Zero sets of solutions of the hyperbolic Darboux equation Текст научной статьи по специальности «Математика»

УДК: 517.444 MSC2010: 44A35, 44A12

ZERO SETS OF SOLUTIONS OF THE HYPERBOLIC DARBOUX

EQUATION

© V. V. Volchkov, Vit. V. Volchkov

Donetsk National University Faculty of Mathematics and Information Technologies Universitetskaya 24, Donetsk, 83001, Ukraine e-mail: [email protected], [email protected]

Abstract. Zero sets of solutions of the hyperbolic Darboux equation Volchkov V. V. and Volchkov Vit. V.

A hyperbolic analog of the generalized Darboux equation is considered. We investigate the structure of zero sets of its solutions for the case where the solution is a radial function of second variable. We show that every solution vanishing on some annulus must be zero in some other annulus containing the first one.

Keywords : Darboux equation, hyperbolic plane, zero sets, uniqueness theorems, transmutation maps

Introduction and statement of main result

Let D = {z E C : |z| < 1}. The partial differential equation

(1 -M2)2^ = (1 - |w|2)2 ^

dz dz dw dw

with f = f (z,w) E C2(D x D) is called the generalized Darboux equation. Equations of type (1) are of considerable interest in their own right, but they are also important for many applications in geometric analysis (see [1], [2]) and integral geometry (see [3]-[5]). In particular, such equations are closely connected with the mean value operators on symmetric spaces (see [1]-[3], [5]).

If f (z,w) is a function of z and |w|, and f (z,w) = h(z,p), p = arth |w|, then equation (1) can be rewritten as

4i(1 = dph + 2cth2pfh. (2)

This relation is called the hyperbolic Darboux equation. Some Euclidean analogs of equation (2) were studied in [1], [4, Part 5] and [6].

In this paper, we study zero sets of solutions of the hyperbolic Darboux equation. The main result of the paper is as follows.

Theorem 1. Let f E C2(D x D) satisfy (1) and assume that f (z,w) is a function of z and |w|. Suppose that R E (0,1) and r E [0, R) are given and the following conditions hold.

(i) f (z, 0) = 0 if r < |z| < R.

(ii) f (z, w) = 0 for all z,w E D, |z| = R.

Then f (z,w) = 0 for all z,w E D such that r < |z| < R and |w| < th (arth |z|—arth r).

Some Euclidean analogs of Theorem 1 can be found in [4, Part 5] and [6]. As regards other aspects of the theory of differential equations on symmetric spaces and their applications, see [2, Ch. 5].

1. Notation

In the paper, we use the following standard notation: R, N, Z, and Z+ denote the sets of real, natural, integer, and non-negative integers, respectively; r is the gamma-function; F(a, b; c; z) is the Gauss hypergeometric function.

We shall consider the disk D as a Riemannian manifold with the Riemannian structure

ds2 =_H2__(3)

= (1 — |z12)2 • (3)

The Laplace-Beltrami operator for (3) is given by

L = 4i(1 — |z|2)2 ^

dz dz

The hyperbolic distance d between the points zi, z2 E D is defined by

u \ ^ 11 — z1z21 + 1 z2 — z1|

d(zi, z2) = - in --=—.-.--•

2 |1 — z 1^2 | — |z2 — zi|

In particular,

1 1 + |z|

d(z, 0) = -in--^ and |z| = thd(z, 0), z E D.

2 1 | z |

The Riemannian measure d^ on D has the form

i dz A dz d^(z) = --

2(1 - |z|2)2'

For R > 0, BR(w) denotes the open non-Euclidean disk of radius R centered at w, i.e.,

BR(w) = {z G D : d(z,w) < R}.

We set BR = BR(0) and SR(w) = {z G D : d(z,w) = R}. Furthermore, let xR be the characteristic function (the indicator) of the disk BR.

We need the following classes of functions and distributions on D: L(D) and Lloc( are the classes of functions integrable and locally integrable on D with respect to the measure dp; D'(D) and E'(D) are the spaces of distributions and compactly supported distributions on D, respectively; D(D) is the space of compactly supported functions infinite differentiable in D.

Let T be a distribution with compact support in R. Its Fourier transform is defined by the relation

T(z) = (T,e-izt), z e C.

For a distribution f, f denotes its complex conjugation, supp f stands for the support of f. The symbol x denotes the convolution of distributions on D in the cases where it exists (see [1, Ch. 2, § 5]). For the convolution of distributions on R, we use the usual symbol "*".

