A Characterization of the Radon Transform’s Range by a System of PDEs
Arp´´ ad Kurusa
Abstract. Letg be a compactly supported function of d-planes in Rn. We prove that then g is in the range of the Radon transform if and only if g satisfies an ultrahyperbolic system of PDEs. We parameterize thed-planes by d+ 1 pointsx0, x1, . . . , xd on them and get the PDE
∂2
∂xki∂xlj − ∂2
∂xli∂xkj
!
g(x0, x1, . . . , xd) Vol{xi−x0}i=1,d
= 0,
wherexki denotes thek−thcoordinate ofxi. At the end we analyze in detail the case ofd= 1.
1. Introduction
In this paper we consider the range of the (d, n) Radon transformRnd, which is defined by
Rndf(ξ) = Z
ξ
f(x)dx,
where f ∈D(Rn),ξis an element ofG(d, n), the set ofddimensional hyperplanes in Rn, 1≤d≤n−2 anddxis the surface measure onξ.
There are many papers about the range of the Radon transform considered on several different spaces (e.g. [2],...,[12]), some of which ([3],[4],[8],[11]) give PDEs to characterize the range of the Radon transform.
The first one as far as I know in which a characterization by a system of PDEs was given is [8]. F. John characterized the range only in the case (1,3) and used in his proof many special properties of the 3 dimensional space and the Asgeirsson’s lemma, but it was very geometrical compared with the others.
AMS Subject Classification(1980): 44A05, 53C65 .
J. Math. Anal. Appl., 161(1991), 218–226. c A. Kurusa´
These later extensions of John’s result used strong tools of analysis like Fourier transforms etc. and a strong not too natural, by my opinion, definition of S(G(d, n)). On the other hand, as Richter also pointed out, Grinberg’s and Gelfand’s proof was incomplete. Other substantial difference between these papers is the parameterization’s method of the d-planes. We will follow John’s method parameterizing the d-planes byd+ 1 points on them. The other possible method, that was used by Richter, Grinberg and Gelfand is to parameterize asy=Ax+c, where Ais an (n−d)×dmatrix andc∈Rn−d.
In the following we characterize the range ofRnd with a system of PDEs on a relatively easy geometrical way deriving Helgason’s moment condition directly from the differential equation.
At the end we analyze in detail the case of (1, n) and give a short proof of John’s main theorem which gives all the solutions of the ultrahyperbolic partial differential equation with four variables. This equation (5) plays part in the Yang- Mills theory [1, pp. 78-81].
2. Preliminaries
To any set ofd+ 1 pointsx0, x1, . . . , xd ∈Rn, which are general position, one can associate a uniquely determined d-plane ξ(x0, x1, . . . , xd) ∈ G(d, n) through them. Using this parameterization forG(d, n) we can write
Rndf(x0, x1, . . . , xd) = Z
ξ
f(x)dx.
It is worthwhile to note thatRndf in this context is interpreted on a principal fibre bundle over G(d, n) with the affine group of Rd as its structure group. The undermentioned functionRdnf /|detUx−1|is definitely interpreted on this principal fibre bundle too. Let Ux be an automorphism of Rn transforming the system {xi−x0}i=1,dinto an orthonormal system. Then a substitution gives
Rndf(x0, x1, . . . , xd) = Z
Rd
f x0+
d
X
i=1
λi(xi−x0)
|detUx−1|dλ,
where λ= (λ1, . . . , λd) ∈ Rd. We note that|detUx−1| is the volume of the par- allelepiped spanned by the vectors {xi−x0}i=1,d. For example, in the case of (1, n) |detUx−1| = |x1−x0| . We will use this remark frequently. Now we can state an easy lemma, which can be proven by applying the differentiation under the integral sign.
Lemma 2.1. If f ∈D(Rn)then
(∂i,k∂j,e−∂i,e∂j,k)Rndf(x0, x1, . . . , xd) Vol{xi−x0}i=1,d
= 0,
where 0≤i, j≤d,1≤k, e≤nand∂i,k denotes the differentiation with respect to the k−thcoordinate xki of thei−thpoint xi.
Now we recall a definition and a statement of Helgason [6].
A functionf on G(d, n) is said to be in DH(G(d, n)) if f is C∞, has com- pact support and satisties the following condition: For each k∈N there exists a homogeneous kth-degree polynomial Pk on Rn such that for each d-dimensional subspace σthe polynomial
Pσ,k(u) = Z
σ⊥
f(x+σ)hx, uikdx foru∈σ⊥,
where σ⊥ is the orthogonal complement ofσin Rn anddxis the surface measure onσ⊥, coincides with the restrictionPk σ⊥.
