Stable self-similar blowup for a supercritical defocusing Schrödinger equation on the torus

Proof. The bracket in the integrand of (2.5) is O(u) at zero, locally uniformly in the parameters; at infinity it and its parameter derivatives grow at most polynomially, with additional powers of \log u. Thus the regularized integral is locally holomorphic for \operatorname{Re}q>-1. The reciprocal gamma function vanishes at zero, which proves the last assertion in (2.6).

For \operatorname{Re}q>0 the expression in braces can be combined into the usual integral. On x>0, substitution u=xv gives

Differentiation and integration by parts prove both the equation and the derivative identity; holomorphic continuation proves them on the stated domain. The branches in the regularized integral are regular on a neighborhood of each nonzero imaginary x, so the boundary values used in the matching problem are genuine holomorphic values.

To obtain (2.7), split the integral at u=|x|/2. Taylor’s formula applies uniformly to (1+u/x)^{m-1-q} on the first piece. Its jth coefficient integrates to (-1)^j(q)_j(1+q-m)_j/j!. On the second piece the exponential weight bounds the integral by any prescribed inverse power of |x|; the same argument applies after symbol differentiation. This proves the full assertion about remainders. For completeness, the other local solution at infinity obtained by reduction of order has the form e^xx^{q-m} times an inverse-power expansion. On either imaginary ray its repeated symbol derivatives acquire arbitrarily high powers of |x|. It therefore cannot be added to (2.7) while preserving all its symbol estimates. This also explains the slow normalization on the rays.□

Proof. Integration by parts gives orthonormality and [t\partial_t^2+(1-t)\partial_t]L_n=-nL_n. Polynomials are complete: a function orthogonal to all powers has a Laplace transform whose derivatives vanish at an interior point of its domain of holomorphy, hence its Laplace transform, and then the function, vanishes. The function G and all its derivatives are smooth at zero and of polynomial growth at infinity by Lemma 2.2. Repeated integration by parts with this Laguerre operator shows that g_n, and the Laguerre coefficients of each derivative of G, decrease faster than every inverse power of n. The boundary terms vanish because the coefficient of the derivative at zero is t, and the weight at infinity is exponential. These arguments are uniform on compact parameter sets, also after holomorphic differentiation.

The generating function gives, initially for 0\le x<1,

Taking its value and first derivative as x\uparrow1 proves (2.8). Since L_k'=-\sum_{n<k}L_n, the coefficients of G' are -B_{n+1}, whereas those of G''-G' are C_n. Rewriting the differential equation as

proves the first recurrence. The other two follow from g_n=B_n-B_{n+1} and the definition of C_n.

Finally insert B_{n+1}=B_n-g_n and C_{n+1}=C_n-B_{n+1} into (2.10). Since s is purely imaginary, the cross term \operatorname{Re}(s|B_{n+1}|^2) vanishes. Use t_ng_n=-MB_{n+1}-sC_n to simplify the remaining terms. The coefficient of |g_n|^2 becomes \operatorname{Re}t_n-M/2=\operatorname{Re}\lambda+n-5/2, which is (2.11).□

Proof. The only zero of (q)_K in the specified half-plane is q=0, and it is simple. It can occur only for \ell=0. At this point G=1, so every g_n with n\ge1 vanishes. Holomorphic division therefore removes the apparent singularity for n\ge K. Local Cauchy estimates preserve the rapid decay established above, justifying division of the sums and continuation of their identities.

The normalized tail cannot vanish identically in its sequence index. Indeed repeated use of (2.6) gives

The right side is nonzero by its leading asymptotic coefficient, also at q=0. If all normalized coefficients with n\ge K vanished, the function [G-\sum_{n<K}g_nL_n]/(q)_K, with its removable value at zero, would vanish. Its Kth derivative would be zero, contradicting (2.14).

Sum (2.11) from K to infinity and divide by |(q)_K|^2, taking the removable value when necessary. Rapid decay makes the limiting form zero. Each weight in the resulting sum is positive because K\ge3 and \operatorname{Re}\lambda\ge-1/32. The preceding nontriviality gives strictness in (2.12). If \widehat C_K were zero the left side would be nonnegative. Dividing by its nonzero squared modulus and completing the square proves (2.13).□

Proof. The recurrences imply the division-free identity F_n(B_{n+1},C_{n+1})^t=t_n(B_n,C_n)^t. Multiplying gives P_K(B_K,C_K)^t=(q)_K(B_0,C_0)^t. Divide by (q)_K and use Lemma 2.4; analytic continuation gives the identity at the removable point. A simultaneous zero of H and H' at x=-s\ne0 would, by uniqueness for its second order equation, make H identically zero, contrary to (2.7).□

Proof. Suppose the determinant vanished. Since the two columns are nonzero, multiply their functions by nonzero constants to arrange H_+(-iZ)=H_-(iZ) and H_+'(-iZ)=H_-'(iZ). Here H_h(x) denotes the resulting multiple of H(\lambda+\ell/2-hi b,M+1,x) and M=\ell+5\ge9. The substitution U_h=e^{-x/2}x^{M/2}H_h gives

Integrate this equation along the conjugate paths

Write \eta_0=\vartheta_0/2 and

In particular L>0 for \eta>0 and

The real part of x tends to +\infty linearly, so the exponential factor in U_h justifies every boundary limit and integration below.

For a bounded real function \alpha, smooth on each side of \eta_0, put B_c=\alpha L+\vartheta'N/2. Pair (2.17) with \overline U_h\,d\eta and take real parts. The kinetic expression after integration by parts is

Completing the square writes this as L|U_h'+(\alpha-hi\vartheta'/2)U_h|^2 -B_c(|U_h|^2)'-L(\alpha^2+\vartheta'^2/4)|U_h|^2. The real lower endpoint terms before this completion sum to \sum_h(M/2)|U_h(0)|^2. To see the cancellation, expand x\overline U_h\partial_xU_h using U_h=e^{-x/2}x^{M/2}H_h: the terms containing x\overline H_hH_h' cancel between the two matched jets, and the real contribution from the gauge is M/2. Since B_c(0)=-Z, integration of the displayed derivative, separately on the two path pieces, now yields

where, with \sigma=\operatorname{Re}\lambda,

There is no missing corner term: it is precisely the jump displayed in ().

On the arc take \alpha=.3. Here u=\sin\vartheta/2, v=Z+(1-\cos\vartheta)/2, and

It follows directly from (2.19), dropping its nonnegative u terms and sine term, that V_c>-3-1/32+.3(10.7)+(3-.09)L>0. At the join 1.13<L(\eta_0)<1.19. On the straight piece, with t=\eta-\eta_0, choose

The jump is -1.5-.3L(\eta_0)-N(\eta_0)>0. The geometry of the ray gives

Set d_0=(.100+.981t)^2+(2.72+.20t)^2 and q_0=1+1.5t. Since M^2\ge81, V_cL_0d_0q_0^2 is bounded below by

The first strict comparison is coefficientwise exact arithmetic; the last two quadratic discriminants are -17519/25 and -174119/10000. Appendix A.4 provides the full polynomial recipe and verifier. Thus all terms of () are nonnegative, the bulk potential is strictly positive, and M/2>Z. It follows that U_h=0, a contradiction.□

Proof. Choose the edges so that t_n,d_n\ne0 for n<8 on them. If the columns became dependent at such a point, their nonzero common initial jet (B_0,C_0) would have two forward images satisfying p_{8,h}\le0. Indeed the backward matrices are invertible there and their inverses are exactly the forward recurrences, up to nonzero scalar factors. For the same initial jet in both channels,

For \sigma>5/2 all increment weights are positive. If B_0\ne0 the first term is positive; if B_0=0 and C_0\ne0, then g_{0,h}=-s_hC_0/d_{0,h}\ne0. This excludes the right edge, which can be taken beyond every possible zero of its divisors.

For the horizontal edges let v=\operatorname{Im}\lambda, C_0=vc, and normalize |B_0|^2+|c|^2=1. Uniformly for \sigma in a bounded interval and n<8,

Consequently (2.21) equals

for large |v|, since 8\sigma+8\ge31/4. This proves the two horizontal exclusions uniformly.

For the left edge set s=iZ and form the forward matrix without its divisors:

Thus the actual forward jet is T(B_0,C_0)^t/\prod_{n<K}d_n. Define

The second summand is the form for h=-1. Each summand evaluates to 2|\prod d_{n,h}|^2p_{K,h} on the common initial pair. We shall prove \mathsf S(v)>0 for every real v, excluding the two simultaneous nonpositive values.

First, its determinant is a real even polynomial of degree at most 4K-2=30 for every real (b,Z) in the parameter box. Hermitian symmetry gives real diagonal entries, and the reflection in (2.23) makes the diagonal entries even and the off-diagonal entry satisfy C(-v)=\overline{C(v)}. To check the degree bound structurally, hold the initial jet fixed and let |v|\to\infty. The forward recurrences give g_{n,h}=O((|B_0|+|C_0|)/|v|) for n<K. Also |\prod d_{n,h}|^2=|v|^{2K}(1+O(1/|v|)). Using (2.11), each form divided by |v|^{2K} has its initial cone form as leading term and an O(v^{-2}) increment. The opposite initial cross terms cancel in their sum. Thus the first diagonal, off-diagonal, and second diagonal entries of \mathsf S/|v|^{2K} have respective orders O(1), O(v^{-1}), O(v^{-2}). This proves the degree bound, independently of any cancellation at a central parameter value. Its first diagonal is positive for sufficiently large |v|.

