SOLUTIONS TO HW 1

TERENCE TAO

1. Exercise 3.28 (H^{s} growth for 1D cubic NLS)

From Proposition 3.23, the solution u is already known to be in S^{0}(I × R) for any compact interval I, and in particular lies in L_{t}^{4}L_{x}^{∞} locally in time. We would like to apply Proposition 3.21, but first we need to show that u is a strong H_{x}^{s}(R) solution^{1}. For this we need a local well-posedness result in H_{x}^{s}(R). This can be done by considering the iteration scheme that places u in L_{t}^{4}L_{x}^{∞} ∩ C_{t}^{0}H_{x}^{s} and |u|^{2}u in L_{t}^{1}H_{x}^{s}. From Lemma A.8 we have

∥|u|^{2}u L_{t}^{1}H_{x}^{s}(I × R) ≲ (∥u∥_{L_{t}^{4}L_{x}^{∞}(I×R)} + ∥u∥_{C_{t}^{0}H_{x}^{s}(I×R)})^{3}

and more generally

∥|u|^{2}u − |v|^{2}v L_{t}^{1}H_{x}^{s}(I × R) ≲ |I|^{1/2}(∥u∥_{L_{t}^{4}L_{x}^{∞}(I×R)} + ∥u∥_{C_{t}^{0}H_{x}^{s}(I×R)} + ∥v∥_{L_{t}^{4}L_{x}^{∞}(I×R)} + ∥v∥_{C_{t}^{0}H_{x}^{s}(I×R)})^{2}
[∥u − v∥_{L_{t}^{4}L_{x}^{∞}(I×R)} + ∥u − v∥_{C_{t}^{0}H_{x}^{s}(I×R)}].

This can be iterated in the usual fashion (together with Strichartz estimates) to show that the equation is locally wellposed in H_{x}^{s}, with time of existence depending only on the H_{x}^{s} norm of the initial data. Furthermore this solution is locally in S^{0}, and thus by the L^{2} uniqueness theory, this local solution must match exactly with the solution constructed in Proposition 3.23. Concatenating the local theory with itself in the usual manner, we see that either the H^{s} solution exists globally in time, or else the H^{s} norm blows up in finite time. But by Proposition 3.21, the only way the latter can happen is if the L_{t}^{2}L_{x}^{∞} norm becomes infinite in finite time. But the S^{0} norm controls the L_{t}^{4}L_{x}^{∞} norm and hence the L_{t}^{2}L_{x}^{∞} norm locally in time, so no blowup occurs and so the global L^{2} solution is in fact a global H^{s} solution.

Now for the bounds. Here we will need to use the L^{2} conservation law

∥u(t)∥_{L_{x}^{2}} = ∥u_{0}∥_{L_{x}^{2}}

1This is a typical, and somewhat annoying, technical issue in this business: before one can obtain quantitative results (such as bounds in the H^{s} norm), one often must first establish qualitative results (such as that the H^{s} norm is finite to begin with). Fortunately, once the local wellposedness theory is in place, the qualitative results are usually quite an easy consequence. In practice it is usually safe to skip the qualitative step and move directly to the quantitative step, which contains the real essence of the problem, but to be totally rigorous one needs to spend a little time on the qualitative results. For instance, the local wellposedness theory combined with a limiting argument often lets us work a priori with Schwartz solutions, and sometimes even global Schwartz solutions, in which case all of the qualititative issues are trivial.

which gives a bound on the L^{2} norm of u(t) which is uniform in t. The local L^{2} theory starting at time t then gives, for each time t, a local interval of existence [t − τ, t + τ] for some τ = τ(∥u_{0}∥_{L_{x}^{2}}) > 0 for which we have the bounds

∥u∥_{S^{0}([t−τ,t+τ]×R)} ≲_{∥u_{0}∥_{L^{2}_{x}}} 1.

The point is that the bounds here (and the time τ) are uniform in t. In particular we have

∥u∥_{L_{t}^{4}L_{x}^{∞}([t−τ,t+τ]×R)} ≲_{∥u_{0}∥_{L^{2}_{x}}} 1.

Pasting together many such intervals, we see that

∥u∥_{L_{t}^{4}L_{x}^{∞}([−T,T]×R)} ≲_{∥u_{0}∥_{L^{2}_{x}}} T^{1/4}

for all T > 1. By Hölder this gives

∥u∥_{L_{t}^{2}L_{x}^{∞}([−T,T]×R)} ≲_{∥u_{0}∥_{L^{2}_{x}}} T^{1/2}.

Applying Proposition 3.21 we obtain the result.

2. Exercise 3.45 (Alternate derivation of interaction Morawetz inequality)

Many of you repeated the derivation I gave in class. Which is fine, but not exactly what I had intended for this question.

If u : R × R^{3} → C is a classical solution to the three-dimensional free Schrödinger equation, direct calculation shows that the function U : R × R^{6} → C, defined by U(t, x, y) := u(t, x)u(t, y) is also a solution to the six-dimensional Schrödinger equation. For notational convenience let us reparameterise

(x, y) = (x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}) = (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}) = x.

