SOLUTIONS TO HW 3

TERENCE TAO

1. Exercise 2.23 (Conformal energy)

Let us first show that the quantity

E(t) := ∥u(t)∥_{Ḣ^{1/2}(R^{d})}^{2} + ∥∂_{t}u(t)∥_{Ḣ^{−1/2}(R^{d})}^{2}

is conserved in time. There are several ways to do this. One is via the explicit solution

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

and hence (noting that ∂_{t} commutes with the spatial Fourier transform)

ˆ∂_{t}u(t, ξ) = −|ξ| sin(t|ξ|)û_{0}(ξ) + cos(t|ξ|)û_{1}(ξ)

which implies that (up to factors of 2π, which are irrelevant here)

∥u(t)∥_{Ḣ^{1/2}(R^{d})}^{2} = ∫_{R_{d}} |û_{0}(ξ)|^{2}|ξ| cos^{2}(t|ξ|)+2Re(û_{0}(ξ)\overline{û_{1}(ξ)}) cos(t|ξ|) sin(t|ξ|)+\frac{|û_{1}(ξ)|^{2}}{|ξ|} sin^{2}(t|ξ|) dξ

and

∥∂_{t}u(t)∥_{Ḣ^{−1/2}(R^{d})}^{2} = ∫_{R_{d}} |û_{0}(ξ)|^{2}|ξ| sin^{2}(t|ξ|)−2Re(û_{0}(ξ)\overline{û_{1}(ξ)}) cos(t|ξ|) sin(t|ξ|)+\frac{|û_{1}(ξ)|^{2}}{|ξ|} cos^{2}(t|ξ|) dξ

and hence

E(t) = ∫_{R_{d}} |û_{0}(ξ)|^{2}|ξ| + \frac{|û_{1}(ξ)|^{2}}{|ξ|} dξ

which is obviously conserved in time. A second approach is to show that ∂_{t}E(t) = 0. One way is via the Fourier representation (again ignoring factors of 2π)

E(t) = ∫_{R_{d}} |û(t, ξ)|^{2}|ξ| + \frac{|∂_{t}û(t, ξ)|^{2}}{|ξ|} dξ

whence

∂_{t}E(t) = ∫_{R_{d}} 2Re(û(t, ξ)\overline{∂_{t}û(t, ξ)})|ξ| + 2\frac{Re(∂_{tt}û(t, ξ)\overline{∂_{t}û(t, ξ)})}{|ξ|} dξ.

But by taking spatial Fourier transforms of the wave equation

∂_{tt}u(t, x) = Δu(t, x)

we have

∂_{tt}û(t, ξ) = −|ξ|^{2}û(t, ξ)

and we thus see that the right-hand side vanishes as claimed.

Alternatively, one can work in physical space, writing (again ignoring factors of 2π)

E(t) = ∥|∇|^{1/2}u(t)∥_{L^{2}(R^{d})}^{2}+∥|∇|^{−1/2}∂_{t}u(t)∥_{L^{2}(R^{d})}^{2} = ⟨u(t), |∇|u(t)⟩_{L^{2}(R^{d})}+⟨∂_{t}u(t), |∇|^{−1}∂_{t}u(t)⟩_{L^{2}(R^{d})}

where we use the fact that the powers of |∇| are self-adjoint. We differentiate in time and conclude (after juggling the powers of |∇| around a bit)

∂_{t}E(t) = ⟨∂_{t}u(t), |∇|u(t)⟩_{L^{2}(R^{d})} + ⟨∂_{t}u(t), |∇|^{−1}∂_{tt}u(t)⟩_{L^{2}(R^{d})}.

But from the wave equation we have

|∇|^{−1}∂_{tt}u(t) = |∇|^{−1}Δu(t) = −|∇|u(t)

(as one can see from a quick appeal to the Fourier transform) and the claim follows. (Note that this proof is really the Fourier proof in disguise).

