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Again put $T=ϵt$. To leading order the oscillator has form $x_0=k-r(T)\cos (t+\varphi (T))$, while the second variable has a slow part $Y(T)$. Averaging its equation over the fast period givesThe oscillatory forcing of the second variable produces only an order-$ϵ$ correction: its leading fast derivative is $-r(T)\cos t$, whose bounded primitive is $-r(T)\sin t$. Thus it creates no secular term.

The oscillator solvability condition is the same as in part (i), with $c$ replaced by $Y(T)$:Introduce accumulated slow timeThe shifted Van der Pol amplitude evolution therefore gives, for $a=k^2-1$,ConsequentlyAs before, uniformity for $t=O({ϵ}^{-1})$ means a fixed bounded slow-time interval on which the envelope stays bounded. For $k^2>4/5$ it fails when $S(T)$ reaches $\log (5-4/k^2)/(k^2-1)$, with limiting value $4$ at $k=1$. There is no bounded leading approximation covering all such slow times for arbitrary positive $k$.

Put $T=ϵt$ and use the method of multiple scales with $x=x_0(t,T)+ϵx_1+\cdots$. At leading order, $x_{0,tt}+x_0=k$, so write $x_0=k-r(T)\cos (t+\varphi (T))$ with $r(0)=k$, $\varphi (0)=0$. The order-$ϵ$ equation isUsing ${\cos }^{2}t\sin t=(\sin t+\sin 3t)/4$, its fundamental sine coefficient is $-2r_T-cr(1-k^2-r^2/4)$, and its fundamental cosine coefficient forces ${\varphi }_T=0$. Removing these resonant forcings is the solvability condition in the method of multiple scales. Thus the shifted Van der Pol amplitude evolution isWith $a=k^2-1$ and $R=r^2$, solve $R_T=caR+c{R}^2/4$ by putting $W=1/R$. This givesThe uniformly valid leading approximation is $x(t)=k-r(ϵt)\cos t+O(ϵ)$ on bounded slow-time intervals for which $r$ remains bounded. The exact initial derivative is recovered by the first-order correction; the displayed leading expression already satisfies the leading initial data.

There is an important limitation to the source's unqualified time-range wording. Starting from $r(0)=k$, the amplitude decays for $k^2<4/5$, is constant at leading order for $k^2=4/5$, and grows for $k^2>4/5$. In the latter case the denominator vanishes atA bounded-amplitude multiple-scale expansion is uniform for $0\le T\le C<T_b$, not through its predicted large-amplitude breakdown. For the decaying case every fixed bounded slow-time interval is available; the leading unstable equilibrium at $k^2=4/5$ also has bounded amplitude over such intervals. The envelope singularity is a breakdown warning, not a proof that the exact original solution has precisely this blow-up time. The coefficient in the PDF is $ϵc$, correcting the converted TeX's duplicated $ϵ$.

Use the Burgers reduction of weakly nonlinear rightgoing acoustics, with $\theta =t-x$, $X=Mx$, ${\partial }_t={\partial }_{\theta }$ and ${\partial }_x=-{\partial }_{\theta }+M{\partial }_X$. The equation of state expands asAt order one, mass conservation and momentum conservation give $({\rho }_0-u_0{)}_{\theta }=0$ and $(u_0-p_0{)}_{\theta }=0$. For the pure rightgoing acoustic wave with no independent leading mean density or velocity shift, ${\rho }_0=u_0=p_0$. This is the background choice implicit in the requested traveling-wave expansion; additional mean components would modify its phase convention. The source's duplicated phrase about expanding $p$ should refer to expanding $\rho$ as well as $u$.

At order $M$, the mass conservation equation becomeswhereas the momentum conservation equation becomesIn the latter equation, the two nonlinear products ${\rho }_0{u}_{0,\theta }$ and $-u_0{u}_{0,\theta }$ cancel. Substitute the equation of state relation into the former equation and add the two equations to eliminate the first corrections. The result isThe same calculation gives $u_1-p_1=-(\gamma +1)p_0^2/4+C(X)$, so the first corrections may be chosen without an explicit secular factor of $\theta$. This is why the retarded-time description can remain uniform for $|\theta |=O(M^{-1})$, provided the leading profile and its relevant derivatives remain bounded over that range and the slow propagation interval.

