Math 567: Assignment 1 Solutions
1. Dispersion relations: find the dispersion relations and group velocities for each:
(a) Rossby waves: for ψ(x), x = (x1 , x2 ) ∈ R2 ,
∂
∂
∂
+c
∆ψ + β
ψ=0
∂t
∂x1
∂x1
Substitute in ψ(x) = ei(ξ·x−ωξ ) , ξ = (ξ1 , ξ2 ) ∈ R2 to get
(−iω + ciξ1 )(−|ξ|2 ) + βiξ1 = 0
so the dispersion relation is
ω = h(ξ) := cξ1 − β
ξ1
|ξ|2
and the group velocity is
vgr
ξ1 ˆ
β
= ∇h(ξ) = c − 2 î + 2β 3 ξ,
|ξ|
|ξ|
ξ
ξˆ :=
.
|ξ|
Remarks: for large frequencies (|ξ| |β/c|), this is essentially transport in the
equatorial (x1 ) direction with velocity c, but for lower frequencies it is different.
ξ2
The component of the group velocity in the x1 direction, î·vgr = c−β |ξ|12 +2β |ξ|14
may be positive or negative, depending on ξ.
(b) (vacuum) Maxwell equations:
~ t = c2 ∇ × B,
~
E
~ t = −∇ × E,
~
B
~ =∇·B
~ = 0,
∇·E
~
~
for vector fields E(x,
t) ∈ R3 (electric), B(x,
t) ∈ R3 (magnetic), x ∈ R3
~ = ~eei(ξ·x−ωt) , B
~ = ~bei(ξ·x−ωt) for ξ ∈ R3 and some constant vectors
Write E
~e = ~e(ξ), ~b = ~b(ξ). The divergence-free conditions give
~ = iξ · ~eei(ξ·x−ωt) ,
0 = ∇E
~ = iξ · ~bei(ξ·x−ωt)
0 = ∇B
1
=⇒
~e ⊥ ξ,
~b ⊥ ξ.
The dynamical equations yield
−iω~e = ic2 ξ × ~b,
−iω~b = −iξ × ~e, =⇒
−ω 2~e = c2 ξ × (ω~b) = c2 ξ × (ξ × ~e)
(and a similar equation for ~b). Since ~e ⊥ ξ, ξ × (ξ × ~e) = −|ξ|2~e, so −ω 2~e =
−|ξ|2 c2~e yielding our dispersion relation
ω 2 = c2 |ξ|2
=⇒
ω = ±c|ξ|
and group velocity
ˆ
vgr = ±∇(c|ξ|) = ±cξ.
Remark: unsurprisingly, this is the same as for the wave equation (light!).
(c) Maxwell in an (uniaxial) anisotropic medium:
~ t = ∇ × B,
~ B
~ t = −∇ × E,
~ ∇ · (eE)
~ =∇·B
~ = 0, e = diag(α, 1, 1)
(eE)
~ = ~bei(ξ·x−ωt) for ξ ∈ R3 and some constant
~ = ~eei(ξ·x−ωt) , B
As above write E
vectors ~e = ~e(ξ), ~b = ~b(ξ). The divergence-free conditions give
~ = iξ·(e~e)ei(ξ·x−ωt) ,
0 = ∇(eE)
~ = iξ·~bei(ξ·x−ωt)
0 = ∇B
=⇒
e~e ⊥ ξ,
~b ⊥ ξ.
The dynamical equations yield
−iωe~e = iξ × ~b,
−iω~b = −iξ × ~e, =⇒
−ω 2 e~e = ξ × (ω~b) = ξ × (ξ × ~e).