Now we turn to the Lorentz invariance. Here I unfortunately discovered that the formula given for the Lorentz transformation in the notes was incorrect (except in one dimension!); sorry for the inconvenience! Let us set v = v_{1}e_{1} for simplicity. Then the Lorentz transformation is given by

u_{v}(t, x_{1}, \underline{x}) := u(\frac{t − v_{1}x_{1}}{\sqrt{1 − v_{1}^{2}}} , \frac{x_{1} − v_{1}t}{\sqrt{1 − v_{1}^{2}}} , \underline{x})

where \underline{x} = (x_{2}, …, x_{d}), and if we set

E_{v}(t) = ∥u_{v}(t)∥_{Ḣ^{1/2}(R^{d})}^{2} + ∥∂_{t}u_{v}(t)∥_{Ḣ^{−1/2}(R^{d})}^{2}

our task is to show that E_{v} is independent of v, thus

E_{v}(t) = E(t).

Let us first demonstrate this directly, via the explicit solution (1). Here, one can save some trouble by converting the sines and cosines to complex exponentials, which will be easier to manipulate. More precisely we have

û(t, ξ) = f_{+}(ξ)e^{it|ξ|} + f_{−}(ξ)e^{−it|ξ|}

where f_{±}(ξ) := \frac{1}{2} û_{0}(ξ) ± \frac{1}{2i|ξ|} û_{1}(ξ). Observe that

û_{0}(ξ) = f_{+}(ξ) + f_{−}(ξ) and û_{1}(ξ) = i|ξ|f_{+}(ξ) − i|ξ|f_{−}(ξ)

and hence after a brief calculation (and ignoring factors of 2π)

E(t) = E(0) = ∑_{±} ∫_{R_{d}} |f_{±}(ξ)|^{2} |ξ|dξ.  (2)

Now we apply the Lorentz transform to see how this transforms f_{±} and then E(t). Since this transform is defined in physical space, let us work in physical co-ordinates (t, x). From the Fourier inversion formula we have

u(t, x) = ∑_{±} ∫_{R_{d}} f_{±}(ξ)e^{ix·ξ±it|ξ|} dξ  (3)

and so upon applying the Lorentz transform

u_{v}(t, x) = ∑_{±} ∫_{R_{d}} f_{±}(ξ)e^{i_{\frac{x_{1}−v_{1}t}{\sqrt{1−v^{2}_{1}}}} ξ_{1}+\underline{x}·\underline{ξ}±i_{\frac{t−v_{1}x_{1}}{\sqrt{1−v^{2}_{1}}}} |ξ|} dξ.

We rearrange this a bit as

u_{v}(t, x) = ∑_{±} ∫_{R_{d}} f_{±}(ξ)e^{ix·Φ_{±}(ξ)±itΨ_{±}(ξ)} dξ

where

Φ_{±}(ξ) := (\frac{ξ_{1} ∓ v_{1}|ξ|}{\sqrt{1 − v_{1}^{2}}}, \underline{ξ})

and

Ψ_{±}(ξ) := \frac{|ξ| ∓ v_{1}ξ_{1}}{\sqrt{1 − v_{1}^{2}}} .

A brief computation shows that

Ψ_{±}(ξ) = |Φ_{±}(ξ)|;

this miracle is connected to the fact that the Lorentz transformation converts solutions of the wave equation to solutions of the wave equation. Changing ξ to ξ_{±} (to add clarity a little later) we thus have

u_{v}(t, x) = ∑_{±} ∫_{R_{d}} f_{±}(ξ_{±})e^{ix·Φ_{±}(ξ_{±})±it|Φ_{±}(ξ_{±})|} dξ_{±}.

Now it is natural to make the change of variables

η_{±} = Φ_{±}(ξ_{±})

which is invertible, indeed a brief computation shows that

ξ_{±} = (\frac{(η_{±})_{1} ± v_{1}|η_{±}|}{\sqrt{1 − v_{1}^{2}}}, \underline{η_{±}})

and we have as Jacobian

dη_{±} = \frac{1 ∓ v_{1}(ξ_{±})_{1}/|ξ_{±}|}{\sqrt{1 − v_{1}^{2}}}dξ_{±} = \frac{Ψ_{±}(ξ_{±})}{|ξ_{±}|}dξ_{±} = \frac{|η_{±}|}{|ξ_{±}|}dξ_{±}

and thus

u_{v}(t, x) = ∑_{±} ∫_{R_{d}} f_{±}(ξ_{±})e^{ix·η_{±}±it|η_{±}|} \frac{|ξ_{±}|}{|η_{±}|}dξ_{±}.