This Inviscid Burgers equation can develop a shock, so the source's uniformity assertion needs a pre-shock qualification. If $p_0(\theta ,0)=F(\theta )$ and $a=(\gamma +1)/2$, its method of characteristics givesThe smooth construction ends when this denominator vanishes; if $\max F^{\prime }>0$, the first such distance is $X_s=[a\max F^{\prime }{]}^{-1}$.

For viscosity write $\beta =M\nu$ with bounded positive $\nu$. Since $u_{0,xx}=p_{0,\theta \theta }+O(M)$, the order-$M$ momentum conservation equation instead reads $(u_1-p_1{)}_{\theta }+p_{0,X}=\nu p_{0,\theta \theta }$. Addition now yields the viscous Burgers equationIts positive diffusion resolves the inviscid steepening rather than supporting an unqualified smooth inviscid expansion through shock formation.

Assume again positive diffusion $\gamma >0$. Remove the drift with $A(x,t)=e^{Ux/(2\gamma )-i\omega t}f(x)$. Differentiating explicitly cancels the first derivative and givesBalance the second derivative with the linear confinement by taking $x=O({ϵ}^{-1/3})$, so the requested exponent is $\sigma =1/3$. More precisely, putThe eigenvalue problem reduces to $f_{\xi \xi }=(\xi -c)f$. Its two solutions are Airy functions. The growing $\mathrm{Bi}(\xi -c)$ is excluded by the condition at infinity: its growth proportional to $e^{2{\xi }^{3/2}/3}$ dominates any linear-in-$x$ drift factor. The decaying $\mathrm{Ai}(\xi -c)$ remains admissible even after multiplication by $e^{Ux/(2\gamma )}$. The wall imposes $\mathrm{Ai}(-c)=0$, so $-c=z_n$, giving the Airy global modes of a linearly confined Ginzburg-Landau equation:The least negative zero, $z_1\simeq -2.338107$, gives the largest temporal growth rate. HenceAt equality the leading global eigenmode is neutral, and below it all these eigenmodes decay. In contrast, freezing the coefficients locally gives absolute hydrodynamic instability wherever ${\mu }_0-ϵ\lambda x>U^2/(4\gamma )$. Such a pocket exists iff ${\mu }_0>U^2/(4\gamma )$ and extends to $x_a=[{\mu }_0-U^2/(4\gamma )]/(ϵ\lambda )$. Thus global instability implies a locally absolutely unstable pocket, while the converse fails: a pocket shorter than $|z_1|(\gamma /(ϵ\lambda ){)}^{1/3}$ does not overcome the wall and confinement losses. The additional threshold is order ${ϵ}^{2/3}$, precisely the loss produced by the Airy function spatial scale.

Use local coordinates with the interface $z=0$, solid below and fluid above, and choose the P-SV displacement potentials convention $u_x={\varphi }_x-{\psi }_z$, $u_z={\varphi }_z+{\psi }_x$. Let the reflected P and SV potential wave amplitudes be $A,B$, while the transmitted P wave amplitude is $T_P$. All four waves have the same tangential slownessThe incident and transmitted retarded times have negative $q_{P}z,q_{f}z$, while elastic waves have positive $q_{P}z,q_{S}z$. At the interface all waveform derivatives therefore have the same argument. displacement field continuity givesWith Lamé parameters $\mu =\rho {\beta }^2$ and $\lambda =\rho ({\alpha }^2-2{\beta }^2)$, the shear and normal stresses follow directly from differentiation:The fluid has no shear traction, so the coefficient of $f^{\prime \prime }$ in ${\sigma }_{xz}$ must vanish:The coefficient of $f^{\prime \prime }$ in the solid stress is $\rho [\cos 2{\theta }_S(1+A)-\sin 2{\theta }_{S}B]$; in the fluid it is ${\rho }_f{T}_P$. ThusThese are the three required matching conditions; tangential displacement field need not be continuous at a solid–inviscid-fluid boundary.

