an introduction to wave diffraction & the wiener–hopf …pcitho17/personal/pdfs/wh_talk.pdfe....
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An introduction to Wave diffraction & the
Wiener–Hopf technique
Ian Thompson
Department of Mathematical Sciences
What is the Wiener–Hopf technique?
• A method for solving problems involving discontinuous boundaryconditions.
• Uses classical complex analysis:
• poles & residues,
• branch points,
• Cauchy & Fourier integrals,
• Liouville’s theorem,
• Jordan’s lemma.
• Surrounded by an aura of fear, thanks largely to some terribleliterature.
1
The Sommerfeld problem
x
y
barrier
shadow region
reflection region
Θ
Incident waves
• There is a fundamental physics problem here, as well as a difficultmaths problem.
2
The Sommerfeld problem (continued)
• Arnold Sommerfeld solved his famous problem in 18961.
• His method doesn’t seem to work for any other problems!
• Around 1931, rigourous work on integral equations by N. Wiener,E. Hopf and others 2 was adapted to solve the Sommerfeld problem3.
• This came to be known as the Wiener–Hopf technique.
• It enables us to solve many otherwise intractable problems.
• Later, D. S. Jones found an easier way to apply the method usingFourier integrals.
• This strips away a lot of unnecessary mathematical apparatus.1Sommerfeld: ‘Optics’ (1964) [In English]2Paley & Wiener: ‘Fourier transforms in the complex domain’ (1934)3Wiener: ‘Collected Works’ 1981 [In English]
3
How it works
• Let Γ be a contour that divides C into upper and lower regions.
• Represent the wavefield φ in the form
φ(x, y) =1
2πi
∫
Γ
φ̂(α; y)e−iαx dα.
• Suppose that φ̂(α; 0) → 0 as |α| → ∞.
• Jordan’s lemma says that for x > 0, we can close Γ in the lowerhalf plane, whereas for x < 0 we can close in the upper half plane.
• This is absolutely crucial:
Different singularity structures above and below Γ givedifferent results for x > 0 and x < 0.
4
The wave equation
• We have (
∇2 − 1
c2
∂2
∂t2
)
Φ(x, y, t) = 0.
• Assume that fields are time-harmonic , i.e.
Φ(x, y, t) = ℜ[φ(x, y)e−iωt
].
• Now we have to solve the Helmholtz equation
(∇2 + k2)φ(x, y) = 0, k = ω/c,
• Note propagation directions, eg. eiky propagates upwards, because
φ = eiky ⇒ Φ = ℜ[
ei(ky−ωt)]
.
5
Setting up the Sommerfeld problem
• Write φt = φi + φs.
• The incident wave is known:
φi = eik(x cos Θ+y sin Θ).
• We need to solve the Helmholtzequation subject to the b.c.
φs(x, 0) = −eikx cos Θ on x > 0.
• Also, ∇φs ∼ O(r−1/2) as
r =√
x2 + y2 → 0 (tip condition).
• φs must not include any contributionsthat are incoming toward the barrier, orthat grow as |y| is increased.
x
y
φt = 0
Θ
Incident waves
6
The first golden rule
• Always look to exploit symmetries.
• Decompose the incident wave into odd and even (in y) functions:
φiS(x, y) =
1
2
[φi(x, y) + φi(x,−y)
], φi
A(x, y) =1
2
[φi(x, y) − φi(x,−y)
].
• φiA(x, 0) = 0, so this satisfies the b.c. on its own.
• In other words, φsA(x, y) ≡ 0, and so φs(x, y) = φs
S(x, y).
• φs(x, 0) must be continuous on x < 0 (where there is no barrier),and it’s an even function, so
∂φs
∂y= 0 y = 0, x < 0.
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The second golden rule
• Exhaust all elementary arguments before resorting to W–H.
• We need to solve for φs(x, y) in y ≥ 0.
• Start by writing
φs(x, y) =1
2πi
∫
Γ
φ̂(α; y)e−iαx dα.
• Apply operator (∇2 + k2) (ok to commute with integral):∫
Γ
[
∂2φ̂
∂y2− (α2 − k2)
]
e−iαx dα = 0.
• This can only be true ∀x if [·] ≡ 0, and hence
φ̂(α; y) = A(α)e−γ(α)y + B(α)eγ(α)y,
where γ(α) = (α2 − k2)1/2.
8
The functions A and B
φs(x, y) =1
2πi
∫
Γ
[
A(α)e−γ(α)y + B(α)eγ(α)y]
e−iαx dα.
• At points on Γ where ℜ[γ(α)] = ℜ[(α2 − k2)1/2] 6= 0 :
If e−γ(α)y → 0 as y → ∞, then |eγ(α)y| → ∞ & vice-versa.
• At points on Γ where ℜ[γ(α)] = ℜ[(α2 − k2)1/2] = 0 :
If e−γ(α)y represents an outgoing contribution then eγ(α)y
represents an incoming contribution & vice-versa.
• One solution is always unphysical—doesn’t matter which becausewe have yet to choose the branch of γ(α).
• It is conventional to take B(α) ≡ 0.
