)}^T$ is $4a^2+b^2$. A first-coordinate error of one costs as much as a second-coordinate error of two. The unit sphere is the ellipse $4a^2+b^2=1$, and orthogonality means $4u_1{v}_1+u_2{v}_2=0$. Thus “closest” and “perpendicular” depend on the chosen inner product. The orthogonal projection residual example shows how this changes a best approximation.

On $(0,L)$, the differential operator $\mathcal{L}={\partial }_x^2+\beta {\partial }_x=e^{-\beta x}{\partial }_x(e^{\beta x}{\partial }_x)$ with homogeneous Neumann boundary conditions is self-adjoint in the weighted inner product with weight $e^{\beta x}$. Its eigenfunctions and nonnegative decay rates areTheir squared norms are $N_0=(e^{\beta L}-1)/\beta$ and $N_k=L[1+{\beta }^2/(4q_k^2)]/2$. The Sturm-Liouville eigenfunction expansion givesThis kernel acts against the measure $e^{\beta y}dy$, not Lebesgue measure alone. It is symmetric in $x,y$; the transition density against Lebesgue measure is $e^{\beta y}{K}_{\beta }(x,y,t)$.

A positive measurable weight $w$ defines the displayed norm on the L2 space for the measure $w(x)dx$, with functions identified when equal almost everywhere. This is the norm induced by the weighted inner product. Changing $w$ changes the contribution of each region to approximation error. If $p_0,\dots ,p_N$ are nonzero mutually orthogonal real functions, completing squares shows that the unique best approximation to real $f$ in their linear span has coefficients $a_j=⟨f,p_j{⟩}_w/\Vert p_j{\Vert }_w^2$.