Put $C=\cos 2{\theta }_S$ and $X=1-A$. From (2), $B=({\beta }^2/{\alpha }^2)(\sin 2{\theta }_P/C)X$. In (1), the coefficient of $X$ simplifies because $C+2({\beta }^2/{\alpha }^2){\sin }^2{\theta }_P=1$. Hence $X=(\alpha C\cos {\theta }_f)/({\alpha }_f\cos {\theta }_P)T_P$. Substitute this into (3) to obtainTherefore the solid-fluid P-wave transmission coefficient isDivision by $C$ was only an elimination device; the final expression extends continuously to $C=0$. The normal-incidence value is $2\rho {\alpha }_f/(\rho \alpha +{\rho }_f{\alpha }_f)$. This is a potential coefficient; longitudinal displacement field transmission additionally multiplies it by $\alpha /{\alpha }_f$. The original PDF's bracket lies in the denominator and contains ${\beta }^2/{\alpha }^2$ and $\cos {\theta }_f$. The converted TeX misplaces the bracket, substitutes ${\theta }^2$ for ${\beta }^2$, and loses the final cosine.

The leading source displacement field isThe second term is one derivative smoother at the leading arrival. At each surface point, use its tangent plane, the angles from (i)–(ii), and the local solid-fluid P-wave transmission coefficient. The incident potential coefficient is $1/r(R)$, so the transmitted leading potential coefficient isTransport this coefficient along the vertical fluid rays. Their transverse area does not change with height, so ray-tube conservation for a P-wave jump gives no further geometrical spreading. The leading displacement field coefficient is consequentlyThis is the transmitted jump at a collimating solid-fluid interface. Radial derivatives of $A_{\varphi }$ multiply $f$, and therefore do not contribute to the first derivative jump carried by $f^{\prime }$. A varying transverse wave amplitude does generate smoother later terms, so this construction is a leading-front result, not a complete global plane-wave field.

Since the source is zero before $t=0$ and is $C^2$ there, $f(0)=f^{\prime }(0)=f^{\prime \prime }(0)=0$. Thus displacement field itself does not jump. If $J=[f^{(3)}(0)]$ exists and is the first nonzero jump, the leading discontinuity is the acceleration jump $[{\partial }_t^2\mathbf{u}]=-A_u(R)J{\mathbf{e}}_z$. More generally, a first jump in $f^{(m)}$ gives $[{\partial }_t^{m-1}\mathbf{u}]=-A_u(R)[f^{(m)}]{\mathbf{e}}_z$. The supplied regularity alone does not guarantee a nonzero discontinuity: a source smooth to all orders may have none. The geometrical transfer factor $A_u$ is nevertheless determined.

At $R=0$, $r=z_0$ and all three angles vanish. HenceFor large $R$, put $C_{\infty }=1-2{\beta }^2/{\alpha }^2$ and retain $s=z_0(\alpha -{\alpha }_f)$, $d=\sqrt{{\alpha }^2-{\alpha }_f^2}$. The relevant limits areWhen $C_{\infty }\ne 0$, the square-bracket factor in the denominator of $T_P$ is $C_{\infty }^2+O(R^{-1})$, soThese are signed waveform coefficients; their absolute values give wave amplitude magnitudes. The leading displacement field decays as $R^{-2}$ rather than $R^{-1}$ because grazing incidence adds a small transmission factor to spherical spreading. The leading coefficient is independent of fluid and solid mass densities, although the exact wave amplitude and its higher-order terms depend on them.

The assumptions do not exclude ${\beta }^2={\alpha }^2/2$, where $C_{\infty }=0$ and the preceding division is invalid. In this special case $\cos 2{\theta }_S={\cos }^2{\theta }_P$ exactly, and the square-bracket factor is $\cos {\theta }_P+O({\cos }^3{\theta }_P)$. Keeping both denominator terms givesIt has no $R^{-2}$ term. This exceptional case matters when requesting an asymptotic formula without an additional restriction on the elastic moduli.

Use the quadratic spectral flux $S_z=-\frac{1}{2}\mathrm{Re}(V^{\ast }T)$. For positive real material mass density and shear modulus, differentiation of the velocity–traction system givesBecause $k$ is real, $\mu p_{\beta }^2=\rho -\mu k^2/{\omega }^2$. If $v=\mathrm{Im}\omega >0$, then $\mathrm{Re}(i{\omega }^{\ast })=v$ and $\mathrm{Re}(-i\omega \mu p_{\beta }^2)=v(\rho +\mu k^2/|\omega {|}^2)$. This proves complex-frequency SH flux monotonicity:Strictness holds for a nontrivial wavefield: if $V,T$ vanished together at one point, uniqueness of the first-order system would make the entire solution zero. The zero solution is the necessary exception to the literal strict inequality. Across a bonded material boundary $V,T$, and hence $S_z$, are continuous; integrating the negative derivative gives strict decrease across any positive-depth interval as well.