• Also, A(α) ∼ O(α−η) with η > 1 as α → ∞ ∈ Γ.
9
The contour Γ
φs(x, y) =1
2πi
∫
Γ
A(α)e−γ(α)ye−iαx dα.
• Recall: Γ divides C into upper and lower regions.
• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .
• γ(α) =√
α2 − k2 for real α with |α| > k.
• Go clockwise around α = −k to reduce theargument from 0 to −π/2.
• Go anticlockwise around = k to change itback to 0.
• D±= {α ± iu : α ∈ Γ, u ≥ 0}.
−k
k ℜ[α]
ℑ[α]
10
The contour Γ
φs(x, y) =1
2πi
∫
Γ
A(α)e−γ(α)ye−iαx dα.
• Recall: Γ divides C into upper and lower regions.
• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .
• γ(α) =√
α2 − k2 for real α with |α| > k.
• Go clockwise around α = −k to reduce theargument from 0 to −π/2.
• Go anticlockwise around = k to change itback to 0.
• D±= {α ± iu : α ∈ Γ, u ≥ 0}.
−k
k ℜ[α]
ℑ[α]
10
The contour Γ
φs(x, y) =1
2πi
∫
Γ
A(α)e−γ(α)ye−iαx dα.
• Recall: Γ divides C into upper and lower regions.
• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .
• γ(α) =√
α2 − k2 for real α with |α| > k.
• Go clockwise around α = −k to reduce theargument from 0 to −π/2.
• Go anticlockwise around α = k to change itback to 0.
• D±= {α ± iu : α ∈ Γ, u ≥ 0}.
−k
k ℜ[α]
ℑ[α]
10
The contour Γ
φs(x, y) =1
2πi
∫
Γ
A(α)e−γ(α)ye−iαx dα.
• Recall: Γ divides C into upper and lower regions.
• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .
• γ(α) =√
α2 − k2 for real α with |α| > k.
• Go clockwise around α = −k to reduce theargument from 0 to −π/2.
• Go anticlockwise around α = k to change itback to 0.
• D±= {α ± iu : α ∈ Γ, u ≥ 0}.
−k
k ℜ[α]
ℑ[α]
10
The contour Γ
φs(x, y) =1
2πi
∫
Γ
A(α)e−γ(α)ye−iαx dα.
• Recall: Γ divides C into upper and lower regions.
• φs contains only decaying (ℜ[γ] > 0) and outgoing (ℑ[γ] < 0)contributions. So, . . .
• γ(α) =√
α2 − k2 for real α with |α| > k.
• Go clockwise around α = −k to reduce theargument from 0 to −π/2.
• Go anticlockwise around α = k to change itback to 0.
• D±= {α ± iu : α ∈ Γ, u ≥ 0}.
−k
k ℜ[α]
ℑ[α]
D+
D−
10
Boundary Condition on x > 0
φs(x, y) =1
2πi
∫
Γ
A(α)e−iαx dα.
• We need to satisfy φs(x, 0) = −eikx cos Θ on x > 0.
• So, if x > 0, the only contribution must come from a simple poleat α = −k cos Θ, otherwise we get the wrong x dependence.
• Γ passes above α = −k cos Θ, and we can write
A(α) =1
α + k cos Θ+ A
−(α),
• A−(α) is unknown, but it is analytic in D−
, and A−(α) → 0 as
α → ∞ ∈ D−.
• Jordan’s lemma applies (close Γ in D−); the BC on x > 0 is
satisfied.
11
Symmetry Condition on x < 0
∂φs
∂y(x, 0) =
−1
2πi
∫
Γ
G(α)e−iαx dα where G(α) = γ(α)A(α).
• We need to satisfy∂φs
∂y(x, 0) = 0 on x < 0.
• If x < 0, there must be no contributions to the integral.
• Hence, G(α) = G+(α), i.e. it’s is analytic in D+
, and G+(α) → 0
as α → ∞ ∈ D+.
• Jordan’s lemma applies again (close Γ in D+); the symmetry
condition on x > 0 is now satisfied.
12
Wiener–Hopf equation
• We have G(α) = γ(α)A(α), and hence
G+(α) = γ(α)
[1
α + k cos Θ+ A−(α)
]
. (∗)
• We can product split γ(α) = (α2 − k2)1/2:
γ+(α) = (α + k)1/2; γ
−(α) = (α − k)1/2.
• Use in (∗): G+(α)
γ+(α)=
γ−(α)
α + k cos Θ+ γ
−(α)A−(α).
• LHS is analytic in D+. RHS is analytic in D− \ {−k cos Θ}.
• Subtractγ−(−k cos Θ)
α + k cos Θfrom both sides. . .
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Wiener–Hopf equation (continued)
G+(α)
γ+(α)− γ
−(−k cos Θ)
α + k cos Θ︸ ︷︷ ︸
=γ−(α) − γ
−(−k cos Θ)
α + k cos Θ+ γ
−(α)A−(α)
︸ ︷︷ ︸
.
Analytic in D+Analytic in D−
• Both sides represent the same function! Call it J(α).
