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The free surface is a material surface. For $F(x,y,t)=y-\eta (x,t)$, the material derivative vanishes on $F=0$. Since the velocity is $\nabla \varphi$, this gives the exact kinematic boundary condition for a free-surface graphAt the surface the pressure equals the constant atmospheric pressure. Absorbing that constant into the time-dependent potential gauge in Unsteady Bernoulli equation yields the exact dynamic boundary condition for an inviscid interfaceThere is no surface-tension term in this model.

Write $\varphi =Ux+\phi$ and linearize in the small perturbation $\phi$ and the surface displacement. The base value is $F=U^2/2$, and evaluation at the displaced surface can be replaced by evaluation at zero to first order. ThusIncompressibility requires Laplace's equation ${\phi }_{xx}+{\phi }_{yy}=0$. For $k>0$, the complex perturbation $b{e}^{ky}{e}^{i(kx-\omega t)}$ satisfies this equation and decays with depth. Put $\eta =\mathrm{Re}\{a{e}^{i(kx-\omega t)}\}$ and $\Omega =\omega -kU$. The linear boundary conditions becomeThey are consistent precisely whenConsequently the two laboratory phase velocities of a deep-water gravity wave areIf propagation of a wave packet is intended, its group velocities are instead $d\omega /dk=U\pm \frac{1}{2}\sqrt{g/k}$.

Neglect end effects, so the electrostatic potential depends only on the cylindrical radius $r$. The radial Laplace equation is $(r{\varphi }^{\prime }{)}^{\prime }=0$, with general solution $A\log r+B$. Imposing the conductor boundary conditions givesThe radial electric field is $E_r=-{\varphi }^{\prime }$. Just inside the middle cylinder it is $-V/[\lambda a\log \lambda ]$, directed towards the inner cylinder; just outside it is $V/[\lambda a\log (2/\lambda )]$, directed towards the outer cylinder. By Gauss's law, its charge per unit length isHence the capacitance per unit length isThe two coaxial cylindrical capacitors contribute in parallel because both connect the middle cylinder to the same grounded potential. This is the coaxial capacitor with grounded inner and outer cylinders construction.

For $\lambda =1+\delta$ with $\delta \to 0^{+}$, $\log \lambda =\delta +O({\delta }^2)$ while $\log (2/\lambda )\to \log 2$. ThereforeThe inner gap has width $a\delta$ and area per unit length approximately $2\pi a$. Its parallel-plate capacitor approximation gives exactly the same leading result; the outer gap contributes only a bounded correction.

For the extremum, set $u=\log \lambda$, $L=\log 2$, so $0<u<L$. ThenThus the unique stationary point is $u=L/2$. Strict convexity and divergence at both endpoints show it is a global minimum:There is no maximum on $1<\lambda <2$.

On a closed compact Riemannian manifold, the heat kernel for the positive Laplace-Beltrami operator is the smooth kernel $K_M(t,x,y)$ of $e^{-t{\Delta }_M}$ for $t>0$. It solves $({\partial }_t+{\Delta }_x)K_M=0$ and tends to the delta kernel as $t\downarrow 0$, so thatFor an orthonormal eigenbasis with eigenvalues repeated according to multiplicity,The exponential factors and elliptic regularity ensure smooth convergence for positive time. If a boundary is present, the same construction uses a specified boundary condition preserved by the covering; the closed-manifold case is the one needed below.

Let $p:N\to M=N/U$ be the finite normal locally isometric covering map. The heat equation for $f\circ p$ on $N$ is the lift of the one for $f$ on $M$. Integrate the lifted solution over a fundamental domain for the deck transformation group $U$, splitting the integral over $N$ into its translates. This givesThere is no averaging factor in this kernel formula: each translate accounts for one lift of the integration variable. To verify it directly, the sum solves the lifted heat equation and its integral against $f$ converges to $f(px)$; uniqueness of the heat solution identifies the kernel. Replacing either lift $x$ or $y$ only permutes the sum, since isometries preserve the heat kernel.

