AN INFORMAL SUMMARY OF GRILLAKIS’S RADIAL CRITICAL NLS RESULT

J. COLLIANDER, M. KEEL, G. STAFFILANI, H. TAKAOKA, AND T. TAO

Abstract. We try to summarize in broad detail Grillakis’ argument for global regularity of H^{1}-critical NLS.

1. Introduction

We consider the quintic defocussing Schrödinger equation in R^{3+1}:

iu_{t} + Δu = |u|^{4}u =: F (u).  (1)

This equation has a conserved Hamiltonian

H(u) = ∫ \frac{1}{2}|∇u|^{2} + \frac{1}{6}|u|^{6}.

We take data which is radial, smooth, and in H^{1}, so that H(u) is bounded. Grillakis’ result is that there is a global smooth solution. This is a somewhat weaker version of Bourgain’s argument, which works for arbitrary H^{1} data and also gives global spacetime bounds (and hence scattering).

Here I’ll try to summarize the basic points of the argument.

2. Step 1: Reduce to L^{∞} bounds

If u(0) is smooth, then the local theory says that we have a smooth solution up to some blowup time 0 < T_{∗} ≤ ∞. What Grillakis does is assume that T_{∗} < ∞ and get a contradiction.

So u(t) is smooth for times t strictly less than T_{∗}. What Grillakis does is show that the L^{∞} norm ∥u(t)∥_{∞} stays bounded as t → T_{∗}; by standard energy estimates (and Strichartz, etc.) this means that the solution can indeed be continued smoothly across the time T_{∗}, a contradiction. This is a standard step and I won’t insult your intelligence by repeating it here.

J.E.C. is supported in part by an N.S.F. Postdoctoral Research Fellowship.

M.K. is supported in part by N.S.F. Grant DMS 9801558.

G.S. is supported in part by N.S.F. Grant DMS 9800879 and by a grant from Hewlett and Packard and the Sloan Foundation.

H.T. is supported in part by J.S.P.S. Grant No. 13740087.

T.T. is a Clay Prize Fellow and is supported in part by grants from the Packard Foundation.

Let S(t)u(0) be the free solution. Since we are assuming u(0) is smooth, we know that S(t)u(0) is globally bounded^{1} in L^{∞}:

∥S(t)u(0)∥_{∞} ≲ 1.

(Our constants are allowed to depend on smooth norms of u(0), as well as on the Hamiltonian H(u). This is one reason why Grillakis can only get regularity and not GWP; of course it is hopeless to try for L^{∞} bounds if the solution is only in H^{1}, as Sobolev embedding clearly indicates).

So from Duhamel we have

|u(t_{0}, x_{0})| ≲ 1 + ∫_{0}^{t_{0}} |S(s)F (u(t_{0} − s))(x_{0})| ds  (2)

for any t_{0}, x_{0}. The idea is to try to control the Duhamel integral in terms of ∥u∥_{∞} and H(u), and thus to get some recursive control on ∥u∥_{∞} if the energy (or some localized version of this energy) is small.

3. Estimating the Duhamel term

We need a small dimensionless parameter 0 < δ ≪ 1 to be chosen later (it depends on u(0) and H(u)).

Let I = [t_{−1}, t_{0}] ⊆ [0, T^{∗}) be a time interval, and let (t_{0}, x_{0}) be a point at the end of the spacetime slab I × R^{3}. Let’s define the normalized sup norm

size(I) := |I|^{1/4}∥u∥_{L_{t,x}^{∞}(I×R^{3})};

note that this expression is dimensionless (u has the units of length^{−1/2}, and |I| has the units of length^{2}). Let’s also define the localized (and normalized) energy^{2}

energy(I) := \frac{1}{|I|} ∫_{0}^{|I|} ∫_{|x|≲δ^{−1}|I|^{1/2}} |u(t_{0} − s)|^{6} dxds.

Note that this energy is localized at the spatial origin; we won’t need to care about energy concentration anywhere else, mostly thanks to Bourgain’s radial-Sobolev estimate.

