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PDE LECTURE NOTES, M ATH 237A-B 185
- Wave Equation on Rn (Ref Courant & Hilbert Vol II, Chap VI §12.) We now consider the wave equation
(14.1) utt − 4 u = 0 with u(0, x) = f (x) and ut(0, x) = g(x) for x ∈ Rn.
According to Section 13, the solution (in the L^2 — sense) is given by
(14.2) u(t, ·) = (cos(t
p − 4 )f + sin(t
g.
To work out the results in Eq. (14.2) we must diagonalize ∆. This is of course done using the Fourier transform. Let F denote the Fourier transform in the x — variables only. Then
u^ ¨ˆ(t, k) + |k|^2 ˆu(t, k) = 0 with ˆu(0, k) = fˆ (k) and uˆ˙(t, k) = ˆg(k).
Therefore
u ˆ(t, k) = cos(t|k|) fˆ (k) + sin(t|k|) |k| ˆg(k).
and so u(t, x) = F−^1
cos(t|k|) fˆ (k) + sin(t|k|) |k| ˆg(k)
(x),
i.e. sin(t
√ −^4 )
g = F−^1
sin(t|k|) |k| gˆ(k)
(14.3) and
cos(t
p − 4 )f = F−^1
h cos(t|k|) fˆ (k)
i
d dt F
− 1
sin(t|k|) |k| ˆg(k)
Our next goal is to work out these expressions in x — space alone.
14.1. n = 1 Case. As we see from Eq. (14.4) it suffices to compute:
sin(t
√ −^4 )
g = F−^1
μ sin(t|ξ|) |ξ| ˆg(ξ)
= lim M →∞
F−^1
μ (^1) |ξ|≤M^ sin(t|ξ|) |ξ| gˆ(ξ)
= lim M→∞
F−^1
μ (^1) |ξ|≤M sin(t|ξ|) |ξ|
(14.5) B g.
This inverse Fourier transform will be computed in Proposition 14.2 below using the following lemma.
Lemma 14.1. Let CM denote the contour shown in Figure 38, then for λ 6 = 0 we have
M^ lim→∞
Z
CM
eiλξ ξ dξ = 2πi (^1) λ> 0.
Proof. First assume that λ > 0 and let ΓM denote the contour shown in Figure
- Then ¯ ¯¯ ¯¯ ¯
Z
ΓM
eiλξ ξ dξ
Z (^) π
0
¯eiλMe iθ ¯¯ ¯ dθ = 2π
Z (^) π
0
dθe−λM^ sin^ θ^ → 0 as M → ∞.
186 BRUCE K. DRIVER†
Therefore
M^ lim→∞
Z
CM
eiλξ ξ dξ = (^) Mlim→∞
Z
CM +ΓM
eiλξ ξ dξ = 2πiresξ=
μ eiλξ ξ
= 2πi.
Figure 38. A couple of contours in C.
If λ < 0 , the same argument shows
lim M→∞
Z
CM
eiλξ ξ dξ = lim M→∞
Z
CM +˜ΓM
eiλξ ξ dξ
and the later integral is 0 since the integrand is holomorphic inside the contour CM + ˜ΓM.
Proposition 14.2. (^) Mlim→∞ F−^1
(^1) |ξ|≤M sin( |ξt||ξ|)
(x) = sgn(t)
√π √ 2 1 |x|<|t|.
Proof. Let
IM =
2 πF−^1
μ (^1) |ξ|≤M sin(t|ξ|) |ξ|
(x) =
Z
|ξ|≤M
sin(tξ) ξ eiξ·xdξ.
Then by deforming the contour we may write
IM =
Z
CM
sin tξ ξ eiξ·xdξ =^1 2 i
Z
CM
eitξ^ − e−itξ ξ eiξ·xdξ
2 i
Z
CM
ei(x+t)ξ^ − ei(x−t)ξ ξ dξ
By Lemma 14.1 we conclude that
lim M→∞
IM =^1
2 i 2 πi(1(x+t)> 0 − (^1) (x−t)> 0 ) = πsgn(t) 1|x|<|t|.
(For the last equality, suppose t > 0. Then x − t > 0 implies x + t > 0 so we get 0 and if x < −t, i.e. x + t < 0 then x − t < 0 and we get 0 again. If |x| < t the first term is 1 while the second is zero. Similar arguments work when t < 0 as well.)
