SOLUTIONS TO FINAL

TERENCE TAO

1. Exercise 1.48 (Virial identity)

Let A(t) be the action

A(t) := ∑_{j=1}^{N} m_{j}x_{j}(t) · ∂_{t}x_{j}(t).

We differentiate in time and obtain

∂_{t}A(t) = ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(t)|^{2} + ∑_{j=1}^{N} m_{j}x_{j}(t) · ∂_{tt}x_{j}(t).

We apply the equation and rewrite this as

∂_{t}A(t) = ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(t)|^{2} − ∑_{j=1}^{N} ∑_{i≠j} x_{j}(t) · G\frac{m_{i}m_{j}}{|x_{i}(t) − x_{j}(t)|^{2}} \frac{x_{j}(t) − x_{i}(t)}{|x_{j}(t) − x_{i}(t)|}.

We can symmetrise the second term by swapping i and j to obtain

∂_{t}A(t) = ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(t)|^{2} − \frac{1}{2} ∑_{j=1}^{N} ∑_{i≠j}(x_{j}(t)−x_{i}(t))·G\frac{m_{i}m_{j}}{|x_{i}(t) − x_{j}(t)|^{2}} \frac{x_{j}(t) − x_{i}(t)}{|x_{j}(t) − x_{i}(t)|}

which simplifies to

∂_{t}A(t) = ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(t)|^{2} − ∑_{1≤i<j≤N} G\frac{m_{i}m_{j}}{|x_{i}(t) − x_{j}(t)|}.

We average this in time from −T to T and obtain

\frac{A(T) − A(−T)}{2T} = \frac{1}{2T} ∫_{−T}^{T} ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(t)|^{2} − ∑_{1≤i<j≤N} G\frac{m_{i}m_{j}}{|x_{i}(t) − x_{j}(t)|} dt.

Taking limits as T → +∞ we obtain that the limit of both sides exists and

lim_{T →+∞} \frac{A(T) − A(−T)}{2T} = 2⟨T ⟩ + ⟨V ⟩

and so it will suffice to show that

lim_{T →+∞} \frac{A(T) − A(−T)}{2T} = 0.

Actually, since the limit is already known to exist, it will suffice to find a sequence of times T_{n} → +∞ for which

lim_{n→∞} \frac{A(T_{n}) − A(−T_{n})}{2T_{n}} = 0.

Now from Cauchy-Schwarz we have

|A(t)| ≤ (∑_{j=1}^{N} m_{j}|x_{j}(t)|^{2})^{1/2}(∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(t)|^{2})^{1/2}.

On the one hand, we know that the x_{j} are uniformly bounded, hence we have

|A(t)| ≲_{N,m,x} (∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(t)|^{2})^{1/2}.

Thus it will suffice to find a sequence of times T_{n} → +∞ for which

lim_{n→∞}[∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(T_{n})|^{2} + ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(−T_{n})|^{2}]/T_{n}^{2} = 0.

Note that we are not initially assuming the velocities to be uniformly bounded. Energy conservation will not directly help us out here because the potential energy can be arbitrarily negative (we have no lower bound on the separation between the masses). However we do know that the average kinetic energy ⟨T ⟩ exists asymptotically. But from the convergence (and hence boundedness) of this average kinetic energy we already know that the averages

\frac{1}{T/2} ∫_{T/2}^{T} ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(t)|^{2} + ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(−t)|^{2} dt

(say) stays bounded as T → ∞. Thus by the pigeonhole principle we can find a sequence of times T_{n} → +∞ for which

∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(T_{n})|^{2} + ∑_{j=1}^{N} m_{j}|∂_{t}x_{j}(−T_{n})|^{2}

is uniformly bounded, and the claim follows.

2. Exercise 1.55 (Stable manifold)

Observe that whenever u : [0, T] → D is a classical solution, we have

∂_{t}∥u∥_{D}^{2} = 2⟨u_{t}, u⟩
≤ 2⟨Lu, u⟩
≤ 0

and so we have ∥u(t)∥_{D} ≤ ∥u_{0}∥_{D}. Combining this with the Picard existence and uniqueness theorem we see that the solution can be continued indefinitely forward in time (since the only way that blowup can occur is if the norm ∥u(t)∥_{D} goes to infinity in finite time, which is not the case. Thus we have a unique classical solution on [0, +∞). (Blowup can occur as t → −∞, but this will not concern us here.)

