The input is even, so its Fourier transform is real and even. Decomposing the cosine into complex exponentials and integrating over its support givesCombining the fractions and using the angle addition formula yieldsThe apparent poles at $k=\pm 1$ are removable singularities; either the integral or the sine-quotient limit gives $\tilde{f}(\pm 1)=a+\sin (2a)/2$.

If $\cos a\ne 0$, expansion of the denominator givesThe oscillatory envelope is therefore $O(|k{|}^{-1})$. At $a=\pi /2+n\pi$, $n=0,1,2,\dots$, the endpoint values vanish and the first term cancels:This gives an $O(|k{|}^{-2})$ envelope. At general $a$ the truncation creates jumps in the function; at these special values it leaves the function continuous but creates jumps in its first derivative. Each additional degree of endpoint smoothness gives another inverse power, as explained by Fourier decay from a derivative jump. The exact example is the Fourier transform of a truncated cosine; isolated zeros of its oscillatory factors do not change the envelope order.

Use the trigonometric identity $2{\cos }^2(ax)=1+\cos (2ax)$ and express the cosine as a sum of two complex exponentials. The Fourier transform of each exponential is a shifted Dirac delta distribution:By linearity,The original function does not decay and is not absolutely integrable on the real line, so this answer is a tempered distribution, rather than an ordinary convergent Fourier integral. Its three lines represent the constant part and the two opposite spatial frequencies.

First take a Schwartz function, so that Fourier inversion and the exchange of integrals are justified. Inserting the inverse transform of the complex conjugate givesFor $f\in L^1\cap L^2$, approximate it by Schwartz functions in both norms. The identity just proved makes their transforms a Cauchy sequence in L2 space, while convergence in L1 space makes the transforms converge uniformly to the defining transform of $f$. Thus the two transform limits agree, and passing to the $L^2$ norms proves the finite Parseval identity. This is the normalization of the Plancherel theorem appropriate to the paper.

There is a qualification to the printed hypothesis: absolute integrability alone does not guarantee a finite square integral. For instance $f(x)=x^{-3/4}$ on $(0,1)$ and zero elsewhere belongs to $L^1$ but not $L^2$. For an $L^1$ function of this kind the transform is not in $L^2$ either, by L2 density from a square-integrable Fourier transform, so both sides are infinite if the identity is interpreted in that extended sense. The finite equality requires square integrability as well.

Insert the source into the Green-function representation and split the integral at its source discontinuity in derivative:Using ${\sin }^{2}x+{\cos }^{2}x=1$, the answer isBoth endpoint values are zero. As a direct check, conjugating by $e^x$ sends $y^{\prime \prime }-2y^{\prime }+2y$ to $e^x(v^{\prime \prime }+v)$, and $v=1-\cos x-\sin x$ gives $v^{\prime \prime }+v=1$. Thus the differential equation and both boundary conditions are satisfied.

The homogeneous characteristic equation is $r^2-2r+2=0$, with roots $1\pm i$. Choose the endpoint-adapted homogeneous solutionsso that $u(0)=0$ and $v(\pi /2)=0$. Their Wronskian iswhich never vanishes. A continuous Green function with the required endpoints has the form $C(\xi )u(x_{<})v(x_{>})$, where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. Its derivative jump is $C(\xi )W(\xi )$. Because the leading coefficient is one, part (b) requires this to equal one. ThereforeIn particular, the branches are $-e^{x-\xi }\sin x\cos \xi$ for $x\le \xi$ and $-e^{x-\xi }\sin \xi \cos x$ for $x\ge \xi$. This is the Dirichlet Green function for a conjugated harmonic oscillator with interval length $\pi /2$, drift parameter one and oscillator wave number one.

For a fixed source point $\xi \in (a,b)$, the Green function is defined bywhere $\delta$ is the Dirac delta distribution and the operator acts on $x$. The corresponding solution isApplying $L_x$ under the integral gives $f(x)$ by the defining delta identity, while the endpoint values vanish. With the stated continuous coefficients, one may define the normalized distributional equation $G_{xx}+(\beta /\alpha )G_x+(\gamma /\alpha )G=\delta (x-\xi )/\alpha (\xi )$; this avoids multiplying a general distributional second derivative by a merely continuous coefficient.

Existence of this inverse Green function additionally requires that the homogeneous two-point boundary value problem have only the zero solution. The assumptions printed in the question alone do not ensure this: $y^{\prime \prime }+y$ on $[0,\pi ]$ has the nonzero zero-endpoint solution $\sin x$. At such resonance there are solvability conditions and no ordinary inverse kernel for arbitrary $f$. The construction below is nonresonant.

For spherical symmetry the angular mode is $\ell =0$. Put $u(t,r)=r\psi (t,r)$. Direct differentiation givesso the radial heat equation in three dimensions becomes the ordinary one-dimensional heat equation $u_t=\kappa u_{rr}$. Since $r_{\pm }>0$, the zero endpoint temperatures become $u(t,r_{\pm })=0$.

Write $L=r_{+}-r_{-}$. The two initial sine modes already satisfy these Dirichlet boundary conditions; their squared wave numbers are $(\pi /L{)}^2$ and $(2\pi /L{)}^2$. Each therefore evolves independently with its corresponding exponential decay. Dividing back by $r$ givesAt $t=0$ the initial profile is recovered, and each sine is zero at both endpoints. The first mode decays four times more slowly in its exponential rate than the second. This is the two-mode instance of the spherical-shell heat equation via radial rescaling; uniqueness follows, for example, by applying the heat equation energy identity to the difference of two solutions.

Seek a nonzero separated solution $\psi =T(t)R(r)\Theta (\theta )$. Substitution into the heat equation and division by the product givesThe left side depends only on time, so set it to $-k^2$. Multiplication by $r^2$ then separates the angular dependence, with separation constant $\ell (\ell +1)$. The three ordinary differential equations areThe temporal solution is $T=T_0{e}^{-\kappa k^{2}t}$. With $q=\cos \theta$, the angular equation is the Legendre differential equation; regularity at both angular endpoints selects $\ell =0,1,2,\dots$ and $\Theta \propto P_{\ell }(q)$, the Legendre polynomial. The radial equation has the Spherical Bessel functions $j_{\ell }(kr)$ and $y_{\ell }(kr)$ for $k>0$. Regularity at $r=0$, if imposed, excludes the singular second solution. For $k=0$ the radial powers are $r^{\ell }$ and $r^{-\ell -1}$. Without specified radial boundary conditions, no discrete set of $k$ is selected; $-k^2$ is a convention for the temporal separation constant.

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: