UniquenessforanInverseProblemwithFormallyDetermined
OsetData
Zachary Bailey
Advisor: Dr. Rakesh
Department of Mathematical Sciences, University of Delaware
Motivation
The problem under consideration arises in the study of geophysics and
medical imaging. Let x = (x1 , x2 , z) be a point in R3 , a = (a1 , a2 , 0) and
h = (h, 0, 0) for some h ≥ 0. We consider an acoustic medium occupying
the half space z ≤ 0 and let q(x) represent some acoustic property of the
medium (e.g. oil, minerals, a tumor).
Source
a−h
Receiver
Inverse Problem: Given measured data
v (a + h, t), is the coecient
unique? I.e. Given two solutions of (6)-(7) v1a and v2a that yield the
same measured data v1a (a + h, t) = v2a (a + h, t), are the corresponding
coecients q1 and q2 equal as well?
a
q(x)
To investigate this, we require the condition in the following theorem:
Theorem 1.
a+h
a
Inverse Problem
then
q1 = q 2
If
v1a (a + h, t) = v2a (a + h, t) for all a ∈ {z = 0} and t ∈ R,
C , independent of z such that
The medium is probed by an acoustic wave, generated at a − h, and the
medium response, U a is measured at the oset boundary location a + h,
for every a on the boundary z = 0. The goal is to recover the acoustic
property q(x) given U a (a + h, t) for all a on z = 0 and for all time t.
Progressing Wave Expansion
Given a real valued q = q(x) on z ≤ 0 representing the acoustic property of
the medium, emit a acoustic wave at x = a − h, characterized by U a (x, t).
It is known that U a satises the following PDE:
Utta − Δx U a − qU a = 0,
∂z U a (x, t) = δ(x − a + h, t),
a
x ∈ R3 , z ≤ 0, t ∈ R
{z = 0}, t ∈ R
3
U (x, t) = 0,
x ∈ R , z ≤ 0, t < 0
(1)
(2)
(3)
We extend U a to an even function in z , called V a . Then
Vtta − Δx V a − qV a = δ(x − a + h, t),
V a (x, t) = 0,
x ∈ R3 , t ∈ R.
x ∈ R3 , t < 0.
(4)
(5)
Using the progressing wave expansion" technique, we get
where va (x, t) = 0 outside of the cone t = |x − a + h| and inside it is the
solution of the Goursat problem:
a
a
− Δx v − qv = 0,
1
v a (x, |x − a + h|) =
8π
(6)
3
x ∈ R , t ≥ |x − a + h|
1
0
q(a − h + s(x − a + h))ds,
x ∈ R3
q
a ∈ {z = 0}, t ∈ R.
(M p)(a, τ ) =
for 0 ≤ t ≤ 2τ , and the PDE (6)-(7).
Goal: Show the coecient q(x) is unique for each va (a + h, 2τ ), i.e. the
map F : q(x) → va (a + h, 2τ ) is injective.
Step 1. Formulate the PDE as a dierence of two solutions with the same
boundary data, W a = V1a − V2a , where p = q1 − q2 . Then derive an
identity for W a .
Step 2. Derive an identity for a mean value operator of p, M (p) in terms
of p and ∇x p.
Step 3. Use steps 1 and 2 to estimate M (p) and ∇x p in terms of p and
p, then use Gronwall's to determine p = 0, i.e. q1 = q2 .
Physical Data:
1
16π 2
R3
p(x)δ(ϕ(x))dx +
E(a,τ )
p(x)k(x)dx,
(9)
where k is smooth and ϕ(x) represents the ellipsoid:
2 ∂
|∇y p(x)|2
√
dSx .
|p(a + σe3 )|2 (M p)(a, τ ) +
∂σ
σ2 − z2
∂E(a,τ )
V1a = V2a .
We rst estimate
2 ∂
(M p)(a, τ ) ∂σ
∂E(a,τ )
|p(x)|2 dSx .
Thus from (10), we get
|p(a + σe)|2 Let P (z) =
R2
∂E(a,τ )
|p(x, z)|2 dx,
|p(x)|2 dSx +
∂E(a,τ )
|∇x p(x)|2
√
dSx
σ2 − z2
(11)
then from the estimates (8) and (11), we get
P (σ) ≤ C
σ
0
P (z)dz
∀σ ∈ (0, 1].
From Gronwall's inequality, this gives that P (σ) = 0, implying p = 0 i.e.
q1 = q2 .
Future Work: A similar problem, but instead of working over the ellip-
soid,
|x − (a − h)| + |x − (a + h)| ≤ 2τ
we work over the hyperboloid
|x − a| − |x| ≤ τ
= {x ∈ R3 : |x − a + h| + |x − a − h| ≤ 2τ }
a+h
(10)
Step 3: Estimates
E(a, τ ) = {0 ≤ ϕ(x)}
a
p(x)
dSx .
|∇ϕ(x)|
∂E(a,τ )
Future Work and References
Let W a = V1a − V2a where V1a and V2a solve (4)-(5). Then
√
v a (a + h, t)
Step 1: The Ellipsoid
1
16π 2
Taking a derivative in σ = τ 2 − h2 , we can extract the value of p at the
north pole of the ellipsoid, a + σe3 , and get the following estimate
Let W a = 0 i.e.
a−h
directly. The physical measurements we can
v a (a + h, t),
the gradient in the rst two coordinates.
√
Forward Problem: Given q(x), determine va (x, t).
Pitfall: Cannot measure
Method
(8)
.References:
(7)
Forward Problem
make are
∇x = e 1 ∂ 1 + e 2 ∂ 2 ,
W a (a + h, 2τ ) =
1 δ(t − |x − a + h|)
V a (x, t) =
+ v a (x, t),
4π
|x − a + h|
a
vtt
where
We can rewrite the rst integral in (9) as a "mean value" operator,
provided there is a constant
∇x (q1 − q2 )(·, z)L2 (R2 ) ≤ C(q1 − q2 )(·, z)L2 (R2 ) , ∀z ∈ (0, 1],
z
Step 2: Mean Value Identity
τ 2 − h2
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pp, (2014).
[3] V. G. Romanov. Integral Geometry and Inverse Problems for Hyperbolic Equations, Springer Tracts in Natural Philosophy, Volume 26,
(1974).