Math 567: Assignment 2 Solutions
1. Space-time dispersive estimates for Airy: recall that the solution operator for
the Airy equation
3
u(x, t) = [e−t(d/dx) u0 ](x)
ut + uxxx = 0, u(x, 0) = u0 (x),
satisfies the decay estimates
3
3
ke−t(d/dx) u0 kL2x = ku0 kL2x ,
ke−t(d/dx) u0 kL∞
≤ Ct−1/3 ku0 kL1 .
x
(a) Use interpolation to establish the family of estimates
3
ke−t(d/dx) u0 kLpx ≤ Cp t−σp ku0 kLp0 ,
x
finding the exponent σp in terms of p (recall
1
p
+
1
p0
= 1).
By the Riesz-Thorin interpolation theorem, for any p ∈ [2, ∞], with
θ
1−θ
∞ = 2 , we have
2
)
−( 31 − 3p
3
ke−t(d/dx) u0 kLpx ≤ (1)1−θ (Ct−1/3 )θ ku0 kLqx = Cp t
1
q
where
=
1−θ
2
+
θ
1
=1−
1
p
1
p
=
1−θ
2
+
ku0 kLqx
and so q = p0 .
(b) By observing that uλ (x, t) := u(λx, λ3 t) also solves the Airy PDE, find the
admissibility relation that must hold between exponents r and p in order for
there to hold a homogeneous space-time estimate of the form
3
ke−t(d/dx) u0 kLrt ([0,∞);Lpx (R) ≤ Cku0 kL2 (R) .
It is straightforward to check that uλ also solves Airy. By simple change of
variables,
kuλ krLr Lpx =
t
Z
∞ Z
0
R
−(3+r/p)
=λ
|u(λx, λ3 t)|p dx
r/p
dt = λ−(3+r/p)
Z
0
kukrLr Lpx .
t
1
∞ Z
R
|u(x, t)|p dx
r/p
dt
Since uλ (x, 0) = u0 (λx) and
ku0 (λ·)k2L2
Z
1
|u0 (λx)|2 dx = λ− 2 ku0 k2L2 ,
=
R
the given space-time estimate for uλ reads
1
λ2
− r3 − p1
kukLrt Lpx ≤ Cku0 kL2 ,
and the only way this can possible hold for all λ > 0 is if
3 1
1
+ = .
r p
2
(c) Use the generalized Young inequality, together with part (a), to establish the
inhomogeneous space-time estimates
Z t
e(s−t)(d/dx)3 f (·, s)ds
≤ Ckf kLr0 Lp0
Lrt Lpx
0
t
x
for a pair (r, p) (2 ≤ p ≤ ∞) satisfying the admissibility criterion you found in
part (b).
Remark: you are not asked to do so here, but the full suite of space-time estimates,
as for the Schrödinger equation, can be obtained following the same arguments.
By part (a),
3
ke(s−t)(d/dx) f (·, s)kLpx ≤ C(t − s)
2
−( 13 − 3p
)
kf (·, s)kLp0 ,
x
and so
Z
k
0
t
3
e(s−t)(d/dx) f (·, s) dskLpx ≤
Z
t
3
ke(s−t)(d/dx) f (·, s)kLpx ds
0
Z t
−( 1 − 2 )
≤C
(t − s) 3 3p kf (·, s)kLp0 ds
0
2
−( 31 − 3p
)
= Ct
x
χ[0,∞) (t) ∗t kf (·, t)kLp0 χ[0,∞) (t)
x
0
2
−( 13 − 3p
)
(convolution of two functions of t). Now if kf (·, t)kLp0 ∈ Lrt , and since t
Lµweak
where µ =
the convolution is
∈
x
2 −1
− 3p
) for p >
ν
in Lt where (using
( 31
2, the generalized Young inequality yields that
the admissibility relation from (b))
1
1
1
1 1
2
4 1 2 1 3
1
= − −
−
=1+
=⇒
µ=r
1+ = 0 + =1− + −
µ
r
µ
r 3 3p
3 r 3 2 r
r
2
and moreover
(s−t)(d/dx)3
p
ke
f
(·,
s)k
Lx Lrt
≤ C 0 kf kLr0 Lp0
t
x
as desired. The remaining case p = 2, r = ∞ not covered by the generalized Young
inequality is elementary: for any t,
Z t
Z t
Z t
(s−t)(d/dx)3
e(s−t)(d/dx)3 f (·, s)ds ≤
kf (·, s)kL2x ds ≤ kf kL1t L2x .
ke
f (·, s)kL2x ds =
0
L2x
0
0
2. Dispersive decay estimate for the wave equation: The basic decay estimate
for solutions of the wave equation
utt = ∆u,
u(x, 0) = 0,
ut (x, 0) = v0 (x)
(with zero initial displacement) for x ∈ Rn with n odd is
|u(x, t)| ≤ Ct−
n−1
2
k(∇)
n−1
2
v0 kL1 (Rn ) .