Integrating the diagonal and using the covering degree gives the heat trace formulaSuppose $U$ is contained in a finite group $T$ of isometries of $N$. For $s\in T$, change variables $x=sy$ and use invariance of both volume and the heat kernel. ThenThus $F_t$ is a class function on $T$, and for its conjugacy classes $\mathcal{C}$,

Consequently, if $U_1,U_2\le T$ are Gassmann equivalent and act freely, their orders are equal and the displayed class-by-class sums give $Z_{N/U_1}(t)=Z_{N/U_2}(t)$ for every $t>0$. Equal heat traces determine equal eigenvalues with multiplicities: the least eigenvalue at which the multiplicities differed would give a nonzero leading exponential in their trace difference as $t\to \infty$, a contradiction. Hence the two smooth quotient manifolds are isospectral. This proves the Sunada theorem, with the free-action hypothesis ensuring that the quotients are manifolds rather than orbifolds. The theorem alone makes no nonisometry claim.

A natural cubic spline interpolant $s$ satisfies $s(x_i)=y_i$, is twice continuously differentiable, and is a polynomial of degree at most three on each interval $[x_i,x_{i+1}]$. Its natural boundary conditions areIf defined outside the data interval, it continues linearly on both exterior intervals. Thus function values and the first two derivatives match at the knots, but the third derivative may jump.

For an explicit construction, let $h_i=x_{i+1}-x_i>0$ and $M_i=s^{\prime \prime }(x_i)$. Set $M_1=M_n=0$ and solveOn $[x_i,x_{i+1}]$, putThis formula interpolates the values and matches second derivatives; the tridiagonal equations enforce first-derivative matching. The homogeneous system has only the zero solution: in a row with maximal nonzero $|M_i|$, its diagonal coefficient strictly exceeds the sum of its neighbour coefficients, making cancellation impossible. Thus the natural interpolant exists uniquely.

With the convention that the residual term is a sum rather than an average, the penalised least squares criterion isMinimize over functions whose first derivative is absolutely continuous and whose second derivative is square-integrable, that is, the usual Sobolev space $H^2[x_1,x_n]$. The cubic smoothing spline ${\hat{g}}_{\lambda }$ is its minimizer, with natural boundary conditions and linear continuation outside the data range. It is a natural cubic spline with possible knots at the observed $x_i$, but usually does not interpolate the noisy $y_i$.

To see why the minimizer is a natural spline, for any $g$ take the natural cubic spline interpolant $s$ through its values $g(x_i)$ and write $h=g-s$. Then $h(x_i)=0$. Integrating $\int s^{\prime \prime }{h}^{\prime \prime }$ by parts on every knot interval twice gives zero: the terms involving $h$ vanish at knots, the terms involving $h^{\prime }$ cancel because $s^{\prime \prime }$ is continuous there, and the outer terms vanish by the natural boundary conditions. Hencewhile their residual sums of squares agree. Replacing $g$ by $s$ never increases the criterion. This proves the minimum roughness property of the natural cubic spline interpolant needed here.

For fitted values $u_i=s(x_i)$, write $\int (s^{\prime \prime }{)}^2=u^{T}Ku$, where $K$ is positive semidefinite and has the sampled affine functions as its two-dimensional nullspace. The finite-dimensional minimization becomes $\Vert y-u{\Vert }^2+\lambda u^{T}Ku$. Its strictly positive definite Hessian is $2(I+\lambda K)$, so it has the unique solutionThis is the roughness-matrix formula for a natural cubic smoothing spline. Larger $\lambda$ penalizes curvature more strongly. As $\lambda \downarrow 0$ the fit approaches interpolation; as $\lambda \to \infty$ it approaches the ordinary least-squares affine line.

