Blowup dynamics for smooth data equivariant solutions to the critical Schrödinger map problem

Abstract

We consider the energy critical Schrödinger map problem with the 2-sphere target for equivariant initial data of homotopy index k=1. We show the existence of a codimension one set of smooth well localized initial data arbitrarily close to the ground state harmonic map in the energy critical norm, which generates finite time blowup solutions. We give a sharp description of the corresponding singularity formation which occurs by concentration of a universal bubble of energy.

Appendices

Appendix A: L ² coercivity estimates

This appendix is devoted to the derivation of L ² weighted coercivity estimates for the operator H and its iterate H ², which generalize related results in [32, 36].

1.1 A.1 Hardy inequalities

Lemma A.1

(Logarithmic Hardy inequalities)

∀R>2, $\forall v\in\dot{H}^{1}_{rad}(\mathbb{R}^{2})$ and γ>0, there holds the following controls:

(A.1)

(A.2)

(A.3)

(A.4)

(A.5)

If $\int_{y\leq1}\frac{|v|^{2}}{y^{2}}<+\infty$, then:

$$ \int_{y\leq2 R} \frac{|v|^2}{y^2}\lesssim \log R \int_{y\leq 2}|v|^2+(\log R)^2\int _{y\leq2R}|\nabla v|^2. $$

(A.6)

Proof

Let v smooth and radially symmetric. First recall from the one dimensional Sobolev embedding H ¹(1≤y≤2)↪L ^∞(1≤y≤2) that

$$\bigl|v(1)\bigr|^2\lesssim\int_{1\leq y\leq2}\bigl(|v|^2+| \partial_yv|^2\bigr). $$

Let $f(y)=-\frac{\mathbf{e}_{y}}{y(1+\mathopen{|}\log y|)}$ so that

and integrate by parts to get:

and (A.1) follows. To prove (A.2), let γ>0 and

$$f(y)=-\frac{\mathbf{e}_y}{y^{\gamma+1}(1+\log y)^2} $$

so that for y≥1:

$$\nabla\cdot f= \frac{1}{y^{\gamma+2}(1+\log y)^2} \biggl[\gamma +\frac{2}{(1+\log y)^3} \biggr] \geq\frac{\gamma}{y^{\gamma +2}(1+\log y)^2}. $$

We then integrate by parts to get:

and (A.2) follows. To prove (A.3), we have: ∀y∈[1,R],

$$\bigl|v(y)\bigr|=\biggl|v(1)+\int_{1}^yv'(r)dr \biggr|\lesssim\bigl|v(1)\bigr|+R \biggl(\int_{1\leq y\leq R} \frac{|\nabla v|^2}{y^2} \biggr)^{\frac{1}{2}}, $$

and (A.3) follows. Similarly,

$$\bigl|v(y)\bigr|=\biggl|v(1)+\int_{1}^yv'(r)dr \biggr|\lesssim\bigl|v(1)\bigr|+ \biggl(\int_{y\leq R} |\nabla v|^2 \biggr)^{\frac{1}{2}}\sqrt{\log R}, $$

and (A.4), (A.5) follow by squaring this estimate and integrating in R. Finally, (A.6) follows from (A.5) by summing over dyadic R-intervals. □

1.2 A.2 Sub-coercivity estimates

In this section we establish weighted sub-coercive estimates for the operators H and H ² which will play a key role in the proof of the coercive estimates under additional orthogonality conditions.

Lemma A.2

(Sub-coercivity for H)

Let u be a function with the property

$$ \int\frac{|u|^2}{y^4{\approx}(1+\mathopen {|}\log y|)^2}+\int\bigl|\partial_y (Au)\bigr|^2+\int\frac{|Au|^2}{y^2(1+y^2)}<+\infty $$

(A.7)

then

(A.8)

(A.9)

Proof of (A.8)

First observe from (2.12) that:

$$\int(Hu)^2=\int\bigl(A^*Au\bigr)^2=\bigl(Au, \tilde{H}(Au)\bigr)\approx\int|\partial_yAu|^2+\int \frac{|Au|^2}{y^2(1+y^2)}. $$

Let now a smooth cut off function χ(y)=1 for y≤1, χ(y)=0 for y≥2, and consider the decomposition:

$$u=u_1+u_2=\chi u+(1-\chi) u. $$

Then from (A.1):

(A.10)

For the first term, we rewrite:

where in the last step we integrated by parts the quantity:

We hence conclude from |Z(y)−1|≲y for y≤1 and the Hardy inequality (A.1) applied to $\frac{u_{1}}{ y}\in H^{1}_{rad}$ from (A.7) that:

$$ \int\frac{|Au|^2}{y^2(1+y^2)}\gtrsim\biggl [\int \frac {|u_1|^2}{y^4(1+\mathopen{|}\log y|)^2}-C\int_{y\leq2}|u|^2 \biggr]. $$

(A.11)

Similarily we estimate:

(A.12)

where we applied the weighted Hardy (A.2) to yu ₂ with γ=4 and integrated by parts for the last step using the bound $|1+Z(y)|\lesssim\frac{1}{y^{2}}$ for y≥1. Equations (A.10), (A.11) and (A.12) imply:

$$ \int\frac{|Au|^2}{y^2(1+y^2)}+\int \bigl|\partial_y (Au)\bigr|^2\gtrsim\biggl[\int\frac{|u|^2}{y^4(1+\mathopen{|}\log y|)^2}-C\int \frac{|u|^2}{1+y^5} \biggr]. $$

(A.13)

This implies using again (A.1):

(A.14)

