Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
7.1 The Radon transform and its inverse
We describe here some uniqueness and stability results of recovery of a function given its integrals over a family of hyperplanes (straight lines inR2or planes in R3).
The Radon transformR f of a compactly supported function f ∈C(R2) is de- fined as the integral
(7.1.1) F(ω,s)=R f(ω,s)=
xãω=s
f.
When f ∈L1(Rn) and compactly supported, the Radon transform is defined by the same formula for almost all (ω,s)∈×R. Due to the rotational invariancy ofR f, we can make use of the expansion of f,Fin spherical harmonics:
(7.1.2) f(x)=
fj,m(r)Yj,m(σ), F(ω,s)=
Fj,m(s)Yj,m(ω).
192
7.1. The Radon transform and its inverse 193 We recall that a spherical harmonicYj,m(σ) is a homogeneous harmonic poly- nomial of degree j restricted to the unit sphere. There are M(j,n) (M(j,2)= 2,M(j,3)=2j+1) such linearly independent polynomials. The functionsYj,m
andYk,l with different j andk areL2-orthogonal on the unit sphere. By apply- ing the standard orthogonalization procedure we can assume thatYj,m andYj,l
are orthogonal whenm=l as well, and that the whole system is orthonormal in L2(Sn−1).
The following result is due to Cormak (n=2) [Cor] and Deans [D].
Theorem 7.1.1. We have (7.1.3) fj,m(r)=cnr2−n
+∞
r
(s2−r2)(n−3)/2C(nj −2)/2(s/r)F(nj,m−1)(s)ds, where Ckj(t)is a normalized Gegenbauer polynomial of degree j,c2= −1/π, c3=π−3/2/2.
To prove the theorem we need the Funk-Hecke formula:
Sn−1
h(σãω)Yj,m(θ)d S(σ)=c(n) 1
−1h(t)C(nj−2)/2(t)(1−t2)(n−3)/2dt Yj,m(ω) (7.1.3)
relating spherical harmonics and Gegenbauer polynomials. Herec(2)=2,c(3)= 2π. Detailed references on this classical result and further properties of Gegenbauer polynomials related to the multiplicative analogue of the Fourier transformation can be found in the book of Natterer [Nat]. Another needed tool is the so-called Mellin transform
M f(s)= ∞
0
f(r)rs−1dr, which plays the same role for the multiplicative convolution
f∗mg(s)= ∞
0
f(r)g(s/r)r−1dr
as does the Fourier transformation for convolutions. Here f,gare bounded measur- able functions with compact supports. We will need the following known properties of the Mellin transform:
M(f∗mg)(s)=M f(s)Mg(s), M f(s+t)=M(xtf(x))(s),
M f(k)(s)=(1−s). . .(k−s)M f(s−k).
(7.1.4)
We observe that by introducing new variablesξ =lnr, η=lns we replace the Mellin transform by the Laplace transform and the multiplicative convolution by the usual one, so properties (7.1.4) are not completely surprising.
Exercise 7.1.2. Prove properties (7.1.4).
Also, we will use in the proof the following known results about the Mellin transform of Gegenbauer polynomials. Letg(r) beC(nj −2)/2(r)(1−r2)(n−3)/2when 0≤r≤1 and 0 when 1<r, and letg1(r) beC(nj −2)/2(1/r)(1−r2)(n−3)/2 when 0≤r≤1 and 0 when 1<r. Then
Mg(s)= 1
2 n−1
2 (s)2−s
j+s+n−1 2
s+1−j 2
,
Mg1(s)= 2s−1(n−2)
s−j 2
n−2+s+j 2
n−2
2 (s+n−2) ,
(7.1.5)
provided that j <s. Here(s) is Euler’s gamma function.
PROOF OFTHEOREM7.1.1. First, we will parametrize the hyperplane{xãω=s}
by pointsσ of the unit sphereaccording to the formulaσ →(s/(σãω))σ and write the integral over the hyperplane as the integral over,
xãω=s
fj,m(r)Yj,m(σ)d(x)=
fj,m(s/(σãω))Yj,m(σ)sn−1/(σãω)nd(σ).
Using the Funk-Hecke theorem (7.1.3) withh(t)= fj,m(s/t)sn−1/tnwhent >0 and 0 otherwise, we obtain
xãω=s
fj,m(r)Yj,m(σ)d(x)
=cn
1
0
fj,m(s/t)sn−1/tnC(nJ−2)/2(t)(1−t2)(n−3)/2dt Yj,m(ω)
=cn
∞
s
C(nj −2)/2(s/r)(1−s2/r2)(n−3)/2fj,m(r)rn−2dr Yj,m(ω), where we have made use of the substitutionr =s/t. By comparing with (7.1.2) we conclude that
(7.1.6) Fj,m(s)=cn
∞
s
C(nj−2)/2(s/r)(1−s2/r2)(n−3)/2fj,m(r)rn−2dr.