The quantity energy(I) is also dimensionless. Since the Hamiltonian H(u(t)) is bounded we have

energy(I) ≲ 1.

later on we will care very much about when this energy is in fact small (yes, we are going to do a standard ”small energy implies regularity” + ”energy does not concentrate” argument).

Grillakis uses (2) to obtain the following estimate, which is fundamental to his argument:

1Note however that this estimate does not scale properly under the H^{1} scaling, because it uses the smoothness of u(0).

2We only include the L^{6} portion of the energy here, as it is all we shall need, although Grillakis actually manages to control the full energy density in his paper.

Lemma 3.1. Let the notation be as above. If |I| ≪ 1 is sufficiently small, then

|I|^{1/4}∥u(t_{0})∥_{∞} ≲ C_{δ} + δsize(I) + C_{δ}size(I)^{3}energy(I)^{0.1}.  (3)

Morally speaking, (3) implies that

size(I) ≲ C_{δ} + δsize(I) + C_{δ}size(I)^{3}energy(I).

If the energy is sufficiently small depending on δ, then one should be able to absorb the latter two terms by the left-hand side and conclude that size(I) is bounded, which will lead to L^{∞} bounds. The rest of the argument then consists of showing that energy does not concentrate (so that energy(I) becomes small as one approaches the blow up time T_{∗}).

Proof Fix an x_{0} ∈ R^{3}. We need to show that

|I|^{1/4}|u(t_{0}, x_{0})| ≲ C_{δ} + δsize(I) + C_{δ}size(I)^{3}energy(I)^{0.1}.

First observe from the Bourgain-radial Sobolev estimate that

u(t_{0}, x_{0}) ≲ |x_{0}|^{−1/2}∥u(t_{0})∥Ḣ^{1} ≲ |x_{0}|^{−1/2}.  (4)

Hence we may assume that

|x_{0}| ≪ |I|^{1/2}  (5)

since the claim is trivial otherwise. In other words, the sup norm only becomes dangerous near the spatial origin (because of the radial symmetry).

From Duhamel (2) it now suffices to show that

|I|^{1/4} ∫_{0}^{t_{0}} |S(s)F (u(t_{0} − s))(x_{0})| ds ≲ C_{δ} + δsize(I) + C_{δ}size(I)^{3}energy(I)^{0.1}. (6)

Let’s understand the expression |S(s)F (x_{0})| more carefully, for arbitrary functions F. We can expand it out as

|S(s)F (x_{0})| = Cs^{−3/2}| ∫ F (x) exp(i|x − x_{0}|^{2}/4s) dx|.

So we clearly have a trivial bound of

|S(s)F (x_{0})| ≲ s^{−3/2} ∫ |F (x)| dx.  (7)

On the other hand, we can also integrate by parts, writing

exp(i|x − x_{0}|^{2}/4s) = s\frac{x − x_{0}}{C|x − x_{0}|^{2}} · ∇_{x} exp(i|x − x_{0}|^{2}/4s)

to obtain

|S(s)F (x_{0})| = Cs^{−1/2}| ∫ exp(i|x − x_{0}|^{2}/4s)∇_{x} · (\frac{x − x_{0}}{|x − x_{0}|^{2}} F (x)) dx|.

Thus we can obtain a bound of

|S(s)F (x_{0})| ≲ s^{−1/2} ∫ \frac{|F (x)|}{|x − x_{0}|^{2}} + \frac{|∇F (x)|}{|x − x_{0}|}.  (8)

Broadly speaking, the first estimate is better when |x − x_{0}| ≲ s^{1/2}, and the second estimate is better when |x − x_{0}| ≳ s^{1/2}.

Now we prove (6). We split the integral ∫_{0}^{t_{0}} into the distant past ∫_{|I|}^{t_{0}} and the recent

past ∫_{0}^{|I|}. The scheme will be as follows:

• Distant past, close to x_{0}: use decay estimates (7), Hamiltonian estimates.

• Distant past, far from x_{0}: use integration by parts (8), radial-Sobolev, Hamiltonian estimates.