The point is that the new solution U also has a stress-energy tensor T, defined by

T_{00}(t, x) := |U(t, x)|^{2}
T_{0j}(t, x) = T_{j0}(t, x) := Im(\overline{U}(t, x)∂_{x_{j}}U(t, x))
T_{jk}(t, x) := Re(∂_{x_{j}}U(t, x)∂_{x_{k}} \overline{U}(t, x)) − \frac{1}{4}δ_{jk}ΔR^{6}(|U(t, x)|^{2})

where j, k now range from 1, …, 6 rather than 1, …, 3, and we still have the pointwise conservation laws (2.35). Here ΔR^{6} denotes the six-dimensional Laplacian. In particular we can use the one-particle second variation formula (3.36) and obtain, for any test function a : R^{6} → R,

∂_{t} ∫_{R_{6}} ∂_{x_{j}}a(x)Im(\overline{U}(t, x)∂_{x_{j}}U(t, x)) dx = ∫_{R_{6}}(∂_{x_{j}}∂_{x_{k}}a(x))Re(∂_{x_{j}}U(t, x)∂_{x_{k}} \overline{U}(t, x))

− \frac{1}{4} ∫_{R_{6}} |U(t, x)|^{2}ΔR^{6}ΔR^{6}a(x) dx.

Both sides are linear in a, so if U is classical one can take distributional limits and obtain the same identity for tempered distributions a (interpreting ∂_{t} in the weak sense if necessary).

Observe that if a is convex, then ∂_{x_{j}}∂_{x_{k}}a(x) is positive definite, and so the first term is non-negative, thus

∂_{t} ∫_{R_{6}} ∂_{x_{j}}a(x)Im(\overline{U}(t, x)∂_{x_{j}}U(t, x)) dx ≥ −\frac{1}{4} ∫_{R_{6}} |U(t, x)|^{2}ΔR^{6}ΔR^{6}a(x) dx.

One then observes the geometrically obvious fact that the function a(x) := |x − y| is a convex function in six dimensions (this is easiest to see in centre-of-mass coordinates z = (x + y)/2, w = (x − y)/2). Also a direct computation shows that ΔR^{6}ΔR^{6}a(x) = −16πδ(x − y), and so

∂_{t} ∫_{R_{6}} ∂_{x_{j}}a(x)Im(\overline{U}(t, x)∂_{x_{j}}U(t, x)) dx ≥ 4 ∫_{R_{6}} |U(t, x)|^{2}δ(x − y) dx.

The right-hand side simplifies to 4 ∫R^{3} |u(t, x)|^{4} dx, and so by the fundamental theorem of calculus

∫_{−T}^{T} ∫_{R_{3}} |u(t, x)|^{4} dxdt ≲ sup_{t=−T,+T} | ∫_{R_{6}} ∂_{x_{j}}a(x)Im(\overline{U}(t, x)∂_{x_{j}}U(t, x)) dx|.

The left-hand side can be bounded using symmetry as

≲ sup_{t=−T,+T} | ∫_{R_{3}} ∫_{R_{3}} \frac{x_{j} − y_{j}}{|x − y|} Im(\overline{u}(t, x)\overline{u}(t, y)∂_{x_{j}}u(t, x)u(t, y)) dxdy|

which we estimate as

≲ sup_{t=−T,+T} ∫_{R_{3}} | ∫_{R_{3}} \frac{x_{j} − y_{j}}{|x − y|} Im(\overline{u}(t, x)∂_{x_{j}}u(t, x)) dx||u(t, y)|^{2}dy.

But by Lemma A.10 and translation invariance

∫_{R_{3}} \frac{x_{j} − y_{j}}{|x − y|} Im(\overline{u}(t, x)∂_{x_{j}}u(t, x)) dx ≲ ∥u(t)∥2 ; Ḣ_{x}^{1/2}

and the claim follows.

The point of doing things this way is that it makes the “N-particle” generalisations of this estimate rather straightforward. For instance, I invite you to now show the four-particle estimate

∫R ∫R |u(t, x)|^{8} dxdt ≲ ∥u(0)∥_{L_{x}^{2}}^{7}∥u(0)∥_{H_{x}^{1}}

whenever u solves the one-dimensional free Schrödinger equation. The relevant weight function a : R^{4} → R is the distance function to the diagonal {(x, x, x, x) : x ∈ R}.

With a defocusing nonlinearity |u|^{p−1}u, things become a bit tricky with this approach, because U does not solve the nonlinear Schrödinger equation directly; instead we have

i∂_{t}U + \frac{1}{2}ΔR^{6}U = V U

where V (t, x, y) := |u(t, x)|^{p−1} + |u(t, y)|^{p−1}. This changes the stress-energy tensor a little bit. More precisely, we have

T_{jk} = Re(∂_{x_{j}}U(t, x)∂_{x_{k}} \overline{U}(t, x)) − \frac{1}{4}δ_{jk}ΔR^{6}(|U(t, x)|^{2}) + δ_{jk}W_{j}|u|^{2}

4 TERENCE TAO

where W_{j} := \frac{p−1}{p+1}|u(t, x)|^{p−1} + |u(t, y)|^{p−1} when j = 1, 2, 3 and W_{j} := |u(t, x)|^{p−1} + \frac{p−1}{p+1}|u(t, y)|^{p−1} when j = 2, 3, 4. The main point however is that the W_{j} remain nonnegative, and so the contribution of the nonlinear terms are similarly nonnegative and thus do not disrupt the above argument.

UCLA Department of Mathematics, Los Angeles, CA 90095-1596.

E-mail address: tao@math.ucla.edu