The crucial point of this condition is the independence ofPσ,k(u) from σ if u∈σ⊥, because it is obviously a homogeneous kth-degree polynomial.
Lemma 2.2. (Corollary 2.28. in [6] ). The(d, n)Radon transform is a bijection of D(Rn)ontoDH(G(d, n)).
This statement characterize the range of our transforms and we will use it as starting point. The following result is a slight extension of Theorem 1.2. of [8] and can be proven by simple calculation.
Lemma 2.3. Suppose that v ∈ C∞(Rn(d+1)) such that it depends only on the d- planes of the d+ 1 points inRn,ξ(x0, x1, . . . , xd)∈G(d, n)and satisfies
(1) (∂i,k∂j,e−∂i,e∂j,k) v(x0, x1, . . . , xd) Vol{xi−x0}i=1,d
= 0, where 0≤i, j≤d,1≤k, e≤n. Then the function
w(x0, x1, . . . , xd) =v(Ax0, Ax1, . . . , Axd)
also satisfies (1) and also depends only on ξ(x0, x1, . . . , xd)∈G(d, n)for any A∈ SO(n), the group of orthogonal automorphisms of determinant1 of Rn.
J. Math. Anal. Appl., 161(1991), 218–226. c A. Kurusa´
3. The main result
Theorem 3.1. Let v ∈D(G(d, n)). Then there exists a function f ∈D(Rn) such that v=Rndf if and only ifv satisfies the system of PDEs (1).
Proof. The necessity is proved by Lemma 2.1. Thus we only have to prove the sufficiency of the condition, i.e. that our condition implies the condition of Helgason in Lemma 2.2. From the definition we have for u∈σ⊥
Pσ,k(u) = Z
σ⊥
v(x, x+σ1, x+σ2, . . . , x+σd)hx, uikdx,
where {σi}i=1,d is an orthonormal basis of σ. We have to show that Pσ,k(u) = Pσ,k¯ (u) ifu∈σ⊥∩σ¯⊥. First we simplify the statement to be proven.
Without any restriction of generality one can suppose that |u| = 1. Let e1, e2, . . . , en be an orthonormal basis of Rn. Let ε = ξ(0, e1, e2, . . . , ed) and let A∈SO(n) such thatσi=Aei andu=Aen. On substitutingx=Aywe get
Pσ,k(u) =PAε,k(en) = Z
ε⊥
v(Ay, A(y+e1), . . . , A(y+ed))hy, enikdy, hence by Lemma 2.3. it is enough to consider the case whenu=enandσi=ei to prove the statement below from which our theorem will follow easily.
(∗)
Ifσ1, σ2, . . . , σd is orthonormal system, ¯σ1 ⊥σ2, . . . , σd, |¯σ1|= 1, σ=ξ(0, σ1, σ2, . . . , σd) and ¯σ=ξ(0,σ¯1, σ2, . . . , σd) thenPσ,k(en)
=Pσ,k¯ (en).
Thus we are going to prove that ifE=α1e1+Pn−1
i=d+1αiei, whereα21+Pn−1 i=d+1α2i = 1, then Pε,k(en) = Pε,k¯ (en), where ¯ε = ξ(0, E, e2, . . . , ed). We will prove it by approximating E in a way that the Pεappr,k(en) does not vary, where εappr = ξ(0, Eappr, e2, . . . , ed).
As a first step, take Eappr = e1sinβ+eicosβ, where i > d. Without any restriction of generality we can assumei=d+ 1. Let
P(β) =Pεappr,k(en)
= Z ∞
−∞
· · · Z ∞
−∞
v(x, x+Eappr, . . . , x+ed)λkndλndλn−1. . . dλd+1,
where x = (−e1cosβ +ed+1sinβ)λd+1+Pn
i=d+2λiei. Differentiating this with respect to β we obtain
P0(β) = Z ∞
−∞
· · · Z ∞
−∞
λd+1
d
X
j=0
(sinβ∂j,1v+ cosβ∂j,d+1v) + + cosβ∂2,1v−sinβ∂2,d+1v
λkndλn. . . dλd+1. To compute this integral, we need a simple fact. Since v is a function onG(d, n) for every ν∈R(6= 0) we have
v(x0, x1, . . . , xd) Vol{xi−x0}i=1,d
=ν v(y0, y1, . . . , yd) Vol{yi−y0}i=1,d
,
where yk=xk except thei-th for whichyi=xj+ν(xi−xj). Then the differenti- ation of this equation with respect toν atν= 1 gives
(2) v(x0, x1, . . . , xd) Vol{xi−x0}i=1,d +
n
X
m=1
(xmi −xmj )∂i,m
v(x0, x1, . . . , xd) Vol{xi−x0}i=1,d = 0.