To distinguish Eulerian and Lagrangian fluid perturbations, use $\mathbf{\xi }=({\xi }_x,{\xi }_y,{\xi }_z)$ for the fluid displacement, $\varphi ={\Phi }^{\prime }$ for the Eulerian potential perturbation, and $\delta \rho$ for the Eulerian mass density perturbation. The equilibrium mass density $\rho (z)$ is the profile from part (a); subscript $z$ below denotes differentiation. Linearized mass conservation and the isothermal equation of state giveThe velocity perturbation is $-i\omega \mathbf{\xi }$. Linearized momentum initially has the formSince $\nabla {\Phi }_0=-c_s^2\nabla \ln \rho$, the pressure and perturbed mass density gravity terms combine into $-\rho \nabla (c_s^2\delta \rho /\rho )$. Define $W=c_s^2\delta \rho /\rho +\varphi$. The closed linearized system isThe $y$ component is a neutral transverse sector; it vanishes for nonzero-frequency modes.

To prove reality rather than assume it, multiply momentum by $\rho {\mathbf{\xi }}^{\ast }$ and integrate over $z$. With $I=\int \rho |\mathbf{\xi }{|}^{2}dz>0$, integration by parts and the conjugate continuity equation giveUsing the conjugate Poisson equation for Newtonian gravity in the last integral givesTake disturbances with finite $I$, finite pressure and gravitational integrals, and vanishing pressure work and gravitational boundary terms. For $k\ne 0$, require the potential to decay at both infinities and the fluid displacement to make $\rho {\xi }_z^{\ast }W\to 0$ as well. In a source-free outer region the decaying potential is proportional to $e^{-|k||z|}$. The isothermal displacement energy of a self-gravitating slab then givesSince $I$ is strictly positive for a nonzero displacement, ${\omega }^2$ is real. Positive values describe oscillations; negative values give pure growth and decay. The result is the energy identity for the pressure-gravity displacement operator, a self-adjoint operator in the mass density-weighted inner product. The identity excludes an oscillatory growing eigenmode under these boundary conditions. At $k=0$ the same identity applies when its boundary terms vanish and its integrals remain finite, with the gravitational integral involving only $|{\varphi }_z{|}^2$.

The solenoidal magnetic-field constraint impliesWith $\delta B/B_0=O(ϵ)$ and ${\partial }_z=O(ϵ{\nabla }_{\perp })$, the right-hand side is $O({ϵ}^2{k}_{\perp }{B}_0)$. Thus the leading, order-$ϵ$ perpendicular field is divergence-free. On a simply connected perpendicular patch, or for the nonzero perpendicular Fourier modes of a periodic domain, it has a stream function. Choose its normalization byConsequently the leading reduced electron magnetohydrodynamics representation isThe error denotes an order-${ϵ}^2$ field relative to $B_0$. The parallel perturbation is an independent scalar and is not removed by leading perpendicular solenoidality.

For clarity, the representation is asymptotic rather than an exact decomposition of an arbitrary three-dimensional solenoidal field. If $\delta B_{\parallel }$ depends on $z$, exact solenoidality can be restored by adding ${\nabla }_{\perp }\chi$ to $\delta {\mathbf{B}}_{\perp }$, where ${\nabla }_{\perp }^2\chi =-{\partial }_z\delta B_{\parallel }$. Its amplitude is $O({ϵ}^2{B}_0)$, so it lies beyond the retained field order. A perpendicular harmonic component can be absorbed into the guide field or fixed separately by boundary conditions.

In the Boussinesq approximation, density departures from a constant reference density ${\rho }_0$ are small enough to neglect in inertia and mass continuity, but their gravitational effect is retained. Write the total buoyancy acceleration as $b=-g(\rho -{\rho }_0)/{\rho }_0$, choose a hydrostatic background $\bar{b}(z)$, and set $\sigma =b-\bar{b}$. Thus $\sigma$ is the buoyancy perturbation, positive for a parcel lighter than the background at its present height. The squared buoyancy frequency isStable stratification has $N^2>0$. A small vertical displacement $\xi$ of a parcel conserving its buoyancy gives $\sigma =-N^2\xi$, explaining the restoring acceleration and the name of $N$.

