WKB Analysis
Slowly varying coefficients/media:
We consider equations of the form
or
utt + ω 2 (t)u = f 0 (u, ut )
∇2 u + k 2 (x̄)u = 0
(1)
(2)
The first equation describes an oscillator with slowly varying frequency, with a possible weakly
nonlinear forcing.
The second is the Helmholtz equation in a slowly varying media. Recall, that the Helmholtz
equation is the equation for spatial dependence of time-harmonic waves
⇒
utt − c2 uxx = 0
u = eiωt v(x)
vxx + k 2 v = 0
If the physical parameters vary slowly in space we get equation (2) above.
It appears obvious that there are 2 scales in the problem, so one would naturally look for a
multiscale solution.
Consider (1), let u = u(t, τ ) = u0 (t, τ ) + u1 (t, τ ),
τ = t.
Then to leading order we get:
u0tt + ω 2 (t)u0 = 0
u0 = A sin ωt + B cos ωt
A = A(τ ), B(τ )
Then the next order contribution has secular terms.
Instead, we have to use a more complicated multi-scale, let
T = f (t)
,
| {z }
fast function of time
τ = t } slow time
Haberman covers this method, but it is relatively inefficient.
Instead, consider the equation (1) again. Let’s make the change of variables s = t, v(s) = u(t)
⇒
2 vss + ω 2 (s)v = f (v, vs )
Looks like a singular perturbation problem – but s is not a spatial variable, we don’t have boundary
conditions in s.
Notice that the “s” is a slow scale; reconsidering the equation on the “t” scale we saw that the
leading order equation gives us a solution A sin ωt where the frequency appears almost constant on
the t scale.
On the “s” scale, that solution appears to oscillate very fast
1
t
or s
t=O(1)
s=O(1)
Then, we look for solutions of the form v = Re iψ(s)/ , where we can use
R ∼ R0 + R1 + . . .
ψ ∼ ψ0 + ψ1 + . . .
Substituting in the equation for v:
#
"
(ψs )2
iψss
2iRs ψs
2
−
R+
R eiψ + Rω 2 (s)eiψ = 0
Rss +
2
p
2
⇒ 0(1)
−ψ0s
+ ω 2 (s) = 0 ⇒ ψ0s = ± ω 2 (s)
(so the solution should really be v = Re iψ/ + c. c.)
Then, the next order equation is:
0() :
ψ0s ψ1s
⇒
ψ0ss
2Rs
=−
ψ0s
R
= 0 ⇒ ψ1 = const
2Rsψ0s + ψ0ss R = 0 ⇒
log ψ0s = −2 log R ⇒
Rs √
Rs √
C2
C1
ω 2 (ŝ)dŝ
ω 2 (ŝ)dŝ
i
−i
p
+
v= p
e
e
4
4
2
2
ω (s)
ω (s)
What about weakly nonlinear problems?
We can write the solution as
iψ
v = Re
| {z } +v1
v0
Then the 0() equation has the form
eiψ (2Rs ψs + ψss R) + 2 v1ss + ω 2 v1 = (g1 (R, Rs )) eiψ/ +
n
X
j=2
Then
2Rs ψs + ψss R = g1 (R, Rs )
X
v1 =
aj eijψ/
For strongly nonlinear problems one cannot use WKB (in general).
What if ω 2 (x) < 0? or if ω 2 (x) = 0 for some x = x0 .
Consider equations of the form
utt + h(t)u = 0
h(t) > 0
for t < t0
h(t) < 0
for t > t0
2
(gj )eijψ/
For t > t0
Rs √
Rs √
C1
C2
−h(ŝ)dŝ
−h(ŝ)dŝ
−
p
u∼ p
+
e
e
4
4
−h(ŝ)
−h(ŝ)
s = t.
For t < t0
Near t ∼ t0
Rs √
Rs √
D1
D2
h(ŝ)ds
h(ŝ)dŝ
i
−i
p
u∼ p
+
e
e
4
4
h(ŝ)
h(ŝ)
h(s) ∼ −A(s − s0 )
(s0 = t0 )
Then solve 2 uss − A(s − s0 )u = 0
γ ξ = (s − s0 )
⇒
Rs
−
C1
s0
e
as ξ → ∞ match to p
4
−h(s)
√
γ = 2/3
uξξ − Aξu = 0
u = Ai A−1/3 ξ + Bi A−1/3 ξ
−h(ŝ) dŝ
Ai decays as ξ → ∞, Bi grows exponentially as ξ → ∞.
−1/4
+1/3 ξ 3/2
1 )
Ai ∼ √ A1/3 ξ
e−2/3(A
2 π
For t > t0
u ∼
C1
√
e
4
+Aγ ξ
=
−
C1
√
e
4
γ
A ξ
−
2/3 ξ
s0 +
R
s0
2/3
R ξ √Az
0
√
A(ŝ−s0 ) dŝ
C1
dz = p
e
4
A2/3 ξ
(A1/3 2/3 ξ)
−2/3
3/2
This gives an equation for C1 (the exponents match)
For t < t0 as ξ → −∞
For t < t0 , t ∼ t0
1
Ai ∼ √ p
sin
π 4 −A1/3 ξ
3/2 π 2
1/3
+
−A ξ
3
4
3/2
3/2
D1
D2
i2/3( −A1/3 2/3 ξ )
−i2/3(−A1/3 2/3 ξ )
p
u∼ p
e
e
+
4
4
−A2/3 ξ
−A2/3 ξ
3