• Recent past, close to x_{0}: use decay estimates (7), localized energy once, and the size bound three times.

• Recent past, far from x_{0}: use integration by parts (8), Hamiltonian estimates, and the size bound once.

Case 1: The contribution of the distant past.

Fix |I| ≤ s ≤ t_{0}. We claim that

|I|^{1/4}|S(s)F (u(t_{0} − s))(x_{0})| ≲ C_{δ}s^{−5/4}|I|^{1/4},  (9)

and so by integrating in s we see that this portion is acceptable.

We have to prove (9). First suppose that F (u(t_{0} − s)) is supported in the region |x − x_{0}| ≲ s^{1/2}. Then we use (7) to estimate the left-hand side of (9) by

|I|^{1/4}s^{−3/2} ∫_{|x−x_{0}|≲s^{1/2}} |u(t_{0} − s)|^{5} dx.

On the other hand, we have an L^{6} bound on u(t_{0} − s), so by Hölder this term is acceptable.

Now suppose that F (u(t_{0} − s)) is supported in the region |x − x_{0}| ≳ s^{1/2}. Then we use (8) to estimate the left-hand side of (9) by

|I|^{1/4}s^{−1/2} ∫_{|x−x_{0}|≲s^{1/2}} \frac{|u(t_{0} − s)|^{5}}{|x − x_{0}|^{2}} + \frac{|u(t_{0} − s)|^{4}|∇_{x}u(t_{0} − s)|}{|x − x_{0}|} dx.

To control the first term, we again use the L^{6} bound and Hölder to bound this by

|I|^{1/4}s^{−1/2}∥u(t_{0} − s)∥_{6}^{5}∥\frac{χ_{|x−x_{0}|≲s^{1/2}}}{|x − x_{0}|^{2}} ∥_{6} ≲ |I|^{1/4}s^{−1/2}s^{−3/4}

as desired. To control the second term, we exploit (5), (4) to obtain

\frac{|u(t_{0} − s)|^{4}|∇_{x}u(t_{0} − s)|}{|x − x_{0}|} ≲ |u(t_{0} − s)|^{3}|∇_{x}u(t_{0} − s)|s^{−3/4}.

But from the control on H(u) and Cauchy-Schwarz we see that |u(t_{0}−s)|^{3}|∇_{x}u(t_{0}− s)| is integrable, and so this term is also acceptable.

One can combine the two cases by a standard application of cutoff functions.

Case 2: The contribution of the recent past.

Fix 0 ≤ s ≤ |I|. We introduce the radius

R(s) := δ^{−1}\sqrt{s}(\frac{|I|}{s})^{0.1}  (10)

and claim that

|I|^{1/4}|S(s)F (u(t_{0} − s))(x_{0})| ≲ δ|I|^{−0.1}s^{−0.9}size(I) + C_{δ}size(I)^{3}|I|^{−1/2}s^{−3/2} ∫_{|x−x_{0}|≲R(s)} |u|^{2} dx. (11)

Suppose (11) was true. Then we could integrate in s and obtain an acceptable contribution so long as we could show that

∫_{0}^{|I|} |I|^{−1/2}s^{−3/2} ∫_{|x−x_{0}|≲R(s)} |u|^{2} dxds ≲ C_{δ}energy(I)^{0.1}.

But by Hölder we have

∫_{|x−x_{0}|≲R(s)} |u|^{2} dx ≲ R(s)^{2}(∫_{|x|≲δ^{−1}|I|^{1/2}} |u|^{6})^{1/3}

since |x − x_{0}| ≲ R(s) and (5) force |x| ≲ δ^{−1}|I|^{−1/2}. The integrand is bounded from Hamiltonian conservation so we may replace the 1/3 power by, say, 1/10. We then use (10) and reduce to showing

∫_{0}^{|I|} |I|^{−0.3}s^{−0.7}(∫_{|x|≲δ^{−1}|I|^{1/2}} |u|^{6})^{1/10} dxds ≲ energy(I)^{0.1}.