Finally, examining the expression

$$\partial_y (A u)=\partial_y \biggl(- \partial_y u+\frac{Z}{y} u\biggr) $$

we also obtain

$$\int\frac{|\partial^2_y u|^2}{(1+\mathopen{|}\log y|)^2} \lesssim \int \bigl|\partial_y (Au)\bigr|^2+ \int\frac {|Au|^2}{y^2(1+y^2)}+\int\frac{|u|^2}{1+y^5} $$

which together with (A.13), (A.14) concludes the proof of (A.8) and of Lemma A.2. □

Lemma A.3

(Weighted sub-coercivity for H)

Let u be a function with the property

(A.15)

then

(A.16)

Proof of Lemma A.3

Let χ(y) be a smooth cut-off function with support in y≥1 and equal to 1 for y≥2. We first consider

We now observe that for k≥0 and y≥1

$$\partial_y^k V(y)= \partial_y^k (1)+ O\bigl(y^{-2-k}\bigr). $$

We may thus apply twice the Hardy inequality with sharp constant (A.8) with γ=6 and get for a sufficiently large universal constant R:

and hence the bound away from the origin:

The control of the third derivative away from the origin follows from:

Near the origin, we first observe from (A.15) that

$$ \int\frac{|Au|^2}{y^2(1+\mathopen{|}\log y|^2)}<+\infty. $$

(A.17)

We now observe from

$$\partial_y\bigl(\log(\varLambda\phi)\bigr)=\frac{Z}{y} $$

that

$$A^*f=\partial_yf+\frac{1+Z}{y}f=\frac{1}{y\varLambda\phi} \partial_y(y\varLambda\phi f), $$

and thus from (A.17):

$$ Au(y)=\frac{1}{y\varLambda\phi(y)}\int_0^y \tau\varLambda\phi(\tau)Hu(\tau)d\tau. $$

(A.18)

We then estimate from Cauchy-Schwarz and Fubini:

and thus:

$$ \int_{y\le1}\frac{|Au|^2}{y^6(1+\mathopen {|}\log y|^2)}\lesssim \int_{y\le 1}\frac{|Hu|^2}{y^4(1+\mathopen{|}\log y|^2)} $$

(A.19)

which implies from A ^∗(Au)=Hu:

$$ \int_{y\le1}\frac{|\partial _yAu|^2}{y^4(1+\mathopen{|}\log y|^2)}\lesssim \int_{y\le1}\frac{|Hu|^2}{y^4(1+\mathopen{|}\log y|^2)}. $$

(A.20)

We now rewrite near the origin:

which implies using (A.19), (A.20),

(A.21)

(A.22)

Let now $\zeta=(1-\chi)^{\frac{1}{2}}$, then ζu satisfies (A.7) from (A.15), (A.19), (A.20), and we thus obtain from (A.8):

(A.23)

Injecting (A.21), (A.22) into (A.23) yields the expected control at the origin. This concludes the proof of (A.16) and Lemma A.3. □

We combine the results of Lemma A.2 and Lemma A.3 and obtain:

Lemma A.4

(Sub-coercivity for H ²)

Let u be a radially symmetric function with

(A.24)

then

(A.25)

1.3 A.3 Coercivity of H ²

We are now in position to derive the fundamental coercivity property of H ² at the heart of our analysis:

Lemma A.5

(Coercivity of H ²)

Let M≥1 be a large enough universal constant. Let Φ _M be given by (4.1). Then there exists a universal constant C(M)>0 such that for all radially symmetric function u satisfiyng (A.24) and the orthogonality conditions

$$(u,\varPhi_M)=0,\qquad(Hu,\varPhi_M)=0, $$

there holds:

(A.26)

Proof of Lemma A.5

We argue by contradiction. Let M>0 fixed and consider a normalized sequence u _n

(A.27)

satisfying the orthogonality conditions

$$ (u_n,\varPhi_M)=0, \qquad(u_n,H \varPhi_M)=0, $$

(A.28)

and

$$ \int\bigl|H^2u_n\bigr|^2\approx \int\frac{|A Hu_n|^2}{y^2(1+y^2)}+\int \bigl|\nabla(A H u_n)\bigr|^2\leq \frac{1}{n}. $$

(A.29)

The normalization condition implies that the sequence u _n is uniformly bounded in $H^{1}_{loc}$. Moreover, as follows from Lemma A.3, for any smooth cut-off function ζ vanishing in a neighborhood of y=0, the sequence ζu _n is uniformly bounded in $H^{3}_{loc}$. As a consequence, we can assume that u _n and ζu _n weakly converge in $H^{1}_{loc}$ and $H^{3}_{loc}$ to u _∞ and ζu _∞ respectively. Moreover, u _∞ satisfies the equation

$$AHu_\infty=0 $$

away from y=0. Integrating the ODE

$$AHu_\infty=-\varLambda\phi\partial_y \biggl(\frac{Hu_\infty }{\varLambda \phi } \biggr)=0 $$

we obtain that Hu _∞(y)=αΛϕ(y) away from y=0. The function u _∞ can be written in the form

Using the condition $u_{\infty}\in H^{1}_{loc}$ we can conclude that γ=0. Passing to the limit in the orthogonality conditions, using that u _n converges to u _∞ weakly in $H^{1}_{loc}$, we conclude that u _∞ satisfies

$$(u_\infty,\varPhi_M)=0,\qquad(u_\infty,H \varPhi_M)=0. $$

We may therefore determine the constants α,β using (4.1), (4.3) which yield α=β=0 and thus u _∞=0.