Let T = {z e C : |z| = 1} and let p, p be the polar coordinates of the point z e D\{0}. Let Hk be the space of spherical harmonics of degree k on T, regarded as a subspace of L2(T) (see [4, Part 1, Ch. 5]). The dimension of Hk is denoted by the symbol ak. We have a0 = 1 and ak = 2 for k > 1. The reader can easily see that L2(T) is the direct sum of the spaces Hk from the L2 theory of Fourier series on the unit circle. Indeed, Hk(T) as a space of functions of the variable e%lf, —n < p < n, is the complex linear span of {etkv, e-ifc^}. From this point of view a Fourier series expansion on T in the same as an expansion into spherical harmonics.

We set

1 n-k

To every function f G L (BR> we assign its Fourier series

тс ak

f(z) - ^^ fkj(p>Y(k)(n>, 0 < p < thR, n = z/e,

fc=0 j=l

where

fk,j(P>= I f (pn> YfV) |dn|

We set

fkj (z) = fkj (P) Y(k\a). Let O(2) be the orthogonal group of R2 with the normalized Haar measure dr. If т is a rotation through angle 9 in R2 then we set t'k i(t) = e-%ke and tkk2(T) = e%ke. In addition,

let t° 1(t) = 1 for all t E O(2). Then one has

f k,j (z) = ak J f (t-1z) (t) dT (4)

O(2)

(cm. [7, Part 2, Ch. 9, formula (9.5)]).

Next, for each f E D'(BR) we define the distribution fk'j E D'(BR) by the formula

<ffcj ,g> = / f, ak J g(T-1z) (t) dA g E D (Br).

* O(2) '

For a set M(Br) C D' (Br) let

Mkj(Br) = {f E M(Br) : f = fkj}, M^(Br) = Mo,1(Br).

2. The functions $A,k)j For the rest of the paper, A E C and

V = v (A) = 1 (iA + 1). For k E Z+, j E {1,..., ak} and z E D\{0} we put

$A;k,j(z) = $A,k(p)lfV), p = |z|, a = z/p,

where

$A,k(p) = r(r)(r+k+1) pk (1 — p2)V F(v + k, v; k + 1; p2). (5)

The equality

$A'kj(z) = 2tt y ' Y(k)(n) ldnl (6)

|z — n|2

T

holds for all A E C and z E D (see [4, Part 2, Ch. 2, formula (2.9)]). Since

1 — |z|2 1 + |z|

I-L12 -rr, z E D, n E T,

|z — n|2 1 — |z|

it follows from (6) that

max

' d \ ai ( d\ a2 .dz dz $A'k'j(z)

O ((1 + |A|)ai+a2 er|ImA|) (7)

for r E (0,1) and a1, a2 E Z+, where the constant in O does not depend on A.

Lemma 1. Let A,^ G C, R G (0,1). Then

R

(A2 - / (1 Фл,к(t) ФМ(t) dt

0

R

(1 - R2)

Proof. We have

(R^'fc(R) - Фм(R)$A,fc(R)

where

Фл,к (t) = g;*? (1 - t2)-k (arth t),

1 + ¿A , ,1 — ¿A , , -, 2

(0 = F ( k + —, k + —k + 1; -sh2 £ ) (10)

(see (5) and [8, Ch. 2.9]). Using now [7, formulae (7.18) and (7.46)] we see that

t

(A2 - ^2WAa(() (0 (0 d£

= Afc(t) Ufc(t)^(t) - (t),

• о m\ 2fc+1

sin 2it

0

= A,(t)

where

Ak(i) = ^

This together with (9) implies (8). □

Equality (9) implies that for all k G Z+, R G (0,1) the function Флк(R) is an even entire function of A. Using [7, Proposition 7.4] we see from Hadamard's theorem [9, Ch. 1, Theorem 13] that Фл,к(R) has infinitely many zeros. Since the function Фл,к(R) is even, the set of these zeros is symmetric with respect to A = 0. It follows from (9), (10), and the expansion of F in a hypergeometric series (see [8, Ch. 2, § 2.1, formula (1)]) that Фл,к(R) > 0 for iA G R.

Lemma 2. All the zeros of Фл,к(R) are real and simple.

Proof. Let Флк(R) = 0 for some A G C. We claim that A G R and | Ф4 k(R) = 0.