The exact coefficient calculation in Appendix A.2 proves \det\mathsf S(v)>0 for all real v and all parameters in \mathcal D. A continuous Hermitian matrix with strictly positive determinant cannot change its signature. Its positive first diagonal at large |v| therefore implies \mathsf S(v)>0 on the whole line. This completes the left-edge exclusion and the proof.□

Proof. Lemma 2.7 and the argument principle preserve total zero multiplicity throughout \rho:1\to0. At \rho=1, Lemma 2.5 identifies this count with the true matching determinant by nonzero holomorphic scalar factors. Individual homotoped columns may vanish at interior singular factors of P_K; these are zeros of a holomorphic determinant and do not affect the boundary argument. No assertion of interior nonvanishing is needed.

We may also move (b,Z) to (B/H_0,S/H_0), where H_0=10^8, B=33477607, S=270506819. This point lies in \mathcal D, and the preceding exclusions hold on the connecting segment. At zero tail ratio, put \lambda=-1/32+X and let (\widetilde x,\widetilde y)^t=P_K(0,1)^t in channel h=1. The real polynomial

is a nonzero constant multiple of the determinant; the bar here conjugates coefficients. Induction in the backward recurrence shows that its degree is 15=2K-1, with unscaled leading coefficient -KZ: \widetilde y has leading term X^K and \widetilde x has leading term -KiZ X^{K-1}. Write

Both real polynomials have degree at most seven.

The rational signs in Appendix A.3 give the following root counts; all roots in the table are simple:

Here is why signs alone suffice. Successive positive separators have alternating one-coordinate sign changes, following the quadrant cycle ++,-+,--,+-. They therefore give the listed distinct positive roots, interlaced between the two polynomials. For \ell=0,1 the additional values at -1/4,-4 supply the listed negative roots. These distinct roots exhaust each degree bound of seven. Hence there are no further roots, none repeated, and no unrecorded axis crossings.

The curve P_\ell(iv) for v>0 crosses each successive axis in the positive rotational direction and finishes in the first quadrant. Let n=5,6,7 be the positive root count of \mathcal R_\ell. Starting with the real value at zero and using the final odd-degree asymptotic, the continuous argument change on [0,\infty) is (n+1/2)\pi. This also follows by counting the consecutive quadrants from the initial signs -+, ++, -- in the three rows. Conjugation doubles that change on the full oriented imaginary axis. A single factor iv-z contributes +\pi if \operatorname{Re}z<0 and -\pi if \operatorname{Re}z>0. Thus

and N_{\mathrm{right}}=(15-(2n+1))/2=2,1,0. The counting boundary has no zeros, so these are also the counts in the closed half-plane. The count transfers back by the homotopies.□

Proof. Use h=1, \ell=\lambda=0, K=34 in (2.15), dividing its nth factor by n+1, and call the resulting product L(b,Z). Set

All matrix indices here are zero-based. The exact rational enclosures of Appendix A.5 give

The same certificate, propagating these estimates by rational inequalities, gives

Since the tail ratio lies in the disk with center -iZ/5 and radius Z/5, the first component L_{00}R_++L_{01} cannot vanish. Lemma 2.5 proves the desired nonvanishing of H before we take its logarithmic derivative.

The second-to-first component ratio is j(b,Z). Subtracting its value at zero tail gives the exact identity

Using |R_+|\le2Z/5 and (2.27) bounds its modulus by

Define

Appendix A.5 gives the exact propagation of the displayed bounds to (2.25), the minimum stretch bound, and |\mathcal L(\eta,\zeta)|\le.283\delta on the disk. Its determinant is negative already from the two enclosing rectangles. The boundary size of the linear map exceeds .048\delta=4.8\cdot10^{-10}; subtracting the error proves the stated boundary margin. Straight homotopy to \mathcal L avoids zero on the boundary, so its degree is -1. Finally .283\delta+2.5\cdot10^{-11}<3\cdot10^{-9} proves the disk bound.□

Proof. Use the forward matrix (2.22) at \lambda=\ell=0, s=iz, K=5, and let

The exact coefficient checks in Appendix A.6 prove

Every b in the disk is positive, so all forward divisors are nonzero. The exact jet has nonpositive terminal cone form, hence J^*\mathsf HJ\le0. If its first component were zero, positivity of d_0 would force its second component to be zero, contradicting Lemma 2.5. Thus j is well defined. Moreover

Taking the largest possible real part and smallest possible imaginary part in this disk, and using (), proves the assertions. All Z in (2.2) exceed 2.704. Finally Q'/Q=(ir/2)j gives (2.28).□

Proof. On these intervals the coefficient 11/r+ir/2 has modulus less than 6, and .334<b<.335. Put u=r-R', so 0\le u\le6\varepsilon. Integrating the two first-order equations in (3.15) gives, by a bootstrap starting at u=0,

For completeness, assuming the first bound up to a first failure gives |F-1|\le.175u^2 and |F''+b|\le6(.35)u+.335(.175)u^2<.0013; integration then gives |F'|<.337u<.35u, excluding that failure. The displayed weaker .003 bound follows as well. Thus

Since |R(Z)-2\sqrt{Z_*}|<10^{-8}, taking R'=R(Z) proves (3.16), including the negligible quadratic error when passing from the complex value to its modulus.

Differentiate the initial-value problem with respect to R'. The derivative G=\partial_{R'}F satisfies the same linear equation and has data G(R')=0, G'(R')=b. Integration, or the same bootstrap, gives |G(r_b)/(r_b-R')-b|<.003. As F(r_b)=1+O((r_b-R')^2),

□

Proof. First take a small enough that h_b^{1/a}\le.3 uniformly on the disk; this also ensures h_b<.5^a. For a continuous radial input A satisfying (3.17), define w,V by the first and second lines of (3.4). The useful nonsingular formula is

Since h_b>.999, r_b^2<11, and a is small, it implies

Indeed c_*h_b^2/12\le w/r\le c_*/(12h_b^2); inserting these bounds into V=b+r^2(1/16-(w/r)^2/4) gives the first assertion for small a. The Lipschitz constant of A\mapsto w/r is at most c_*(2h_b^{-3}+2h_b^{-2})/12<2.01; multiplying by r_b^2(\sup w/r)/2 gives a constant less than 10 for V.

For this prescribed V, solve the scalar Dirichlet problem

The constants h_b and .5^a are respectively a lower and an upper solution. One can obtain H by monotone iteration: choose L>0 so that t\mapsto(L+V(r))t-t^{1+1/a} is increasing on [h_b,.5^a], and at each step invert -\Delta+L with the given boundary value. This inverse is positive by the minimum principle. Its existence, if needed directly in the radial class, follows by minimizing the strictly convex functional \int(|\nabla u|^2+L|u|^2-2fu)/2 for zero boundary data; the radial equation then reads (r^{11}u')'=r^{11}(Lu-f) and gives the classical solution. Iteration from h_b is increasing and stays below .5^a. For fixed a its right sides are uniformly bounded; radial integration gives compactness in C^1, and its limit solves (3.21).

We record a comparison argument that also applies below. The radial Dirichlet inverse of -\Delta is

Thus the positive geometric series defines T=(-\Delta-.5)^{-1}=\sum_{n\ge0}(.5K)^nK with zero boundary values. Also \psi=1+K(1+.5\psi) has a positive solution and satisfies (-\Delta-.5)\psi=1, \psi(r_b)=1. For every q\ge0 and V\le.5,

If a function with nonnegative boundary values violated the comparison principle for -\Delta-V+q, its ratio to \psi would have a negative interior minimum, where the product rule gives a strictly negative value for that operator. This is a contradiction. Taking q to be the divided difference of t^{1+1/a} proves uniqueness in (3.21) and comparison with lower solutions.

The function B(r)=.2^a-.02(r-r_0)_+^2 is such a lower solution. For small a it is positive, B(r_b)<h_b by (3.16), B^{1/a}\le.2, and

At r_0 the first derivative is continuous, so no interface measure occurs in this inequality. Comparison proves (3.18) for every output H.

Let H_1,H_2 be outputs for two inputs. Their difference solves

The barrier implies q\ge.2/a on r\le r_0, whereas q\ge0 everywhere. A positive supersolution for unit forcing is

On the inner ball, the constant term contributes at least 2-5a and the second term contributes a nonnegative quantity. On the shell, the second term contributes at least 2, while the constant term is bounded below by -5a. Thus the comparison principle applies, including at the shell interface. To bound S, radial integration gives

Here the second inequality uses (1-.5\|K\|)^{-1}<2. Consequently the input-output map satisfies

Our already fixed shell width makes the second contribution less than 1/4; taking a small makes the whole factor less than 1/2. The closed interval (3.17) in the radial continuous functions is complete, so successive iteration gives its unique fixed point. This proves the coupled inner problem.

Finally 0<P\le.5 and (3.20) give

These are the asserted uniform bounds. For fixed a, convergent parameter sequences have C^1 convergent subsequences; the equation and uniqueness identify their limits. This proves continuous dependence, including when the boundary value h_b varies.□

Proof. The preceding bounds give C^1 compactness of the amplitudes. Every limit A_0 is at most 1 and, by (3.18), equals 1 on [0,r_0]. Formula (3.19) and the last equation in (3.4) give C^1 convergence of w, including at zero. On every compact interval where A_0<1, the pressure tends uniformly to zero. There the limit satisfies \Delta A_0=-V_0A_0, with V_0\ge.3.