Let $P=p^{\prime }/{\rho }_0$ be pressure with the background hydrostatic pressure removed, $\mathbf{u}=(u,v,w)$, and $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ the material derivative. The ideal, nonrotating Boussinesq equations areThe last equation is equivalent to $Db/Dt=0$. There is neither viscosity nor buoyancy diffusion. The reference density occurs in the pressure normalization, while the small density anomaly occurs only through buoyancy.

For the two-dimensional disturbance, linearization about rest givesA plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$ therefore satisfiesThese internal-wave polarization relations and the dispersion relation determine the individual responses in parts i and ii. A radiating response has no incoming internal gravity wave from $x=+\infty$; an evanescent response is bounded there. The harmonic solutions describe the response after excluding arbitrary initial transients.

For the localized oscillatory boundary, define $\alpha =\omega /\sqrt{N^2-{\omega }^2}$ and $I(z)=ϵ{\int }_0^{\infty }{\hat{u}}_0(m)e^{imz}dm=f(z)+ig(z)$. Both signs of vertical wavenumber must radiate into $x>0$. Their horizontal wavenumber is $k=\alpha |m|$, so the positive-$m$ contribution involves $Z_1=z+\alpha x$ and the negative-$m$ contribution involves $Z_2=z-\alpha x$. Consequently the complex horizontal velocity isTaking its real part provesThe positive-$m$ internal-wave polarization has $w=-\alpha u$ and the negative-$m$ one has $w=+\alpha u$. HenceIn particular, $2f(z)=ϵu_0(z)$, so the boundary condition is recovered exactly. This is two-beam radiation from a localized oscillating boundary.

Take a smooth localized profile, for example $u_0(z)=e^{-z^2/H^2}$. Its effective support is a neighbourhood of width a few $H$; increasing that neighbourhood makes the tails arbitrarily small. The boundary displacement is $ϵu_0(z)\sin \omega t/\omega$ to first order, so at $t=0$ it is flat and moving at maximum speed. At that instant,There are two narrow oblique beams centered on $z=-\alpha x$ and $z=+\alpha x$. Away from their overlap, the velocity is parallel to the relevant beam, in the directions $(1,-\alpha )$ and $(1,+\alpha )$. The figure uses $\alpha =3/4$ and measures velocities in units of $ϵ$.

At other phases the $g$ terms contribute. For the positive-frequency convention above, $g$ is the Hilbert transform of $f$; it generally has longer tails than $f$. Thus even when the boundary is nearly stationary, the interior need not be stationary or confined to the bright cores shown here. Half a period later all velocities reverse. The beam directions remain the same, but the instantaneous velocity profile and its quadrature component change.

A short pulse has a broad frequency spectrum. Frequencies below $N$ produce outgoing internal gravity wave packets with frequency-dependent ray slopes $\pm \omega /\sqrt{N^2-{\omega }^2}$; frequencies above $N$ contribute an evanescent near-boundary response. Moreover the group velocity depends on wavelength as well as direction. The radiated packet therefore spreads into a dispersive fan or wake, with different frequencies traveling at different angles and different scales separating along those directions. After the forcing ends the propagating packet travels away, rather than maintaining the two permanently driven monochromatic beams.

Define the quasi-geostrophic streamfunction by $p^{\prime }/{\rho }_0=f_0\psi$. Leading horizontal geostrophic balance, $f_0\hat{\mathbf{z}}\times {\mathbf{u}}_g=-{\nabla }_{H}P$, gives $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$. The hydrostatic approximation gives $P_z=\sigma$, henceThus $D_g$ is the material derivative along the leading geostrophic flow, and the vertical derivative of pressure determines the buoyancy perturbation. This also gives thermal wind: $u_{gz}=-{\sigma }_y/f_0$ and $v_{gz}={\sigma }_x/f_0$.