188 BRUCE K. DRIVER†
Now u solves (∂t − ∂x)u = v, i.e. ∂tu = ∂xu + v. Therefore
u(t, x) = et∂x^ u(0, x) +
Z (^) t
0
e(t−τ^ )∂x^ v(τ, x)dτ
= u(0, x + t) +
Z (^) t
0
v(τ, x + t − τ )dτ
= u(0, x + t) +
Z (^) t
0
v(0, x + t| {z } − 2 τ s
)dτ
= u(0, x + t) +
Z (^) t
−t
v(0, x + s)ds
= f (x + t) +^1 2
Z (^) t
−t
(g(x + s) − f 0 (x + s))ds
= f (x + t) −
2 f^ (x^ +^ s)
s=t s=−t
Z (^) t
−t
g(x + s)ds
f (x + t) + f (x − t) 2
Z (^) t
−t
g(x + s)ds
which is equivalent to Eq. (14.8).
14.2. Solution for n = 3. Given a function f : Rn^ → R and t ∈ R let
f^ ¯ (x; t) :=
Z
S^2
f (x + tω)dσ(ω) =
Z
Z
|y|=|t|
f (x + y)dσ(y).
Theorem 14.4. For f ∈ L^2
R^3
sin
−∆t
f = F−^1
sin |ξ|t |ξ| fˆ (ξ)
(x) = t f¯ (x; t)
and
cos
−∆t
g = d dt
t f¯ (x; t)
In particular the solution to the wave equation (14.1) for n = 3 is given by
u(t, x) = ∂ ∂t (t f¯ (x; t)) + t g(x; t)
=
4 π
Z
|ω|=
(tg(x + tω) + f (x + tω) + t∇f (x + tω) · ω)dσ(ω).
PDE LECTURE NOTES, M ATH 237A-B 189
Proof. Let gM := F−^1
h sin |ξ|t |ξ| 1 |ξ|≤M
i , then by symmetry and passing to spher- ical coordinates,
(2π)^3 /^2 gM (x) =
Z
|ξ|≤M
sin |ξ|t |ξ| eiξ·xdξ =
Z
|ξ|≤M
sin |ξ|t |ξ| ei|x|ξ^3 dξ
Z M
0
dρρ^2
Z (^2) π
0
dθ
Z (^) π
0
dφ sin ρt ρ eiρ|x|^ cos^ φ^ sin φ
= 2π
Z M
0
dρ sin ρt eiρ|x|^ cos^ φ −i|x|
π 0
= 2π
Z M
0
dρ sin ρt eiρ|x|^ − e−iρ|x| i|x|
4 π |x|
Z M
0
sin ρt sin ρ |x| dρ.
Using
sin A sin B =
[cos(A − B) − cos(A + B)]
in this last equality, shows
gM (x) = (2π)−^3 /^2 2 π |x|
Z M
0
[cos((t − |x|)ρ) − cos((t + |x|)ρ)]dρ
= (2π)−^3 /^2 π |x| hM (|x|)
where
hM (r) :=
Z M
−M
[cos((t − r)α) − cos((t + r)α)]dα,
an odd function in r. Since
F−^1
sin |ξ|t |ξ| fˆ (ξ)
= (^) Mlim→∞ F−^1 (ˆgM (ξ) fˆ (ξ)) = (^) Mlim→∞(gM B f )(x)
we need to compute gM B f. To this end
gM B f (x) =
μ 1 2 π
π
Z
R^3
|y| hM (|y|)f (x − y)dy
μ 1 2 π
π
Z ∞
0
dρ hM (ρ) ρ
Z
|y|=ρ
f (x − y)dσ(y)
μ 1 2 π
π
Z ∞
0
dρ hM^ (ρ) ρ 4 πρ
2
Z
|y|=ρ
f (x − y)dσ(y)
2 π
Z ∞
0
dρ hM (ρ)ρ f¯ (x; ρ) =
4 π
Z ∞
−∞
dρ hM (ρ)ρ f¯ (x; ρ)
where the last equality is a consequence of the fact that hM (ρ)ρ f¯ (x; ρ) is an even function of ρ. Continuing to work on this expression suing ρ → ρ f¯ (x; ρ) is odd
PDE LECTURE NOTES, M ATH 237A-B 191