Next, we integrate the above monotonicity formula in time to obtain

∥u(T)∥_{D}^{2} − ∥u_{0}∥_{D}^{2} ≤ 2 ∫_{0}^{T} ⟨Lu(t), u(t)⟩ dt

and hence on taking limits as T → ∞

∫_{0}^{T} |⟨Lu(t), u(t)⟩| dt ≤ ∥u_{0}∥_{D}^{2}.

Now we know that u(t) stays bounded as t → ∞, which by the equation u_{t} = Lu + Nu (and the continuity of L and N) shows that u_{t} is also bounded, thus u is uniformly Lipschitz and bounded. This implies that ⟨Lu(t), u(t)⟩ → 0 as t → +∞. This (via a standard compactness argument) shows that u(t) must converge to the space V = {u ∈ D : ⟨Lu, u⟩ = 0} as claimed. Note that V is in fact a vector space, as can be seen by considering the bilinear positive semi-definite inner product (u, v) := −⟨Lu, v⟩ − ⟨u, Lv⟩.

3. Exercise 2.52 (Weighted Sobolev spaces)

First let us give the Fourier-analytic proof. Let us normalise ∥f∥_{H_{x}^{k,k}(R^{d})} = 1, so that ∥⟨x⟩^{j}f∥_{H_{x}^{k−j}(R^{d})} ≤ 1 for all 0 ≤ j ≤ k, and thus

∥∇^{i}(⟨x⟩^{j}f)∥_{L_{x}^{2}(R^{d})} ≲_{k,d} 1

whenever i, j ≥ 0 are such that i + j ≤ d. By using the Leibnitz rule

∇^{i}(⟨x⟩^{j}f) = ⟨x⟩^{j}∇^{i}f + l.o.t.

and an induction on j we see that

∥⟨x⟩^{j}∇^{i}f∥_{L_{x}^{2}(R^{d})} ≲_{k,d} 1  (1)

whenever i, j ≥ 0 are such that i + j ≤ d.

Our task is then to show that

∥⟨x⟩^{j}e^{itΔ/2}f∥_{H_{x}^{k−j}(R^{d})} ≲_{k,d} ⟨t⟩^{k}

for all 0 ≤ j ≤ k. This in turn would follow if we could show

∥∇^{l}(⟨x⟩^{j}e^{itΔ/2}f)∥_{L_{x}^{2}(R^{d})} ≲_{k,d} ⟨t⟩^{k}

for 0 ≤ l ≤ k − j. Applying the Leibnitz rule repeatedly, noting that ∇^{i}⟨x⟩^{j} = O_{i,j,d}(⟨x⟩^{j−i}), we reduce to showing that

∥⟨x⟩^{j−i}∇^{l−i}e^{itΔ/2}f∥_{L_{x}^{2}(R^{d})} ≲_{k,d} ⟨t⟩^{k}

for all 0 ≤ i ≤ l. Note that ∇^{l−i} commutes with e^{itΔ/2}. We thus reduce to showing that

∥x^{a}e^{itΔ/2}∇^{b}f∥_{L_{x}^{2}(R^{d})} ≲_{k,d} ⟨t⟩^{k}

for all multi-indices a, b with |a| + |b| ≤ k. We apply Fourier transforms and reduce to showing

∥∇^{a}[e^{−it|ξ|^{2}/2}ξ^{b}f̂]∥_{L_{x}^{2}(R^{d})} ≲_{k,d} ⟨t⟩^{k}.

We observe that each derivative in ∇^{a} either falls on f̂, falls on ξ^{b} (thus lowering the multiindex b by one), or falls on e^{−it|ξ|^{2}/2} (which increases the multiindex ξ^{b} and also gains a t). At the end of the day we see that we obtain a finite linear combination of terms of the form

t^{c}e^{−it|ξ|^{2}/2}ξ^{e}∇^{g}f̂

where c ≥ 0 and e, g are multiindices such that 0 ≤ c ≤ k, and 0 ≤ |e| + |g| ≤ k. (Some further estimates are available but we will not need them). It thus suffices to show that

∥ξ^{e}∇^{g}f̂∥_{L_{x}^{2}(R^{d})} ≲_{k,d} 1.

We invert the Fourier transform and reduce to

∥∇^{e}(ξ^{g}f̂)∥_{L_{x}^{2}(R^{d})} ≲_{k,d} 1.

Applying the Leibnitz rule and using (1) we obtain the claim.