The sub-coercivity bound (A.25) together with (A.29) ensures:

Coupling this with the normalization condition we obtain that

$$\int\zeta\biggl[\frac{|H u_n|^2}{1+y^5}+\int\frac{(\partial_y u_n)^2}{1+y^8}+ \int \frac{|u_n|^2}{1+y^{10}} \biggr]\ge c $$

for some positive constant c>0 and a smooth cut-off function ζ vanishing for y<ϵ and y>ϵ ⁻¹. The size of ϵ depends only on the universal constant C. Since u _n weakly converges to u _∞ in H ³ on any compact subinterval of y∈(0,∞), we can pass to the limit to conclude

$$\int\zeta\biggl[\frac{|H u_\infty|^2}{1+y^5}+\int\frac{(\partial_y u_\infty)^2}{(1+y^8)} + \int \frac{|u_\infty|^2}{(1+y^{10})} \biggr]\ge c. $$

This contradicts the established identity u _∞≡0 and concludes the proof of Lemma A.5 modulo the additional bound for

$$\int\frac{|\partial^4_y u|^2}{(1+\mathopen{|}\log y|^2)} $$

claimed in (A.26). To control this term we simply observe that for y≥1:

$$H^2 u =H\biggl(-\varDelta +\frac{V}{y^2}\biggr) u = \partial_y^4 u + O \Biggl(\sum _{i=0}^3 \frac{|\partial_y^i u|}{y^{4-i}} \Biggr) $$

and the desired estimate easily follows from the already established bounds for lower derivatives. For y≤1 we write

We further note that

The estimate for y≤1 now follows from the bounds for ∂ _y Hu, Hu and the coercivity estimate (A.19) for Au. □

1.4 A.4 Coercivity of H

We complement the coercivity property of the operator H ², established in the previous section, by the corresponding statement for the operator H, which follows from standard compactness argument. A complete proof⁵ is given in [32]:

Lemma A.6

(Coercivity of H)

Let M≥1 fixed. Then there exists c(M)>0 such that the following holds true. Let u∈H ¹ with

$$(u,\varPhi_M)=0 $$

and

$$ \int\frac{|u|^2}{y^4(1+\mathopen{|}\log y|)^2}+\int \bigl|\partial_y (Au)\bigr|^2<+\infty, $$

(A.30)

then:

Appendix B: Interpolation estimates

We derive interpolation bounds in the bootstrap regime of Proposition 4.2 and in the regime of parameters described by Remark 4.3. We recall the notation:

$$\mathbf{w}=\left| \begin{array}{@{}l@{}}\alpha\\\beta\\\gamma \end{array} \right.= \mathbf{w}^\perp+\gamma e_z $$

and the norms $\mathcal{E}_{1}, \mathcal{E}_{2}, \mathcal{E}_{4}$, introduced in (4.7), together with their bootstrap bounds (4.13), (4.14), (4.15):

Note that the $\mathcal{E}_{2}$ bound is better than the interpolated bound between $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$. This bound will be essential to derive suitable interpolation estimates, Lemma B.1, which are used to close the control of the nonlinear terms in Lemma 4.8, see Appendix C.

2.1 B.1 Regularity at the origin

The use of the coercivity bounds in Appendix A and the interpolation estimates below requires establishing a priori regularity of the Schrödinger map u, expressed in Frenet basis, at the origin. We will show that the smoothness of the map $u(t): {\mathbb{R}}^{2}\to{\mathbb {S}}^{2}\subset{\mathbb{R}}^{3}$ implies boundedness of the quantities

$$\biggl|\frac{\mathbf{w}}{y}\biggr|, \ |{ \partial_y \mathbf{w}} |,\ \biggl|\frac{A\mathbf{w}^\perp}{y} \biggr|, \ \biggl|\frac{H \mathbf{w}^\perp}{y}\biggr|, \ \biggl |\frac{AH \mathbf{w}^\perp}{y}\biggr|, \ \bigl|{\partial_y A H\mathbf{w}^\perp} \bigr|. $$

We first consider the expression $\nabla u = (\partial_{y} u, \frac{1}{y} \partial_{\theta}u)$ which, as long as u is smooth, is bounded at the origin. Using Lemma 2.1 we compute this in terms of the coordinates $(\hat{\alpha},\hat{\beta},\hat{\gamma})$ in the Frenet basis of u:

This immediately implies boundedness of

$$\biggl|\frac{\mathbf{w}}{y}\biggr|, \ |{ \partial_y \mathbf{w}} |. $$

Similarly, computing the expression Δu+|∇Q|² u from (2.19)

gives us the boundedness of |H w ^⊥| and |Δγ|. Using cartesian coordinates x=(x ₁,x ₂) on ℝ² we examine the expressions

$$\lim_{x_2=0, x_1\to0} \nabla_{x_1} u= \lim_{x_1=0, x_2\to0} \nabla_{x_1} u. $$

Computing this in polar coordinates and relative to the Frenet basis we immediately obtain the relations

$$\partial_y\hat{\alpha}=\frac{1}{y} \hat{\alpha},\qquad \partial_y\hat{\beta}=\frac{1}{y} \hat{\beta} $$

in the limit as y→0. This immediately gives the boundedness of

$$\biggl|\frac{A\alpha}{y}\biggr|, \ \biggl| \frac{A\beta}{y} \biggr|. $$

The remaining bounds can be shown by similar arguments. We omit the details.

2.2 B.2 Interpolation bounds for w

We now turn to the proof of interpolation estimates for w in the bootstrap regimes which are used all along the proof of Proposition 4.2.