• This is analytic in D+ ∪ D−= C, i.e. it’s entire.
• We need to determine its behaviour as |α| → ∞.
• From the LHS, J(α) → 0 as α → ∞ ∈ D+.
• It’s not clear what happens to γ−(α)A
−(α) as α → ∞ ∈ D−
.
14
Tauberian Theorems
• Suppose that
f(x) =1
2πi
∫
Γ
F−(α)e−iαx dα,
where F−(α) → 0 as α → ∞ ∈ D−
, so that f(x) = 0 for x > 0.
• If f(x) ∼ O(xη) as x → 0−
(with η > −1) then F−(α) ∼
O(α−(η+1)) as α → ∞ ∈ D−
• We have to apply this to φt, because φt(x, 0) = 0 for x > 0.
φt(x, 0) =1
2πi
∫
Γ
⌣
[1
α + k cos Θ+ A
−(α)
]
e−iαx dα
(the residue at α = −k cos Θ is eikx cos Θ = φi(x, 0)).
• ∇φt ∼ O(r−1/2) as r → 0, so φt(x, 0) ∼ O(1) as x → 0−
(at
most). =⇒ A−(α) ∼ O(α−1) as α → ∞ ∈ D−
(at most).
15
Solving the Wiener–Hopf equation
• At last, we have everything we need. . .
G+(α)
γ+(α)− γ
−(−k cos Θ)
α + k cos Θ︸ ︷︷ ︸
=γ−(α) − γ
−(−k cos Θ)
α + k cos Θ+ γ
−(α)A−(α)
︸ ︷︷ ︸
= J(α).
→ 0 as α → ∞ ∈ D+ → 0 as α → ∞ ∈ D−
• J(α) → 0 as α → ∞ in any direction, and it’s an entire function.
• Liouville’s theorem says that J(α) ≡ 0.
• Now we can complete the solution using either G+
or A−:
G+(α) = γ(α)A(α) =⇒ A(α) =
γ−(−k cos Θ)
γ−(α)(α + k cos Θ).
16
What does it all mean?
φs(x, y) =1
2πi
∫
Γ
γ−(−k cos Θ)
γ−(α)(α + k cos Θ)e−γ(α)|y|e−iαx dα.
• This satisfies all of the required conditions.
• In this (very special) case, the integral can be evaluated exactly(!)
φs(r, θ) =eikr
2
{
w[
eiπ/4√
2kr sin(
θ+Θ2
)]
+ w[
eiπ/4√
2kr sin(
θ−Θ2
)]}
.
• w(z) is the scaled complex error function; w(z) = e−z2[1− erf(z)].
• This is Sommerfeld’s original solution.
• The error function rapidly but continuously (de)activates the planewave terms as the optical boundaries are crossed.
• There is also a circular wave that is O[(kr)−1/2] in the far field,away from the boundaries.
17
k = 1.0, Θ = π/4
-
−20−20
20
202.52.5
(Plane wave terms included)
17
k = 1.0, Θ = π/4
-
−20−20
20
202.52.5
(Plane wave terms excluded—diffraction only)
17
k = 1.0, Θ = π/4
-
−20−20
20
202.52.5
(Plane wave terms excluded—diffraction only)
17
One more thing. . .
• To solve the Sommerfeld problem, we had to write
γ(α) = γ+(α)γ
−(α).
• This is called a kernel factorisation.
• It was easy to find γ±(α); just ‘share out’ the singularities:
γ(α) = (α2 − k2)1/2, γ±(α) = (α ± k)1/2.
• It doesn’t take much to make this hard, eg.
f(α) = 1 + γ(α), f±(α) =?
18
Cauchy integrals
• Suppose that K(α) → 1 as α → ∞ ∈ Γ. Define
I+(α) =
1
2πi
∫
Γ
⌣log[K(z)]
z − αdz, α ∈ D+
,
I−(α) =
−1
2πi
∫
Γ
⌢log[K(z)]
z − αdz, α ∈ D−
.
• Then K±(α) = exp[I
±(α)], because
K+(α)K
−(α) = exp[I+(α) + I
−(α)] = exp[log[K(α)]].
• We use functions with known factorisations to make K(α) → 1:
f(α) = 1 + γ(α), K(α) =1 + γ(α)
γ(α), f
±(α) = γ
±(α)K
±(α).
• It is usually very hard to evaluate these integrals, even numerically.
19
References
• The Sommerfeld problem appears in lots of books, eg.
• Linton & McIver (2001) ‘Handbook of Mathematical techniques for wave/structure interactions’,
• Billingham & King (2000) ‘Wave motion’,
• Crighton et al. (1992) ‘Modern Methods in Analytical Acoustics’.
• Jones (1964) ‘The Theory of Electromagnetism’,
• Linton & McIver solve one more problem via the Wiener–Hopfmethod.
• Only the dreaded
• Noble (1988) ‘Methods based on the Wiener–Hopf technique’.
seems to contain further examples.
• Don’t read this after dark when there is a full moon.
• Prof. David Abrahams is in the process of writing a new book. . .
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