Use the solution formulas in the cases n = 1 and n = 3 to prove this estimate.
Remark: notice a couple of things here: 1. for n = 1 there is no decay, as the equation
is not at all dispersive, while for n = 3 the weak dispersion coming from the variation
of the group velocity with the direction (but not the magnitude!) of the frequency
vector does result in some decay (albeit slower than for Schrödinger). 2. this decay
estimate is also weaker than that of (say) Schrödinger or Airy in that we must take
also take some derivatives of the initial data: ‘loss of derivative’.
The n = 1 case is easy, using d’Alembert’s formula:
Z
Z
1 x+t
1
1
u(x, t) =
v0 (y)dy
=⇒
|u(x, t)| ≤
|v0 (y)|dy = kv0 kL1 .
2 x−t
2 R
2
The formula for n = 3 is
1
u(x, t) =
4πt
Z
v0 (y)dS(y).
|y−x|=t
One way to re-express this surface integral as a bulk integral is to write, using spherical
coordinates (r, Ω) centred at y = x (so the sphere is r = t) and the fundamental
theorem of calculus (and assuming v0 decays to zero at spatial infinity),
Z Z ∞
Z ∞
1
(v0 )r (r, Ω)dr t2 dΩ
v0 (t, Ω) = −
(v0 )r (r, Ω)dr
=⇒
u(x, t) = −
4πt
S2 t
t
and so since r ≥ t in the integral,
Z Z ∞
Z Z ∞
Z
1
1
1
2
2
|u(x, t)| ≤
|(v0 )r (r, Ω)|r dr dΩ ≤
|∇v0 (r, Ω)|r dr dΩ =
|∇v0 |.
4πt S 2 t
4πt S 2 0
4πt R3
3
3. Linearized water waves: linearize the water wave system
− h ≤ x1 ≤ η(x0 , t), x0 ∈ R2
ηt + ∇x0 φ · ∇x0 η − φx1 = 0
x1 = η(x0 , t), x0 ∈ R2
φt + 12 |∇φ|2 + gη = 0
∆φ = 0
x1 = −h, x0 ∈ R2
φ x1 = 0
and obtain the dispersion relation ω 2 = g|ξ| tanh(h|ξ|) for the result.
The linearized system is
− h ≤ x1 ≤ 0, x0 ∈ R2
η t − φ x1 = 0
x1 = 0, x0 ∈ R2
φt + gη = 0
∆φ = 0
x1 = −h, x0 ∈ R2
φx1 = 0
(where note we have replaced the perturbed surface by the flat one for evaluating the
boundary condition). Separating variables φ(x1 , x0 , t) = A(x1 )B(x0 , t),
A00
∆x0 B
=−
= λ2
A
B
(λ > 0) we arrive at the single-variable problem
0 = A00 B + A∆x0 B
=⇒
A00 = λ2 A,
A0 (−h) = 0
whose solutions are proportional to cosh(λ(x1 +h)), so φ(x1 , x0 , t) = B(x0 , t) cosh(λ(x1 +
h)). Taking η = ei(ξ·x−ωt) , on the upper boundary x1 = 0, we have
ω
ei(ξ·x−ωt) .
φ x1 = η t
=⇒ λB(x0 , t) sinh(λh) = −iωei(ξ·x−ωt)
=⇒
B = −i
λ sinh(hλ)
Then
−λ2 =
∆ x0 B
= −|ξ|2
B
shows λ = |ξ|. So again at x1 = 0,
gη = −φt =
ω 2 cosh(h|ξ|)
η
|ξ| sinh(h|ξ|)
=⇒
ω 2 = g|ξ| tanh(h|ξ|).
4. Local existence for the Hartree equation: a variant of the NLS arising in Quantum Mechanics (many-body theory) is the Hartree equation, which has a convolutiontype nonlinearity:
iut + ∆u = (V ∗ |u|2 )u,
u(x, 0) = u0 (x),
where V : R3 → R is the potential (describing the inter-particle interaction).