The linear transformation ξ × (ξ × ·) is given (in the standard basis) by the
matrix ξ ⊗ ξ − |ξ|2 I (where (ξ ⊗ ξ)jk = ξj ξk ) and so the condition for a non-zero
solution ~e of the last equation is
0 = det[ξ ⊗ ξ − |ξ|2 + ω 2 e] = ω 2 (ω 2 − |ξ|2 )(αω 2 − (αξ12 + ξ22 + ξ32 ))
(after some simplification), yielding a dispersion relation with two ‘branches’:
ω 2 = |ξ|2 ,
ω 2 = ξ12 +
1 2
(ξ + ξ32 )
α 2
(the second one reflecting the anisotropy) with corresponding group velocities
ˆ
vgr = ±cξ,
(ξ1 , α1 ξ2 , α1 ξ3 )
vgr = ± q
ξ12 + α1 (ξ22 + ξ32 )
.
2
2. Fundamental solution of the 1D transport equation: the dispersion relation
for ut + cux = 0 (n = 1) is h(ξ) = cξ. Show directly that the fundamental solution
Z ∞
1
1 h −ith(ξ) iˇ
(x) := lim
eixξ e−itcξ e−|ξ| dξ = δ(x − ct)
Φt (x) := √
e
→0+
2π
2π
−∞
in the ‘sense of distributions’. That is, for any function φ(x) which is smooth (C ∞ )
and compactly supported (vanishing outside a bounded interval), show
Z ∞
Z ∞
1
ixξ −itcξ −|ξ|
φ(x)
lim
e e
e
dξ dx = φ(ct).
→0+ −∞
2π −∞
We have
Z ∞
Z 0
Z ∞
1
ixξ −itcξ −|ξ|
(i(x−ct)+)ξ
e e
e
dξ =
e
dξ +
e(i(x−ct)−)ξ dξ
:=
2π −∞
−∞
0
1
1
1
1
=
+
=
2π + i(x − ct) − i(x − ct)
π 2 + (x − ct)2
1
x − ct
1 1
= η
,
η(y) :=
.
π 1 + y2
R∞
Notice that −∞ η(y)dy = 1. So, using the change of variables y = (x − ct)/,
Z ∞
Z ∞
Φt (x)φ(x)dx − φ(ct) =
η(y)[φ(ct + y) − φ(ct)]dy.
Φt (x)
−∞
−∞
Since φ is a bounded function, the integrand is bounded (independently of ) by the
integrable function Cη(y). Moreover, for each y, the integrand converges to 0 as
→ 0. So by the Lebesgue Dominated Convergence Theorem, this integral → 0 as
→ 0. (If you prefer to avoid the use of the LDCT, just observe that since φ0 is
bounded, |φ(ct + y) − φ(ct)| ≤ C|y| by the mean value theorem, and go from there.)
3. Fundamental solution of the free Schrödinger equation: the dispersion relation for iut = ∆u (with x ∈ Rn ) is h(ξ) = |ξ|2 . Show directly that
Z
|x|2
1 h −ith(ξ) iˇ
1
1
ix·ξ −it|ξ|2 −|ξ|2
i 4t
Φt (x) :=
e
(x)
:=
lim
e
e
e
dξ
=
e
n
n
→0+ (2π)n Rn
(2π) 2
(4πit) 2
Q
Qn
2
2
2
n
R
Since (2π)−n eix·ξ e−it|ξ| e−|ξ| = nj=1 (2π)−1 eixj ξj −(it+)ξj , we have ΦR
t (x) =
j=1 Φt (xj )
and it suffices to compute the n = 1 case. Completing the square,
Z
Z
ix
x2
1
1
−(+it)(ξ− 2(+it)
)2 − 4(+it)
−(+it)ξ 2 +ixξ
e
dξ =
e
dξ,
2π R
2π R
3
1
1
√
2π + it
√
+ it(ξ − ix/(2( + it))), we obtain
Z
x2
2
− 4(+it)
e
e−η dη.
and making the change of variables η =
Note that since this is a complex change of variables, this integral is now over a
straight-line contour in the complex plane, and a standard complex-integration argument must be given (details omitted here) that the contour canRbe moved to the real
√
2
axis. That done, we are left with a standard Gaussian integral R e−η dη = π, and
so the above is
2
ix2
1
1
− x
p
e 4(+it) →→0 √
e 4t
4πit
4π( + it)
as required.
