CC BY
5УДК: 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: valeriyvolchkov@gmail.com, volna936@gmail.com
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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