Applying this in our situation for j= 1 andj= 0 our integral becomes P0(β) =
Z ∞
−∞
· · · Z ∞
−∞
(cosβ∂2,1v−sinβ∂2,d+1v)λkndλn. . . dλd+1. Then by partial integration with respect to λn we have
(k+ 1)P0(β)
= Z ∞
−∞
· · · Z ∞
−∞
d dλn
(cosβ∂2,1v−sinβ∂2,d+1v)λk+1n dλn. . . dλd+1. But dλd
n =Pd
j=0∂j,n in our case therefore (1) gives that (k+ 1)P0(β) =
Z ∞
−∞
· · · Z ∞
−∞
−d
dλd+1∂2,nv λk+1n dλn. . . dλd+1.
Since n > d+ 1 the integration with respect to λd+1 shows P0(β) ≡ 0, i.e.
Pεappr,k(en) =Pε,k(en). One can easily see that with at most n−1 steps of this kind we can reach the generalE, therefore (∗) is proved.
Now let σ1, σ2, . . . , σd and ¯σ1,σ¯2, . . . ,σ¯d be orthonormal systems and σ = ξ(0, σ1, σ2, . . . , σd), ¯σ=ξ(0,σ¯1,σ¯2, . . . ,¯σd). We prove Pσ,k(en) =P¯σ,k(en) step by
J. Math. Anal. Appl., 161(1991), 218–226. c A. Kurusa´
step showing a sequence of the ddimensional subspaces σr such thatPσ,k(en) = Pσ1,k(en) =· · ·=Pσm,k(en) =Pσ,k¯ (en), where all these equations are true by (∗).
If ¯σi is at most for one index j not perpendicular to σj it can be substi- tuted into the place of σj by (∗) on such a way thatPσ,k(en) does not changei.e.
Pσ,k(en) = Pσ1,k(en). Let us continue this replacements with the last obtained σr−1 as far as possible. This procedure can stop at theσr if and only if there are at least two elements of{σir}i=1,d, which are not perpendicular to ¯σi.
In this case letσrj andσrkbe two vectors not perpendicular to ¯σi. Transforming these two vectors as
σjr+1=σrjcosα+σrksinα σkr+1=σjrsinα−σrkcosα,
where tanα = h¯σi, σrki/h¯σi, σrji, and leaving σsr+1 = σsr for other indexes s we obtain the σr+1 = ξ(0, σr+11 , σ2r+1, . . . , σdr+1) subspace, which satisfies obviously Pσr+1,k(en) = Pσr,k(en) and has more vectors perpendicular to ¯σi. Continuing this procedure as far as possible we will get a subspace the orthonormal spanning system of which will become treatable by the previous method. As a result of this line of reasoning finally we shall obtain the desired sequence, which proves the theorem.
4. The (1, n) case
From now on we concentrate on the special case of (1, n) Radon transform, when we integrate over the lines. We parameterize the lines G(1, n) by Pl¨ucker coordinates.
Letξ= (ξ1, . . . , ξn) andη= (η1, . . . , ηn) be two different points of the lineg.
The Pl¨ucker coordinates ofg are pi,k= det
ξi ξk ηi ηk
and qj= det ξj 1
ηj 1
,
where 1≤i < k≤nand 1≤j ≤n. As it is well known ratios of the coordinates pi,n andqj are unchanged under replacingξandη by two different points ofgand determine the line uniquely. Below for brevity we simply write pi forpi,n.
By using the Pl¨ucker coordinates now we define a bijection between functions onG(1, n) and the functions onR2n−2. To any functionvonG(1, n) let us associate a functionuonR2n−2 by the equation
(3) v(ξ, η) =
n
X
i=1
qi
qn
2!1/2
u x1
qn,x2
qn,x3
qn, . . . ,x2n−2 qn
,
wherexi=Pn−1
j=1 (λi,jpj+δi,jqj) andλi,j, δi,j ∈R(1≤i≤2n−2). The function v can be recovered fromuif and only if the matrices
Λ = [λi,j]i=1,2n−2
j=1,n−1 and ∆ = [δi,j]i=1,2n−2
j=1,n−1
are chosen on the way that the (2n−2)×(2n−2) matrix Γ = [Λ,∆] is invertible.