4=Л

_2 _

Assume that A G R; then A2 = A , since iA G R. Putting ^ = A in (8) and taking into

account that Фл k(R) = 0, we infer that

R

/ (1 -tt2)2 |Фл,k(t)|2dt = 0, (11)

0

which is impossible. Now assume that dt k(R)l = 0. Letting p ^ A in (8) we

t=A

obtain (11) once again. Hence, all the zeros of $A;k(R) are real and simple. □

Let Nk (R) be the set of positive zeros A of the function $A;k(R). Lemma 2 shows that Nk(R) has the form Nk(R) = {Ai,A2,...}, where Am = Am(R,k) is the sequence of all positive zeros of $A;k(R) numbered in the ascending order. Owing to [9, Ch. 1, Theorem 6], we have

E Am'-

m < ^

m=1

for any £ > 0.

Lemma 3. Let A E Nk (R) and

R

I (A ) = / (r-tt2)2 |$A,k (t)|2 dt. 0

Then I (A) > CA-4, where C > 0 does not depend on A. Proof. Because of (10), for t > 0 one has

t

r(k + lHsh2t)2k f

i)

2)

.A,M-^ + )2k-3 J (ch2t - ch2^)k-

■ F f 2k, 0; k + 1 cos A£ d£

V 2 2 ch t J S S

(see [10, equality (2.21)]). Using now (9) and repeating the arguments in [4, Part 2, the proof of Lemma 2.7] we arrive at the desired statement. □

3. The spherical transform

For any m E Z we consider the differential operator dm defined on C'(0,1) as follows:

tm d f /1 \m \

(M® =(1-^4(1-1) f(t))• f E C1(0'1).

Let Lk = L - 4(k - 1)kI, where I is the identity operator. A simple calculation shows that

(Lk f)(z) = (dk-id-ku)(p)Y(k)(a) (12)

if f E C2(Br) has the form f (z) = u(p)Yj(kV).

Using (5) and [8, formulae 2.8(25), 2.8(26))] we easily obtain

(dk$A,k)(p) = (iA - 2k - 1)$A,k+i(p), (13)

(d-k-i$A,k+i) (p) = (iA + 2k + 1)$A,k (p). (14)

In what follows we assume that all functions that are defined and continuous in a punctured neighbourhood of zero in C and admit continuous extension to 0 are defined at 0 by continuity. The functions $A,k,j admit continuous extension to the point z = 0, becoming real-analytic functions on D. Formulae (13), (14) and (12) imply that

(L + (A2 + 1)l)($A,kJ) =0. (15)

Formula (15) with k = 0 implies that $A,0(|z|) coincides with the elementary spherical function pA on D (see [8, Ch. 4, §4.2]). The spherical transform /(A) of a distribution f E E' (D) is defined by

7(A) = <f,PA>. (16)

By (16) and (15) we conclude that

Lmf(A) = (-1)m(A2 + 1)m/(A), m E Z+. This together with (7) shows that for f E « n C2m)(D)

f(A) = O(|A|-2m), A ^ro, A e R, (17)

where the constant of the symbol O is independent of A.

Lemma 4. Let T E E'(D), f E C2(D) and Lf = -(A2 + 1)f. Then

(f x T)(z) = T(A)f (z), z E D. (18)

In particular,

($A,fcj X Xr)(z) = n sh 2r $A,1Vthr) $A,fcj (z) for any r > 0.

'19)

Proof. The first equality follows from the mean value theorem for the eigenfunctions of the operator L (see [8, Ch. 4, § 2.2]). Next one has

thr

Xr(A) = J $A,o(|z|)dMz) = 2K J ^^pprdp.

Br 0

Combining this with (14), we obtain

Xr (A) = n sh2r—!-.

Thus the second equality in the lemma follows from the first with T = xr. □

Lemma 5. Let t > 0 and k G Z+. Then there exists Ut)k G E'(R) such that supp Ut)k C [—t, t] and

Uk(A) = $A;fc(tht), A G C. (20)

l(v + k)

Proof. As already pointed out in § 3, for each t > 0 the function p^k) (tht) is an even entire function of A. Moreover, it follows from (7) that

(tht)| < cet|ImA|, (21)

where the constant c > 0 does not depend on A. Now the Paley-Wiener theorem (see [11, Theorem 7.3.1]) completes the proof. □

We now define an operator allowing the reduction of several problems for convolution in D to the one-dimensional case. As usual, we set p = |z|, a = z/p. For f G Lloc(D), Z G D \ {0}, z G D \ {0} we set

1 — P2

Ac(f)(z) = (f X Xarth|z|)(Z)Y(1) (a). (22)

Lemma 6. Let a > 0. There exists a linear homeomorphism : D'kj(Ba) ^ D'(—a, a) such that the following assertions hold.