Each component of \{A_0<1\} has a left endpoint R' where A_0(R')=1, A_0'(R')=0. On this component (r^{11}A_0')'=-r^{11}V_0A_0<0, so A_0'<0 and the component cannot have a right endpoint below r_b. As A_0(r_b)=h_b<1, there is exactly one such component, (R',r_b], and R'\in[r_0,r_b). On the plateau (3.19) gives w_0=r/2 and hence \phi_0=0. Reconstruction gives the free complex equation with data 1,0 at R'. Its boundary modulus is h_b, so Lemma 3.2 forces R'=R(Z).

The pressures are bounded by .5. Any weak-star limit P_0 satisfies in distributions

There is no surface measure at R because A_0 and A_0' are continuous. On the plateau the right side is b, and on its exterior it is zero. Since A_0 is positive, this identifies P_0 uniquely. The preceding identification applies to every subsequence; compactness therefore proves all convergence assertions. Reconstruction by \phi'=(w-r/2)/2 gives the asserted C^1 convergence of the complex profiles as well.□

Proof. Writing \nu=-2a+2ib and Q=r^\nu f transforms the equation into

Set N_0=0. Choose a small fixed neighborhood of the compact family (m_0(b,Z),b) such that the corresponding free functions f stay in an annulus

This is possible because at a=0 the equation is complex linear in the coefficient m=\lim f. Enlarge the annulus slightly, still keeping its upper radius below 1. On this enlarged annulus every fixed real derivative of N_a tends uniformly to zero as a\downarrow0: its norm is bounded by C_j a^{-j}\rho_1^{1/a} for some \rho_1<1 (adjusting C_j and the power of a does not affect this conclusion). Thus all fixed differentiability orders needed below are uniformly controlled.

For prescribed m,b, construct a formal expansion f_N=\sum_{q=0}^Nm_qr^{-2q} with m_0=m. Substituting in (3.25) determines m_{q+1} from the coefficient of r^{-2q}: its coefficient in (ir^2/2)Df_N is -i(q+1)m_{q+1}. All the other terms at that order involve only m_0,\ldots,m_q and derivatives of N_a at m. This proves existence and uniqueness of the coefficients, boundedness at every fixed order, continuous dependence down to a=0, and the same bounds after a fixed number of derivatives in the real and imaginary parts of m and in b. The residual of f_N is O(r^{-2N}), with its fixed symbol and parameter derivatives.

Here is a convergent construction behind this expansion. Put t=\log r and Y=(e,De) for e=f-f_N. The error equation has the form

where the constant matrix \mathcal B has bounded norm, \mathcal N(t,0)=0, its Lipschitz constant on the annulus is bounded uniformly in a,m,b, and |\mathcal R_N(t)|\le C_Ne^{-2Nt}. The propagator \Phi(t,s) of the displayed diagonal part is unitary. The backward integral equation is

On t\ge t_* use the norm \|Y\|_N=\sup e^{2Nt}|Y(t)|. If C bounds the combined Lipschitz constant of the two terms involving Y, their integral has norm at most C/(2N). Choose one integer N_* with this ratio below 1/2, and then choose t_* large enough that the corresponding ball around f_{N_*} remains in the enlarged annulus. The residual term has norm at most C_{N_*}/(2N_*), so iteration in a ball of twice this radius converges geometrically to a solution. No decay estimate for an oscillatory propagator has been used; unitarity and the weight make the integral absolutely convergent.

The same construction works at any larger truncation order. Two solutions so obtained differ by O(r^{-2N_*}) in both entries after subtracting their common expansion through order N_*. Their difference solves (3.28) without the residual, on a common sufficiently long tail. The contraction estimate forces it to vanish there, and uniqueness for the finite-interval ODE makes the solutions identical. This proves an expansion to every order for the single solution already constructed. Symbol derivatives cause no loss in this conclusion. Indeed (3.27) bounds the next derivative of an error by at most C r^2 times its two entries and the residual; differentiating it J times loses at most 2J powers of r. Constructing at an order greater than N+J+2 first therefore gives the remainder estimate of (3.10) through the prescribed order J. This also explains why the constants need only be uniform at each fixed order.

Differentiating the integral equation with respect to m or b gives the same contraction for its derivative, with a bounded inhomogeneous term. Difference quotients converge to that solution, proving real C^1 dependence; repeated differentiation works at every fixed order. Continuity at a=0 follows directly from the uniform contraction and convergence of its coefficients. Continue the solution from the fixed large radius back to r_b. On this finite interval the right side of the first-order form of (3.25) and all its fixed derivatives are bounded and converge on the enlarged annulus. The integral equation on finitely many short intervals, followed by Gronwall’s inequality for differences, keeps the solution close to its free counterpart and proves existence and parameter convergence down to r_b. This also proves convergence of all fixed radial derivatives and the first parameter derivatives there. In view of (3.26), the final functions satisfy \kappa\le|f|\le\rho<1 for fixed \kappa,\rho, uniformly in small a and the chosen parameter neighborhood.

It remains to choose m so that Q(r_b)=h_b. At a=0 the map m\mapsto Q(r_b) is complex multiplication by a nonzero factor. Its real two-dimensional inverse is uniformly bounded over the compact parameter disk. The just proved C^1 convergence implies that, on a fixed small neighborhood of each m_0, the equation for m is a contraction after multiplying by that limiting inverse. Its residual at m_0 tends to zero uniformly, so it has a unique root near m_0. These roots agree on overlapping neighborhoods and depend continuously on (a,b,Z). This imposes the required boundary value. All tail bounds extend from the initially fixed large radius to r_b by the compact interval bounds, completing the proof.□

Proof. Let Q^{\rm in}_a be reconstructed with phase zero at the origin, and define the continuous complex-valued matching map

Lemmas 3.4 and 3.5 give uniform convergence on \mathcal D to M_0=F_{\rm in}'(r_b)/F_{\rm in}(r_b)-G_b'(r_b)/G_b(r_b). Both denominators are nonzero.

The Wronskian \mathcal W=F_{\rm in}G_b'-F_{\rm in}'G_b satisfies \mathcal W'=-(11/r+ir/2)\mathcal W. At R=2\sqrt Z, G_b'(R)=(iR/2)j(b,Z)G_b(R), and hence

The factor K is continuous and never zero on the entire disk. Contracting its parameter argument to the disk center gives a homotopy through nonzero factors to a constant. Complex multiplication by that constant preserves orientation. Proposition 2.1 therefore gives

In particular M_0 is separated from zero on the boundary. Uniform convergence preserves the boundary homotopy and the degree for every sufficiently small a, so M_a has a zero.

At such a zero the values have equal modulus h_b. Rotate the outer profile to match the value of the inner one; equality of logarithmic derivatives then matches the first derivatives as well. The ODE gives smoothness across r_b, since the profile is nonzero. To verify smoothness through the origin without a regular-singular shortcut, write q(x)=Q(\sqrt x) for x\ge0. Integrating the complex radial equation gives

The right side is continuous at x=0. It first proves q\in C^1 and then, by successive differentiation, q\in C^\infty because q is bounded away from zero on this compact interval. Thus Q(y)=q(|y|^2) is smooth at the origin. The normalization at zero was already imposed. Finally, a convergent sequence of matching parameters satisfies M_0(b,Z)=0, so (3.29) gives j(b,Z)=0. At such a root the free initial-value solution equals G_b/G_b(R), proving (3.7).□

Proof. First consider the inner amplitude. On every component of \{P<.2\},

At a positive left endpoint A'\le0, since the amplitude is entering the sublevel set A<.2^a; at the origin A'=0. Thus A'<0 at every positive radius in this sublevel set. If M=\max_{[0,r_b]}(A')_+ is positive, it is attained at a point r_m\in(0,r_b): the derivative is zero at the origin and strictly negative at r_b for small a by Lemma 3.4. At this point P\ge.2, A''=0, and A'''\le0.

Let q=w(r_m)/r_m. The bounds A\le.5^a, A(r_m)\ge.2^a and (3.19) give

The fixed lower bound follows from h_b>.999 and small a; it need not approach 1/2 at any prescribed rate. Differentiating V and using (3.4) gives the exact identity

The polynomial q\mapsto1/8-c_*q/2+11q^2/2 is increasing for q\ge.49 and small a, and its value at q_+ is O(a), since its value at (a,q)=(0,1/2) is zero. Thus at r_m,

Differentiate the amplitude equation. At this maximum,

so

It follows that M\le Ca^2. Since w\ge0 and w/A is bounded on the inner ball,

which proves (3.12).

In the exterior, (3.9) gives P/a=r^{-2}|f|^{1/a}/a. Repeated Euler differentiation and \kappa\le|f|\le\rho<1 bound every fixed derivative by C_j a^{-N_j}\rho^{1/a}r^{-2} for some finite integer N_j. This tends to zero uniformly after multiplication by r^2. Also e^{2i\phi}=r^{4ib}f/\overline f, so all its fixed Euler derivatives are uniformly bounded. As w/r is uniformly bounded in the exterior, these facts give (3.11) and the exterior bound on W(P/a). Together with P\le.5 in the interior they give the uniform upper bound on its positive part. By taking a smaller if necessary, the exterior formula also gives P\le1 everywhere. The same bounds for f, and (3.17), prove the two-sided comparison for \mu in (3.6).

