SOLUTIONS TO HW 6

TERENCE TAO

1. Exercise 2.55 (Local smoothing for Airy equation)

There is a typo in the question: the conservation law should read

∂_{t}(u^{2}) = −∂_{xxx}(u^{2}) + 3∂_{x}(u_{x}^{2}).

Using the product rule repeatedly, the right-hand side can be evaluated as −2uu_{xxx}, while the left-hand side is 2uu_{t}, so the claim follows from the Airy equation u_{t} + u_{xxx} = 0.

Integrating the conservation law in space, observing that the right-hand side is zero, yields

∂_{t} ∫R u(t, x)^{2} dx = 0

for Schwartz solutions u, so we have L^{2} norm conservation:

∫R u(t, x)^{2} dx = ∫R u(0, x)^{2} dx.

Now let η : R → R be a smooth non-decreasing function to be selected later. Multiplying the conservation law against η(x/R) and integrating in space, we obtain

∂_{t} ∫R u(t, x)^{2}η(x/R) dx = ∫R −∂_{xxx}(u^{2}(t, x))η(x/R) + 3∂_{x}(u_{x}^{2}(t, x))η(x/R) dx.

Integrating by parts repeatedly we obtain

∂_{t} ∫R u(t, x)^{2}η(x/R) dx = ∫R \frac{1}{R^{3}} u^{2}(t, x)η′′′(x/R) − \frac{3}{R}u_{x}^{2}(t, x)η′(x/R) dx.

We integrate this in time as well and obtain

∫R u(t, x)^{2}η(x/R) dx|_{t=0}^{T} = ∫_{0}^{T} ∫R \frac{1}{R^{3}} u^{2}(t, x)η′′′(x/R) − \frac{3}{R}u_{x}^{2}(t, x)η′(x/R) dxdt.

Now choose η so that η, η′′′ are O(1) and that η′ is at least 1 (say) for |x| ≤ 1 and non-negative elsewhere. Then by L^{2} norm conservation the LHS has magnitude O(∫R u(0, x)^{2} dx), while the first term on the RHS has magnitude O(T/R^{3}). We conclude that

\frac{1}{R} ∫_{0}^{T} ∫R u_{x}^{2}(t, x)η′(x/R) dxdt ≤ O(1) + O(T/R^{3})

and then by the assumptions on η′ the claim follows.

2. Exercise 2.64 (Morawetz inequality for wave equation)

There is another typo in the notes; the identity should read

∂_{α}(T^{αβ}X_{β}) = \frac{|∇ / u|^{2}}{|x|} − \frac{1}{2|x|}□(|u|^{2}).

From the notes we have

∂_{α}(T^{αβ}X_{β}) = \frac{1}{2}T^{αβ}π_{αβ}

where π is the deformation tensor

π_{αβ} = ∂_{α}X_{β} + ∂_{β}X_{α}.

By definition, we have X_{i} = x_{i}/|x| for i = 1, 2, 3 and X_{0} = 0. So π_{αβ} = 0 when α = 0 or β = 0 (this reflects the fact that the radial vector field does not distort the geometry in the time variable). To compute the spatial components π_{ij} of the tensor for i, j = 1, 2, 3 we compute

π_{ij} = ∂_{i} \frac{x_{j}}{|x|} + ∂_{j} \frac{x_{i}}{|x|}

= 2(\frac{δ_{ij}}{|x|} − \frac{x_{i}x_{j}}{|x|^{3}} )

and hence

∂_{α}(T^{αβ}X_{β}) = \frac{T_{ii}}{|x|} − \frac{T_{ij}x_{i}x_{j}}{|x|^{3}} .

On the other hand, we have

T^{αβ} = Re(∂^{α}u\overline{∂^{β}u}) − \frac{1}{2}g^{αβ}Re(∂^{γ}u∂_{γ}u) = Re(∂^{α}u\overline{∂^{β}u}) − \frac{1}{4}g^{αβ}□(|u|^{2})

from the product rule and the equation □u = 0, so we have

∂_{α}(T^{αβ}X_{β}) = \frac{Re(u_{i}\overline{u_{i}})}{|x|} − \frac{Re(u_{i}\overline{u_{j}})x_{i}x_{j}}{|x|^{3}} − \frac{1}{4}[\frac{g_{ii}}{|x|} − \frac{g_{ij}x_{i}x_{j}}{|x|^{3}} ]□(|u|^{2}).