The balanced regime is favored by small Rossby number, a shallow aspect ratio, and a Burger number $N^2{H}^2/(f_0^2{L}^2)$ of order one. Equivalently the stratified Froude number $U/(NH)=\mathrm{Ro}/\sqrt{\mathrm{Bu}}$ is small. On a beta plane, require $\beta L/|f_0|=O(\mathrm{Ro})$, and take slow, initially balanced motion away from the equator. Under the traditional approximation, the horizontal Coriolis force generated by vertical velocity is small because vertical velocity is small, while the vertical rotational force is small compared with the leading hydrostatic pressure and gravity terms. Strong stratification and small aspect ratio justify retaining only the vertical component of Earth's rotation in this regime; this argument is not uniform near the equator.

Put $A(z)=f_0^2/N^2(z)$. The buoyancy equation gives $w=-f_0{D}_g{\psi }_z/N^2$. Substitution into the vertical vorticity equation yieldsThere is an important commutator cancellation. With $J_{xy}(a,b)=a_x{b}_y-a_y{b}_x$,Also $D_{g}f=\beta {\psi }_x$ for $f=f_0+\beta y$. Adding terms therefore proves potential-vorticity conservation:This is the three-dimensional quasi-geostrophic potential vorticity.

At an impermeable stationary lower boundary $z=b(x,y)$, the kinematic condition is $w={\mathbf{u}}_H\cdot {\nabla }_{H}b=D_{g}b$. For small elevation and slope, evaluate this at the reference level $z=0$ to leading order. Combining it with the buoyancy equation gives the rigid-boundary buoyancy condition for quasi-geostrophic wavesThe quantity in parentheses is the boundary displacement plus its buoyancy-equivalent displacement.

Now take constant $N,f$ and the mean quasi-geostrophic streamfunction $\bar{\psi }=-(U_0+\Lambda z)y$. Its mean buoyancy perturbation is $-f\Lambda y$, consistent with thermal wind, and its interior potential vorticity is constant. In the boundary condition, the perturbation velocity $v^{\prime }={\psi }_x^{\prime }$ advects the mean boundary buoyancy gradient. Keeping this contribution givesIn particular the $-\Lambda {\psi }_x^{\prime }$ term must not be omitted.

The linear interior potential-vorticity equation is $({\partial }_t+U(z){\partial }_x)Q^{\prime }=0$. Starting with zero interior perturbation potential vorticity, and with forcing only through the boundary, it remains zero. For a steady solution this is also the regular zero-interior-PV choice, continued through any isolated zero of $U$. The inversion equation for a horizontal mode isThe printed convention $K>0$ takes $f>0$; for either hemisphere the decay rate is defined using $|f|$. Boundedness aloft selects the decaying mode. Writewith real part understood. The linear boundary condition becomesFor $K{U}_0+\Lambda \ne 0$, a steady quasi-geostrophic mountain wave is thereforeOn $y=0$, its real meridional velocity isFor the positive-parameter convention $k,ϵ,f,U_0>0$ and $K{U}_0+\Lambda >0$, the velocity minimum lies one quarter wavelength downstream of each topographic maximum; its maximum lies one quarter wavelength downstream of each topographic minimum. If the velocity prefactor changes sign, maxima and minima interchange. The condition $K{U}_0+\Lambda >0$ alone does not fix that prefactor if $U_0$ is allowed to be negative. The upper panel shows the positive-current convention.

At $K{U}_0+\Lambda =0$, the free boundary Eady edge wave has zero frequency in the ground frame. If $kϵU_0\ne 0$, a steady amplitude cannot satisfy the forced amplitude equation. With $A(0)=0$, integration givesConsequentlyThe symbol multiplying $i$ here is the horizontal wavenumber $k$, as in the PDF. Taking the real part and differentiating in $x$ givesFor $U_0=-\Lambda /K>0$, positive $ϵ,f$, and $t>0$, the meridional-velocity minima coincide with topographic maxima, and its maxima with topographic minima. This is the lower panel. The change from quadrature to opposition is the phase of a resonantly forced topographic edge wave.