Lemma B.1

(Interpolation estimates for w ^⊥)

There holds:

(B.1)

(B.2)

(B.3)

(B.4)

(B.5)

(B.6)

(B.7)

(B.8)

(B.9)

(B.10)

(B.11)

(B.12)

(B.13)

Proof of Lemma B.1

The estimate (B.6) follows from the $\mathcal{E}_{1}$ bound:

$$\bigl\|\mathbf{w}^\perp\bigr\|_{L^{\infty}}^2\lesssim \bigl\| \partial_y \mathbf{w}^\perp\bigr\|^2_{L^2}+ \biggl\|\frac{\mathbf{w}^\perp}{y}\biggr\|^2_{L^2}\lesssim\delta \bigl(b^* \bigr). $$

The estimate (B.1) follows from H ² coercivity of Lemma A.5 and the $\mathcal{E}_{4}$ bound. For i=0,1,2 (B.3) is easily implied by Lemma A.6 and the $\mathcal{E}_{2}$ bound. For i=3 we split the integral at $y= B_{0}^{20}$ and estimate the inner contribution using Lemma A.5 and the $\mathcal{E}_{4}$ bound, and the outer by interpolating between the $\mathcal{E}_{4}$ and $\mathcal{E}_{2}$ bounds:

Similarly, the estimate (B.4) follows by splitting the integral at y=B ₀ and using the $\mathcal{E}_{4}$ and $\mathcal{E}_{2}$ bounds for the inner and outer regions respectively.

To obtain (B.5) we write

$$\partial_y H\alpha= -A H\alpha+ \frac{Z}{y} H\alpha, $$

which, using the $\mathcal{E}_{2}$ and $\mathcal{E}_{4}$ bounds, implies with B∈[B ₀,2B ₀]

To prove (B.11) and (B.12) let a∈[1,2] such that

$$\bigl|\partial_y\mathbf{w}^\perp(a)\bigr|^2\lesssim\int _{1\leq y\leq 2}|\partial_y\mathbf{w}|^2\le C(M) \mathcal{E}_4, $$

then for y≤1:

$$\bigl|\partial_y\mathbf{w}^\perp\bigr|\lesssim\bigl|\partial_y \mathbf{w}^\perp(a)\bigr|+\int_y^a\bigl| \partial_{yy}\mathbf{w}^\perp\bigr|dy\le C(M) \sqrt{ \mathcal{E}_4}\lesssim b^2, $$

and for y≥1:

where in the last step we split the integral at $y=B_{0}^{2}$ and use $\mathcal{E}_{2}$ bound for the inner and $\mathcal{E}_{1}$ for the outer parts. Next, we have from w ^⊥(0)=0:

$$\biggl\|\frac{\mathbf{w}^\perp}{y}\biggr\|_{L^{\infty}(y\leq 1)}\lesssim\bigl\|{ \partial_y\mathbf{w}^\perp}\bigr\|_{L^{\infty}(y\leq1)} $$

and

$$ \biggl\|\frac{\mathbf{w}^\perp}{y}\biggr \|^2_{L^{\infty}(y\geq 1)}\lesssim\int_{y\geq1} \frac{|\partial_y\mathbf{w}^\perp|^2}{y^2}+\int_{y\geq1} \frac {|\mathbf{w}^\perp|^2}{y^4}. $$

(B.14)

The estimates (B.11) and (B.12) now easily follow.

The estimate (B.8) follows directly from (A.19).

For (B.7):

$$\|A\alpha\|_{L^{\infty}}^2\lesssim\biggl(\int\frac{|A\alpha |^2}{y^2} \biggr)^{\frac{1}{2}} \biggl(\int|\partial_yA\alpha|^2 \biggr)^{\frac{1}{2}} \lesssim b^2\mathopen{|}\log b|^9, $$

where in the last step we used the coercivity of H of Lemma A.6, split the first integral at $y=B_{0}^{2}$ and used the $\mathcal{E}_{2}$ bound for inner and the $\mathcal{E}_{1}$ for the outer parts.

For (B.9), we recall from (A.18):

$$H\alpha=A^*A\alpha\quad \mbox{and thus}\quad A\alpha=\frac {1}{y\varLambda \phi}\int _0^y z \varLambda\phi(H\alpha) dz. $$

This yields for y≤1:

Similarily,

$$AH\alpha=\frac{1}{y\varLambda\phi}\int_0^y z\varLambda \phi\bigl(H^2\alpha\bigr) dz $$

yields for y≤1:

$$ \bigl|\tilde{H}A\alpha(y)\bigr|= \bigl|AH\alpha(y)\bigr|\lesssim\frac{1}{y^2} \biggl(\int|H^2(\alpha)|^2 \biggr)^{\frac{1}{2}} \biggl( \int_0^y \frac {z^4}{z}dz \biggr)^{\frac{1}{2}}\lesssim\sqrt{\mathcal{E}_4}. $$

(B.15)

This implies for y≤1:

Similar estimates can be shown for β and the last term in (B.9). Let now a∈[1,2] be such that

$$|H\alpha|^2(a)\lesssim\int_{1\leq y\leq2}|H \alpha|^2\lesssim\mathcal{E}_4, $$

then from

$$ Af=-\varLambda\phi\partial_y \biggl( \frac{f}{\varLambda\phi} \biggr), $$

(B.16)

there holds for y≤1:

$$\bigl|H\alpha(y)\bigr|\leq\varLambda\phi(y) \biggl[ \frac {|H\alpha |(a)}{\varLambda \phi(a)}+\int _y^a\frac{|AH\alpha|}{\varLambda\phi}dz \biggr]\lesssim |y|\bigl(1+\mathopen{|}\log y|\bigr)\sqrt{\mathcal{E}_4}, $$

where in the last step we used the coercivity of A ^∗. The bound (B.10) follows.