It remains to prove positive deformation. The two eigenvalues of the derivative of a radial field are w' and w/r. For every limiting matched profile they equal 1/2 in the core. On the exterior, A_0'<0, w_0>0, and w_0/r<.543, so

Integrating from zero also gives w_0/r\ge.027. The compact C^1 convergence of A and formula (3.19), followed by the last equation of (3.4), give convergence of both eigenvalues on every fixed compact interval, including zero. For the remaining tail, the uniform expansion of Lemma 3.5 gives

uniformly in small a. Choose a large fixed radius to make these tail bounds positive, and only then take a small enough for the compact convergence. A subsequence contradiction, using compactness of the parameter disk, makes this choice uniform for all matched profiles. For example c=.01 is admissible after reducing a_0. This proves (3.6).□

Completion of Proposition 3.1. The inner construction, exterior construction and degree argument give existence for every 0<a<a_0 after finitely many reductions of a_0. The order of choices is fixed: the disk and \varepsilon first, then the core comparison estimates, the exterior annulus and a single large truncation order and starting radius, and finally sufficiently small a. Any later fixed derivative order is obtained by taking a higher auxiliary truncation of the same exterior solution; it does not require changing that solution or the already fixed geometric parameters.

The limiting statements on the inner ball are Lemma 3.4, and on [r_b,\infty) follow from Lemma 3.5 and matching. On a compact subset of (R,r_b), the limiting amplitude is strictly below 1; uniform C^1 convergence therefore places the profiles in a fixed annulus bounded away from both 0 and 1. All derivatives of z\mapsto|z|^{1/a}z tend to zero there. The complex radial ODE, successively differentiated, then proves C^\infty convergence and P/a\to0 on that compact set. The weak-star pressure assertion on any larger fixed ball follows by joining the inner assertion to this exterior convergence. The C^1 convergence of \mu,w has already been proved from their amplitude formulas.

Finally, radial Euler estimates imply Cartesian symbol estimates away from zero: an ordinary derivative of order j is a linear combination of r^{-j} times Euler derivatives with bounded smooth angular coefficients. The fixed-a smoothness on a compact ball supplies the remaining bounds in (3.13); their constants are not asserted to be uniform as a\downarrow0. As a-ib+\nu/2=0,

and (3.10) gives (3.14). All parts of the proposition follow.□

Proof. Suppose an off-block entry is b_{jk}=O(r^{-q}). A change with entry -r^{-2}b_{jk}/(\Lambda_j-\Lambda_k) cancels it by commutation with r^2\Lambda. Differentiating this change or multiplying it by the bounded part gives O(r^{-q-2}); its products with the other changes have the same or better order. The denominator and all its fixed Euler derivatives are uniformly bounded by the spectral gap. Perform this operation simultaneously on every off-block entry, and iterate a finite number of times. The product of the changes is invertible for sufficiently large r by its I+O_m(r^{-2}) bound. Differentiating the construction proves the symbol estimates. All operations are algebraic operations and differentiation with nonzero denominators, proving the last assertion.□

Proof. Use (v_h,v_h'/r) for each channel. The coefficient of r^2 in its D equation is

Diagonalize by the eigenvectors with first component one, group the two zero eigenvalues, and apply Lemma 4.2. Writing

the slow block is \operatorname{diag}(\nu_h)+O(r^{-2}), and the fast entries are -hi r^2/2-12-\nu_h+O(r^{-2}). For example, before diagonalization the bounded contribution to the second row is -12(v_h'/r)-(b+hi(a+\lambda))v_h; this verifies both constant exponents.

Factor out r^{\nu_h} in a slow coordinate and e^{-hi r^2/4}r^{-12-\nu_h} in a fast coordinate. The diagonal errors are integrable in d r/r. The off-block errors are also integrable if m>|12+2\operatorname{Re}\nu_h| with a fixed positive margin on the parameter neighborhood. The factored equation is thus an integrable perturbation of a constant vector. On a sufficiently late half-line, its integral from infinity is a contraction on bounded vectors; prescribing its four limits gives four independent solutions. Continuation gives a basis on the entire exterior. This construction is holomorphic on a slightly smaller parameter neighborhood because the integrable majorants are uniform there.

Setting both fast limits to zero gives exactly the outgoing subspace. In fact the fast coordinates forced by slow coordinates are O(r^{\operatorname{Re}\nu_h-m}) before factoring, by the backward integral. Taking m larger before each fixed number of differentiations gives (4.1). Increasing the reduction order does not change the limiting normalization: the additional transformations have arbitrarily high order relative to the prescribed normalization, and uniqueness of the backward integral applies on a common tail. Uniform exterior symbol bounds for the profiles and their convergence give convergence of these integral equations by dominated convergence; ordinary ODE continuation then proves the asserted local convergence. At a=0 the slow columns are, up to their nonzero normalizations,

in their separate channels, by Lemma 2.2.

If the fast limit d_h in channel h is nonzero, reconstruction and k radial differentiations give the leading fast term

The other channel’s fast contribution is smaller by r^{-2}; this follows from the off-block reconstruction. The same formula follows directly by successively applying D to the reduced fast equation, whose leading multiplier is -hi r^2/2. For \operatorname{Re}\lambda\ge-1/32 and k\ge7, 2k>12+2\operatorname{Re}\nu_h. Consequently the first term in (4.6) dominates the slow term and its square is not integrable against r^{11}\,dr. Thus every fast limit vanishes. For any particular finite \lambda,\ell, the argument applies in a bounded neighborhood of those parameters; it does not require a uniform bound on the parameters of the mode being tested.□

Proof. Put V_a=P/a and \sigma=\operatorname{Re}\lambda. Test (4.3) against (\lambda+W)X, and take real parts. The J term has zero real part. Integration by parts, using (4.4), gives

Here the matrix \nabla w acts on spatial indices and the expression is summed over the two components. Boundary terms vanish: after division by A, the slow bound is X=O(r^{-2\sigma}), with its symbol derivatives, and \sigma\ge4 gives all needed integrability.

For completeness, the weighted Hardy inequality used here is uniform for small a:

Indeed, for \rho=\langle r\rangle^{-4a}, \operatorname{div}(\rho y/r^2)\ge(10-4a)\rho/r^2. Integration against |f|^2 and Cauchy–Schwarz prove the inequality with \rho; comparability of \rho and \mu proves (4.8). Cutoffs justify the calculation by density. Proposition 3.1 gives \nabla w\ge cI and W(V_a)\le C a V_a on the inner ball, while its exterior positive part is o(1)r^{-2}. Since \sigma-c_*/2\ge1, these two negative contributions in (4.7) are absorbed, respectively, by the pressure and gradient terms. A nonzero X is impossible.□

Proof. For r\ge E use (v_h,v_h'/r) again. Diagonalize the leading matrix

Its eigenvalues are

The square root stays uniformly positive on r\ge E. Eigenvectors with first entry one give a uniformly invertible symbol change. Group the two plus signs as the outgoing block; take the two minus signs as separate blocks. These blocks have uniformly positive gaps. The bounded remainder has norm at most a fixed C_0, with uniform fixed symbol derivatives. The leading diagonal is purely imaginary.

The order of choices is important. First choose C>C_0+2, enlarged if necessary to bound the Hermitian parts of the initial full and block systems. Next choose one fixed integer m>2C+4. Apply Lemma 4.2 to this order. Finally pass far enough along the sequence that its finitely many O_m(r^{-2}) corrections have norm at most one on r\ge E. Both the full reduced evolution and each block evolution then have propagator bound

This is Gronwall in \log r after using skewness of the leading diagonal. No separation inside the outgoing block is required.

For each individual outgoing solution, all reduced coordinates have symbol bounds O(r^\rho), where \rho=-2(a+\sigma)<1; their constants may depend on that solution and its parameters. In an incoming equation the r^2 leading coefficient is invertible. Solving for that coordinate and differentiating gives a gain of two powers of r; iteration gives the same gain for each fixed symbol derivative, up to the O_m(r^{-m}) forcing from the outgoing block. Thus the incoming coordinate is o(r^{-C-1}), with m as chosen. This qualitative bound is used only to remove its terminal value in variation of constants at infinity.

After that terminal value has vanished, the quantitative estimate uses only (4.13). If Y denotes all reduced coordinates, its incoming part at E is at most

Undoing the changes gives

The full coordinate size is bounded by C|v(E)|, since the incoming coordinates are small and the first entry of each outgoing eigenvector is one. Multiplication by the scalar defining q_h proves (4.12); the 11/(2E) term and the difference between L_\ell and \ell(\ell+10) are within its error.□

Proof. The derivative F'=r/8+2L_\ell/r^3 is positive, so there is at most one zero. In the turning cases its defining equation gives

It follows that k_0^2\asymp S, r_0\to\infty, r_0<E/2, and d_0\to0. For the h=-1 turn in Case I, for example, r_0^2\asymp \ell^2/(\omega+\ell), which tends to infinity precisely because \ell/\sqrt{1+\omega}\to\infty. The other turn has r_0^2\asymp\omega+\ell. In the no-turn case, necessarily the negative channel in Case II, F\asymp S throughout [R,E].