4
(a) Use the Hölder and/or Young inequalities to show that if V is a bounded function, then the nonlinearity u 7→ (V ∗ |u|2 )u is locally Lipshitz in L2 (R3 ) (and so
it follows from the standard argument given in class that for u0 ∈ L2 (R3 ) we
can construct a local solution u ∈ C([0, T ]; L2 (R3 ))).
Denote F (u) := (V ∗ |u|2 )u so that by triangle inequality, Hölder’s inequality,
Young’s inequality (actually just the L∞ (i.e. Hölder) version of Young), and
||u| − |v|| ≤ |u − v|,
kF (u) − F (v)k2 ≤ k(V ∗ |u|2 )(u − v)k2 + k(V ∗ (|u|2 − |v|2 ))vk2
≤ kV ∗ |u|2 k∞ ku − vk2 + kV ∗ ((|u| − |v|)(|u| + |v|))k∞ kvk2
≤ kV k∞ k|u|2 k1 ku − vk2 + kV k∞ k(|u − v|)(|u| + |v|)k1 kvk2
≤ kV k∞ kuk22 ku − vk2 + kV k∞ (kuk2 + kvk2 )kvk2 ku − vk2
≤ C (kV k∞ , kuk2 , kvk2 ) ku − vk2 ,
which is the desired local Lipshitz condition.
(b) Physically important potentials such as the Coulomb potential V (x) = ±1/|x|
are not bounded. Show that the Coulomb potential lies in the class L∞ (R3 ) +
L3/2 (R3 ) (that is, it can be written as a sum of two terms, one lying in the first
space, the other in the second).
±
1
1
1
= ± χ|x|≤1 ±
χ
|x|
|x|
|x| |x|>1
where χE denotes the characteristic (indicator) function of a set E. The second
term is bounded (hence in L∞ ) while the first term is in L3/2 since
1
3/2
k±
χ
k =
|x| |x|≤1 3/2
Z
−3/2
|x|
Z
dx = 4π
|x|≤1
1
r
0
−3/2 2
Z
r dr = 4π
1
r1/2 dr < ∞.
0
(c) Use the Hölder and/or Young and/or Sobolev inequalities (more precisely for
the latter, kukL6 (R3 ) ≤ ckukH 1 (R3 ) ) to show that if V ∈ L∞ (R3 )+L3/2 (R3 ), then
the Hartree nonlinearity is locally Lipshitz in H 1 (R3 ) (and so for u0 ∈ H 1 (R3 ),
the standard argument gives a local solution u ∈ C([0, T ], H 1 (R3 )) for this class
of potentials, which includes the Coulomb).
We dealt with V ∈ L∞ above, so if suffices now to take V ∈ L3/2 (though we
only estimated L2 above, this easily extends to H 1 by taking the derivative and
estimating L2 again). Similiar to the above, but now using also the Sobolev
5
inequality kuk6 ≤ CkukH 1 , we have
kF (u) − F (v)k2 ≤ kV ∗ |u|2 k∞ ku − vk2 + kV ∗ (|u|2 − |v|2 )k3 kvk6
≤ kV k3/2 k|u|2 k3 ku − vk2 + kV k3/2 k|u|2 − |v|2 k3/2 kvk6
≤ kV k3/2 kuk26 + k|u| + |v|k6 kvk6 ku − vk2
≤ C kV k3/2 , kukH 1 , kvkH 1 ku − vk2 .
For the derivative, we use the property ∇(f ∗ g) = f ∗ ∇g = (∇f ) ∗ g of the
convolution, so that
∇F (u) = (V ∗ |u|2 )∇u + (V ∗ 2Re(ū∇u))u.
So there are many terms in F (u) − F (v), which we can estimate as follows:
•
•
•
•
•
k(V
k(V
k(V
k(V
k(V
∗ |u|2 )∇(u − v)k2 ≤ kV k3/2 kuk26 k∇(u − v)k2
∗ (|u|2 − |v|2 )∇uk2 ≤ kV k3/2 k|u| + |v|k6 ku − vk6 k∇uk2
∗ |u||∇u|)(u − v)k2 ≤ kV k3/2 kuk6 k∇uk2 ku − vk6
∗ (|u|∇(u − v)|)uk2 ≤ kV k3/2 kuk6 k∇(u − v)k2 kuk6
∗ (|u − v||∇u|))uk2 ≤ kV k3/2 ku − vk6 k∇uk2 kuk6 .
and putting all of this together we arrive at the desired local Lipshitz estimate:
kF (u) − F (v)kH 1 ≤ C kV k3/2 , kukH 1 , kvkH 1 ku − vkH 1 .