For (B.13):

Now

The argument for the other terms is similar, and (B.13) is proved. □

Lemma B.2

(Interpolation bound for γ=w ^z)

There holds:

(B.17)

(B.18)

(B.19)

(B.20)

(B.21)

(B.22)

(B.23)

(B.24)

(B.25)

(B.26)

Proof of Lemma B.2

Recall the normalization relation:

$$(1+\hat{\gamma})^2+\hat{\alpha}^2+\hat{ \beta}^2=1. $$

We expand to get:

$$ 2\gamma=- \bigl[\bigl(2{\tilde{\gamma}}_0+ \tilde{\beta}_0^2\bigr)+\tilde{\alpha}_0^2+ \alpha(\alpha+2\tilde{\alpha}_0)+\beta(\beta+\tilde{ \alpha}_0) \bigr]-(\gamma+{\tilde{\gamma}}_0)^2. $$

(B.27)

Note also by construction (3.4) that the leading order $(2{\tilde{\gamma}}_{0}+\tilde{\beta}_{0}^{2})$ contribution to γ is cancelled on y≤B ₁, and we obtain:

from (B.6) and (B.11) for w ^⊥ and the corresponding bounds for $\widetilde{\mathbf{w}}_{0}$. The remaining estimates in (B.17), (B.18), (B.19), (B.21), (B.22) and (B.23) are obtained similarly from (B.27) and the corresponding statements for w ^⊥. We omit the details.

The estimate (B.20) for y≥1 follows from (B.2) and the challenge here is the behavior of the integrand at the origin. To treat we write

$$A \partial_y \gamma= -\partial_y^2 \gamma+ \frac{Z}{y} \partial_y \gamma. $$

Therefore for y≤1:

The most important aspect of this formula is the last term containing A w ^⊥ and providing the necessary vanishing at the origin from Lemma B.1:

We now turn to the proof of (B.24). We first estimate in brute force from (3.15), (3.16) and the relation HT ₁=ΛQ:

(B.28)

and similarly:

$$\int\frac{| \widetilde{\mathbf{w}}_0|^2}{1+y^4} \lesssim b^2+\int _{y\leq2B_1} \biggl|\frac{b\mathopen{|}\log y|}{1+y}+\frac{b^3 y^3}{(1+y^2)\mathopen{|}\log b|}\biggr|^2 \lesssim b^2\mathopen{|}\log b|^3. $$

We now use this and the estimates of Lemma B.1 and the properties of the profile $\widetilde{\mathbf{w}}_{0}$ to estimate:

We then estimate from (B.12), (B.28):

$$\int\bigl|\nabla\mathbf{w}^\perp\bigr|^2|\nabla\widetilde {\mathbf {w}}_0|^2\lesssim\bigl\|\partial_y \mathbf{w}^\perp\bigr\|_{L^{\infty}}^2\sqrt{b}\mathopen{|}\log b|^C\lesssim b^2\mathopen{|}\log b|^2, $$

and using (B.2):

$$\int\bigl|\varDelta \mathbf{w}^\perp\bigr|^2|\widetilde{ \mathbf{w}}_0|^2 \lesssim\sqrt{b}b^2\mathopen{|}\log b|^C\lesssim b^2\mathopen{|}\log b|^2, $$

and thus arrive to the bound:

$$\int|\varDelta \gamma|^2 \lesssim b^2\mathopen{|}\log b|^2+ \int\bigl|\varDelta \mathbf{w}^\perp\bigr|^2\bigl|\mathbf{w}^\perp \bigr|^2+ \int\bigl|\nabla\mathbf{w}^\perp\bigr|^4. $$

From (B.1) we easily see that

$$\int_{y\le1} \bigl|\varDelta \mathbf{w}^\perp\bigr|^2 \bigl| \mathbf{w}^\perp\bigr|^2+ \int_{y\le 1} \bigl|\nabla \mathbf{w}^\perp\bigr|^4\le C(M) \mathcal{E}_4\le b^4. $$

For y≥1 we write

$$\bigl|\varDelta \mathbf{w}^\perp\bigr|^2\lesssim|H\alpha|^2+|H \beta|^2 + \frac {|\mathbf{w}^\perp|^2}{y^4} ,\qquad \bigl|\partial_y \mathbf{w}^\perp\bigr|^4\lesssim \bigl|A\mathbf{w}^\perp\bigr|^4+ \frac{|\mathbf{w}^\perp|^4}{y^4} $$

and estimate

(B.29)

where we used (B.6) and the Gagliardo-Nirenberg inequality

Hence (B.24) follows from:

$$ \int_{y\geq1} \frac{|\mathbf{w}^\perp |^4}{y^4}\lesssim b^2+\delta\bigl(b^*\bigr)\mathcal{E}_2. $$

(B.30)

Indeed, let a cut-off function with ψ(y)=0 for y≤1 and ψ(y)=1 for y≥2. We compute:

where we used that $|Z(y)+1|\lesssim\frac{1}{y^{2}}$ for y≥1 and (B.4). We now use Hölder and Sobolev inequalities to derive:

and (B.30) follows.

The L ^∞ bound (B.25) follows from:

$$|\varDelta \gamma|^2_{L^{\infty}(y\geq1)}\lesssim\biggl(\int _{y\geq 1}\frac{|\partial_y \varDelta \gamma|^2}{y^2} \biggr)^{\frac{1}{2}} \biggl(\int _{y\geq1} |\varDelta \gamma|^2 \biggr)^{\frac{1}{2}} \lesssim b^3\mathopen{|}\log b|^8. $$

It remains to prove (B.26). Using (B.27), we treat the most delicate quadratic term, other terms are treated similarly and are easier to handle. We claim:

$$ \int|\alpha|^2\bigl|\varDelta ^2\bigl( \alpha^2\bigr)\bigr|^2\lesssim\delta\bigl(b^*\bigr) \biggl( \mathcal{E}_4+\frac{b^4}{\mathopen{|}\log b|^2} \biggr). $$

(B.31)

To treat the singularity at the origin, we write

Taking another Laplacian and multiplying by α now yields sufficient vanishing at the origin to close the L ² estimate (B.31) near the origin using the estimates of Lemma B.1. Far out, we write

$$\varDelta \bigl(\alpha^2\bigr) = 2|\partial_y \alpha|^2-2\alpha H\alpha+2 \biggl(\frac {\alpha}{y} \biggr)^2+2(V-1) \biggl(\frac{\alpha}{y} \biggr)^2. $$

The contribution of the last term is easily estimated using the extra decay $|V-1|\lesssim\frac{1}{1+y^{2}}$. We next compute:

Next, we write:

The contribution of all other terms can be treated in a fashion similar to the previous argument.

