The initial displacement lies in an invariant subspace of axial motions. Let the three opposite vertex pairs have positionsFor a vertex on the positive $x$ axis, the two springs towards the $y$ vertices contribute $-k(q_x+q_y)$ in the $x$ direction; the two towards the $z$ vertices contribute $-k(q_x+q_z)$. Transverse components cancel in pairs. Cyclically,The matrix is $I+11^{\mathsf{T}}$. It has eigenvalue $4$ on the breathing vector $(1,1,1)$, and eigenvalue $1$ on the orthogonal plane $q_x+q_y+q_z=0$. This identifies the axial normal modes of an octahedral spring framework without a large characteristic determinant.

Put ${\omega }_0=\sqrt{k/m}$ and decompose the initial vector asSince all initial velocities are zero, only cosine time factors occur. The complete motion to first order in the displacements iswith the vertex positions given by the first displayed formula. No translation or rotation is excited: the initial displacements are orthogonal to those zero modes and their initial velocities vanish. The two checks ${ϵ}_1={ϵ}_2$ and ${ϵ}_1+2{ϵ}_2=0$ respectively isolate pure breathing and pure traceless axial oscillation. Nonlinear corrections beyond the small-displacement approximation have been omitted.

For a breathing displacement take ${\mathbf{u}}_i=q(t){\mathbf{r}}_i$. Every spring has equilibrium length $\sqrt{2}$ in the given length units, so its linear extension is $\sqrt{2}q$. At the vertex ${\mathbf{r}}_i$, the four adjacent equilibrium vertices sum to zero. Therefore the spring force there isIt remains radial and proportional to the original displacement at every vertex, proving that this symmetric displacement is a normal mode. Newton's second law givesThe first is angular frequency and the second is cycles per unit time. For $k$ in $\mathrm{N}{\mathrm{m}}^{-1}$ and $m$ in $\mathrm{kg}$, these have units $\mathrm{rad}{\mathrm{s}}^{-1}$ and $\mathrm{Hz}$ respectively. An energy check gives $T=3m{\dot{q}}^2$ and $V=12k{q}^2$, leading to the same ratio ${\omega }_b^2=24k/(6m)$.

Label the equilibrium vertices in the order $+x,-x,+y,-y,+z,-z$, and write their displacements as ${\mathbf{u}}_i(t)\in {\mathbb{R}}^3$. Their eighteen Cartesian components are generalized coordinates. Linearize the axial Hookean springs about the unstressed equilibrium. For an edge with unit direction ${\mathbf{n}}_{ij}$, its first-order extension is ${\mathbf{n}}_{ij}\cdot ({\mathbf{u}}_i-{\mathbf{u}}_j)$, so the quadratic potential energy isThree zero modes are translations: for each Cartesian unit vector ${\mathbf{e}}_{\alpha }$, take ${\mathbf{u}}_i={\mathbf{e}}_{\alpha }$ at every vertex. Three more are infinitesimal rotations ${\mathbf{u}}_i={\mathbf{e}}_{\alpha }\times {\mathbf{r}}_i$. Explicitly, in the stated vertex order, their displacement vectors areTranslations leave each displacement difference zero. For rotations,because the equilibrium edge vector is parallel to ${\mathbf{n}}_{ij}$. Hence every spring has zero first-order extension in all six patterns, and the stiffness matrix annihilates them.

The translations have nonzero summed displacement, while all the rotations have zero sum. The three rotational patterns are themselves independent because the vertices do not lie on a single line. ThusThese are the rigid-motion zero modes of a spring framework. The rotational displacements describe infinitesimal rigid rotations; an arbitrary finite rotation would require its nonlinear displacement field.

Let $n\ge 2$ be even and integrate $f(z)=1/(1+z^n)$ around an upper-half-plane semicircle of radius $R>1$ with the real segment traversed from $-R$ to $R$. There are no real poles, and the $n/2$ upper-half-plane poles areEach is simple. Since $z_j^n=-1$,On the semicircle the integral magnitude is at most $\pi R/(R^n-1)$, which tends to zero. The real integral is absolutely convergent because its tail is of order $|x{|}^{-n}$. ThusEvaluate the finite geometric series, writing $\theta =\pi /n$:ConsequentlyFor $n=2$ this reduces to $\pi$, as a check on the contour orientation and residue signs.

If $f$ is analytic on and inside a positively oriented simple closed contour except for finitely many isolated poles $z_j$ strictly inside it, then the residue theorem statesThere must be no singularity on the contour. More generally each residue is multiplied by the contour's winding number about its point.