Comparing this with (3), and letting f_{±}^{(v)} be the counterpart of f_{±} for u_{v}, we conclude that

f_{±}^{(v)}(η_{±}) = f_{±}(ξ_{±})\frac{|ξ_{±}|}{|η_{±}|}

and thus by (2)

E_{v}(t) = ∑_{±} ∫_{R_{d}} f_{±}(ξ_{±})^{2} \frac{|ξ_{±}|^{2}}{|η_{±}|^{2}} |η_{±}| dη_{±}

and then by undoing the change of variables

E_{v}(t) = ∑_{±} ∫_{R_{d}} f_{±}(ξ_{±})^{2}|ξ_{±}| dξ_{±} = E_{0}(t)

as desired.

Let us give a second, more “local” way to demonstrate the Lorentz invariance, taking advantage of a group law. Let S(v_{1}) denote the Lorentz transform operator with velocity v_{1}, S(v_{1})u := u_{v}, or in other words

S(v_{1})u(t, x) := u(\frac{t − v_{1}x_{1}}{\sqrt{1 − v_{1}^{2}}} , \frac{x_{1} − v_{1}t}{\sqrt{1 − v_{1}^{2}}} , \underline{x}).

Observe the velocity addition law

S(v_{1})S(w_{1})u = S(\frac{v_{1} + w_{1}}{1 + v_{1}w_{1}} )u.  (4)

This law can be clarified by using the hyperbolic substitution v_{1} = tanh α, w_{1} = tanh β,in which case

S(v_{1})u(t, x) := u(cosh αt − sinh αx_{1}, cosh αx_{1} − sinh αt, \underline{x})

and the velocity addition law is basically reflecting the fact hyperbolic tangent addition formula

tanh(α + β) = \frac{tanh α + tanh β}{1 + tanh α tanh β} .

Now let E[u] denote the conformal energy of u. We have to show that

E[S(v_{1})u] = E[S(0)u] = E[u]

and so by the fundamental theorem of calculus it suffices to show that

\frac{d}{dv_{1}} E[S(v_{1})u] = 0

for all v_{1}.

Actually we can use the velocity addition law to reduce to the case v_{1} = 0. If we differentiate both sides of (4) with respect to v_{1} and then set v_{1} = 0 we obtain

S′(0)S(w_{1})u = (1 − w_{1}^{2})S′(w_{1})u

and hence by the chain rule

\frac{d}{dw_{1}} E[S(w_{1})u] = \frac{1}{1 − w_{1}^{2}} \frac{d}{dv_{1}} E[S(v_{1})S(w_{1})u]|_{v_{1}=0}.

Since S(w_{1})u also solves the wave equation, it thus suffices to show that

\frac{d}{dv_{1}} E[S(v_{1})u]|_{v_{1}=0} = 0

for any (classical) solution u to the wave equation. Now we have (ignoring factors of 2π)

E[S(v_{1})u] = ⟨S(v_{1})u(0), |∇|S(v_{1})u(0)⟩_{L_{x}^{2}(R^{d})}+⟨S(v_{1})∂_{t}u(0), |∇|^{−1}S(v_{1})∂_{t}u(0)⟩_{L_{x}^{2}(R^{d})}

so on differentiating and rearranging a bit (noting that S(0) is just the identity)

\frac{d}{dv_{1}} E[S(v_{1})u]|_{v_{1}=0} = 2⟨S′(0)u(0), |∇|u(0)⟩_{L_{x}^{2}(R^{d})}+2⟨∂_{t}S′(0)u(0), |∇|^{−1}∂_{t}u(0)⟩_{L_{x}^{2}(R^{d})}.

Now a brief calculation shows that

S′(0)u = −x_{1}∂_{t}u − t∂_{x_{1}}u

and hence

S′(0)u(0) = −x_{1}∂_{t}u and ∂_{t}S′(0)u(0) = −x_{1}∂_{tt}u − ∂_{x_{1}}u = −x_{1}Δu − ∂_{x_{1}}u

where we have used the wave equation. Thus we have

\frac{d}{dv_{1}} E[S(v_{1})u]|_{v_{1}=0} = −2⟨x_{1}∂_{t}u(0), |∇|u(0)⟩_{L_{x}^{2}(R^{d})}
− 2⟨∂_{x_{1}}u(0), |∇|^{−1}∂_{t}u(0)⟩_{L_{x}^{2}(R^{d})}
− 2⟨x_{1}Δu(0), |∇|^{−1}∂_{t}u(0)⟩_{L_{x}^{2}(R^{d})}.