This concludes the proof of Lemma B.2. □

2.3 B.3 Interpolation bounds for w ₂

We recall that

$$\mathbf{w}_2=\widehat{J}\mathbb{H}\mathbf{w}, \quad \widehat {J}=(e_z+\widehat{\mathbf{w}})\wedge $$

and the decomposition from (4.37):

$$ \mathbf{w}_2=\mathbf{w}_2^0+ \mathbf{w}_2^1, \qquad \mathbf{w}_2^0=R_z \mathbb{H}\mathbf{w}^\perp, \qquad \mathbf{w}_3=R_z \mathbb{A}\mathbf{w}_2^0. $$

(B.32)

From the explicit definition (2.26) of ℍ, we have the formula:

$$ \mathbf{w}^1_2=\mathbf{w}_2^2+ \widehat{\mathbf{w}}\wedge\mathbb{H}\mathbf{w}, \qquad \mathbf{w}_2^2=R_z \left| \begin{array}{@{}l@{}}-2(1+Z)\partial_y\gamma\\0\\0 \end{array} \right.. $$

(B.33)

Lemma B.3

(Interpolation bounds for w ₂)

There holds:

(B.34)

(B.35)

(B.36)

(B.37)

(B.38)

(B.39)

(B.40)

(B.41)

(B.42)

Proof of Lemma B.3

Step 1 :

Estimates for w ₂.

Observe that

$$ \mathcal{E}_2=\int \bigl|\mathbf{w}_0^2\bigr|^2, \qquad\mathcal{E}_4=\int\bigl|R_z\mathbb{H} \mathbf{w}_2^0\bigr|^2=\int\bigl|\mathbb{A}^* \mathbf{w}_3\bigr|^2. $$

(B.43)

We then estimate from (B.2), (B.23), (B.4), (B.19):

which yields (B.35). Now from (B.19):

$$\int\bigl|\mathbf{w}_2^1\bigr|^2 \lesssim\int \frac{|\partial_y\gamma |^2}{1+y^4}+\delta\bigl(b^*\bigr)\|\mathbb{H}\mathbf{w} \|_{L^2}^2\lesssim b^2\mathopen{|}\log b|^2+ \delta\bigl(b^*\bigr)\mathcal{E}_2 $$

which together with (B.43) concludes the proof of (B.34).

From the explicit definition of ℍ:

and (B.36) follows from (B.1) and (B.17). For (B.37) we simply observe

$$\bigl|\mathbf{w}^0_2\bigr|=|R_z\mathbb{H}\mathbf{w}| \le|\mathbb{H}\mathbf{w}|. $$

On the other hand,

$$\bigl|\mathbf{w}_2^1\bigr|\le|\widehat{\mathbf{w}}\wedge H \mathbf{w}|\lesssim|\widehat{\mathbf{w}}|\, |\mathbb{H}\mathbf{w}| $$

and (B.38) follows from the bound $\|\widehat{\mathbf{w}}\| _{L^{\infty}}\le\delta(b^{*})$, see (B.6) and (B.21).

Step 2 :

Estimates for ℍw ₂.

We now turn to the H ⁴ estimates (B.39) and (B.41). Recalling the decomposition (B.32) we first examine the quantity $\mathbb{H}\mathbf{w}_{2}^{0}$:

$$\bigl|\mathbb{H}\mathbf{w}_2^0\bigr|\le\bigl|H^2 \alpha\bigr|+\bigl|H^2\beta\bigr|+\frac{1}{1+y^2} \biggl(| \partial_y H\beta|+\frac{|H\beta|}{y}\biggr). $$

The estimate

$$\int\bigl|\mathbb{H}\mathbf{w}_2^0\bigr|^2\lesssim C(M) \mathcal{E}_4 $$

now easily follow from the definition of $\mathcal{E}_{4}$, the coercivity bounds of Lemma A.5. To obtain the improved bound for $\widehat{J}\mathbb{H}\mathbf{w}_{2}^{0}$ we simply note that

$$\bigl|R_z \mathbb{H}\mathbf{w}_2^0\bigr| \le\bigl|H^2\alpha\bigr|+\bigl|H^2\beta\bigr| $$

and

$$\bigl|\hat{\mathbf{w}}\wedge\mathbb{H}\mathbf{w}_2^0\bigr|\le \delta \bigl(b^*\bigr)\bigl|H \mathbf{w}_2^0\bigr| $$

from which

$$\int\bigl|\widehat{J}\mathbb{H}\mathbf{w}^0_2\bigr|^2 \lesssim\mathcal{E}_4+\frac{b^4}{\mathopen{|}\log b|^2} $$

and thus (B.40) follows from the first part of (B.41).