14.3. Du Hamel’s Principle. The solution to
utt = 4 u + f with u(0, x) = 0 and ut(0, x) = 0
is given by
(14.9) u(t, x) =
4 π
Z
B(x,t)
f (t − |y − x|, y) |y − x| dy =
4 π
Z
|z| 192 BRUCE K. DRIVER†
Then U solves Utt = 1 rn−^1 ∂r(rn−^1 Ur)
with
U(0, r) =
Z
∂B(0,1)
u(0, x + rω)dσ(ω) = f¯ (x; r)
Ut(0, r) = g(x; r). Proof. This has already been proved, nevertheless, let us give another proof which does not rely on using integration over O(n). To this hence we compute
∂r U(t, r) = ∂r
Z
∂B(0,1)
u(t, x + rω)dσ(ω)
Z
∂B(0,1)
∇u(t, x + rω) · ωdσ(ω)
σ (Sn−^1 ) rn−^1
Z
|y|=r
∇u(t, x + y) · ydσˆ (y)
σ (Sn−^1 ) rn−^1
Z
|y|≤r
∆u(t, x + y)dy
σ (Sn−^1 ) rn−^1
Z (^) r
0
dρ
Z
|y|=ρ
∆u(t, x + y)dσ(y)
so that
1 rn−^1 ∂r(r
n− (^1) Ur ) = 1 rn−^1 ∂r
σ (Sn−^1 )
Z (^) r
0
dρ
Z
|y|=ρ
∆u(t, x + y)dσ(y)
σ (Sn−^1 ) rn−^1
Z
|y|=r
∆u(t, x + y)dσ(y)
=
Z
|y|=r
∆u(t, x + y)dσ(y)
Z
|y|=r
utt(t, x + y)dσ(y) = Utt.
We can now use the above result to solve the wave equation. For simplicity, assume n = 3 and let V (t, r) = r u(t, x; r) = r U (t, r). Then for r > 0 we have
Vrr = 2Ur + r Urr = r(Urr +
r Ur) = r Utt = Vtt.
This is also valid for r < 0 because V (t, r) is odd in r. Indeed for r < 0 , let v(t, r) = V (t, −r), then Vrr(t, r) = Vrr(t, −r) = Vtt(t, −r) = Vtt(t, r). By our solution to the one dimensional wave equation we find
V (t, r) =^1 2 (V (0, t + r) + V (0, r − t)) +^1 2
Z^ r+t
r−t
Vt(0, y)dy.
194 BRUCE K. DRIVER†
Proof. First recall that d dr
Z
B(x,r)
f dx = d dr
Z (^) r
0
dρ
Z
|y−x|=ρ
f (y)dσ(y) =
Z
∂B(x,r)
f dσ.
Hence
e ˙(t) = d dt
Z
B(x,R−t)
{| u˙(t, y)|^2 + |∇u(t, y)|^2 }dy
Z
∂B(x,R−t)
(| u˙|^2 + |∇u|^2 )dσ +
Z
B(x,R−t)
[ u˙ u¨ + ∇u · ∇ u˙] dm
Z
∂B(x,R−t)
(| u˙|^2 + |∇u|^2 )dσ +
Z
B(x,R−t)
[ u˙ ∆u + ∇u · ∇ u˙] dm
Z
∂B(x,R−t)
(| u˙|^2 + |∇u|^2 )dσ + 2
Z
∂B(x,R−t)
u ˙ ∂u ∂n dσ
Z
∂B(x,R−t)
{ 2 u˙ (∇u · n) − (| u˙|^2 + |∇u|^2 )}dσ ≤ 0
wherein we have used the elementary estimate,
2 (∇u · n) u˙ ≤ 2 |∇u| | u˙| ≤ (| u˙|^2 + |∇u|^2 ).
Therefore e(t) ≤ e(0) = 0 for all t i.e. e(t) := 0.
Corollary 14.9 (Uniqueness of Solutions). Suppose that u is a classical solution to the wave equation with u(0, ·) = 0 = ut(0, ·). Then u ≡ 0.