Using integration by parts to put everything off of the ∂_{t}u(0) term, we see that we have to show that

x_{1}|∇|u(0) + |∇|^{−1}∂_{x_{1}}u(0) + |∇|^{−1}[x_{1}Δu(0)] = 0.

This is easiest seen by taking Fourier transforms:

i∂_{ξ_{1}}[|ξ|û(0)] + |ξ|^{−1}iξ_{1}û(0) − |ξ|^{−1}i∂_{ξ_{1}}[|ξ|^{2}û(0)] = 0,

which is verified by direct calculation.

Now we give a third proof proceeding via the spacetime Fourier transform

ũ(τ, ξ) := ∫R ∫_{R_{d}} e^{−ix·ξ−itτ} u(t, x) dtdx

of the solution u. This transform is only defined in a distributional sense. The wave equation

−∂_{tt}u + Δu = 0

becomes

(τ^{2} − |ξ|^{2})ũ = 0.

Morally speaking, this means that ũ takes the form

ũ(τ, ξ) = F (τ, ξ)δ(τ^{2} − |ξ|^{2})

for some function F, where δ is the Dirac mass, thus δ(τ^{2}−|ξ|^{2}) is the distributional limit of ψ_{ε}(τ^{2} − |ξ|^{2}) as ε → 0, where ψ_{ε} is your favourite approximation to the identity. To actually show this rigorously is somewhat tricky, however, because of the double zero that τ^{2} − |ξ|^{2} has at the origin (τ, ξ) = (0, 0). One can fix this in a number of ways; one way is to approximate the solution u to another solution to the wave equation whose Fourier transforms (both spatial and spacetime) vanish near the origin, and then take limits at the end. Another way is via the representation (3), which we can rewrite as

u(t, x) = ∑_{±} ∫R ∫_{R_{d}} f_{±}(ξ)e^{ix·ξ+itτ} δ(τ ∓ |ξ|) dξdτ

which on comparison with the spacetime Fourier inversion formula yields

ũ(τ, ξ) = ∑_{±} f_{±}(ξ)δ(τ ∓ |ξ|).

Now, observe the distributional identity

δ(τ^{2} − |ξ|^{2}) = \frac{1}{2|ξ|}δ(τ + |ξ|) + \frac{1}{2|ξ|}δ(τ − |ξ|)  (5)

which is basically another way of saying that

lim_{ε→0} ∫_{R×R^{d}} ψ_{ε}(τ^{2} − |ξ|^{2})ϕ(τ, ξ) dτdξ = ∑_{±} ∫_{R_{d}} ϕ(±|ξ|, ξ)\frac{dξ}{2|ξ|}

for all test functions ϕ. (Note that when τ is close to ±|ξ|, then τ^{2} − |ξ|^{2} is approximately equal to ±2|ξ|(τ ∓ |ξ|).) So we have

ũ(τ, ξ) = F (τ, ξ)δ(τ^{2} − |ξ|^{2})

for some F, such as

F (τ, ξ) := 2|ξ|f_{sgn(τ)}(ξ).

(Note that only the values of F on the light cone τ^{2} − |ξ|^{2} = 0 are relevant.) From (2) and (5) we observe that

E(t) = \frac{1}{2} ∫ |F (τ, ξ)|^{2}δ(τ^{2} − |ξ|^{2}) dτdξ.

This is a formulation of the conformal energy which is more manifestly Lorentzinvariant. Indeed, if u_{v} is the Lorentz transform of u as before, we can write u_{v}(t, x) = u(L(t, x)), where L is the linear transformation

L(t, x) := (\frac{t − v_{1}x_{1}}{\sqrt{1 − v_{1}^{2}}} , \frac{x_{1} − v_{1}t}{\sqrt{1 − v_{1}^{2}}} , \underline{x}).

This transform is unimodular (i.e. det L = 1), so the spacetime Fourier transform is given by

ũ_{v}(τ, ξ) = u_{v}((L^{∗})^{−1}(τ, ξ)).

Since L preserves the form t^{2} − |x|^{2}, L^{∗} (and hence (L^{∗})^{−1}) will preserve the dual form τ^{2} − |ξ|^{2}. So

ũ_{v}(τ, ξ) = F ((L^{∗})^{−1}(τ, ξ))δ(τ^{2} − |ξ|^{2})

and hence by the preceding formula

E_{v}(t) = \frac{1}{2} ∫ |F ((L^{∗})^{−1}(τ, ξ))|^{2}δ(τ^{2} − |ξ|^{2}) dτdξ.