Where the SH input impedance is finite, $S_z=\frac{1}{2}\mathrm{Re}Z|V{|}^2$. If $\mathrm{Re}Z(z_1)>0$, then $S_z(z_1)>0$ and all shallower fluxes are larger and positive. They cannot have $V=0$, since that would make $S_z=0$. Thus $\mathrm{Re}Z(z)>0$ for every $z<z_1$. Likewise, negative real SH input impedance at $z_0$ gives negative flux throughout $z>z_0$, excludes velocity zeros there and implies $\mathrm{Re}Z(z)<0$ for every $z>z_0$.

In a uniform region let $q=\mu p_{\beta }$ on the causal vertical wavenumber branch. The constant Riccati solutions are $Z=\pm q$, and integrating the one-way equation givesSince $\mathrm{Re}q>0$, the plus solution carries positive downward flux and the minus solution negative upward flux. Writing $k_{\beta }=a+ib$ with $b>0$, their moduli are proportional to $e^{-bz}$ and $e^{bz}$ respectively: each decays in its direction of energy flow.

For complex frequency the displayed quadratic form is the spectral flux diagnostic specified in the question; a genuinely periodic time average is recovered on the real-frequency axis. The decrease here comes from the upper-half-plane convergence parameter, not from material dissipation. At real frequency in a lossless layer the corresponding flux is constant.

First let $n\ge 0$. The ordinary Leibniz rule, applied to a test function $h$, givesEquivalently, the composition of differential operators isFor completeness, this formula follows by induction: applying $\partial$ differentiates either the coefficient or the remaining derivative of $h$, and the two contributions combine by $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$. The initial case is multiplication by $f$.

For all integers $n$, define $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n(n-1)\cdots (n-r+1)/r!$. The formal pseudodifferential composition rule isFor $n=-s<0$, $\left(\genfrac{}{}{0ex}{}{-s}{r}\right)=(-1{)}^r(\genfrac{}{}{0ex}{}{s+r-1}{r})$, so this is usually an infinite series. In particularApplying $\partial$ on the left gives $m_f$: all later coefficient derivatives cancel in pairs. This uniquely determines the normal-ordered inverse expansion. Repeating the inverse construction, or using the same binomial recurrence downward in $n$, gives every negative power. More explicitly, applying $\partial$ to the proposed $n$-series yields coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)+\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ at $f^{(r)}{\partial }^{n+1-r}$, so it is consistent with the already determined $(n+1)$-series. Negative powers here are formal; differentiation on all of $C^{\infty }(\mathbb{R})$ has a nontrivial kernel and has no genuine two-sided inverse there.

Write $P={\sum }_{j\le \alpha }{p}_j(x){\partial }^j$, setting unused coefficients to zero. Split the Adler trace into its fixed leading part and variable parts:Cyclicity from part (i) givesThus this Hamiltonian trace functional on monic differential operators has the required formTo make each coefficient explicit, expand ${\partial }^l{p}_j$ by the Leibniz rule. A term with $k$ derivatives of $p_j$ contributes to the residue exactly when $l-k+j=-1$. Consequently,These are smooth functions depending on $P$, not on the variable coefficients of $L$. In particular $a_0=p_{-1}$ and $a_1=p_{-2}+p_{-1}^{\prime }$. The coefficient $a_0$ must be included: it appears in the displayed sum even though the printed coefficient list starts at $a_1$.

The monic operators form an affine space, not a vector space. Accordingly $l_P$ is affine in the coordinates $u_l$, with constant $c$, and its differential is linear in variations. Its variational derivatives are $\delta l_P/\delta u_l=a_l$. It is the restriction of a linear trace pairing on the full operator space, which explains the terminology without discarding the constant term.