Now we give the integration by parts proof. We write u_{0} := f and u(t) := e^{itΔ/2}f, so that u solves the Schrödinger equation with initial data u_{0}. We also normalise ∥u_{0}∥_{H_{x}^{k,k}(R^{d})} = 1. Our task is then to show that

∥u(t)∥_{H_{x}^{k,k}(R^{d})} ≲_{k,d} ⟨t⟩^{k}.

After repeated application of the Leibnitz rule, we see that it suffices to show that

∥⟨x⟩^{a}∇^{b}u(t)∥_{L_{x}^{2}(R^{d})} ≲_{k,d} ⟨t⟩^{a}

for all a ≥ 0 and multi-indices b with a + |b| ≤ k. The exponent of a on the right emerges rather naturally from the inductive argument (or from considering the k = 1 case), as we shall see. Squaring, we reduce to showing that

∫_{R_{d}}⟨x⟩^{2a}|∇^{b}u(t, x)|^{2} dx ≲_{k,d} ⟨t⟩^{2a}.

We induct on a. When a = 0 the claim follows from H^{s} norm conservation, so assume a > 0 and that the claim has already been proven for all smaller values of a. We apply (2.39) from the notes followed by the triangle inequality and conclude that

|∂_{t} ∫_{R_{d}}⟨x⟩^{2a}|∇^{b}u(t, x)|^{2} dx| ≲_{k,d} ∫_{R_{d}}⟨x⟩^{2a−1}|∇^{|b|}u(t, x)||∇^{|b|+1}u(t, x)| dx.

By induction hypothesis we have

∫_{R_{d}}⟨x⟩^{2a−2}|∇^{|b|+1}u(t, x)|^{2} dx ≲_{k,d} ⟨t⟩^{2a−2}

so by Cauchy-Schwarz we have

|∂_{t} ∫_{R_{d}}⟨x⟩^{2a}|∇^{b}u(t, x)|^{2} dx| ≲_{k,d} ⟨t⟩^{a−1}(∫_{R_{d}}⟨x⟩^{2a}|∇^{|b|}u(t, x)|^{2})^{1/2}.

We can morally cancel the integrals here, but there is a slight issue because of ∇^{b} versus ∇^{|b|}. So we shall integrate and obtain

∫_{R_{d}}⟨x⟩^{2a}|∇^{b}u(t, x)|^{2} dx ≲_{k,d} 1 + ∫_{0}^{t}⟨t′⟩^{a−1}(∫_{R_{d}}⟨x⟩^{2a}|∇^{|b|}u(t′, x)|^{2})^{1/2} dt′

and hence

∫_{R_{d}}⟨x⟩^{2a}|∇^{|b|}u(t, x)|^{2} dx ≲_{k,d} 1 + ⟨t⟩^{a} sup_{0≤t′≤t}(∫_{R_{d}}⟨x⟩^{2a}|∇^{|b|}u(t′, x)|^{2})^{1/2}

and hence

sup_{0≤t′≤t}(∫_{R_{d}}⟨x⟩^{2a}|∇^{|b|}u(t, x)|^{2} dx) ≲_{k,d} 1 + ⟨t⟩^{a} sup_{0≤t′≤t}(∫_{R_{d}}⟨x⟩^{2a}|∇^{|b|}u(t′, x)|^{2})^{1/2}.

We solve for this to obtain

sup_{0≤t′≤t}(∫_{R_{d}}⟨x⟩^{2a}|∇^{|b|}u(t, x)|^{2} dx) ≲_{k,d} ⟨t⟩^{2a}

and the claim easily follows.

4. Exercise 3.9. (Localised blowup of NLW)

All of you got this question correctly (I realise now that the hint was a bit too much of a giveaway!), so I won’t replicate the solution here.

5. Exercise 3.18. (L^{2} critical wellposedness of NLS)

Fix p, μ, d, R, u_{∗} as in the theorem, and let ε_{0} be chosen later. Let I, u_{0} be as in the theorem. There are several ways to proceed. Here we will follow the hint and employ the norm

∥u∥_{S} := δ∥u∥_{S^{0}(I×R^{d})} + ∥u∥_{L_{t,x}^{2(d+2)/d}(I×R^{d})}

where 0 < δ ≤ 1 is a small quantity to be chosen later. Let u_{0} ∈ B. We first locate a fixed point of the functional

Φ(u) := e^{itΔ/2}u_{0} − i ∫_{0}^{t} e^{i(t−t′)Δ/2}F (u(t′)) dt′

where F (z) := μ|z|^{p−1}z, in the ball B_{S} := {u ∈ S : ∥u∥_{S} ≤ C_{0}ε_{0}}, where C_{0} is a big constant to be chosen later..