Fix a large number M and remove the interval |r-r_0|<Md_0. On either remaining interval choose \operatorname{Re}p>0 with p^2=\operatorname{sign}(F)(F+hi\gamma). Use the two approximate solutions

All integrals in this paragraph are over the outer intervals. Direct differentiation of (4.18) gives

Here and below the constants multiplying M^{-3/2} do not depend on M. To verify the estimates, on [r_0/2,2r_0] use |F|\asymp k_0^2|r-r_0|/r_0, |F'|\le Ck_0^2/r_0, |F''|\le Ck_0^2/r_0^2 and d_0^3=r_0/(2k_0^2). Integrating the powers |r-r_0|^{-3/2} and |r-r_0|^{-5/2} from Md_0 gives the stated error. Off this interval |F|\gtrsim k_0^2, with relative derivative bounds C/r,C/r^2, and E\lesssim k_0; splitting at r_0/2 and 2r_0 gives a vanishing error and the bounded integral in (4.19). For no turn the same computations use F\asymp S and give o(1) in (4.20) on all of [R,E].

Here is a transfer justification retaining the error that will be needed below. Each approximate solution has residual potential

The approximate pair has constant nonzero Wronskian. Its transfer matrix in the N norms is at most C\exp(|H_f(r)-H_f(t)|): K\asymp|p|, and \int|p-\sqrt{|F|}|\,dr\le C by (4.19). Variation of constants for the residual followed by Gronwall bounds the difference between exact and approximate transfer by

Indeed the residual divided by K has integral bounded by (4.20); when r,s,t occur in order, monotonicity of H_f gives |H_f(r)-H_f(s)|+|H_f(s)-H_f(t)|=|H_f(r)-H_f(t)|. This removes the exponential factor from the integral inequality. With no turn the error coefficient is o(1).

At E, the prescribed data select the outgoing approximate branch up to an error tending to zero. On the positive outer interval, (4.19) bounds its normalized amplitude above and below, because the absolute imaginary action is bounded. Consequently

This proves all claims in the no-turn case.

In a turning case set \xi=(r-r_0)/d_0 and Y(\xi)=d_0^{-1/2}U(r_0+d_0\xi). Since F'(r_0)d_0^3=1 and \gamma d_0^2\to0, its equation converges on every fixed \xi interval to

The corresponding transfer on [-M,M] is bounded in the scaled data norm, with a bound depending only on fixed M. Figure 2 records these three regions.

Choose one fixed M_0 so large that the error in (4.22) is below its lower amplitude margin. At \xi=M_0, the scaled Cauchy data are bounded and their flux

has sign h and magnitude bounded below. A subsequence of these data converges, and continuous dependence of the scaled ODE gives one solution Y_\infty on every fixed interval. Its flux is nonzero and constant, since (4.23) has real coefficients.

We spell out why this gives the required relative margin on the forbidden side. With t=-\xi, the real equation is y''-ty=0. On a sufficiently late half-line it has a positive growing solution comparable to t^{-1/4}\exp(2t^{3/2}/3). To construct one, prescribe the growing approximate data at a fixed large T and use the residual transfer estimate with p=\sqrt t: the total error relative to the growing scale is at most CT^{-3/2}. The exact solution and its derivative are positive when T is large. Its logarithmic slope z=y'/(\sqrt t\,y) satisfies

This equation implies z\to1: for every fixed \epsilon>0, its derivative points into [1-\epsilon,1+\epsilon] at both endpoints for sufficiently large t, and outside that interval it points inwards with magnitude bounded below by a positive multiple of \sqrt t. A decaying independent solution is obtained by multiplying the growing one by \int_t^\infty y(s)^{-2}\,ds. The growing comparison bound makes that integral finite and shows that this second solution and its derivative are negligible relative to the first. These real solutions have nonzero Wronskian. A complex multiple of the decaying solution alone has zero flux. Thus the coefficient of the growing solution in Y_\infty is nonzero.

As M\to\infty its growing contribution dominates its decaying one, and

In the approximate basis referenced at r_-=r_0-Md_0, the coefficient of growth to the left therefore has magnitude at least a fixed positive fraction of N_{r_-}(U), for every sufficiently large fixed M and subsequently late sequence indices. That fraction can be chosen independently of M: the normalized basis vectors have comparable size one at their reference point, and the displayed slope is separated from the decaying slope +1. For each such fixed M, N_{r_-}(U) is also bounded below by a positive constant.

Transfer to R now gives a growing term of size at least cN_{r_-}(U)e^{H_f(R)-H_f(r_-)}; the decaying term is negligible since the action difference tends to infinity. The transfer error (4.21), relative to this growing term, is still CM^{-3/2}+o(1) with C independent of M, precisely because of the relative coefficient margin. For the value lower bound, choose one sufficiently large fixed M and retain it; then H_f(r_-)=O_M(1) and (4.16) follows with a positive constant. For the slope take an arbitrary sufficiently large fixed M, obtain error CM^{-3/2}+o(1), and then let M\to\infty. This proves (4.17). The transfer estimate follows from the two outer estimates and the scaled central transfer, using one fixed M. Both channels can use a common extracted subsequence. Finally, all estimates allow a moving R in the fixed shell, where |F_h(R)|\asymp n_*^2. If uniformity there failed, the same argument at a sequence of offending endpoints would contradict the estimates.□

Proof. For each scalar channel let D_h(R)=0, D_h'(R)=K_h(R)^{1/2}. Lemma 4.6 gives

where H_h=H_f in that channel. The Green kernel with left Dirichlet data and the right Robin condition of U_h is the product of D_h on the left and U_h on the right, divided by the Wronskian. It obeys the following two different action bounds:

Each exponential is at most one, by monotonicity within its own channel. There is no comparison of H_+ and H_-. A right Robin derivative error \beta_h, with zero left data, similarly gives N_r\le CK_h(E)^{-1/2}|\beta_h|; this follows either from the same Wronskian or by using D_h for the correction.

Let \kappa=\min_{h,r\in[R,E]}K_h(r). Then \kappa\to\infty: at a turn it is d_0^{-1}, and in the no-turn case it is comparable to \sqrt S. Put \mathcal N(q)=\max_h\sup_rN_r(q_h). The matrix forcing in (4.11) has Green norm at most

Also K_h(E)\asymp E for both channels, so the right error (4.12) has norm at most CE^{-2}\mathcal N(q). The diagonal outgoing extension q_h(R)U_h(r)/U_h(R) has norm at most Cn_*^{1/2}|q(R)|, since K_h(R)\asymp n_* and H_h(r)\le H_h(R). Variation of constants and absorption of (4.26) and the Robin error give

Taking the derivative at R costs K_h(R)^{1/2}\asymp n_*^{1/2}. Together with (4.17), this is (4.24). Every estimate used is homogeneous in the left data and uniform over the shell.□

Proof. Use (4.9) and Lemma 4.7. All volume pairings in this proof have weight \mu, and [\cdot] denotes its surface integral at the current sphere, including the factor r^{11}. On a fixed ball \mu is uniformly bounded above and below, and the coefficients of W and L+\Delta are uniformly bounded.

In Case I pair (4.3) with X on the ball r<B_1 and take real parts. The second-order term is

The boundary term has nonpositive real part, by (4.24) and both negative signs of F_h. More precisely its real part is \frac12\operatorname{Re}\sum_h\overline q_hq_h'+O(|q|^2), so its leading value is -c n_*|q|^2 and absorbs its errors. The absolute value of the remaining real terms is at most C\|X\|_2\|\nabla X\|_2+C(1+\omega)\|X\|_2^2. The angular barrier

and \omega=o(\ell^2) absorb these terms into an arbitrarily small fraction of the gradient term. Nonnegative pressure then gives a contradiction.

In Case II normalize

Write the shell error in (4.24) as \varepsilon n_*|q|, with \varepsilon\to0. Its leading diagonal has modulus at least c n_*, so n_*|q|\le C|q'| on the shell. There is a point in its first half at which both |q'| and n_*|q| are bounded, by the shell integral bound furnished by (4.27). For the channel h=-1, the energy |q_-'|^2+F_-|q_-|^2 propagates this bound to B_1. Indeed |F_-'|/F_-\le C, F_-\asymp n_*^2 on the shell, and the forcing in its equation has bounded integral, since it is at most C|q| there. Differentiating the square root of this energy and applying Gronwall proves

from that point to B_1. The positive channel then obeys, in the upper-derivative sense,

The distance from the chosen point to B_1 is at least 1/2. Integrating this inequality with \sqrt{-F_+}\ge c n_* gives

We next remove the other boundary channel by a flux identity. The imaginary part of the pairing of (4.3) with X is

To check the first-order term, integration by parts and (4.4) give

Also the pressure-free equation Lg=(\lambda+W)f, tested against g, gives

Its integrated gradient term, W term, and boundary term are bounded by the normalization and the boundary bounds already obtained. The change to circular coordinates gives the exact cancellation

Multiplying (4.29) by n_*\asymp\sqrt\omega, and using (4.30), makes its right volume term tend to zero. The positive channel contributes o(1) by (4.28). The outgoing relation for the other channel gives \operatorname{Im}(\overline q_-q_-')=-\sqrt{F_-}|q_-|^2+ o(n_*)|q_-||q|. Its sign and F_-\asymp n_*^2 imply

Rellich compactness on the ball, including the full angular fields, gives a strongly L^2 convergent subsequence of f. The equation Lg-Wf=\lambda f has uniformly bounded left side in H^{-1} by the normalization. Division by \lambda, whose modulus tends to infinity, shows that the limit is zero. Thus

Finally put S_d=W+c_*/2 and test (4.3) against S_dX. The real part of the second-order and pressure terms is

All boundary terms vanish by (4.31); the tangential derivatives cost \ell\le n_*, and P/a is bounded on this exterior sphere. The JS_dX pairing has zero real part. For I=\int(S_dX)^*JX\,\mu, (4.32) and one integration by parts in the term containing Wf give I\to0. Further,

Hence \omega\operatorname{Im}I\to0, since \omega=O(n_*^2) and the boundary values are o(n_*^{-1}). The remaining pairing (\lambda-c_*/2)I therefore has real part tending to zero. We conclude that the displayed bulk integral tends to zero. The profile estimates \nabla w\ge cI and W(P/a)\le C, together with (4.32), force \|\nabla X\|_2\to0. The vanishing boundary trace and Poincaré’s inequality then force \|X\|_2\to0, contradicting (4.27).□

Proof. Rellich compactness gives the volume part of j_E; in a fixed harmonic the trace has only two scalar entries and is bounded, hence compact. Uniform upper and lower bounds on \mu give uniform coercivity of (4.33). Inverses exist by the Hilbert-space coercive form argument, also on the closed subspace E_0.

