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Move the differential terms to the left in the sign convention of part (a). Then $\tilde{a}=1+x$, $\tilde{c}=0$, and $\tilde{b}=(3+2x)-{\tilde{a}}^{\prime }=2(1+x)$. The integrating factor is therefore $w=e^{2x}$, up to a positive constant. The Sturm-Liouville problem isThe trial $y=x{e}^{-x}$ is admissible because $y(0)=0$ and $y^{\prime }=e^{-x}(1-x)$ vanishes at one. Its Rayleigh quotient has numeratorand denominatorConsequently the Rayleigh-Ritz variational principle givesThe numerator is positive for every nonzero admissible function, so the lowest eigenvalue is positive. The trial is not an exact eigenfunction: substituting it would require $x^2+2x-1=\lambda x$ for every $x$. Thus the upper inequality is actually strict. This example underlies the exponential-polynomial trial space for a mixed Sturm-Liouville problem.

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.

Expand the two operators before comparing coefficients:Equality $L=w\tilde{L}$ requires $a=\tilde{a}$, $c=\tilde{c}$, and $(aw{)}^{\prime }=w(a^{\prime }+\tilde{b})$. Thus $a{w}^{\prime }=\tilde{b}w$, giving the positive integrating factorWith $p=aw>0$, the result is the Sturm-Liouville form $L=-(pd/dx{)}^{\prime }-wc$. The arbitrary positive multiplier $C$ rescales both sides of the eigenvalue problem and does not change its eigenvalues. The coefficient ${\tilde{a}}^{\prime }$ is already present in the divergence term of $\tilde{L}$; it must not be counted a second time in the exponential.

For a nondegenerate surface take $R>0$, and abbreviate $C=\cosh R$, $H=\sinh R$. Close the upper oblate spheroid by its equatorial disk $D$, obtaining a half-ellipsoid $V$. The divergence isThe integral of $2y$ vanishes by reflection symmetry. At height $z$, the cross-section is a disk of radius $q=C\sqrt{1-z^2/H^2}$. Using planar polar coordinates,ConsequentlyThe outward unit normal vector on $D$ is $-{\mathbf{e}}_z$, so its flux is $-2\pi C^2$. Applying the divergence theorem to the closed surface and subtracting this disk contribution givesThis is a special case of polynomial flux through a half-ellipsoid. Including the closing disk is essential: the constant vertical component has zero divergence but nonzero flux through the open curved surface.

Put $s^2=x^2+y^2$. For $\rho >0$, elimination of $u$ givesThese are oblate spheroids with equatorial radius $\cosh \rho$ and polar radius $\sinh \rho$, flattened along the $z$ axis. At $\rho =0$ the limiting surface is the disk $z=0$, $s\le 1$.

For $0<|u|<\pi /2$, elimination of $\rho$ instead givesThese are one-sheet hyperboloids of revolution, restricted to their upper half for $u>0$ and lower half for $u<0$. The restriction matters because $\rho \ge 0$. At $u=0$, the surface degenerates to the exterior equatorial plane $z=0$, $s\ge 1$; at $u=\pm \pi /2$ it degenerates to the corresponding half of the $z$ axis.

Finally, fixed $v$ specifies a vertical half-plane through the $z$ axis, with horizontal direction $(\cos v,\sin v)$. The two values $v$ and $v+\pi$ together give a full plane. These surfaces form the mutually orthogonal families of the hyperbolic parametrization of oblate spheroidal coordinates.

Differentiate the position vector with respect to the three coordinates:Their pairwise dot products vanish. For example, ${\mathbf{r}}_{\rho }\cdot {\mathbf{r}}_u=-\sinh \rho \cosh \rho \sin u\cos u+\sinh \rho \cosh \rho \sin u\cos u=0$. Thus these are orthogonal coordinates wherever the coordinate tangents are nonzero. Their scale factors of orthogonal coordinates areThe absolute Jacobian determinant is their product, givingThe ordered coordinate system is left handed: ${\mathbf{r}}_{\rho }\cdot ({\mathbf{r}}_u\times {\mathbf{r}}_v)=-h_{\rho }{h}_u{h}_v$. The volume measure uses the absolute value. The axis and the focal circle contain coordinate degeneracies, as expected for oblate spheroidal coordinates. This is the hyperbolic parametrization of oblate spheroidal coordinates. The local TeX has an OCR substitution ${\sinh }^{2}u$ in the volume formula; the original PDF correctly prints ${\sin }^{2}u$.

Let $V\subset {\mathbb{R}}^3$ be a bounded region with piecewise smooth closed boundary $\partial V$, let $\mathbf{n}$ be its outward unit normal vector, and let $\mathbf{A}$ be a continuously differentiable vector field on a neighbourhood of its closure. The divergence theorem statesHere $dV$ is the positive volume measure, $dS$ is the surface area measure, $\nabla \cdot \mathbf{A}={\partial }_x{A}_x+{\partial }_y{A}_y+{\partial }_z{A}_z$, and the right side is the total outward flux. All boundary components must be included, with their normals pointing away from $V$.

In Cartesian coordinates, write ${\partial }_i=\partial /\partial x_i$ and use the Levi-Civita symbol. The contraction of two Levi-Civita symbols givesHere the Kronecker delta selects the indicated indices, and continuous second partial derivatives allow their order to be interchanged. Identifying the gradient, divergence and componentwise Laplacian proves the curl of the curl identity:

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 $g$.