5. Persistence of regularity: consider a nonlinear Schrd̈inger equation with pure
power nonlinearity
iut + ∆u = ±up
u(x, 0) = u0 (x),
with p a positive integer. Recall that if u0 ∈ H s (Rn ) where s > n/2 is (for
simplicity) an integer, we constructed a ‘classical solution’ u ∈ C([0, T max ); H s )
with ku(·, t)kH s → ∞ if T max < ∞. Prove that if T max < ∞, then in fact
kukLp−1 L∞ (Rn ×[0,T max )) = ∞.
t
x
Remark: one implication of an estimate like this one, is ‘persistence of regularity’ –
namely that the constructed H s solutions for different s will all have the same lifespan.
More precisely, an H s2 solution is also an H s1 solution if n/2 < s1 < s2 , but the H s2
norm cannot blow up before the H s1 norm does: clearly T max (H s1 ) ≥ T max (H s2 ), but
in fact we cannot have T max (H s1 ) > T max (H s2 ). For if so, the boundedness of kukH s1
on [0, T max (H s2 )] implies (by Sobolev imbedding) kukLp−1 L∞ (Rn ×[0,T max (H s2 )]) < ∞,
x
t
contradicting what was proved in this exercise.
6
A slick way to proceed is to use the Sobolev imbedding inequality
p−1
kup kH s ≤ CkukL
∞ kukH s
for s > n/2 (you can look this up, or else prove it by differentiating up s times, and
using some Sobolev-type interpolation inequalities) and the easy ‘Strichartz’ estimate
Z t
it∆
s
≤ ku0 kH s + kf kL1t Hxs ([0,T ])
ei(t−s)∆ f (·, s)dskL∞
ke u0 + i
t Hx ([0,T ])
0
(which just uses keit∆ φkH s = kφkH s , which is easy to see by taking the Fourier
p−1
transform). Suppose now kukLp−1 L∞ [0,T max ) < ∞. This means that kukL
∞ in an
x
x
t
integrable function on t on [0, T max ), and thus we can break this interval up into
finitely many sub-intervals on which its integral is sufficiently small:
[0, T max ) = [T0 = 0, T1 )∪[T1 , T2 )∪· · ·∪[TN −1 , TN = T max ),
kukp−1
p−1
Lt
L∞
x ([Tj ,Tj+1 )
≤
1
2C
where C is the constant in the above Sobolev imbedding inequality. Now writing the
NLS in Duhamel form
Z t
it∆
ei(t−s)∆ up (·, s)ds,
u(t) = e u0 ± i
0
we apply the above Strichartz estimate on the first sub-interval [0, T1 ) to get
s
kukL∞
≤ ku0 kH s + kup kL1t Hxs ([0,T1 ])
t Hx ([0,T1 ])
Now use the Sobolev inequality and Hölder’s inequality (in t) to see
kup kHxs ≤ Ckukp−1
kukHxs
L∞
x
=⇒
kup kL1t Hxs ([0,T1 ]) ≤ Ckukp−1
p−1
Lt
L∞
x ([0,T1 ])
s
kukL∞
t Hs ([0,T1 ])
1
s
≤ kukL∞
t Hx ([0,T1 ])
2
and so we may move this term to the left side of the previous estimate to get
s
kukL∞
≤ 2ku0 kH s
t Hx ([0,T1 ])
and so in particular ku(·, T1 )kH s ≤ 2ku0 kH s . Now we may carry out the same estimate on the next sub-interval [T1 , T2 ], with u(·, T1 ) playing the role of initial data, to
find
s
kukL∞
≤ ku(·, T1 )kH s ≤ 2ku0 kH s .
t Hx ([T1 ,T2 ])
Repeating this on each sub-interval, we get
s
kukL∞
≤ 2j ku0 kH s
t Hx ([Tj ,Tj+1 ])
7
and so
N +1
max )) ≤ 2
s
ku0 kH s ,
kukL∞
t Hx ([0,T
contradicting ku(·, t)kH s → ∞ as t → T max .
Remark: this argument can be done a bit more efficiently using ‘Gronwall’s inequality’, if you know it.
8