If we then make a change of variables (τ′, ξ′) := (L^{∗})^{−1}(τ, ξ) and once again use the fact that L^{∗} preserves the form τ^{2} − |ξ|^{2} and is unimodular, we obtain the desired invariance.

2. Exercise 2.25 (Integration in time yields regularity in space)

There are both Fourier space and physical space proofs of this result. The Fourier space result proceeds via the Fourier representation (1). Integrating this on the time interval I and using elementary calculus we have

| ; ∫_{I} u(t) dt(ξ)| ≲ \frac{1}{|ξ|}|û_{0}(ξ)| + \frac{1}{|ξ|^{2}} |û_{0}(ξ)|

where we bound the sines and cosines arising from the integral by 1. Multiplying by |ξ|^{2} and taking L_{ξ}^{2} norms gives the result.

Alternatively, we can observe (from Fourier analysis) that the Ḣ_{x}^{2} norm of a function f is comparable to the L_{x}^{2} norm of Δf. Thus

∥ ∫_{I} u(t) dt∥_{Ḣ_{x}^{2}(R^{d})} ∼ ∥ ∫_{I} Δu(t) dt∥_{L_{x}^{2}(R^{d})}.

But from the wave equation we have Δu(t) = ∂_{tt}u(t). From the fundamental theorem of calculus we thus have

∫_{I} Δu(t) dt = ∂_{t}u(t_{1}) − ∂_{t}u(t_{0})

where t_{0}, t_{1} are the endpoints of I. On the other hand from energy conservation (or (1)) we have

∥∂_{t}u(t_{0})∥_{L_{x}^{2}} ≲ ∥u(0)∥Ḣ_{x}^{1} + ∥∂_{t}u(0)∥_{L_{x}^{2}}

and similarly for t_{1}, and the claim follows.

3. Exercise 2.26 (Gaussian integrals)

We can complete the square and write the Gaussian integral as

∫_{−∞}^{∞} e^{−α(x−β/2α)^{2}} dx = \sqrt{π/α}.

Making the change of variables z = x − β/2α we wish to show

∫_{−∞−β/2α}^{∞−β/2α} e^{−αz^{2}} dz = \sqrt{π/α}.

We can shift the contour, noting that the hypothesis Re(α) > 0 ensures that e^{−αz^{2}} decays very fast on any horizontal strip, and reduce to showing

∫_{−∞}^{∞} e^{−αz^{2}} dz = \sqrt{π/α}.

Both sides are analytic on the half-plane Re(α) > 0, so it suffices to verify the identity when α is real (note that non-zero analytic functions have isolated zeroes). But then a change of variables y = \sqrt{α}x reduces one to the standard integral ∫_{−∞}^{∞} e^{−x^{2}} dx = \sqrt{π}.

There is a typo in the second part of the problem, as h(ξ) is equal to −\frac{ℏ}{2m}|ξ|^{2}. We thus need to show that

lim_{ε→0} \frac{1}{(2π)^{d}} ∫_{R_{d}} e^{i(x·ξ− \frac{ℏ}{2m} t|ξ|^{2})}e^{−ε|ξ|^{2}} dξ = \frac{1}{(2πiℏt/m)^{d/2}} e^{im|x|^{2}/(2ℏt)}.

Writing s := ℏt/2m this becomes

lim_{ε→0} \frac{1}{(2π)^{d}} ∫_{R_{d}} e^{i(x·ξ−s|ξ|^{2})}e^{−ε|ξ|^{2}} dξ = \frac{1}{(4πis)^{d/2}} e^{i|x|^{2}/4s}.

Splitting x and ξ into components and using Fubini’s theorem we see that we only have to show this for one dimension:

lim_{ε→0} \frac{1}{2π} ∫R e^{i(x·ξ−sξ^{2})}e^{−εξ^{2}} dξ = \frac{1}{(4πis)^{1/2}} e^{ix^{2}/4s}.

But the left-hand side is

lim_{ε→0} \frac{1}{2π} \sqrt{\frac{π}{is + ε}}e^{(iξ)^{2}/4(is+ε)}

8 TERENCE TAO

and the claim follows.

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

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