The equality (7.1.6) can be considered as an Abel-type integral equation with respect to fj,m, which can be solved explicitly using the Mellin trans- form. Letting f(r)= fj,m(r)rn−1 and defining g as above, we conclude that M Fj,m(s)=M f(s)Mg(s) because of the first property (7.1.4). To make use of the second formula (7.1.5) we have to replace s bys−1. So by applying the second property (7.1.4) witht = −1 we will have
M(r−1f)(s)=M(f)(s−1)M Fj,m(s−1) Mg(s−1) .
7.1. The Radon transform and its inverse 195 Comparing the two formula (7.1.5) we conclude that
1
Mg(s−1) =dn(s+n−2)
(s−1) Mg1(s),
dn = (n−2)
((n−1)/2)((n−2)/2)(1/2),
(d2=1/2). By using the second and third properties (7.1.4) witht =n−2 and k=n−1 we obtain
M(r−1f)(s)=(−1)n−1dnM(rn−2F(nj,m−1))(s)M(g1)(s).
Returning to the functions from their Mellin transforms by using the first property (7.1.4), we finally will have
rn−2fj,m=(−1)n−1dng1∗m(rn−1F(nj,m−1)),
which is formula (7.1.3) by the definitions of the multiplicative convolution∗m
and the functiong1.
This completes the proof.
We recall that the (normalized) Gegenbauer polynomialCkj(t) of degree j can be defined as
(7.1.7) −1(k)
(−1)m(j−m+k)/(m!(j−2m)!)(2t)j−2m
(the sum is overm< j/2). Whenk=0 the Gegenbauer polynomial is the Cheby- shev polynomial Tj(t)=cos(jarccost), and when k=1/2 it is the Legendre polynomial Pj. By using properties of these polynomials we will obtain later stability estimates for inversion of the Radon transform in the situation of the following corollary.
Corollary 7.1.3(The “hole” Theorem). Let K be a convex compact set inRn. Let f ∈L1(Rn)with compact support.
If R f(ω,s)=0 for (almost) all hyperplanes {xãω=s}not intersecting K , then f =0outside K .
PROOF. Formula (7.1.6) says thatF(ω,s) given for alls>Runiquely determines f(rσ) whenr >R. In other words, this corollary is true whenKis the ballB(0;R).
By using scaling and translations we obtain the result also for any ballB.
In the general case letxbe outsideK. Then, due to the convexity ofK one can find a ballBcontainingK and not containingx. By applying the result forBwe get f(x)=0.
The proof is complete.
For the inversion of the complete Radon transform one can give sharp stability estimates. We introduce the norm
F•,α =
|σ|=1
R(1+τ2)α|Fˆ(σ, τ)|2dτd S(σ) 1/2
,
whereFˆ(σ, τ) is the (partial) Fourier transform of the functionF(σ,s) with respect tos.
The next sharp stability result was obtained by Natterer (n=2) and by Smith, Solmon, and Wagner [SmSW] in the general case.
Theorem 7.1.4. There is a constant C=C(n, α, ρ)such that (7.1.8) C−1f(α)≤ R f•,α+(n−1)/2≤Cf(α)
for all functions f in C0∞(B(0;ρ)).
PROOF. Observe that Fˆ(σ, τ)=
Re−iτs
xãσ=s
f(x)d(x)
ds =
Re−iτxãσf(x)d x= fˆ(τσ).
Therefore,
F2•,α+(n−1)/2 =cn
|σ|=1
R(1+τ2)α+(n−1)/2|fˆ(τσ)|2dτd(σ)
=cn
(1+ |ξ|2)α+(n−1)/2|ξ|−n+1|fˆ(ξ)|2dξ.
(7.1.9)
To obtain the last equality we letξ =τσ and used thatdξ =cnτn−1dτd(σ).
Since (1+ |ξ|2)1/2|ξ|−1≥1, we have the inequality F2•,α+(n−1)/2≥cn
(1+ |ξ|2)α|fˆ(ξ)|2dξ.
The last integral isf2(α). Therefore, we have the first inequality (7.1.8).