Proof. Proposition 14.8 shows 1 2
Z
B(x,T −t)
| u˙(t, y)|^2 + |∇u(t, y)|^2
dy = EB(x,T )(0) = 0
for all 0 ≤ t < T and x ∈ Rn. This then implies that u˙(t, y) = 0 for all y ∈ Rn^ and 0 ≤ t ≤ T and hence u ≡ 0.
Remark 14.10. This result also applies to certain class of weak type solutions in x by first convolving u with an approximate (spatial) delta function, say u(t, x) = u(t, ·) ∗ δ(x). Then u satisfies the hypothesis of Corollary 14.9 and hence is 0. Now let ↓ 0 to find u ≡ 0.
Remark 14.11. Proposition 14.8 also exhibits the finite speed of propagation of the wave equation.
14.6. Wave Equation in Higher Dimensions.
14.6.1. Solution derived from the heat kernel. Let
pnt (x) :=
(2πt)n/^2
e−^ 21 t |x|^2
and simply write pt for p^1 t. Then
2
Z ∞
0
cos ωt pλ(t)dt =
Z
R
eitωpλ(t)dt = e−λ∂ t^2 /^2 eitω|t=0 = e−λω (^2) / 2 .
PDE LECTURE NOTES, M ATH 237A-B 195
Taking ω =
−∆ and writing u(t, x) := cos
−∆t
g(x) the previous identity gives
2
Z ∞
0
u(t, x)
√^1
2 πλ
e−^ 21 λ t^2 dt = 2
Z ∞
0
u(t, x) pλ(t)dt
= eλ∆/^2 g(x) =
Z
Rn
pnλ(y)g(x − y)dy
=
Z
Rn
(2πλ)n/^2
e−^ 21 λ |y|^2 g(x − y)dy
(2πλ)n/^2
Z ∞
0
dρe−^ 21 λ ρ^2
Z
|y|=ρ
g(x − y)dσ(y)
σ(Sn−^1 ) (2πλ)n/^2
Z ∞
0
dρe−^ 21 λ ρ^2 ρn−^1 ¯g(x; ρ),
and so Z (^) ∞
0
u(t, x)e−^21 λ^ t 2 dt =
r πλ 2
σ(Sn−^1 ) (2πλ)n/^2
Z ∞
0
dρe−^21 λ^ ρ 2 ρn−^1 ¯g(x; ρ)
r π 2
σ(Sn−^1 ) (2π)n/^2
λ−(n−1)/^2
Z ∞
0
e−^ 21 λ t^2 tn−^1 ¯g(x; t)dt.
Suppose n = 2k + 1 and let cn :=
p (^) π 2
σ(Sn−^1 ) (2π)n/^2 ,^ then the above equation reads Z (^) ∞
0
u(t, x)e−^ 21 λ t^2 dt = cnλ−k
Z ∞
0
e−^ 21 λ t^2 t^2 k^ ¯g(x; t)dt
= cn
Z ∞
0
μ −
t ∂t
¶k e−^ 21 λ t^2 t^2 k^ g¯(x; t)dt
I.B.P. = c n
Z ∞
0
e−^ 21 λ t^2 (∂tMt− 1 )k^
t^2 k^ ¯g(x; t)
dt.
By the injectivity of the Laplace transform (after making the substitution t →
t, this implies
cos
−∆t
g(x) = u(t, x) = cn (∂tMt− 1 )k^
t^2 k^ ¯g(x; t)
= cn (∂tMt− 1 ∂tMt− 1... ∂tMt− 1 )
t^2 k^ ¯g(x; t)
= cn∂t
z k−^1 }|^ times { Mt− 1 ∂tMt− 1... Mt− 1 ∂t
£t^2 k−^1 ¯g(x; t)¤
= cn∂t
μ 1 t ∂t
¶k− (^1) £ t^2 k−^1 g¯(x; t)
Hence we have derived the following theorem.
Theorem 14.12. Suppose n = 2k + 1 is odd and let cn :=
p (^) π 2
σ(Sn−^1 ) (2π)n/^2 ,^ then
cos
−∆t
g(x) = cn∂t
μ 1 t ∂t
¶k− (^1) £ t^2 k−^1 ¯g(x; t)
PDE LECTURE NOTES, M ATH 237A-B 197
Making the substitution, u = s (^12)
1 + |x|
2 t^2
in the previous integral shows
Qt(x) = (2π)−^1 /^2
2 πt^2
¢−n/ 2
Ã
|x|^2 t^2
!#− n+1 2 Z (^) ∞
0
s n+1 2 e−s^ ds s
= (2π)−^1 /^22 n+1 2 (2π)−n/^2 t
t^2
¢− n+1 2
Ã
|x|^2 t^2
!− n+1 2 Γ
μ n + 1 2
n+1 2 (2π)−^
n+1 2 Γ
μ n + 1 2
t ³ t^2 + |x|^2
´ n+1 2
μ n + 1 2
t π n+1^2
t^2 + |x|^2
´ n+1 2.
