SOLUTIONS TO HW 5

TERENCE TAO

1. Exercise 2.34 (Asymptotic behaviour of Schrödinger)

We first obtain the bound |u(t, x)| ≲_{N} ⟨x⟩^{−N} when |x| ≥ 5|t| and N ≥ 0. Without loss of generality we can take N to be an integer. By time reversal symmetry we can take t > 0 (the case t = 0 is trivial).

The function û_{0} is smooth and is supported on the ball |ξ| ≤ 1. From the Fourier representation

u(t, x) = \frac{1}{(2π)^{d}} ∫_{R_{d}} e^{ix·ξ}e^{it|ξ|^{2}/2}û_{0}(ξ) dξ

we obtain the bound

|u(t, x)| ≲_{u_{0}} 1.  (1)

This is already enough when |x| = O(1), so we take |x| ≫ 1. By rotation symmetry we can take x = (x_{1}, 0, …, 0), thus

u(t, x) = \frac{1}{(2π)^{d}} ∫_{R_{d}} e^{ix_{1}ξ_{1}}e^{it|ξ|^{2}/2}û_{0}(ξ) dξ.  (2)

If one integrates by parts N times by integrating e^{ix_{1}ξ_{1}} in the ξ_{1} direction, and differentiating the other two terms, one obtains

u(t, x) = (−1)^{N} \frac{1}{(2π)^{d}} ∫_{R_{d}} \frac{1}{(ix_{1})^{N}} e^{ix_{1}ξ_{1}}(\frac{∂}{∂_{ξ_{1}}} )^{N}[e^{it|ξ|^{2}/2}û_{0}(ξ)] dξ.

Repeated application of the chain rule (using the fact that û_{0} is smooth) shows that

|(\frac{∂}{∂_{ξ_{1}}} )^{N}[e^{it|ξ|^{2}/2}û_{0}(ξ)]| ≲_{N,u_{0}} ⟨t⟩^{N}

which thus leads to a bound

|u(t, x)| ≲_{N,u_{0}} ⟨t⟩^{N}/|x|^{N}.

This bound suffices when t = O(1), but does not quite work when |t| ≫ 1 (even with the assumption |x| ≥ 5|t|, although a stronger assumption such as |x| ≥ |t|^{1+ε} would suffice). To handle the case |t| ≫ 1 one needs to be a bit smarter with the integration by parts, integrating both phase terms e^{ix_{1}ξ_{1}}e^{it|ξ|^{2}/2} and differentiating only û_{0}. The simplest way to do this is by using the identity

e^{ix_{1}ξ_{1}}e^{it|ξ|^{2}/2} = (\frac{1}{i(x_{1} + tξ_{1})}∂_{ξ_{1}})e^{ix_{1}ξ_{1}}e^{it|ξ|^{2}/2}

2 TERENCE TAO

and hence on iteration

e^{ix_{1}ξ_{1}}e^{it|ξ|^{2}/2} = (\frac{1}{i(x_{1} + tξ_{1})}∂_{ξ_{1}})^{N}e^{ix_{1}ξ_{1}}e^{it|ξ|^{2}/2}.

The transpose^{1} of (\frac{1}{i(x_{1}+tξ_{1})}∂_{ξ_{1}})^{N} is (−∂_{ξ_{1}} \frac{1}{i(x_{1}+tξ_{1})})^{N} and so by inserting these facts back into (2) we obtain

u(t, x) = \frac{1}{(2π)^{d}} ∫_{R_{d}} e^{ix_{1}ξ_{1}}e^{it|ξ|^{2}/2}(−∂_{ξ_{1}} \frac{1}{i(x_{1} + tξ_{1})})^{N} û_{0}(ξ) dξ.  (3)

One can expand out (−∂_{ξ_{1}} \frac{1}{i(x_{1}+tξ_{1})})^{N} û_{0}(ξ) using the product rule. Each one of the ∂_{ξ_{1}} derivatives ultimately either lands on û_{0}, or on one of the N \frac{1}{i(x_{1}+tξ_{1})} factors. If j ∂_{ξ_{1}} derivatives lands on a \frac{1}{i(x_{1}+tξ_{1})} factor, it becomes O_{j}(\frac{|t|^{j}}{|x_{1}+tξ_{1}|^{j+1}} ) = O_{N}(|x|^{−1}) since |x| = x_{1} ≥ 5|t|. Since û_{0} is infinitely differentiable, we conclude that

(−∂_{ξ_{1}} \frac{1}{i(x_{1} + tξ_{1})})^{N} û_{0}(ξ) = O_{u_{0},N}(|x|^{−N}).