We prove a stronger sequential statement. Let F_a\rightharpoonup F be any bounded weakly convergent sequence in E', and put X_a=T_aF_a. Coercivity gives a bounded sequence in E and

After extraction, X_a\rightharpoonup X in E and j_EX_a\to j_EX strongly. Uniform boundedness of P_a permits replacing |f_a|^2 by |f|^2 in the last integral, with error o(1). Its weak-star limit b\mathbf1_{r<R} therefore gives b\int_{r<R}|f|^2=0. This proves X\in E_0 without any pointwise lower bound on pressure near its moving transition.

Test first with \Theta\in E_0 whose first component is supported a fixed positive distance outside r=R. On that support the limiting amplitude is strictly below one, so P_a/a\to0 uniformly. The other coefficients converge uniformly on the ball. Weak convergence in E, and weak convergence of the inputs against this fixed test, therefore give the limiting inverse equation. Such tests are dense in E_0. Indeed the first component has zero outer trace at R; cutting it off across [R+\epsilon,R+2\epsilon] costs, in addition to its shrinking collar norm, at most

The radial weight is comparable to a positive constant on this collar. The second component needs no cutoff. Density and uniqueness identify X=T_0F. Every subsequence has the same possible strong limit after j_E, proving j_ET_aF_a\to j_ET_0F for weakly varying inputs.

If (4.34) failed, choose unit witnessing inputs and extract a weakly convergent subsequence in the Hilbert dual. The sequential assertion just proved and compactness of j_ET_0 would make both images converge strongly to j_ET_0F, a contradiction. This proves exactly the stated norm convergence; no operator-norm convergence into the full H^1 space is asserted.□

Proof. Lemma 4.3 and its large-radius normalization show that the values of a slow basis at B_2 form an invertible matrix, uniformly in a small neighborhood of the candidate parameter and for sufficiently small a. Thus the outgoing condition is

with holomorphic matrices converging locally uniformly to \mathcal M_0. Define K_a on volume values and boundary entries by its pairing with \Theta\in E:

The boundary bracket is at B_2. This is the weak equation obtained by integrating -\Theta^*JW X by parts. It uses only the two types of data in j_EX. Its coefficients converge uniformly, and hence K_a:\mathscr Z\to E' converges locally uniformly in operator norm. Lemma 4.9 proves the asserted convergence of \mathcal C_a, and compactness follows by factoring through E. For a kernel vector z its lift is X=T_aK_a z, with j_EX=z. For positive a, smooth elliptic regularity gives a regular solution through the origin; its Robin condition extends it uniquely into the slow space. Conversely any regular outgoing mode satisfies the weak equation. These two maps are inverse on the kernels.□

Proof. The form domain gives f=0 in the core and continuity of both traces at R. Testing the second equation without a constraint on its test component gives Lg=(\lambda+W)f across the interface. Its right side is locally L^2 and has no surface delta, since f is H^1. Radial integration therefore makes \mu r^{11}g' continuous. Positivity and continuity of \mu give continuity of g'. Inside the core \mu=1 and f=0, so regularity and harmonicity give g=c r^\ell. Piecewise ODE regularity supplies the one-sided derivatives involved. The first equation may contain a constraint reaction at R; the form domain does not impose any condition on f'(R+). In particular we do not impose the generally false condition f'(R+)=0.

Since Q(R)=1 and Q'(R)=0, the exterior circular conditions are

Using (4.5), the derivative of the argument -hi r^2/4 is -hi r/2. Consequently these two equations, after nonzero row and column scalings and a sign change in one unknown coefficient, have columns

This is exactly the determinant of Proposition 2.1. The columns are nonzero by Lemma 2.5.□

Proof. For every positive a the first circular components iQ, \partial_jQ, and (a-ib+D/2)Q solve the linearized equation at 0,1/2,1, respectively, by differentiating phase, center, and blow-up time. These identities can also be checked without referring to a family of solutions. The first-component generator is

Phase covariance gives \mathcal A(iQ)=0; differentiating the stationary equation gives \mathcal A(\partial_jQ)=\partial_jQ/2. Commuting D with \Delta and using homogeneity gives \mathcal A(DQ/2)=i\Delta Q and \mathcal A(aQ)=-i|Q|^{1/a}Q, whose sum equals (a-ib+D/2)Q by the stationary equation. The term -ibQ is a phase multiple and contributes zero.

At a=0 the same exterior identities hold with the power term deleted. The limiting phase, time, and translation fields satisfy (4.37). For the time mode its first circular component at R is -ib, while its derivative is -Rb/2, which is real; for a translation it has value zero and derivative -b times the appropriate first harmonic. These facts follow from the exterior profile equation Q''(R)=-b. The phase mode is immediate. They are slow by the free expansions; in the time mode the leading tail exponent cancels. They are nonzero: b>0, and the exterior profile is nonconstant. Proposition 2.1 gives exactly two, one, and zero roots with multiplicity in degrees 0,1, and \ell\ge2. Thus the displayed distinct roots exhaust the counts and are simple.

Suppose the compact pencil has a first-order chain, that is, z_0\ne0 and

Lift it by T_0K_0, including the derivative term in the lift of z_1. Both first components vanish in the core. Differentiating the second equation shows that both g components are harmonic there, since its new forcing is the leading f, which vanishes in the core. Both obey the interface trace and derivative requirements. At B_2 the correction satisfies the differentiated Robin condition

The analytic slow basis extends these data to a first-order exterior solution. Therefore its two coefficients satisfy D_0c_0=0 and D_0c_1+D_0'c_0=0, where D_0(\lambda) is the matrix in (4.38). A simple determinant zero has rank one and nonzero derivative pairing between a null vector and a conull vector. Applying that conull vector to the second equation is a contradiction. Kernel dimension one follows from the same matrix and the unique harmonic continuation into the core.□

Proof of Theorem 4.1. First work in a fixed harmonic and a compact spectral neighborhood. The compact pencil is Fredholm of index zero. At a limiting simple root split its domain into the kernel and a closed complement, and its range into the image and a one-dimensional complement. The complement block is invertible, so elimination reduces the local kernel problem to a scalar holomorphic function. Its derivative is nonzero: vanishing would produce exactly the first-order chain excluded by Lemma 4.12. Locally uniform norm convergence in Lemma 4.10 gives locally uniform convergence of this scalar function and its derivative, by Cauchy’s formula. The argument principle on a small circle therefore gives one zero counted with multiplicity for small a. The exact symmetry mode places that zero at 0,1/2, or 1, so it remains simple. Away from the limiting roots invertibility persists by a Neumann series.

If the classification failed for arbitrarily small a, select offending modes with a\to0. Lemma 4.4 bounds their real parts, and Lemma 4.8 bounds their angular and imaginary parameters. Extract a fixed harmonic, a convergent spectral parameter, and convergent profile parameters. The preceding compact-pencil argument gives a contradiction. This also gives the assertion uniformly over all the profiles in Proposition 3.1. Lemma 4.3 then proves the homogeneous top-derivative assertion.

Finally suppose a regular vector v_1 with the top-derivative condition satisfies (\mathcal A-\lambda_0)v_1=v_0, where v_0 is one of the surviving modes. Choose an analytic exterior slow family v(\lambda) with v(\lambda_0)=v_0. Differentiating its mode equation gives (\mathcal A-\lambda_0)\partial_\lambda v=v_0 on the exterior. Its differentiated slow tail has bounds

For example, Cauchy’s formula on parameter circles of radius (1+\log r)^{-1} gives these bounds from the locally uniform analytic slow construction. This tail is square integrable in order k\ge7 at the three nonnegative eigenvalues. Subtracting it from v_1 leaves a homogeneous exterior solution with the same top-derivative condition, so Lemma 4.3 makes that difference outgoing. The original v_1 consequently satisfies the differentiated Robin condition and produces a forbidden first-order chain of the finite-a compact pencil. This proves the claimed algebraic simplicity and completes the proof.□

Proof. For the continuous Fourier transform, Cauchy–Schwarz uses

Thus \|\widehat v\|_1\le C\|v\|_Y. This identifies the completion in (5.3) with functions vanishing at infinity, rather than with distributions modulo polynomials. In the periodic case the corresponding inverse-weight sum, after division by (2\pi L)^{12}, is uniformly bounded. Indeed its portion |n|\le L is bounded by

and the portion |n|>L is bounded using k>6. The convolution inequality |\xi+\zeta|^s\le C_s(|\xi|^s+|\zeta|^s), and its version for (L^{-2}+|\xi|^2)^{s/2}, now prove the algebra estimates by Young’s inequality. The volume factors in (5.4) cancel against the Fourier \ell^1 estimate just proved.