Theorem 14.13. Let
cn :=
¡ (^) n+ 2
π n+1^2
(14.11) Qt(x) = cn t ³ t^2 + |x|^2
´ n+1 2
then
(14.12) e−t
√−∆ f (x) =
Z
Rn
Qt(x − y)f (y)dy.
Notice that if u(t, x) := e−t
√−∆ f (x), we have ∂^2 t u(t, x) =
u(t, x) = −∆u(t, x) with u(0, x) = f (x). This explains why Qt is the same Poisson kernel which we already saw in Eq. (9.36) of Theorem 9.31 above. To match the two results, observe Theorem 9.31 is for “spatial dimension” n − 1 not n as in Theorem 14.13. Integrating Eq. (14.12) from t to ∞ then implies √^1 −∆
e−t
√−∆ f (x) = √−^1 −∆
e−τ^
√−∆ f (x)|∞ τ =t
=
Z ∞
t
e−τ^
√−∆ f (x)dτ
=
Z
Rn
Z ∞
t
dτ Qτ (x − y)f (y)dy.
Now Z (^) ∞
t
Qτ (x − y)dτ = cn
Z ∞
t
τ ³ τ 2 + |x|^2
´ n+1 2 dτ^ =^
cn 1 − n
τ 2 + |x|^2
´ 1 − 2 n |∞ τ =t
cn n − 1
t^2 + |x|^2
´− n− 21
and hence
√^1 −∆ e−t
√−∆ f (x) =
Z
Rn
cn n − 1
t^2 + |y|^2
´− n− 21 f (x − y)dy
198 BRUCE K. DRIVER†
and by analytic continuation,
1 √ −∆
e(it−)
√−∆ f (x) =
e−(−it)
√−∆ f (x)
cn n − 1
Z
Rn
( − it)^2 + |y|^2
´− n− 21 f (x − y)dy
= cn n − 1
Z
Rn
|y|^2 − (t − i)^2
´− n− 21 f (x − y)dy
and hence
√^1 −∆
sin
t
f (x) = c^0 n lim ↓ 0
Z
Rn
Im
|y|^2 − (t − i)^2
´− n− 21 f (x − y)dy.
Now if |y| > |t| then
lim ↓ 0
|y|^2 − (t − i)^2
´− n− 21
|y|^2 − t^2
´− n− 21
is real so
lim ↓ 0 Im
|y|^2 − (t − i)^2
´− n− 21 = 0 if |y| > |t|.
Similarly if n is odd lim↓ 0
|y|^2 − (t − i)^2
´− n− 21
|y|^2 − t^2
´− n− 21 ∈ R and so
lim ↓ 0 Im
|y|^2 − (t − i)^2
´− n− 21
is a distribution concentrated on the sphere |y| = |t| which is the sharp propagation again. See Taylor Vol. 1., p. 221— 225 for more on this approach. Let us examine here the special case n = 3,
Im
Ã
|y|^2 − (t − i)^2
= Im
Ã
|y|^2 − t^2 + ^2 + 2it
− 2 t ³ |y|^2 − t^2 + ^2
so
I := lim ↓ 0
Z
Rn
Im
Ã
|y|^2 − (t − i)^2
f (x − y)dy
= lim ↓ 0
Z
Rn
− 2 t ³ |y|^2 − t^2 + ^2
f (x − y)dy
= 4π lim ↓ 0
Z ∞
0
ρ^2 − 2 t (ρ^2 − t^2 + ^2 )^2 + 4^2 t^2
f^ ¯ (x; ρ)dρ
= ct lim ↓ 0
Z ∞
0
ρ^2
(ρ^2 − t^2 + ^2 )^2 + 4^2 t^2
f^ ¯ (x; ρ)dρ.