Put every formal pseudodifferential operator in normal order, with all coefficient functions on the left. Its formal pseudodifferential residue is the coefficient of ${\partial }^{-1}$. The Adler trace isHere and below, use periodic coefficients and integrate over a period, or impose convergence and vanishing boundary terms on the real line. Equivalently, the algebraic integration functional must annihilate total derivatives. This convention is needed for the requested trace property; arbitrary elements of $C^{\infty }(\mathbb{R})$ do not supply it automatically.

It suffices to compute with monomials $A=a{\partial }^p$, $B=b{\partial }^q$. Write $r=p+q+1$. If $r<0$, no term in either product has power ${\partial }^{-1}$, so both residues are zero. If $r\ge 0$, the formal pseudodifferential composition rule givesSince $q=r-1-p$, reversal of the factors in the numerator provesAfter $r$ integrations by parts, $\int a{b}^{(r)}dx=(-1{)}^r\int b{a}^{(r)}dx$. Hence the two product traces agree. More explicitly, for $r\ge 1$ their residue difference is the total derivativeFor $r=0$ the residue difference is $ab-ba=0$. In products of operators bounded above in order, only finitely many monomial pairs can contribute to a residue: the required $r$ is nonnegative, so the two downward coefficient indices have a bounded sum. Summing the monomial result therefore proves

The boundary convention is substantive. Without it, take $A={\partial }^2$, $B=(\tanh x){\partial }^{-2}$. Both individual residues vanish, but $\mathrm{res}[A,B]=2(\tanh x{)}^{\prime }=2{\mathrm{sech}}^{2}x$ and its integral over $\mathbb{R}$ is $4$, not zero. The primitive has unequal limits at the two ends. Also a generic smooth residue need not have a convergent integral at all. Thus the proof supplies the intended cyclic trace under its necessary analytic or formal integration convention.

Use the extended-value convention for a normal family, which is essential here: every sequence in the holomorphic family has a subsequence converging uniformly on compact subsets either to a finite holomorphic function or to infinity. Convergence to infinity means ${\inf }_K|f_n|\to \infty$ for every compact $K\subset D$. Equivalently one uses local uniform convergence in the chordal metric. For holomorphic functions a nonconstant meromorphic limit cannot acquire a pole: near a supposed pole, the reciprocals of sufficiently large-index functions are holomorphic and nowhere zero, and converge to a holomorphic function with a zero. Hurwitz's theorem would force that reciprocal limit to vanish identically, giving the constant infinite limit instead.

Restriction proves that global normality implies normality near each point. For the converse, choose a countable cover by connected discs $U_j$ on which the family is normal, with smaller discs covering $D$ and having compact closure inside the corresponding $U_j$. Such a cover exists from the countable rational-disc basis and local normality. Starting with any sequence, extract a subsequence converging on $U_1$, then a further subsequence converging on $U_2$, and so on. The diagonal sequence converges on each $U_j$. On an overlap the limits agree; a finite holomorphic limit and infinity cannot both occur there. The overlap graph of this cover is connected, since otherwise its components would partition the connected domain into disjoint open subsets. Therefore all local limits are of the same type and patch to one holomorphic function or to infinity throughout $D$. Every compact set has a finite cover by the smaller discs, so convergence on their compact closures gives uniform convergence on that compact set. This proves that normality is a local property.

For the counterexample take the unit disc and the family $f_n(z)=n(1+z^2/2)$. Since $|1+z^2/2|\ge 1-|z{|}^2/2>1/2$, $f_n$ tends uniformly to infinity. Any sequence from this family either has a constant subsequence or has indices tending to infinity, so the family is normal. Its derivatives are $nz$. No subsequence of these derivatives can have a finite locally uniform limit, because their values at any fixed nonzero point grow without bound; nor can they tend locally uniformly to infinity, since every derivative vanishes at zero. Thus a holomorphic normal family need not have a normal derivative family.

Now impose the point bound at $z_0$. For any sequence of derivatives choose corresponding functions from the original family. Normality supplies a locally uniformly convergent subsequence, and the point bound excludes the infinite limit. Let its finite holomorphic limit be $f$. The Cauchy integral formula gives convergence of derivatives uniformly on compact subsets. Explicitly, for compact $K\subset D$ choose $\rho >0$ so its closed $\rho$-neighborhood lies in a compact $L\subset D$; thenEvery sequence of derivatives therefore has a locally uniform subsequential limit. This pointwise anchoring of a holomorphic normal family proves the derivative family is normal under the stated bound.