To prove the second we will split the integration domain in the last integral in (7.1.9) into|ξ| ≥1 and|ξ|<1. In the first integral
(1+ |ξ|2)(n−1)/2|ξ|−n+1≤2(n−1)/2,
so this integral is bounded byCf(α). The second integral is bounded byCfˆ2∞. Using that f is compactly supported and applying the H¨older inequality, one has
fˆ∞≤Cf2(B(0;ρ))≤Cf(α).
the proof is complete.
Theorem 7.1.5. Let n=2or3. Assume that|f|n−1(BR)≤ M. Let AR,ρ = {ρ <
s<R}.
Then there is a constant C=C(R, ρ)such that
f2(BR\Bρ)≤C M2/|lnF2(AR,ρ)|.
PROOF. As a preliminary step we obtain a bound on Gegenbauer polynomials by using (7.1.7) withk=0(n =2) ofk=1/2(n=3). By the basic property of the
7.1. The Radon transform and its inverse 197 gamma function,
(j−m+1/2)=(j−m−1/2). . .3/21/2(1/2)≤(j−m)!(1/2).
therefore,
|Ckj(x)| ≤
2m<j
(j−m)!/(m!(j−2m)!)(2x)j−2m≤
2m<j
j!/((2m)!(j−2m)!)(2x)j−2m≤(1+2x)j (7.1.10)
according to the binomial theorem.
First we will bound the integral R
ρ fj2,mrn−1dr.
We will start with the three-dimensional case, which is easier technically. Ac- cording to Theorem 7.1.1, this integral equals
C R
ρ
R r
C1/2j (s/r)F(2)j,m(s)ds 2
dr
≤C(1+2R/ρ)2j R
ρ
R r
|Fj(2),m(s)|ds 2
dr
≤C(1+2R/ρ)2j R
ρ (R−r) R
r
(F(2)j,m(s))2ds)dr
≤C(1+2R/ρ)2j(R−ρ)2 R
ρ (Fj(2),m(s))2ds.
In the second inequality we made use of (7.1.10), and then of Fubini’s theorem and H¨older’s inequality. Using the interpolation theorem for spacesH(k)(ρ,R) and then the definition of the norm •,2and orthonormality of spherical harmonics on the unit sphere, we obtain
R
ρ (Fj(2),m(s))2ds≤C R
ρ (F(3)j,m(s))2ds
1/3 R
ρ (Fj,m(s))2ds 2/3
≤C
|σ|=1
R(Fj(3),m(s)Yj,m(σ))2ds dσ 1/3
F4/32 (A)
≤CF2/3•,3F4/32 (A)
≤Cf2(2)/3F42/3(A)≤C M2/3F42/3(A) by Theorem 7.1.4. We conclude that
(7.1.11)
R
ρ fj2,mrn−1dr ≤Ce2L jM2−λFλ2(A), L =ln(1+2R/ρ) whenn =3 withλ=4/3.
This integral in the two-dimensional case is R
ρ
R r
(s2−r2)−1/2C0j(s/r)Fj(1),m(s)ds 2
dr
≤Ce2L j R
ρ
R r
(s2−r2)−3/4ds 4/3
dr R
ρ (F(1)j,m(s))3ds 2/3
≤Ce2L j R
ρ (F(1)j,m(s))3ds 2/3
,
where as above we have used (7.1.10) and the H¨older inequality for the integral with respect tos. The last integral by the embedding theorem (withq =3,k= 1,m=76,n=1,p=2) is less than
CFj,m2(7/6)(ρ,R)≤CFj,m143//29Fj,m4(0)/9(ρ,R)≤CF14•,3//92F2(0)/3(A)
≤Cf14(1)/9F4(0)/9(A)≤C M14/9F42/9(A),
where in the second inequality we have used the interpolation theorem withθ = 2/9,s1=3/2,s2 =0. Thus, we obtain the bound (7.1.11) whenn=2.
Now we can complete the proof as in Section 6.1. Observe that
j,m
R
0
j2f2j,mrn−1dr ≤Cf2(1)(B)≤C M2. By orthonormality and completeness of spherical harmonics,
f22(A)=
m,j<J
R
ρ fj2,mrn−1dr+
m,J≤j
R
ρ fj2,mrn−1dr
≤Ce2L JM2−λFλ2(A)+C J−2M2,
where we have made use of (7.1.11) and of the previous observation. Let us choose J from the inequalities 2J≤ −δlnF2(A)≤2J+2, where positiveδis to be determined later. Then the previous inequality yields
f22(A)≤C M2(F−δ2 L+λ(A)+1/(−δ/2 lnF2(A)−1)).
Choosing δ < λ/L and using that the logarithm grows more slowly than any
positive power, we complete the proof.