But this follows from Hölder, since s^{−0.7} is in L_{t}^{10/9}.

We now prove (11). First suppose that F (u(t_{0} − s)) is supported in the region |x − x_{0}| ≲ R(s). As before we use (7) to estimate the left-hand side of (11) by

|I|^{1/4}s^{−3/2} ∫_{|x−x_{0}|≲R(t_{0}−s)} |u(t_{0} − s)|^{5} dx,

which we estimate crudely by

|I|^{−1/2}s^{−3/2}size(I)^{3} ∫_{|x−x_{0}|≲R(s)} |u(t_{0} − s)|^{2} dx

which is acceptable.

Now suppose that F (u(t_{0} − s)) is supported on the region |x − x_{0}| ≳ R(s). Then we use (8) to obtain a bound of

|I|^{1/4}s^{−1/2} ∫_{|x−x_{0}|≳R(s)} \frac{|u(t_{0} − s)|^{5}}{|x − x_{0}|^{2}} + \frac{|u(t_{0} − s)|^{4}|∇_{x}u(t_{0} − s)|}{|x − x_{0}|}.

We estimate one factor of |u(t_{0} − s)| by |I|^{−1/4}size(I), and one denominator of |x − x_{0}| by R(s), to obtain

size(I)s^{−1/2} ∫_{|x−x_{0}|≳R(s)} \frac{|u(t_{0} − s)|^{4}}{|x − x_{0}|^{2}} + \frac{|u(t_{0} − s)|^{3}|∇_{x}u(t_{0} − s)|}{R(s)}.

Now we use Hölder and the fact that u is in L^{6} and ∇u is in L^{2} to obtain a bound of

size(I)s^{−1/2}R(s)^{−1}

which is acceptable by (10).

One can combine the two cases by a standard application of cutoff functions. This completes the proof of (6), and hence (3).

4. Small energy implies regularity

Now we use the above estimate (3) to show that one can keep size under control as long as energy is bounded. We need a large number K = K(δ) ≫ 1. The following lemma shows that size does not blow up as long as the localized energy remains small:

Corollary 4.1. Suppose I = [t_{0} + R, t_{0} + 4R] ⊆ [0, T_{∗}) is an interval with |I| ≪ 1. Write I = I_{1} ∪ I_{2}, where I_{1} := [t_{0} + R, t_{0} + 2R], I_{2} = [t_{0} + 2R, t_{0} + 4R]. Assume that

size(I_{1}) ≤ K

and

energy(I_{1} ∪ I_{2}) ≲ ε

for some 0 < ε ≪ 1 depending on K, δ, H(u), and u_{0}. Then

size(I_{2}) ≤ K/2.

Proof Let Ω ⊆ I_{2} denote the set

Ω := {t ∈ I_{2} : |I_{2}|^{1/4}∥u∥_{L_{x,t}^{∞}([t_{0}+2R,t_{0}+4R]×R)} ≤ K/2.

It suffices to show that the right endpoint of I_{2} lies in Ω. Clearly the left endpoint lies in this set, and Ω is also closed, so it suffices to show that Ω is open.

Let t be any time in Ω, and let I_{t} be the interval I_{t} := [t_{0} + R, t]. Then from hypothesis and definition of Ω we have size(I_{t}) ≲ K. From Lemma 3.1 we thus have

|I|^{1/4}∥u(t)∥_{∞} ≲ C_{δ} + δK + C_{δ}K^{3}ε^{0.1}.

If δ is small enough, K is large enough (depending on δ) and ε is small enough (depending on K, δ), we see that |I|^{1/4}∥u(t)∥_{∞} ≤ K/4. Thus Ω is open, as desired.

We can iterate this:

Corollary 4.2. Let K and ε be as in the previous lemma. Suppose t_{0} ∈ [0, T^{∗}) is a time such that T^{∗} − t_{0} ≪ 1 and

energy([t_{0}, t]) ≲ ε

for all t_{0} < t < T^{∗}. Then

sup_{\frac{t_{0}+T^{∗}}{2} <t<T^{∗}} ∥u(t)∥_{∞} ≲ K|T_{∗} − t_{0}|^{−1/4}.