Now Re(u_{i}\overline{u_{i}}) = |∇u|^{2}, Re(u_{i}\overline{u_{j}})x_{i}x_{j} = |x · ∇u|^{2}, g_{ii} = 3, and g_{ij}x_{i}x_{j} = |x|^{2}, hence

∂_{α}(T^{αβ}X_{β}) = \frac{|∇u|^{2} − |\frac{x}{|x|} · ∇u|^{2}}{|x|} − \frac{1}{2|x|}□(|u|^{2})

which gives the desired identity (at least outside of the time axis {x = 0}. The identity in fact holds in a distributional sense, but we shall just proceed classically here).

We integrate the identity on [−T, T] × (R^{3}\B(0, ε)) for some small ball B(0, ε) and use Stokes’ theorem to conclude that

∫_{R^{3}\B(0,ε)} T^{0β}X_{β} dx|_{t=−T}^{t=T} −∫_{−T}^{T} ∫_{∂B(0,ε)} T^{iβ}X_{β} dSdt = ∫_{−T}^{T} ∫_{R^{3}\B(0,ε)} \frac{|∇ / u|^{2}}{|x|} −\frac{1}{2|x|}□(|u|^{2}) dxdt;

taking limits as ε → 0 and using the local integrability of everything in sight we obtain

∫_{R_{3}} T^{0β}X_{β} dx|_{t=−T}^{t=T} = ∫_{−T}^{T} ∫_{R_{3}} \frac{|∇ / u|^{2}}{|x|} − \frac{1}{2|x|}□(|u|^{2}) dxdt;

Since X is bounded, and the stress-energy tensor is bounded by the energy-density T^{00}, we have

∫_{R_{3}} T^{0β}X_{β} dx = O(∫_{R_{3}} T^{00} dx) = O(E)

and so

∫_{−T}^{T} ∫_{R_{3}} \frac{|∇ / u|^{2}}{|x|} − \frac{1}{2|x|}□(|u|^{2}) dxdt = O(E).

We split □(|u|^{2}) = −∂_{tt}|u|^{2} + Δ(|u|^{2}). For the Laplacian term, we recall that Δ\frac{−1}{4π|x|} = δ and hence

∫_{−T}^{T} ∫_{R_{3}} \frac{1}{2|x|}Δ(|u|^{2}) dxdt = −2π ∫_{−T}^{T} |u(t, 0)|^{2} dt

so to finish the proof it suffices to show that

∫_{−T}^{T} ∫_{R_{3}} \frac{1}{2|x|}∂_{tt}|u|^{2} dxdt = O(E).

By the fundamental theorem of calculus the left-hand side is

∫_{R_{3}} \frac{1}{2|x|}∂_{t}|u|^{2} dx|_{t=−T}^{T} .

But we can use Cauchy-Schwarz to write

∫_{R_{3}} \frac{1}{2|x|}∂_{t}|u|^{2} dx ≤ ∫_{R_{3}} \frac{|u||u_{t}|}{|x|} dx ≤ (∫_{R_{3}} |u_{t}|^{2} dx)^{1/2}(∫_{R_{3}} \frac{|u|^{2}}{|x|^{2}} dx)^{1/2}.

Hardy’s inequality (Lemma A.2) gives

(∫_{R_{3}} \frac{|u|^{2}}{|x|^{2}} dx)^{1/2} ≤ (∫_{R_{3}} |∇u|^{2} dx)^{1/2}

and so

∫_{R_{3}} \frac{1}{2|x|}∂_{t}|u|^{2} dx = O(E).

The claim follows.

An alternate way to proceed (that does not require Hardy’s inequality) is by integrating by a smooth cutoff function χ(t/T) rather than from −T to T. The argument proceeds broadly as before, until one is faced with the expression

∫R ∫_{R_{3}} χ(t/T)\frac{1}{2|x|}∂_{tt}|u|^{2} dxdt.

We can now integrate by parts twice in time to obtain

T^{−2} ∫R ∫_{R_{3}} χ′′(t/T)\frac{1}{2|x|}|u|^{2} dxdt.

To deal with this, we obtain some uniform bounds on u. From the Fourier transform formula

û(t, ξ) = cos(t|ξ|)û_{0}(ξ) + \frac{sin(t|ξ|)}{|ξ|}û_{1}(ξ)

and the assumption that u is Schwartz in space, we see that û(t) lies in L_{ξ}^{1} and L_{ξ}^{2} uniformly in t, which means that u(t) lies in L_{x}^{∞} and L_{x}^{2} uniformly in t. From

4 TERENCE TAO

this one easily shows that ∫R^{3} \frac{|u|^{2}}{|x|} dx is bounded uniformly in t (by some quantity depending on u). This gives

T^{−2} ∫R ∫_{R_{3}} χ′′(t/T)\frac{1}{2|x|}|u|^{2} dxdt = O_{u,χ}(T^{−2}T).

Now let T → ∞ to obtain the claim.

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

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