For j<s_0, split at frequency R^{-1}. The low-frequency part has jth derivative bounded pointwise by CR^{-a-j}\|v\|. This follows by the same inverse-weight integral or sum; it includes the periodic zero mode. Multiplication by the square root of the volume of the ball gives CR^{s_0-j}\|v\|. On higher frequencies, Plancherel gives that same bound because |\xi|^j\le R^{s_0-j}|\xi|^{s_0} there. For s_0\le j\le k, the global derivative norm is bounded by the two defining Sobolev norms. This proves (5.5).

For a localized symbol, the integer derivative norms of order m are O(R^{6-\beta-m}), with the same bounds for derivatives falling on \psi_R. Fourier interpolation gives the fractional order in (5.6). The additional periodic term is bounded by

since the first norm in (5.4) is uniformly equivalent to \|\cdot\|_{\dot H^{s_0}}+L^{-s_0}\|\cdot\|_2. For \beta=2a the first exponent in (5.6) is -a; both exponents are negative. Summing dyadic pieces proves the last assertions, and also convergence of smooth compactly supported approximations to Q in Y.□

Proof. All coefficient derivatives have the bounds in (5.2), uniformly for the cutoff potentials. On an annulus of radius R, a lower Leibniz term with j<k derivatives on v has norm at most

Every exponent is negative: it is -2-k+s_0 for j<s_0 and at most -3 otherwise. Dyadic summation therefore makes these tail terms arbitrarily small.

For the fractional part, use a smooth dyadic partition and apply global Fourier interpolation to the actual localized product z_R=\psi_R\mathcal V_*v. Leibniz’ rule, including derivatives of \psi_R, gives

In the second estimate, terms with at most five derivatives on v are actually O(R^{-2-a}); the sixth derivative term is O(R^{-2}). Since s_0=5a+6(1-a), interpolation yields the explicit bound

The identical proof uses Fourier sums on a torus. Its remaining regularized term is bounded separately by

with a harmless geometric constant when R\le C_0L_s. Both exponents are summable over dyadic annuli; indeed -2+a-a^2\le-7/4 for 0<a<1. Summation uses the Sobolev triangle inequality, not a locality assertion for fractional derivatives.

It remains to estimate a fixed interior region. Its low-order product is controlled by a local H^6 norm, and the lower top-order Leibniz terms by a local H^{k-1} norm. Choose a slightly larger fixed ball and a cutoff equal to one on the region. The usual interpolation inequality on these two balls gives, for each m<k,

For completeness, this follows from compactness of the embedding H^k(B_{2R})\to H^m(B_R): a failure after normalizing the left side would give a bounded sequence whose L^2 norm tends to zero but whose H^m norm does not. Local extension and Fourier truncation give this compactness. By (5.5), the H^k norm on the fixed larger ball is bounded by C_R\|v\|. Choose the tail radius first and \delta next. This proves (5.14), including the interior contribution of the regularized torus term.□

Proof. The two squared free norm rates, on both kinds of space, are

For the first periodic norm this also uses L_s^{-2}=e^{-s}L^{-2}; thus the zero mode has exactly the same first rate. At order k, the leading potential term contributes at most 2B\|\nabla^kv\|_2^2. This constant does not grow with k, because the same pointwise operator acts on each ordered derivative component. All combinatorial constants occur only in the lower terms covered by Lemma 5.2. The entire fractional potential term is also covered there. Consequently, writing N_0,N_k for the two norm components, one obtains

Choose \varepsilon after the fixed k, and apply Young’s inequality. Condition (5.13) leaves a positive c.

This computation does not demand two additional derivatives of the initial datum. On a short step pull the mild solution back by its free propagator. Strong continuity and the bounded potential give first variation -i\mathcal V_*v in the relevant space, while the two free squared norms acquire exactly the exponential factors in (5.20). Taking the norm variation gives the displayed inequality for arbitrary mild solutions. Equivalently one can first use smooth data, integrate the inequality, and pass to general data by the uniform mild-solution bounds. This proves (5.19) as well.□

Proof. We first obtain a statement about the bounded time-one operator T=S(1), rather than deducing decay from a generator spectral gap. If v_j\rightharpoonup0 in Y and \|v_j\|\le1, then for every fixed s the functions S(s)v_j converge strongly to zero in L^2(B_{R_0}). Indeed they converge weakly in Y, and the restriction from Y to that L^2 space is compact by (5.5) and local Sobolev compactness. Slab bounds and dominated convergence in (5.19) give

The same conclusion holds on the complexification, by adding the two real energy inequalities.

Let F_N be increasing finite-dimensional orthogonal projections converging strongly to the identity. Formula (5.24) implies \limsup_N\|T(I-F_N)\|\le e^{-c/2}: otherwise unit vectors in the complements of F_N would give a weakly null counterexample. For some N we therefore have

Outside |z|\le q, invert z-B_0 and write K_0=UV through a finite-dimensional space. The resolvent of T is determined by the finite matrix I-V(z-B_0)^{-1}U. Its determinant is analytic and tends to one as |z|\to\infty. Thus the spectrum there consists of isolated eigenvalues with finite-dimensional generalized eigenspaces; there are only finitely many outside any larger disk. This also proves the usual spectral decomposition in this region by the finite-matrix resolvent formula.

Choose r with

and with |z|=r disjoint from the spectrum. Every spectral subspace of T outside this circle is preserved by S(s), because S(s) commutes with the time-one resolvent and its contour integrals. The restricted continuous group on a finite-dimensional space is a matrix exponential. Each of its generator eigenvalues \lambda satisfies \operatorname{Re}\lambda=\log|z|>\log r>-1/32 for the corresponding time-one eigenvalue z.

Its generator equations are the distributional equations for (5.1), as follows by differentiating (5.10) against compactly supported smooth tests. Local elliptic regularity gives smoothness of eigenvectors and chain vectors. Projection onto any spherical harmonic gives regular radial solutions of the mode equation in Theorem 4.1. The unweighted radial kth derivative is in L^2(r^{11}dr): along each ray \partial_r^k is a contraction of the ordered tensor \nabla^k, and angular projection is bounded on L^2 of each sphere. Smoothness of the projection at the origin follows by projecting the Taylor polynomials, using the harmonic decomposition of homogeneous polynomials. The top-derivative criterion in Theorem 4.1 excludes the fast exterior branches, including in a generalized eigenvector after subtraction of the differentiated slow basis. That theorem therefore leaves precisely the symmetry vectors (5.21) and excludes all nontrivial Jordan chains. Completeness of spherical harmonics gives the statement for the original vectors; a nontrivial chain would have a nonzero harmonic projection of its leading vector and hence would contradict the same classification.

Conversely all columns of E lie in Y by (5.6); the time column even has symbol order -2a-2, since its leading slow power cancels. They are generator domain vectors with the asserted eigenvalues. To check the domain claim directly, multiply a column by a smooth cutoff at radius R. Both the discarded symbol tail and the commutator

tend to zero in Y by the same dyadic estimates. The smooth compactly supported approximations thus converge in the generator graph norm. It follows that the entire spectral subspace outside |z|=r is \operatorname{ran}E, with the action in (5.22). Complex conjugation in the complexification preserves its projection, so this projection is the complexification of a bounded real projection E\pi.

On its complement, the spectrum of T lies in |z|\le r. Choose r<r_1<1. The contour formula for powers on |z|=r_1 gives

Bounded propagation for the remaining time in [0,1] proves (5.23), with \eta=-\log r_1>0 and an enlarged C_S. This derivation supplies actual semigroup decay, not only a description of the generator spectrum.□

Proof. We give the compactness and uniqueness argument in detail. Fix M and a sequence L_j\to\infty with initial data v_j(0) bounded in Y_{L_j}. Write L_{j,s}=L_je^{s/2}. Uniform slab bounds and (5.5) give boundedness in local H^k, uniformly for 0\le s\le M. On each fixed ball the equation also bounds \partial_sv_j in H^{k-2}; the drift coefficient is bounded on that ball. Local compactness, interpolation, and Arzelà–Ascoli therefore give a subsequence converging in C([0,M];L^2(B_R)) for every fixed R, to a function v. One may obtain the stronger convergence in local H^{k-1}, but it is unnecessary here.

The localized functions J_{L_{j,s}}v_j(s) are bounded in Y by (5.8). At each s, every weakly convergent subsequence has the same distributional limit v(s), since \chi(y/L_{j,s})=1 on each fixed ball for all large j. Thus

The limit is weakly continuous in Y: boundedness and its local continuity identify the weak limit along every converging time sequence. As Y is separable, it is strongly measurable and bounded on this slab. The torus coefficients agree with the whole-space coefficients on each fixed ball eventually, so local passage in the equation gives \partial_sv=\mathcal Av in distributions.

This distributional solution is the whole-space mild solution. To see this without a propagation assertion, test against Schwartz functions transported by the free adjoint. Formula (5.9) preserves Schwartz functions, uniformly on compact time intervals. The global L^\infty bound supplied by Y, and the at-most-linear growth of the drift, justify removal of spatial cutoffs in these tests. Integration in time gives

as tempered distributions. The integral is a Bochner integral in Y by boundedness of \mathcal V and strong measurability. The distributional identity is therefore an identity in Y and is equivalent to (5.10). Uniqueness gives

Finite dimensionality, (5.29), and C_L\to I now imply convergence of the initial and final coordinates to \pi v(0) and \pi S(M)v(0), respectively. Their difference in (5.27) tends to zero by (5.22). This proves convergence in operator norm: failure would furnish a sequence of unit initial vectors witnessing a fixed positive lower bound, to which exactly the preceding subsequence argument applies.