Factor the denominator as $(z-z_0)(z+z_0)$. Since $z_0\ne 0$, its zeros are distinct, and the potential simple poles and their residues areFor example, the first follows by taking ${\lim }_{z\to z_0}(z-z_0)f(z)$. If the numerator vanishes at either denominator zero, the singularity there is removable and the corresponding residue is zero. Thus the data identify the possible poles, not an unconditional pole at a point canceled by $g$.

The Cauchy integral formula for derivatives, with $z_0=0$, givesThe derivative cycles of sine and cosine give ${\sin }^{(n)}(0)=\sin (n\pi /2)$ and ${\cos }^{(n)}(0)=\cos (n\pi /2)$. ThereforeIn particular the sine integral vanishes for even $n$, and the cosine integral vanishes for odd $n$. For the remaining cases their signs alternate. The radius is irrelevant provided $R>0$, since the numerator is entire and the origin is enclosed once counterclockwise.

Denote the unshifted differential expression by $L_{0}y=(p{y}^{\prime }{)}^{\prime }+qy$. The supplied Green's function satisfies $L_{0}G(x,\xi )=\delta (x-\xi )$, whereas the inhomogeneous equation is $L_{0}y=\lambda wy+wf$.

Multiply the latter equation by $G$ and the Green equation by $y$, subtract and integrate in $x$:The left-hand side is the boundary termIt vanishes because $G$ and $y$ satisfy the same homogeneous Robin boundary conditions. For example, where $\alpha \ne 0$, substituting $y^{\prime }=-\beta y/\alpha$ and $G_x=-\beta G/\alpha$ cancels the endpoint expression; where $\alpha =0$, both values vanish. The other endpoint works identically. HenceThe Dirac delta function contribution is valid for an interior evaluation point $\xi$; equivalently, $G$ is continuous there and has flux-derivative jump $[p{G}_x{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$.

This uses the existence of the unshifted Green's function stated in the question. Such a function is not automatic for every choice of $p,q$ and endpoint conditions: a nontrivial zero mode of $L_0$ would obstruct an ordinary inverse and would require a compatibility condition and a modified Green's function. Under the stated inverse assumption, the differential equation and the displayed integral equation are equivalent for functions in the boundary-condition domain.

For nonzero $b$, the coefficients satisfy $a_n\sim -b/n^2$, so the analytic power series has radius of convergence $1$. It converges absolutely also at $x=\pm 1$, by comparison with $\sum n^{-2}$.

The real logarithm in the prescribed solution requires $x>0$. ConsequentlyThe series and the function itself have finite limits at $x=1$; in fact the logarithmic companion tends to $1$. Its derivatives are singular there, so convergence of its values at the endpoint does not make it a classical solution through that singular point of a differential equation. At $x=0$ the logarithmic branch diverges. The independent analytic solution $C(1-x)$ is a polynomial and has no such convergence restriction.

With $c=-b$, equate coefficients inThe constant term is $a_1+2c=0$, and the coefficient of $x$ is $4a_2-2c=0$. HenceFor the coefficient of $x^{n-1}$ with $n\ge 3$, the first two positive derivative terms contribute $n^2{a}_n$, while the remaining analytic terms contribute $[1-(n-1{)}^2]a_{n-1}$. ThusIteration gives $a_n=-b/[n(n-1)]$ for $n\ge 2$. Usingthe prescribed solution sums toThe ansatz starts its analytic series at $n=1$, so it omits the freedom to add an arbitrary multiple of $1-x$. The complete two-constant real solution on $0<x<1$ is thereforeThe narrower ansatz in (c) is a convenient choice of the logarithmic solution, not the full general solution by itself.

Let $A(x)={\sum }_{n=1}^{\infty }{a}_n{x}^n$ denote the analytic part of the prescribed ansatz. Its logarithmic part has derivativesAfter substitution, the full differential equation reduces toThe first terms are real analytic at zero and cannot cancel a nonzero logarithm. ThereforeThe cancellation of the apparent $1/x$ terms is already built into this calculation.

For a Frobenius exponent $r$, put $p_0={\lim }_{x\to 0}xP(x)=1$ and $q_0={\lim }_{x\to 0}{x}^{2}Q(x)=0$. The indicial equation issoA fundamental pair has the repeated-root Frobenius formon a positive-$x$ interval. In this case a real analytic solution is particularly simple: $y_1=1-x$ satisfies the equation exactly. A convenient logarithmic companion, obtained below, is $y_2=x+(1-x)\ln [x/(1-x)]$ for $0<x<1$. It is independent of $y_1$ because of its logarithmic divergence at zero.