We now claim:

$$ \int\bigl|\mathbb{H}\mathbf{w}_2^2\bigr|^2 \lesssim\delta\bigl(b^*\bigr) \biggl(\frac {b^4}{\mathopen{|}\log b|^2}+\mathcal{E}_4 \biggr). $$

(B.44)

Indeed, we compute:

$$\mathbb{H}\mathbf{w}_2^2=-2 \bigl[H\bigl((1+Z) \partial_y\gamma\bigr) \bigr]e_y $$

and split the integral:

$$\int\bigl|\mathbb{H}\mathbf{w}_2^2\bigr|^2\leq\int _{y\leq1} \bigl|\mathbb{H}\mathbf{w}_2^2\bigr|^2+ \int_{y\geq1}\bigl|\mathbb{H}\mathbf{w}_2^2\bigr|^2. $$

The outer integral is easily estimated using the extra decay $|1+Z|\lesssim\frac{1}{1+y^{2}}$ and the estimates (B.17) of Lemma B.2. For the inner integral, we use |Z−1|≲|y|² to estimate:

$$\int_{y\leq1} \bigl|\mathbb{H}\mathbf{w}_2^2\bigr| \lesssim\int_{y\leq1} |H\partial_y\gamma |^2+ \delta\bigl(b^*\bigr) \biggl(\frac{b^4}{\mathopen{|}\log b|^2}+ \mathcal{E}_4 \biggr). $$

We then observe near the origin that for any function f:

$$Hf=-\partial_{yy}f-\frac{\partial_y}{y}f+\frac{1}{y^2}f+ \frac {V-1}{y^2}f=-\partial_{yy}f+\frac{Af}{y}+ \frac{V-1+Z-1}{y^2}f. $$

Since |V−1|+|Z−1|≲y ², the estimates of Lemma B.2 imply:

$$\int_{y\leq1} \bigl|\mathbb{H}\mathbf{w}_2^2\bigr|^2 \lesssim\int_{y\leq1} \frac{|A\partial_y\gamma|^2}{y^2}+ \delta \bigl(b^*\bigr) \biggl(\frac{b^4}{\mathopen{|}\log b|^2}+\mathcal{E}_4 \biggr). $$

The bound for the remaining term is given by (B.20).

Step 3 :

Quadratic term.

We claim:

$$ \int\bigl|\mathbb{H}(\widehat{\mathbf {w}}\wedge \mathbb{H} \mathbf{w})\bigr|^2\lesssim\delta\bigl(b^*\bigr) \biggl( \frac{b^4}{\mathopen{|}\log b|^2}+\mathcal{E}_4 \biggr). $$

(B.45)

Indeed, first compute:

$$ \widehat{\mathbf{w}}\wedge\mathbb{H}\mathbf {w}=\left| \begin{array}{@{}l@{}} \hat{\beta} [-\varDelta \gamma +2(1+Z)(\partial_y\alpha+\frac{Z}{y}\alpha) ]-\hat{\gamma}H\beta \\ \noalign{\vspace{3pt}} -\hat{\alpha} [-\varDelta \gamma+2(1+Z)(\partial_y\alpha+\frac {Z}{y}\alpha ) ]+\hat{\gamma} [H\alpha-2(1+Z)\partial_y\gamma ]\\ \noalign{\vspace{3pt}} \hat{\alpha} H\beta-\hat{\beta}(H\alpha-2(1+Z)\partial_y\gamma) \end{array}. \right. $$

(B.46)

We now apply the ℍ operator again and estimate all terms. We sketch the proof for the most delicate terms.

$$ \int\bigl|H(\hat{\beta} \varDelta \gamma)\bigr |^2\lesssim \int|\varDelta \gamma|^2|H\hat{\beta}|^2+\int| \partial_y\varDelta \gamma|^2|\partial_y\hat{ \beta}|^2+\int|\hat{\beta}|^2\bigl|\varDelta ^2 \gamma\bigr|^2. $$

(B.47)

The last integral is the most delicate term estimated from (B.26). For the other terms, we estimate using (B.10), (B.24) and (B.25):

On the other hand, since

$$\partial_y \varDelta \gamma= \partial_y^3 \gamma- \frac{1}{y} A\partial_y\gamma+\frac {Z-1}{y} \partial_y\gamma $$

and |Z−1|≲y ² we can estimate from (B.11), (B.17) and (B.20)

$$\int_{y\le1} |\partial_y\varDelta \gamma|^2|\partial_y\hat{\beta}|^2\lesssim\| \partial_y\hat{\beta}\|_{L^\infty(y\le1)} \int_{y\le1} |\partial_y\varDelta \gamma|^2\le b^6. $$

For y≥1 we can interpolate between (B.25) and (B.26) to obtain

$$\int_{y\ge1} |\partial_y\varDelta \gamma|^2|\partial_y\hat{\beta}|^2\lesssim b^5\mathopen{|}\log b|^C+b^{\frac{10}{3}} \|\partial_y \beta\|_{L^\infty(y\ge1)} \le b^5. $$

For the term involving the last coordinate in (B.46), we compute:

Terms near the origin are easily estimated using Lemma B.2. Far out, the first two terms are easily treated and for the third one, we estimate from (B.13):

$$\int_{y\geq1} |\hat{\alpha}|^2\frac{|H\beta|^2}{y^4} \lesssim b^4\mathopen{|}\log b|^C\sqrt{b}+ \biggl\|\frac{\alpha}{1+y^2} \biggr\|_{L^\infty}^2 \int|H\beta|^2\lesssim\delta\bigl(b^* \bigr) \frac{b^4}{\mathopen{|}\log b|^2}. $$

This concludes the proof of (B.45). The second part of (B.41) and (B.42) can be obtained in a similar fashion.

We omit the details.

This concludes the proof of Lemma B.3. □

Appendix C: Proof of Lemma 4.8

This appendix is devoted to the proof of Lemma 4.8 which is the key to handle the quasilinear structure of the problem. The proof is mostly algebraic and makes an implicit use of the interpolation estimates of Appendix B.

Step 1 :

Gain of two derivatives.