Furthermore this quantity is clearly supported on the set where |ξ| ≤ 1. We conclude from (3) that

|u(t, x)| ≲_{u_{0},N,d} |x|^{−N}

as desired. Note that by doubling N and using the fact that |x| ≥ 5|t|, we can improve this bound “for free” to

|u(t, x)| ≲_{u_{0},N,d} ⟨x⟩^{−N}⟨t⟩^{−N}

whenever |x| ≥ 5|t|, if desired. When |x| < 5|t|, the dispersive inequality gives

|u(t, x)| ≲_{d} |t|^{−d/2}∥u_{0}∥_{L^{1}} ≲_{d,u_{0}} |t|^{−d/2}

which when combined with (1) gives

|u(t, x)| ≲_{d,u_{0}} ⟨t⟩^{−d/2}.

This is already enough to get upper bounds on ∥u(t)∥_{L_{x}^{p}(R^{d})}. For p = ∞ we clearly have ∥u(t)∥_{L_{x}^{∞}(R^{d})} ≲_{d,u_{0}} ⟨t⟩^{−d/2}. For p = 1 we have

∫_{R_{d}} |u(t, x)| dx ≤ ∫_{|x|≤5|t|} |u(t, x)| dx + ∫_{|x|>5|t|} |u(t, x)| dx
≲_{d,u_{0},N} ∫_{|x|≤5|t|}⟨t⟩^{−d/2} dx + ∫_{R_{d}}⟨t⟩^{−N}⟨x⟩^{−N} dx
≲_{d,u_{0},N} ⟨t⟩^{d/2} + ⟨t⟩^{−N}
≲_{d,u_{0},N} ⟨t⟩^{d/2}

if we choose N large enough (N > d will do). Thus ∥u(t)∥_{L_{x}^{1}(R^{d})} ≲_{d,u_{0}} ⟨t⟩^{d/2}.

1We use the transpose rather than the adjoint because we are not conjugating the second factor û_{0} in the inner product.

We can control intermediate L^{p} norms by the same method, or simply invoke Hölder’s inequality:

∥u(t)∥_{L_{x}^{p}(R^{d})} ≤ ∥u(t)∥_{L_{x}^{1}(R^{d})}^{1/p}∥u(t)∥_{L_{x}^{∞}(R^{d})}^{1/p′} ≲_{d,u_{0}} ⟨t⟩^{d( \frac{1}{p} − \frac{1}{2} )}.

On the other hand, from L^{2}-norm conservation and the hypothesis that u_{0} is nonzero

∥u(t)∥_{L_{x}^{2}(R^{d})} = ∥u_{0}∥_{L_{x}^{2}(R^{d})} ∼_{u_{0}} 1

and by Hölder’s inequality again

∥u(t)∥_{L_{x}^{2}(R^{d})} ≤ ∥u(t)∥_{L_{x}^{p}(R^{d})}^{1/2}∥u(t)∥_{L_{x}^{p′} (R^{d})}^{1/2}

for any 1 ≤ p ≤ ∞. Combining these three estimates we see that we can complement the upper bound with a lower bound:

∥u(t)∥_{L_{x}^{p}(R^{d})} ≳_{d,u_{0}} ⟨t⟩^{d( \frac{1}{p} − \frac{1}{2} )}.

(We thus see here that Hölder’s inequality can be used in two equivalent ways when dealing with the L^{p} norms of a single function, in this case u(t). Firstly, it shows that upper bounds on two L^{p} norms imply upper bounds on any intermediate norm. Contrapositively, an upper bound on one L^{p} norm combined with a lower bound on another L^{p} norm gives further lower bounds beyond the second L^{p}.)

2. Exercise 2.50 (Local near-conservation of mass)

By time reversal symmetry we can concern ourselves only with positive times t > 0. Let M(t) be the quantity

M(t) := ∫_{R_{d}} ϕ^{2}(\frac{x}{R})|u(t, x)|^{2} dx

where ϕ is as in the hint. This quantity is smooth in t and from (2.36) and the triangle inequality we have

|∂_{t}M(t)| ≤ \frac{2}{R} ∫_{R_{d}} ϕ(\frac{x}{R})|∇ϕ(\frac{x}{R})||u(t, x)||∇u(t, x)| dx.

By Cauchy-Schwarz we conclude that

|∂_{t}M(t)| ≤ \frac{2}{R}M(t)^{1/2}(∫_{R_{d}} |∇ϕ(\frac{x}{R})|^{2}|∇u(t, x)|^{2} dx)^{1/2}.

On the other hand, from energy conservation we have

∫_{R_{d}} |∇u(t, x)|^{2} dx = 2E

and hence

∫_{R_{d}} |∇ϕ(\frac{x}{R})|^{2}|∇u(t, x)|^{2} dx ≲_{ϕ} E.

We conclude the differential inequality

|∂_{t}M(t)| ≲_{ϕ} \frac{E^{1/2}}{R} M(t)^{1/2}.