The stationary relief continuously excites a boundary Rossby wave whose intrinsic propagation exactly cancels mean advection. The forcing therefore remains coherent with that wave; with no damping or radiation loss for this trapped mode, its linear amplitude grows secularly. This is a resonant topographic quasi-geostrophic wave. Unlimited growth is a conclusion of the linear idealization: eventually the small-disturbance approximation fails.

The assertion that resonance must be unsteady needs nonzero forcing. If $k=0$, then $b_x=0$ and the boundary current does not cross the relief; if $U_0=0$, the forcing also vanishes. In either case the zero disturbance is a steady solution even if $K{U}_0+\Lambda =0$. For example, choose $k=0$, $\ell \ne 0$, and $\Lambda =-K{U}_0$; this is a concrete exception to an unconditional resonance claim. When both horizontal wavenumbers vanish, $K=0$ and the stated decaying-wave ansatz itself does not apply.

Distinguish the full velocity potentials ${\Phi }_i=U_{i}x+{\phi }_i$ from their perturbations ${\phi }_i$. The linear boundary conditions use the perturbations. This distinction reconciles the notation in the printed conditions with the full potentials written in the mode ansatz. All equations below neglect products of perturbations. Assume positive densities and no surface tension.

The rigid-rotation velocity field is $\mathbf{v}=(-\Omega y,\Omega x,0)$, with spatially uniform $\Omega (t)$. Its divergence is zero, since ${\partial }_x(-\Omega y)={\partial }_y(\Omega x)=0$. It therefore satisfies the incompressible continuity equation. On the cylindrical wall, $\mathbf{v}\cdot (x,y,0)=0$, so it has zero normal velocity; on either end cap its vertical component is zero. These are the impermeability boundary conditions for an inviscid flow. Tangential slip is permitted; a no-slip condition is not part of this inviscid model.

The necessary formulation uses a real Hilbert space $V$ of functions satisfying the essential boundary conditions, a continuous symmetric bilinear form $a$, and a continuous linear functional $\ell$. Assume constants $C,c>0$ withThe latter is coercivity, the positive-definiteness condition needed in the chosen complete space. On the domain of the strong differential operator, integration by parts and the boundary conditions identify $a(v,w)=⟨\mathcal{L}v,w⟩$; this form extends to $V$. Let $\ell (w)=⟨f,w⟩$ for admissible forcing. DefineWhen $u$ has the regularity to belong to the strong operator domain, this is $\mathcal{L}u=f$ with its boundary conditions. For complex spaces use a Hermitian form and $-2\mathrm{Re}\ell (v)$ to make the functional real.

Here is an existence proof by minimization. Continuity of $\ell$ and coercivity give $I(v)\ge c\Vert v{\Vert }_V^2-2\Vert \ell {\Vert }_{V^{\ast }}\Vert v{\Vert }_V$, so $m={\inf }_{V}I$ is finite and a minimizing sequence $v_n$ is bounded. The quadratic parallelogram identity isSince its left side is at least $m$,Coercivity makes $v_n$ Cauchy in $V$. Completeness gives $v_n\to u\in V$, and continuity of the form and forcing gives $I(u)=m$.

For every $w\in V$, differentiation along $u+tw$ yieldsThus the Euler-Lagrange equation is exactly the stated weak solution equation. Conversely, if $u$ satisfies it, expansion givesThis proves that the weak solution is the unique global minimum and that two weak solutions coincide. It proves the symmetric coercive case underlying the Lax-Milgram theorem directly, rather than taking existence as a theorem name. The framework is symmetry and coercivity in quadratic energy minimization.