(i) For each A G C,

r(v + k)

Afcj($Aifcj-)(t) = r(;y)(r+k+)l^ cos At. (23)

(ii) If f G L)°j(Ba), t G (0, a), and Z G St, then

2 Ai,i(Ac(f)) =r(k + l) (Afcj(f) * Utifc) (24)

in D'(t — a, a — t).

Proof. According to [7, Theorem 10.21], there exists a linear homeomorphism Afcj : Dj(B) ^ D'(—a, a) satisfying (23). Let us prove (24). First of all, we note that the set Lin{$A,fcj, A G C} is a dense subset of D'kj(Ba) (see [7, Proposition 9.9]). Let a > 0, n G D \ {0}. Consider a sequence fn G Lin {$A,fcj, A G C} such that fn ^ f

in the space L(Ba(n)). Let ^ G D(D), supp ^ C B and let

l p2

^(z) = ^^(z)Yi(1) (a).

Using (22) we obtain

|<A(/ra),p>-<An(f),p>j < < sup I |fn - f |dp / |^|dp <

z€Ba J J

Bd(0,z)(n) B

< f |fn - f |dp/ |^|dp.

Ba (n) Ba

Since fn ^ f in L(Ba(n)) this implies that the sequence {An(fn)} converges to An(f) in D'(D). Therefore, to prove Lemma 6 we can assume without loss of generality that f = $A,k,j, A E C. Next one has

2 Ai,i(Ac($A,k,j))(£) = $A,k,j (Z) cos A£ (25)

(see (19)). On the other hand,

r(k + 1) (A kj ($A,k,j ) * Ut,k) (£) = V^ $A,k(Z) cos A£ (26)

because of (23) and (20). Comparing (25) with (26) we arrive at (24). □

4. Proof of Theorem 1

We now proceed to the proof of Theorem 1. Let f G C2(D X D) and suppose that this function satisfies conditions (i)-(ii) in Theorem 1. For each t G (0,1), Asgeirsson's mean value theorem (see [1, Ch. 2, § 5.6, Theorem 5.28]) yields

J f (Z, 0) d^(Z) = / f (z,Z) d^(Z), z G D,

St (z) St (0)

|dZ I

where d^(Z) = 1 _ |Z|2.

This equality and condition (i) in Theorem 1 show that

f(z-w) = m / f0) dw(Z> (27)

St(z)

for all z G D, w G St(0). Let R = arth R, a = 2R' - arth r. Now define u(z) = f (z, 0) for r < |z| < a and u(z) = 0 for |z| < r. Relation (4) and property (ii) imply that

ufcj (x) = 0 in B

R

for all k G Z+, j G {1,..., ak}. In addition, by property (ii) and (27),

(ufcj X Xi)(C) = 0, Z G Sr(0), for each t G (0, R' — arth r). Hence

Akj(«kJ) * Un^k = 0 in (R' — a, a — R') because of Lemma 6. Using now Lemma 3 and [4, Part 3, Theorem 1.3] we see that

Akj(uk,j)(t)= £ CA.k.j cos At, (29)

AeNfc (n)

where cA;kj G C,

CA,k,j = O(AY), A ^^ (30)

for some 7 > 0, and the series in (29) converges in the space D'(—a, a). Owing to (23), this means that

uk,j (z)= £ CA,k,j $A,k,j (z), (31)

AeNfc (R)

where the series converges in D'(Ba). Let G D^(D) and suppC B£, e G (0,a). In view of Lemma 4, we conclude from (31) that

(wkJ X pe)(z) = CA,k,j ^£(A)$A)k,i(z), z G Ba_£. (32)

AeNfc (R)

Together with (30), relation (17) yields

' 1

cA,k,j ¥?£(A) = O (A&) , A

—> OO

for each b > 0. Taking (21) into account we see that the series in (32) converges uniformly on compacts. Therefore, we obtain

i \

-1

CA,k,j ^e(A) =

(wk'j x ^e)(z)$A,k,j(z) d^(z)

I |$A)fc,j (Z )|2 d^(z)

/ Br,

(see Lemma 1). Letting e ^ 0, for a suitable sequence one has

/ \-1

cA,k,7 =

|$A,k,j (z)|2 d^(z)

uk,j(z) $A,kj(z) d^(z).

\Bfl' / Br,

Hence cA;kj = 0 for all A,k, j because of (28). Now we know that the functions ukj and u vanish in In view of (27) this gives us the assertion of Theorem 1.

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