There are a number of ways to exploit this inequality. One is by the comparison principle, comparing M(t) with a solution to the ODE ∂_{t}F (t) = C_{ϕ} \frac{E^{1/2}}{R} F (t)^{1/2} with the same initial data. Another is by making the substitution M(t) = m(t)^{2} and formally obtaining the inequality

|∂_{t}m(t)| ≲_{ϕ} \frac{E^{1/2}}{R}

as a consequence. Strictly speaking there is a problem because m(t) need not be smooth when M(t) vanishes. (The square root of a non-negative smooth function need not be smooth; look at the square root of t^{2}, for instance.) One can fix this by working with m_{ε}(t) := (ε + M(t)^{1/2} first and then taking limits as ε → 0; one checks that

|∂_{t}m_{ε}(t)| ≲_{ϕ} \frac{E^{1/2}}{R}

for all ε > 0, so in particular

m_{ε}(t) ≤ m_{ε}(0) + O_{ϕ}(\frac{E^{1/2}|t|}{R}).

Taking limits as ε → 0 we obtain

m(t) ≤ m(0) + O_{ϕ}(\frac{E^{1/2}|t|}{R});

since

(∫_{|x|≤R} |u(t, x)|^{2} dx)^{1/2} ≤ m(t); (∫_{|x|≤2R} |u(0, x)|^{2} dx)^{1/2} ≥ m(0)

the claim follows. An alternate approach is to work with the integral form of the inequality, concluding that

sup_{0≤t≤T} M(t) ≤ M(0) + ∫_{0}^{T} |∂_{t}M(t)| dt ≤ M(0) + O_{ϕ}(\frac{E^{1/2}T}{R} [ sup_{0≤t≤T} M(t)]^{1/2})

which leads (using for instance the Cauchy-Schwarz inequality ab ≤ \frac{1}{2}[εa^{2} + \frac{1}{ε}b^{2}]) to the slightly weaker estimate

sup_{0≤t≤T} M(t) ≲_{ϕ} M(0) + (\frac{E^{1/2}T}{R} )^{2}

which almost proves the desired claim except for the loss of a constant.

3. Exercise 2.51 (Local near-conservation of mass II)

Again take t > 0. From energy conservation we have

∫_{0}^{t} ∫_{R_{d}} |∇u(t, x)|^{2} dx = 2tE

and hence by the pigeonhole principle we can find 1 ≤ j ≤ n such that

∫_{0}^{t} ∫_{2^{j}R≤|x|≤2^{j+1}R} |∇u(t, x)|^{2} dx = 2tE/n.

Now we run the argument of the preceding section but now take

M_{j}(t) := ∫_{R_{d}} ϕ^{2}(\frac{x}{2^{j}R})|u(t, x)|^{2} dx

As before we have

|∂_{t}M_{j}(t)| ≤ \frac{2}{2^{j}R}M_{j}(t)^{1/2}(∫_{R_{d}} |∇ϕ(\frac{x}{2^{j}R})|^{2}|∇u(t, x)|^{2} dx)^{1/2}.

Now we take advantage of the fact that ∇ϕ(y) is only non-zero when 1 ≤ |y| ≤ 2, ths

|∂_{t}M_{j}(t)| ≲_{ϕ} \frac{1}{2^{j}R}M_{j}(t)^{1/2}(∫_{2^{j}R≤|x|≤2^{j+1}R} |∇u(t, x)|^{2} dx)^{1/2}.

We formally square root this, setting m_{j}(t) := M_{j}(t)^{1/2}; to be more rigorous one should work with regularised functions such as m_{j,ε}(t) = (ε + M_{j}(t))^{1/2} but we will not do so here. We obtain

|∂_{t}m_{j}(t)| ≲_{ϕ} \frac{1}{2^{j}R}(∫_{2^{j}R≤|x|≤2^{j+1}R} |∇u(t, x)|^{2} dx)^{1/2}

and hence

m_{j}(t) ≤ m_{j}(0) + O_{ϕ}(\frac{1}{2^{j}R} ∫_{0}^{t}(∫_{2^{j}R≤|x|≤2^{j+1}R} |∇u(t′, x)|^{2} dx)^{1/2} dt′).

From Cauchy-Schwarz we have

∫_{0}^{t}(∫_{2^{j}R≤|x|≤2^{j+1}R} |∇u(t′, x)|^{2} dx)^{1/2} dt′ ≤ t(2E/n)^{1/2}

and by discarding the denominator 2^{j} we thus obtain

m_{j}(t) ≤ m_{j}(0) + O_{ϕ}(\frac{E}{Rn^{1/2}} )

whence the claim.

Note that the last step (throwing away the 2^{j}) is extremely wasteful. Indeed, instead of applying the pigeonhole principle we could simply set j = n, foregoing the 1/n gain coming from the pigeonhole principle and taking the 2^{j} = 2^{n} gain instead, to obtain the result with n^{1/2} replaced by 2^{n}. (Indeed, one could simply apply the previous exercise with R replaced by 2^{n}R to achieve this). However, the pigeonhole trick is worth remembering because it comes in very handy in other situations in which such a gain in n cannot be achieved by any other means.

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

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