4. Fundamental solution of the wave equation: show that
Kt (x) := (2π)
−n/2
sin(ct|ξ|) ˇ
(x) =
c|ξ|
1
2c χ[−ct,ct] (x)
1
δ
(x)
4πc2 t |x|=ct
n=1
n=3
indirectly – that is, by computing the Fourier transform of the right-hand side to get
the left-hand side. Use this to write solution formulas for the initial value problem
for the wave equation in dimensions n = 1 and n = 3.
For n = 1,
√
2π
ˆ
Z
1 1 −iξx ct
1
1 ct −iξx
1
sin(ct|ξ|
e
dx =
χ[−ct,ct] (x) (ξ) =
e
|−ct =
sin(ctξ) =
.
2c
2c −ct
2c −iξ
cξ
c|ξ|
The solution formula for initial displacement u0 and initial velocity v0 is then
Z
∂
1
1
1 x+ct
χ[−ct,ct] ∗ v0 + χ[−ct,ct] ∗ u0 =
v0 (y)dy+ [u0 (x − ct) + u0 (x + ct)] .
u(x, t) =
2c
∂t
2c x−ct
2
For n = 3,
3/2
(2π)
ˆ
Z
1
1
δ
e−iξ·x dS(x).
(ξ) =
4πc2 t |x|=ct
4πc2 t |x|=ct
The integral will not depend on the direction of ξ, so we may take ξ = (0, 0, |ξ|).
Then using spherical coordinates, we get
Z 2π Z π
1
t 1 −i|ξ|ct cos φ π
sin(ct|ξ|)
e−i|ξ|(ct) cos φ sin(φ) (ct)2 dφ dθ =
e
|0 =
.
4πc2 t 0
2
i|ξ|ct
c|ξ|
0
4
The solution formula for initial displacement u0 and initial velocity v0 is then
1
∂ 1
δ
∗ v0 +
δ
∗ u0
4πc2 t Z|x|=ct
∂t 4πc2 t |x|=ct
Z
∂ 1
1
v0 (y)dS(y) +
u0 (y)dS(y)
=
4πc2 t |y−x|=ct
∂t 4πc2 t |y−x|=ct
u(x, t) =
(this is probably the simplest way to leave it).
5. Group velocity: if f (x) is a smooth function on Rn with compact support, and φ(x)
is a smooth (real-valued) function on Rn with only one critical point x0 (∇φ(x0 ) = 0)
in the support of f , the method of stationary phase yields the asymptotics:
Z
eitφ(x) f (x)dx = Ct−n/2 eitφ(x0 ) f (x0 ) + O(t−(n/2+1) ), t → ∞
Rn
(where C is a constant (depending on D2 φ(x0 ))). Recall the solution formula
Z
−n/2
u(x, t) = (2π)
ei(x·ξ−th(ξ)) û0 (ξ)dξ
Rn
for the dispersive PDE ut = −ih(−i∇x )u. Now, for a given velocity v ∈ Rn , use
the above to find the large t asymptotics of u(vt, t), the solution as observed while
moving out with velocity v. (You may assume h smooth and û0 smooth with compact
support). Determine the group velocity from your computation.
For a given v ∈ Rn , we have
u(vt, t) = (2π)
−n/2
Z
eit(v·ξ−h(ξ)) û0 (ξ)dξ.
Rn
Now we apply the stationary phase method with φ(ξ) = v·ξ−h(ξ), so that ∇φ(ξ) = v−
∇h(ξ), to conclude that if there is no solution ξ in the support of û0 (ξ) to ∇h(ξ0 ) = v,
then u(vt, t) = O(t−(n/2+1) ), while if there is one such ξ, then
u(vt, t) = Ct−n/2 eit(v·ξ−h(ξ)) û0 (ξ) + O(t−(n/2+1) )
(and if there are several, the leading contribution would be a sum of such terms). In
other words, supposing û0 (ξ) is concentrated around a single frequency ξ0 ∈ Rn , the
largest contribution to the solution travels with velocity v = ∇h(ξ0 ). This is the group
velocity.