It is realistic that one can derive stability estimates from a singular value de- composition for the Radon transform given in the most advanced form by Quinto [Q].
Lissianoi [Lis] showed the one cannot improve logarithmic estimates of Theo- rem 7.1.4 to H¨older even when looking for f in{R+ε <|x|}whileF is given on {R<|s|}. This is contrast to interior estimates for the Cauchy problem in differential equations and shows that it is not possible to apply to integral geom- etry over straight lines and planes reduction to hyperbolic equations and use of Carleman-type estimates. The spherical means tranform defined as integrals of f
7.1. The Radon transform and its inverse 199 over a family of spheres centered atγwith radius changing from 0 toT has better stability properties: indeed it is closely related to the Cauchy problem for the wave equation with the data on the lateral surfaceγ×(0,T). This Cauchy problem is discussed in sections 3.2, 3.4. There is a renewed interest to the spherical transform due to applications in ultrasound detection [FPR].
Despite the sharp estimates of Theorem 7.1.4, the Radon transform is not an invertible map fromH(α)ontoH•,α+(n−1)/2because its range is substantially smaller than this space. There is a complete description of the range given by the following Helgason-Ludwig consistency conditions:
RsmF(σ,s)dsis a homogeneous polynomial of degreeminσ, F(σ,s)=F(−σ,−s).
(7.1.12)
A similar characterization was given by Kuchment and Lvin in 1990 for the atten- uated Radon transform in the plane when the attenuation coefficient is constant, and we refer for some generalizations to the recent paper of Aguilar, Ehrenpreis, and Kuchment [AEK]. It claims that Fàis the exponential Radon transform of a function f ∈C0∞(R2) if and only ifFà∈C0∞(RìS1) and
Fˆà(σ(s),i s)=Fˆà(σ(−s),−i s),
whereσ(s) isσ rotated clockwise by the angle arcsins(s2+à2)1/2. We define the weighted Radon transform of f (with weightρ) as (7.1.13) Rρf(ω;ωãx)=
(y−x)ãω=0ρ(y,x;ω)f(y)d(y)
and observe that the constant attenuation in R2 with the attenuation coefficient à >0 means that
ρ(y,x, ω)=exp(à(y−x)ã(−ω2, ω1)).
Another inversion tool is the averaging operatorSwith respect toω:
S F=
|ω|=1F(ω)dω.
Lemma 7.1.6.
(7.1.14) S Rρf(x)=
|x−y|−1ρ•(y,x)f(y)d y, where
ρ•(y,x)=
|ω|=1,(x−y)ãω=0ρ(y,x;ω)dω.
PROOF. By definition, S Rρf(x)=
|ω|=1
(y−x)ãω=0ρ(y,x;ω)f(y)d(y)d S(ω)
By introducing the polar coordinatesy=x+rν, νãω=0 in the hyperplane (y− x)ãω=0, we write the interior integral as
|ν|=1,νãω=0
+∞
0
rn−2ρ(x+rν,x;ω)f(x+rν)dr d S(ν).
By interchanging the order of integration with respect toωandνwe obtain S Rρf(x)
=
|ν|=1
+∞
0
r−1
×
|ω|=1,ωãν=0ρ(x+rν,x;ω)
d S(ω)f(x+rν)rn−1dr d S(ν)
. Returning from polar coordinates to y=x+rν and using that d y= rn−1dr dν,r= |x−y|, andωãν=0 is equivalent to (x−y)ãω=0, we obtain
the formula (7.1.14).
Whenρ =1 (the classical Radon transform) we have S R1f(x)=cn
|x−y|−1f(y)d y.
In particular, ifn=3, then the right side is the volume potential of f; so applying the Laplace operator to the both parts, we will find that
(7.1.15) f(x)=cx
|ω|=1R1(x, ω)dω,
whereccan be shown to be−1/(8π2). This formula was known already to Radon.
In a more general situation we can reduce the problem of reconstruction of f to a Fredholm integral equation and obtain Schauder-type estimates for f.
Corollary 7.1.7. Let n=3andρ∈C3(××). Then the inversion of the attenuated Radon transform can be obtained as a solution of the following Fred- holm integral equation in C():
(7.1.16) c f(x)+
K(x,y)f(y)d y=(−)S F(x), where
K(x,y)=2∇x|x−y|−1ã ∇xρ•(x,y)+ |x−y|−1xρ•(x,y), We have the Schauder-type estimate
|f|λ()≤C(|S F|2+λ()+ |f|0()).
PROOF. By using that−x|x−y|−1=4πδ(x), we get from (7.1.14) c f(x)+
K(x,y)f(y)d y=(−)S F(x).