In particular u(t) does not blow up at time T^{∗}.

Proof Since u is smooth at time t_{0}, we see that

size([t_{0} + s, t_{0} + 2s]) → 0 as s → 0.

From the previous Corollary we see that

size([t_{0} + s, t_{0} + 2s]) ≤ K ⟹ size([t_{0} + 2s, t_{0} + 4s]) ≤ K

whenever t_{0} < t_{0} + 4s < T_{∗}. From this we see that

size([t_{0} + s, t_{0} + 2s]) ≤ K

for all such s. The claim follows.

Thus we can obtain regularity as long as we can show energy does not concentrate.

5. Energy does not concentrate

We have to find a time 0 < t_{0} < T^{∗} such that T_{∗} − t_{0} ≪ 1 and

energy([t_{0}, t]) ≲ ε for all t_{0} < t < T_{∗}.

Actually it will suffice to find a time 0 < t_{1} < T_{∗} such that T_{∗} − t_{1} ≪ 1 and

\frac{1}{T_{∗} − t_{1}} ∫_{t_{1}}^{T_{∗}} ∫_{|x|≲δ^{−1}(T_{∗}−t_{1})^{1/2}} |u|^{6} dxdt ≲ ε.

To see this, consider the quantity a(t) defined for t_{1} ≤ t ≤ T_{∗} by

a(t) := ∫_{t_{1}}^{t} ∫_{|x|≲δ^{−1}(T_{∗}−t_{1})^{1/2}} |u|^{6} dxdt − Cε(t − t_{1}).

Then a(t_{1}) = 0, and if C is sufficiently large we have a(T_{∗}) < 0 by hypothesis. Also, since the Hamiltonian (and hence L^{6} norm) is bounded, a is Lipschitz. Thus a(t) attains its maximum at some t_{1} ≤ t_{0} < T_{∗}. the claim then follows since a(t_{0}) − a(t) < 0 for all t_{0} < t < T_{∗}.

To summarize, it will suffice to prove the non-concentration estimate

lim inf_{t_{1}→T_{∗}} \frac{1}{T_{∗} − t_{1}} ∫_{t_{1}}^{T_{∗}} ∫_{|x|≲δ^{−1}(T_{∗}−t_{1})^{1/2}} |u|^{6} dxdt = 0.

As always, this will be achieved by a Morawetz estimate. We are accustomed to controlling the Morawetz action by the H^{1/2} norm; what Grillakis does is estimate instead by the geometric mean of the Hamiltonian and the L^{2} norm, and show a certain non-concentration of the L^{2} norm. To make this work we need some facts about L^{2} norm propagation.

Let η(x) be any non-negative spatial bump function which equals 1 for |x| ≤ 1 and equals 0 for |x| > 2, and consider the quantity

L(R, t)^{2} := R^{−2} ∫ η(x/R)^{2}|u(t, x)|^{2} dx,

8 J. COLLIANDER, M. KEEL, G. STAFFILANI, H. TAKAOKA, AND T. TAO

measuring the (normalized) L^{2} norm on a ball of radius R centered at the origin at time t. From the L^{6} control and Hölder we see that L(R, t) is bounded; now let’s see how it changes in time. We differentiate in time. From the mass conservation identity

∂_{t}|u|^{2} = C∇_{x} · Im(\overline{u}∇u)

and integration by parts, we have

|∂_{t}L(R, t)^{2}| ≲ R^{−3} ∫ η(x/R)|∇η|(x/R)|u||∇u|;

by Cauchy-Schwarz and the Hamiltonian control we thus have

|∂_{t}L(R, t)^{2}| ≲ R^{−3}RL(R, t)

and hence

|∂_{t}L(R, t)| ≲ R^{−2}.  (12)

In particular L(R, t) has a limit as t → T_{∗}, and we can meaningfully talk about L(R, T_{∗}). Lemma 5.1. Let 0 < s < T_{∗} and 0 < ε ≪ 1. If L(ε^{−1}s^{1/2}, T_{∗}) ≲ ε^{2}, then

\frac{1}{s} ∫_{T_{∗}−s}^{T_{∗}} ∫_{|x|≲δ^{−1}s^{1/2}} |u|^{6} dxdt ≲ C_{δ}ε.