First we need to verify that Φ maps B_{S} to itself for Schwartz functions (the general case will follow by a limiting argument that we present shortly). If u ∈ B_{S}, we compute

∥Φ(u)∥_{S} ≤ ∥e^{itΔ/2}u_{0}∥_{S} + ∥ ∫_{0}^{t} e^{i(t−t′)Δ/2}F (u(t′)) dt′∥_{S}

≤ δ∥e^{itΔ/2}u_{0}∥_{S^{0}} + ∥e^{itΔ/2}u_{∗}∥_{L_{t,x}^{2(d+2)/d}(I×R^{d})} + ∥e^{itΔ/2}(u_{0} − u_{∗})∥_{L_{t,x}^{2(d+2)/d}(I×R^{d})} + ∥ ∫_{0}^{t} e^{i(t−t′)Δ/2}F (u(t′)) dt′∥_{S}

≤ δ∥e^{itΔ/2}u_{0}∥_{S^{0}} + ε + ∥e^{itΔ/2}(u_{0} − u_{∗})∥_{S^{0}} + ∥ ∫_{0}^{t} e^{i(t−t′)Δ/2}F (u(t′)) dt′∥_{S}

and hence by the Strichartz inequality

∥Φ(u)∥_{S} ≤ O_{d}(δR) + ε_{0} + O(ε_{0}) + O_{d}(∥F (u)∥_{L_{t,x}^{2(d+2)/(d+4)}(I×R^{d})}).

But from Hölder’s inequality we have

∥F (u)∥_{L_{t,x}^{2(d+2)/(d+4)}(I×R^{d})} ≤ ∥u∥_{L_{t,x}^{2(d+2)/d}(I×R^{d})}^{p} ≤ ∥u∥_{S}^{p} ≲_{d} C_{0}^{p}ε_{0}^{p}

and so

∥Φ(u)∥_{S} ≤ O_{d}(δR) + O(ε_{0}) + O_{d}(C_{0}^{p}ε_{0}^{p}).

If C_{0} is sufficiently large depending on d, ε_{0} is sufficiently small depending on C_{0}, d, and then δ is sufficiently small depending on ε_{0}, R, we can make the RHS less than C_{0}ε_{0}, and we obtain boundedness.

With a similar argument we can show that Φ is a contraction. Indeed, if u, v ∈ B_{S} are Schwartz we have

∥Φ(u) − Φ(v)∥_{S} ≤ ∥ ∫_{0}^{t} e^{i(t−t′)Δ/2}[F (u(t′)) − F (v(t′))] dt′∥_{S}
≲_{d} ∥F (u) − F (v)∥_{L_{t,x}^{2(d+2)/(d+4)}(I×R^{d})}).

Now from the pointwise estimate

|F (u) − F (v)| ≲_{p} |u − v|(|u| + |v|)^{p−1}

and Hölder we have

∥F (u) − F (v)∥_{L_{t,x}^{2(d+2)/(d+4)}(I×R^{d})}) ≲ ∥u − v∥_{L_{t,x}^{2(d+2)/d}}[∥u∥_{L_{t,x}^{2(d+2)/d}} + ∥v∥_{L_{t,x}^{2(d+2)/d}}]^{p−1}
≲ ∥u − v∥_{S}[∥u∥_{S} + ∥v∥_{S}^{p−1}]
≲_{d} ε_{0}^{p−1}∥u − v∥_{S}

and hence

∥Φ(u) − Φ(v)∥_{S} ≲_{d} ε_{0}^{p−1}∥u − v∥_{S}.

If ε_{0} is sufficiently small we obtain that Φ is a contraction, say of Lipschitz constant 1/2. One can now extend the previous bounds on Φ to general functions in B_{S} by a limiting argument. We can then run the contraction mapping theorem and obtain a unique solution u ∈ B_{S} for each u_{0} ∈ B. A similar argument to that given in the notes or in class then also shows that the map u_{0} ↦ u is Lipschitz continuous. The last thing is to show that u is in fact unique in the entire class S (not just in the ball B_{S}), but this can be achieved by a continuity argument similar to that done in class.

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

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