Let

$$a=\alpha e_r+\beta e_\tau+\gamma Q,\quad \boldsymbol{ \varGamma}=\left| \begin{array}{@{}l@{}}\alpha\\\beta\\\gamma \end{array} \right. $$

be a decomposition of the vector a relative to the Frenet basis of Q with (α,β,γ)=(α(y),β(y),γ(y)) functions of the radial variable y, and α ²+β ²+(1+γ)²=1. Then from (2.19):

$$ \varDelta a+|\nabla Q|^2a=-\mathbb{H} {\boldsymbol{ \varGamma}}. $$

(C.1)

We compute the action of derivatives

$$ \begin{aligned}[c] &\partial_ya=\partial_y \boldsymbol{\varGamma}+M\boldsymbol{\varGamma}, \qquad M\boldsymbol{\varGamma }:=(1+Z)e_y \wedge{\boldsymbol{\varGamma}}, \\ &\frac{1}{y} \partial_\theta a = N\boldsymbol{\varGamma},\qquad N\boldsymbol{ \varGamma}:= \biggl(\frac{Z}{y} e_z-(1+Z) e_x \biggr)\wedge{ \boldsymbol{\varGamma}}. \end{aligned} $$

(C.2)

We recall the double wedge formula:

$$a\wedge(b\wedge c)=(a\cdot c)b-(a\cdot b)c. $$

Let also

$$u=\widehat{\mathbf{w}}+e_z, \quad |u|^2=1 $$

be a unit vector. The proof of Lemma 4.8 is based on two computations.

The first one relies on the action of the Laplace operator in the Frenet basis:

$$\int a\cdot u\wedge \varDelta a = \int \varDelta a\cdot(a\wedge u)=\int a\cdot \varDelta (a \wedge u)=\int a\cdot[\varDelta a\wedge u+2\nabla a\wedge\nabla u ] $$

and thus

$$ \int a\cdot u\wedge \varDelta a =\int a\cdot(\nabla a\wedge\nabla u). $$

(C.3)

We now compute from (C.1):

$$a\cdot[u\wedge \varDelta a ]=a\cdot\bigl[u\wedge\bigl(\varDelta a+|\nabla Q|^2a\bigr) \bigr]=-\boldsymbol{\varGamma}\cdot\bigl[(e_z+ \widehat{\mathbf{w}})\wedge\mathbb{H}\boldsymbol{\varGamma} \bigr ]=-\boldsymbol{ \varGamma}\cdot\widehat{J}\mathbb{H} \boldsymbol{\varGamma}. $$

Hence from (C.2), (C.3):

The second computation uses the normalization of $\widehat{\mathbf{w}}$:

and the structure of the operator M:

This generates a two derivatives gain:

We now observe from

$$Me_z=(1+Z)e_x, \qquad Ne_z=(1+Z)e_y $$

the cancellation:

We have thus arrived at the formula:

(C.4)

Step 2 :

Leading order b term.

Let us write

$$\widetilde{\mathbf{w}}=\widetilde{\mathbf{w}}_0+\widetilde{\mathbf {w}}_1, \qquad \widetilde{\mathbf{w}}_0=b \tilde{T}_1 e_y. $$

We compute:

$$M\widetilde{\mathbf{w}}_0=0,\qquad N\widetilde{\mathbf{w}}_0=-b \tilde{T}_1 \biggl(\frac{Z}{y} e_x+(1+Z)e_z \biggr). $$

This yields in particular the cancellation:

$$N\widetilde{\mathbf{w}}_0\cdot\biggl(-(1+Z)e_x+\frac{Z}{y}e_z \biggr)=0. $$

We now compute the leading order contribution of $\widetilde{\mathbf {w}}_{0}$ to (C.4). First,

where we used the estimates of Lemma B.1. Next:

thanks to Lemma B.1. Next,

Therefore we obtained

(C.5)

Step 3 :

Upper bound on the quadratic term.

We now claim that

$$ \forall y\geq0, \quad 0\leq(1+Z)A\tilde{T_1} \leq1-d_1 $$

(C.6)

for some universal constant

$$0<d_1<1. $$

We prove the inequality for T ₁, the claim for $\tilde{T}_{1}$ follows immediately. From

$$\frac{(\varLambda\phi)'}{\varLambda\phi}=\frac{Z}{y}, $$

there holds:

$$A^*(AT_1)=\frac{1}{y\varLambda\phi}\frac{\partial}{\partial y} (y\varLambda\phi AT_1 )=\varLambda\phi $$

and thus,

$$AT_1=\frac{1}{y\varLambda\phi}\int_0^y \tau(\varLambda\phi)^2d\tau. $$

Now

$$1+Z=\frac{2}{1+y^2}=\frac{\varLambda\phi}{y}. $$

Therefore,

$$ J(y)=(1+Z)A(T_1)=\frac{1}{y^2}\int _0^y\tau(\varLambda\phi)^2d\tau\geq0. $$

(C.7)

Let now

$$f(y)=\int_0^y\tau(\varLambda\phi)^2d \tau-y^2, $$

then

$$f'(y)=y(\varLambda\phi)^2-2y=\frac{4y^3}{(1+y^2)^2}-2y= \frac {1}{y} \bigl[4y^3-2y\bigl(1+2y^2+y^4 \bigr) \bigr]\leq0 $$

and thus

$$f(y)<f(0)=0 \quad \mbox{for} \ y>0. $$

Hence J(y)<1 for y>0. Now J(y)→0 as y→+∞. It therefore attains its maximum at some y ₀≥0 with J(y ₀)<1, unless y ₀=0 which is ruled out since J(0)=0, and (C.6) is proved.

Step 4 :

Conclusion.

To control the remaining nonlinear terms in (C.5) we use the estimates of Lemma B.1. The first three terms are easily controlled:

The last term requires an additional cancellation to handle a singularity at the origin. Indeed,

Combining these estimates together with (C.6) and (C.5) concludes the proof of (4.49) and of Lemma 4.8.