Since|ξ−η| = Pn
i=1q2i1/2
derivation of v(ξ, η)/|ξ−η|with respect to ξk and ηe, equations (1) and (3) result in
(4)
2n−2
X
i=1
(λi,eδi,k+λi,kδi,e)∂i2u+
+
2n−2
X
1≤i<j
(λi,eδj,k−λi,kδj,e+λj,eδi,k−λj,kδi,e)∂i∂ju= 0.
This means that relation (3) defines an equivalence between the systems of PDEs (1) and (4) ifk, e < n.
Lemma 4.1. Relation (3) gives a bijection between C2(G(1, n)) and C2(R2n−2).
For a pair of functions u ∈ C2(R2n−2) and v ∈ C2(G(1, n)) related by (3) the systems of PDEs (4) and(1)are equivalent.
Proof. The only non trivial thing to prove here is that if usatisfies (4) and v is defined by (3) then v satisfies (1) not only fork, e < nbut also for e =n. From (2) we know that
v(ξ, η)
|ξ−η|+
n
X
e=1
(ξe−ηe) ∂
∂ξe
v(ξ, η)
|ξ−η| = 0.
On differentiating with respect toηk it follows that
n
X
e=1
(ξe−ηe) ∂2
∂ηk∂ξe − ∂
∂ξk + ∂
∂ηk
!v(ξ, η)
|ξ−η| = 0.
We add to this equation the one obtained by replacing ξ and η and get in this manner
n
X
e=1
(ξe−ηe) ∂2
∂ηk∂ξe− ∂2
∂ξk∂ηe
v(ξ, η)
|ξ−η| = 0.
Since v satisfies (1) for all 1≤k, e≤n−1 finally we obtain the desired formula fore=n.
Now we can state one of the main theorems of [8] as a simple consequence of our results.
J. Math. Anal. Appl., 161(1991), 218–226. c A. Kurusa´
Theorem 4.2. The function u:R4 → R is a Cc∞ solution of the ultrahyperbolic PDE
(5) (∂12+∂22−∂32−∂42)u= 0
if and only if there exists a Cc∞ function f:R3→Rsuch that
R31f(ξ, η) =
3
X
i=1
qi
q3
2!1/2 u
p1+q2
q3
,−p2+q1
q3
,p1−q2
q3
,−p1−q1
q3
,
where pi andqi (1≤i≤3) are the Pl¨ucker coordinates of the straightline through ξ andη.
Proof. In (2) let us choose the following matrix
1 0 0 1
0 −1 1 0
1 0 0 −1
0 −1 −1 0
as Γ. Then the relationship betweenv andudefined by (3) is one-to-one and (4) gives only one equation for unamely (5). By Lemma 4.1. and Theorem 3.1. this gives our theorem.
To find such a nice formula for higher dimension it would be necessary that the PDE (4) does not depend onkande. This condition gives a lot of linear equations for the elements of the matrix Γ from which one can conclude by counting these equations that for n ≥ 4 to find matrix Γ that gives only one PDE in (4) is impossible.
In Theorem 3.1. we essentially characterized the range of the (d, n) Radon transformRnd by the system of PDEs (1). The fact that the functionv(ξ, η) depends only on the straightline throughξandηplays a role of an additional condition there.
The following corollary shows how this unessential condition can be removed by making certain transformation on the range of Rn1. At the same time this result may prove useful in solving partial differential equations.
Corollary 4.3. The function u:R2(n−1) → R is a Cc∞ solution of the system of ultrahyperbolic PDEs
(6)
∂2
∂xk∂ye − ∂2
∂yk∂xe
u= 0(1≤e, k≤n−1),
where u=u(x, y), x= (x1, . . . , xn−1)∈ Rn−1 and y = (y1, . . . , yn−1)∈Rn−1, if and only if there exists a Cc∞ function f:Rn→Rsuch that
Rn1f(ξ, η)
=
n
X
i=1
qi qn
2!1/2 u
p1+q1 qn
, ...,pn−1+qn−1 qn
,p1−q1 qn
, ...,pn−1−qn−1 qn
,
where pi and qi (1 ≤ i ≤ n−1) are the Pl¨ucker coordinates of the straightline through ξ andη.
Proof. In (2) let us choose the following matrix U U
U −U
as Γ, whereU is the unit (n−1)×(n−1) matrix. One can conclude the proof as in the previous theorem.
The author would like to thank Z.I. Szab´o for proposing the problem of this paper and L. Feh´er for making valuable suggestions on the form and content.
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A. K´ URUSA, Bolyai Institute, Aradi v´ertan´uk tere 1. 6720 Szeged, Hungary; e-mail:
kurusa@math.u-szeged.hu