The qualifications matter if positive definiteness is read more weakly. For a nonsymmetric operator the derivative of $⟨\mathcal{L}v,v⟩$ uses $(\mathcal{L}+{\mathcal{L}}^{\ast })/2$, so it need not give $\mathcal{L}u=f$. For example, on smooth periodic real functions $\mathcal{L}=1+{\partial }_x$ has $⟨\mathcal{L}v,v⟩=\Vert v{\Vert }_2^2$, but the functional's equation is $u=f$, not $u+u_x=f$. Even a symmetric operator can be strictly positive without being coercive: on l2 sequence space, $(Lv{)}_n=v_n/n$ with $f_n=1/n$ would require a solution $u_n=1$, which is not square summable. The associated functional has values $-{\sum }_{n=1}^{N}1/n$ on vectors with the first $N$ entries equal to one, so it has no minimum in that space. This is strict positivity without coercivity can fail variational solvability. Under the usual symmetric coercive meaning of positive definiteness, the proof above supplies every assertion requested.

Retaining the nonlinear factor $f$ determines the order-$ϵ$ forcing and the matching constants. Write $\delta =\sqrt{ϵ}$. The inner leading equation and its boundary condition at $r=1$ give $f_0=C(1-r^{-1/2})$. Matching to the far-field value fixes $C=1$.

The tail $-r^{-1/2}$ becomes $-\delta x^{-1/2}$ under $x=ϵr$, so a regular expansion in integer powers of $ϵ$ cannot suffice. This is a matched expansion with square-root and logarithmic corrections. DefineThe exponential integral here is normalized by the displayed definite integral. Integration by parts also gives $E(x)=2e^{-x}/\sqrt{x}-2\sqrt{\pi }\mathrm{erfc}\sqrt{x}$, using the complementary error function.

For the outer asymptotic expansion, put $F(x)=f(x/ϵ)$ andThe transformed equation gives $L{F}_1=0$ and $L{F}_2=-F_1{F}_1^{\prime }$, with both corrections vanishing at infinity. Matching the leading inner tail determines $F_1=-E/2$. Multiplying the equation for $F_2$ by its integrating factor yieldsFix the particular solution $G$ unambiguously byThen $(x^{3/2}{e}^x{G}^{\prime }{)}^{\prime }=E$, $G(\infty )=0$, and $G$ has no $x^{-1/2}$ term at zero. The general outer correction is $F_2=\beta E+G/4$, whereThe logarithm in this asymptotic expansion explains why the inner expansion needs an $ϵ\log ϵ$ term.

For the inner asymptotic expansion, all corrections larger than order $ϵ$ solve the homogeneous leading equation and vanish at $r=1$. At order $ϵ$, the forcing isIntegration gives the particular solution $1-\sqrt{r}+\log r$, which vanishes at $r=1$. Thus writeExpress this inner expansion in $x=ϵr$ and compare the terms that survive in the overlap $ϵ\ll x\ll 1$. The order-$\delta$ constant gives $A=\sqrt{\pi }$. At order $ϵ$, the term $C\log ϵ+\log r$ becomes $\log x+(C-1)\log ϵ$, so $C=1$. The singular term $-ϵA/\sqrt{x}$ matches $2ϵ\beta /\sqrt{x}$, giving $\beta =-\sqrt{\pi }/2$. Finally, the regular constants give $B+1=\pi +\mu /4$.

The requested expansions, including every correction through order $ϵ$, are thereforeThe inner formula satisfies the boundary condition at $r=1$ term by term; the outer formula satisfies the condition at infinity. Each formula has its own stated region of validity, and the matched asymptotic expansion relates them in their overlap.

The constant can also be made explicit. Integration by parts in the definition of $G$ gives $G=-EI-{\int }_x^{\infty }E(t{)}^{2}dt$, and $E(x)I(x)\to 8$. Squaring the complementary error function expression for $E$ separates the integral into three elementary pieces:The first piece is $-4\log x-4\gamma -4\log 2+o(1)$, where $\gamma$ is the Euler--Mascheroni constant. Substitution $u=\sqrt{t}$ makes the second integral tend to $\sqrt{\pi }/2$, because $(\mathrm{erfc}u{)}^{\prime }=-2e^{-u^2}/\sqrt{\pi }$. Integration by parts in the third integral gives $1/2-1/\pi$. ConsequentlyEquivalently, defining $G$ by its convergent integral already determines the same matching constant without needing this closed form.