For the contraction, take such a sequence with \|v_j(0)\|_{Y_{L_j}}=1 and \pi_{L_j}v_j(0)=0. The limit satisfies \pi v(0)=0, so (5.23) gives

The constants in this estimate are independent of the fixed slab length M. Apply the integrated observation inequality (5.19) before passing to a local limit. The full initial norm is still one in its first term. Local uniform convergence passes only the observation integral to the limit, giving

Here C_1 uses only the fixed observation embedding, C_J, C_S, and the energy constant; it does not contain the slab bound C_M. Any norm that escaped to spatial infinity remains charged in the first term e^{-cM}. In particular a zero local limit causes no loss in this estimate.

The quantity D(M) tends to zero as M\to\infty; when c=2\eta its integral is Me^{-cM}, and otherwise it is (e^{-2\eta M}-e^{-cM})/(c-2\eta). Let C_P uniformly bound \|I-P_L\| for sufficiently large L. Choose M_0 so that C_P\sqrt{D(M)}<1/16 for every M\ge M_0. For each such fixed M, a failure of (5.28) along arbitrarily large L would supply unit stable vectors contradicting (5.31). This proves the required order of choices: first M, then L. The stated bound on the map from E_L follows from the uniform step and projection bounds. Restarting a step uses the same formulas with its current scale as the initial scale.

Finally identify v\in Y_L with v(Lx) on \mathbb T^{12}. On a compact range of positive L, these norms are equivalent to H^k, continuously in L. The free step on this fixed torus has Fourier phases \exp(-i|n|^2L^{-2}(1-e^{-s})) and scalar factor e^{(-a+ib)s}. Dominated convergence of the weighted Fourier sum proves strong continuity for continuously varying inputs. The cutoff coefficients are smooth and continuous in L on each finite slab; successive approximation in the integral equation transfers this continuity to the full propagator. Dilation is strongly continuous on Y, which gives the same assertion for J_L,E_L and then for \pi_L,P_L by (5.26). This completes the proof.□

Proof. Identify the expanding spaces by dilation and use the free Fourier propagator from Section 5. On a fixed slab its norm is bounded uniformly in the starting scale, as are the linearized multiplication operators. On sufficiently short substeps, the integral equation is a contraction on a small slab-norm ball by (6.2); the forcing is bounded by (6.1). The short-substep length can be chosen independently of L. Iteration and Gronwall’s inequality give the displayed slab bound for sufficiently small \delta and large L_0.

At zero initial perturbation the same integral equation gives O_M(L^{-2-a}), proving (6.4). Differences of two solutions are at most C_M times their initial difference. Subtract the linearized solution from this difference. Its remaining source is \mathcal E_s(v)-\mathcal E_s(\widetilde v); use (6.2) and the slab bound. Integration over the fixed slab gives (6.5). These arguments also prove uniqueness.□

Proof. For a constant K\geq1 to be fixed, use the sequence norm

Define a map by the right side of (6.6) for w_{n+1}, resetting its zeroth coordinate to the prescribed w_0, and by

for the coordinate sequence. The geometric weights make this sum convergent even in the phase coordinate, where D_* equals one.

For differences of sequences, the linear terms in the forward map cost at most 2(1/8+C_M/K) in the sequence norm. Choose K so that this is less than 1/2. The weighted sum in (6.9) costs at most 2K times the weighted supremum of the forcing difference. By (6.5), the remaining Lipschitz costs tend to zero as the sequence ball is made small and then L is made large. The same is true of e_j, uniformly in j. Consequently the complete map has a contraction constant strictly below one, uniformly for all sufficiently large L.

At the zero sequence the residual forcing is O_M(L_n^{-2-a}). Condition (6.3) bounds its weighted sequence norm by C_M L^{-2-a}. Thus a fixed small ball is invariant if w_0 is small, and the contraction theorem yields a unique sequence in that ball. Its norm is at most C(\|w_0\|+L^{-2-a}). The fixed point satisfies (6.7), since consecutive backward sums telescope. Define G_L(w_0)=u_0. This proves (6.8) and the decay at step endpoints. Lemma 6.1 gives the decay between endpoints.

We spell out the continuity assertion because the free Schrödinger flow is not continuous in operator norm as its time varies. At every fixed index, and after dilation to a fixed torus, the free Fourier formula is strongly continuous in the starting scale and in the input. The polynomial integral equation preserves this continuity on each finite slab. Cutoffs and coordinate maps are likewise strongly continuous: prove this first for smooth functions, then use density and their locally uniform bounds. The finite matrix normalizing \pi_L stays invertible and varies continuously.

Start the sequence contraction at zero. Each iterate is continuous at every fixed coordinate: in (6.9), the summands are continuous and the geometric tail is uniform. The iterates converge uniformly in the sequence norm on the allowed parameter set. Their zeroth coordinate is therefore continuous. This proves the assertion without differentiating the graph or taking an operator-norm limit of Schrödinger propagators.□

Proof. Substitution of (6.10) into (6.11) gives

Taylor expansion of the smooth reference profile yields

The profile symbol bounds and their dyadic norm estimates apply to its first two parameter derivatives: translations preserve the tail bounds, and the dilation derivative y\cdot\nabla preserves their order. Hence the Taylor remainder is O_{Y_L}(|d'|^2) uniformly in L. The translated cutoff differs by O(\sqrt{T_0}|\zeta|), with the same scaled derivative estimates on its annulus; multiplication by the profile gives the additional stated error. For small parameters the supports remain strictly inside one physical cell.

The spectral time vector is (a-ib+y\cdot\nabla/2)Q. The missing ibQ contributes b\beta to the phase coordinate, proving (6.12). On the compact parameter ball with T_0 fixed, translation and rescaling from physical H^k to Y_L have uniformly bounded norms. They are strongly continuous, which proves the remaining assertions. Only the smooth reference has been differentiated; no extra derivative is required of nearby H^k data.□

Proof of Theorem 1.1. Choose a and k as above. We seek a zero of the continuous map

The matrix C in Lemma 6.3 is invertible. At every current scale, \pi_LE_L=I and (I-P_L)E_L=0 exactly. Consequently the stable argument of G_L contains only the error in that lemma; varying L introduces no leading stable component. By (6.8), on |d'|\leq\rho we have

where \varepsilon_{\mathrm{data}}=C(T_0) \|u_{\mathrm{in}}-u_{\mathrm{ref}}\|_{H^k}.

Let c_C>0 be the smallest singular value of C. First choose \rho small enough for the quadratic term in (6.14) to be less than c_C\rho/4 and for the graph domain requirements to hold. Next choose T_0 sufficiently small that all scales are in the range of Proposition 6.2, and the two T_0 terms together are less than c_C\rho/4. Finally choose a positive physical H^k radius so that the data term is less than c_C\rho/4. The graph is then defined throughout the closed parameter ball, and on its boundary

The straight-line homotopy to C has no boundary zero. Brouwer degree therefore gives a zero of (6.13) for every datum in this open H^k ball.

At such a zero the initial perturbation lies on the graph. The resulting similarity solution exists for all s\geq0 and its perturbation tends uniformly to zero, by Proposition 6.2 and the uniform Y_{L_s}\to L^\infty embedding. Undoing the similarity variables gives the physical solution on 0\leq t<T:

On each finite similarity slab, the norms are equivalent after dilation to the physical H^k norm. The integral equation and polynomial algebra therefore give u\in C([0,T);H^k)\cap C^1([0,T);H^{k-2}). The usual local uniqueness follows directly from the physical Schrödinger integral equation and its H^k difference estimate. At the center, Q_s(0)=Q(0)\ne0, and hence

Since H^k(\mathbb T^{12}) embeds continuously into C^0, the solution cannot extend continuously in H^k through T. This proves the theorem.□

Proof of Corollary 1.2. Fix the pair p,k supplied by Theorem 1.1 and take \alpha>k+6. Independence and the finite Gaussian second moment give

up to the fixed Fourier normalization. Thus the series converges as an H^k-valued Gaussian random variable. Its finite Fourier segment has a strictly positive density on the whole finite-dimensional complex coefficient space.

For completeness, let B_{H^k}(f,\varepsilon) be any nonempty ball. Choose a finite Fourier projection P_N such that \|(I-P_N)f\|_{H^k}<\varepsilon/4 and

Markov’s inequality gives positive probability that the random tail has norm less than \varepsilon/4 after increasing N further if necessary. Independently, the finite segment belongs to the ball of radius \varepsilon/2 about P_Nf with positive probability. The intersection lies in B_{H^k}(f,\varepsilon), so the law has full support in H^k.

Apply this to the open blowup neighborhood in Theorem 1.1. It has positive probability for every \alpha>k+6. Since 8<k<\alpha-6, the exponent m=k lies in the interval in the universal assertion of Corollary 1.2. The local solution fails its required global continuation with positive probability. The fixed pair also satisfies s_{\mathrm{cr}}=6-2/(p-1)>1. Hence no finite threshold for this pair, and therefore no universal assertion of the stated form, is possible.□