Proof Let R := ε^{−1}s^{1/2}. From the Lipschitz bound (12) we have

L(R, t) ≲ ε^{2}  (13)

for all T_{∗} − s ≤ t ≤ T. Define the localized Morawetz action M_{0}(t) by

M_{0}(t) := Im ∫ η(x/R)\overline{u(t)}u_{r}(t).

From (13), the Hamiltonian bound, and Cauchy-Schwartz we have the estimate

|M_{0}(t)| ≲ C_{δ}ε^{2}R

for all T_{∗} − s ≤ t ≤ T. A standard computation (see e.g. Bourgain’s book) shows that

∂_{t}M_{0}(t) ≳ ∫_{|x|≲R} |u|^{6}/|x| dx + O(∫_{|x|∼R}(|∇u|^{2} + |u|^{6})/|x|).

The error term is O(1/R) by the boundedness of the Hamiltonian. Thus

∫_{T_{∗}−s}^{T} ∫_{|x|≲R} |u|^{6}/|x| dxdt ≲ s/R + C_{δ}ε^{2}R ≲ C_{δ}εs^{1/2}

and thus

∫_{T_{∗}−s}^{T} ∫_{|x|≲δ^{−1}s^{1/2}} \frac{|u|^{6}}{δ^{−1}s^{1/2}} dxdt ≲ εs^{1/2},

and the claim follows.

GRILLAKIS’S NLS ARGUMENT 9

From this lemma we see that it will suffice to show that

lim inf_{R→0} L(R, T_{∗}) = 0.

In fact we have the stronger estimate

∑_{n≥0} L(2^{−n}, T_{∗})^{2} < ∞.

This is because for every 0 < t ≤ T_{∗} we have

∑_{n≥0} L(2^{−n}, t)^{2} ≲ ∫ |u(t)|^{2}/|x|^{2} dx ≲ ∥u(t)∥2 ; Ḣ^{1} ≲ H(u)

by Hardy’s inequality (see below); the claim then follows from Fatou’s lemma since we know that L(R, t) is Lipschitz, hence continuous, in t. Just for reference, here is a quick proof of Hardy’s inequality. Lemma 5.2. For any u ∈ Ḣ^{1}(R^{3}), we have

∫ |u|^{2}/|x|^{2} ≲ ∥u∥2 ; Ḣ^{1}.

Proof Split u = ∑_{N} u_{N}, where N ranges over powers of two and u_{N} has Fourier support in the annulus |ξ| ∼ 2^{N}. We bound the left-hand side by

∑_{M} M^{2} ∫_{|x|≲M^{−1}} |u|^{2},

which by the triangle inequality is bounded by

∑_{M} M^{2}(∑_{N} ∥u_{N}∥_{L^{2}(|x|≲M^{−1})})^{2}.

We can estimate the norm in two different ways. First of all we have

∥u_{N}∥_{L^{2}(|x|≲M^{−1})} ≤ ∥u_{N}∥_{2} ≲ N^{−1}∥u_{N}∥Ḣ^{1}

or we have

∥u_{N}∥_{L^{2}(|x|≲M^{−1})} ≤ M^{−3/2}∥u_{N}∥_{∞} ≲ M^{−3/2}N^{1/2}∥u_{N}∥Ḣ^{1}.

Thus the previous sum is bounded by

∑_{M}(∑_{N} min(M/N, (N/M)^{1/2})∥u_{N}∥_{2})^{2};

by Young’s inequality (since the sequence min(2^{n}, 2^{−n/2}) is summable) this is bounded by

∑_{N} ∥u_{N}∥_{2}^{2}

as desired.

10 J. COLLIANDER, M. KEEL, G. STAFFILANI, H. TAKAOKA, AND T. TAO

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