6. Dispersive estimates for the Airy equation: let
Z R
1 h −itξ3 iˇ
1
3
Φt (x) = √
e
(x) =
lim
ei(xξ−tξ ) dξ
2π R→∞ −R
2π
be the fundamental solution of the Airy equation.
5
(a) Show that Φt (x) = t−1/3 Φ1 (t−1/3 x).
Changing variables in the integral as η = t1/3 ξ, we get
1
Φt (x) =
lim
2π R→∞
Z
R
e
−R
−1/3
= t−1/3 Φ1 (t
i(xξ−tξ 3 )
1
dξ =
lim
2π R→∞
Z
R
−1/3
Z
R
t
−1/3 xη−η 3 )
ei(t
dη
−R
−R
x).
(b) Show that Φ1 (x) is a bounded function over the interval x ∈ (−∞, −1].
Let φ(ξ) := xξ − ξ 3 be the phase. Then integration by parts gives
Z
R
e
iφ(ξ)
Z
R
dξ =
−R
−R
[eiφ ]0
eiφ R
dξ
=
| +
iφ0
iφ0 −R
Z
R
−R
φ00 iφ
e dξ.
i(φ0 )2
Now for x ≤ −1, φ0 (ξ) = x − 3ξ 2 ≤ −(1 + 3ξ 2 ), so |φ0 (ξ)| ≥ 1 and
iφ
e R |−R ≤ 1.
iφ0
Moreover φ00 (ξ) = −6ξ, so
|φ00 |
|ξ|
≤6
0
2
|φ |
(1 + 3ξ 2 )2
so |
∞.
RR
−R e
iφ dξ|
=⇒
Z
R
−R
Z ∞
φ00 iφ |ξ|
e dξ ≤ 6
dξ < ∞,
0
2
2 2
i(φ )
−∞ (1 + 3ξ )
is bounded by a constant (independent of R). Thus supx≤−1 |Φt (x)| <
(c) (Bonus): use more clever integration-by-parts arguments to show Φ1 (x) is
bounded over all x ∈ R.
If x ∈ (−1, 0], just bound the integral over ξ ∈ [−1, 1] by 2, and treat the remaining integrals (over [−R, −1] and
p [1, R] exactly as above. If x > 0, then
the phase has critical points ξ = ± x/3 Let’s treat just the integral over [0, R]
(the [−R, 0] part
p can be treated
p identically). For some fixed M > 1, bound the
integral over [ x/3 − M/2, x/3 + M/2] ∩ [0, R] simply by the (maximum)
length M of this interval. Integrate by parts as above on the remaining in2
tegral(s). The boundary terms
p can be seen to be bounded by (const)M . If
non-empty, the√remaining [0, x/3 − M/2] integral can be seen to be bounded by
p
R x/3
(const)(1/x) 0
ξdξ ≤ const, while the remaining [ x/3 + M/2, R] integral
R∞
is bounded by (const) 0 M 2ξ+ξ2 dξ ≤ const.
6
(d) Given that Φ1 (x) is bounded, conclude |Φt (x)| ≤ Ct−1/3 , and hence for the solution u(x, t) = (Φt ∗ u0 )(x) of the initial value problem: |u(x, t)| ≤ Ct−1/3 ku0 kL1 .
Since Φ1 is bounded, part (a) gives |Φt (x)| ≤ Ct−1/3 and so
Z
Z
|u(x, t)| = | Φt (x − y)u0 (y)dy| ≤ sup |Φt | |u0 (y)|dy ≤ Ct−1/3 ku0 kL1 .
R
R
7