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2009
Establishment of Weak Conditions for DarbouxGoursat-Beudon Theorem
Mekki Terbeche
University of Oran Es-senia
Broderick O. Oluyede
Georgia Southern University, [email protected]
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Terbeche, Mekki, Broderick O. Oluyede. 2009. "Establishment of Weak Conditions for Darboux-Goursat-Beudon Theorem."
International Journal of Mathematical Analysis, 3 (25): 1229-1237. source: http://www.m-hikari.com/ijma/ijma-password-2009/ijmapassword25-28-2009/oluyedeIJMA25-28-2009-2.pdf
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Int. Journal of Math. Analysis, Vol. 3, 2009, no. 25, 1229 - 1237
Establishment of Weak Conditions
for Darboux-Goursat-Beudon Theorem
Mekki Terbeche
Department of Mathematics
University of Oran Es-senia, Oran, Algeria
[email protected]
Broderick O. Oluyede
Department of Mathematical Sciences
Georgia Southern University, Statesboro, GA 30460, USA
[email protected]
Abstract
Abstract. This paper is devoted to the establishment of weak conditions for Darboux-Goursat-Beudon (DGB) theorem in order to improve
analogous results in [1, 7]. By adapting a technique proposed by [7] in
another setting [8] via majorant method, we obtain the generalization
of DGB theorem.
Mathematics Subject Classifications: 35A10, 58A99
Keywords: Cauchy-Kovalevskaya theorem, connected open neighborhood,
support of function
1
Introduction
It is well known that the classical Cauchy and Goursat theorems [1, 2, 5, 6, 7, 8]
play an important role in the theory of differential equations and their solutions. As a result, the study of Darboux-Goursat-Beudon problem is attaining
more prominence. In particular, during the last two decades many useful and
interesting contributions have been made in the investigation of existence and
uniqueness of the solution. The extended DGB problem has been initiated in
[1] and [7] and the theory of analytic function of several variables [3] and [4]
has been applied to ensure the existence and the uniqueness of the generalized
DGB problem. In this paper, we shall continue this study and investigate the
1230
M. Terbeche and B. O. Oluyede
existence and uniqueness of the solution of DGB problem with weak hypotheses in this framework.
2
Utility notions and basic results
We start by presenting some basic notations. Let Rn+1 be the (n + 1)dimensional Euclidean space, R+ the set of real numbers ≥ 0, Rn+1
be the
+
n+1
be the (n + 1)-dimensional
set of all r = (r0 , r1 , ..., rn ) with rj ∈ R+ and C
complex space with variables z = (z0 , z1 , ..., zn ) and Ω be an open subset of
C n+1 containing the origin. We use the standard multi-index notation. More
precisely, let Z be the set of integers, > 0 or ≤ 0, and Z+ be the set of integers
is the set of all α = (α0 , α1 , ..., αn ) with αj ∈ Z+ for each
≥ 0. Then Zn+1
+
j = 0, 1, ..., n. The length of α ∈ Zn+1
is |α| = α0 + α1 + ... + αn ; α ≤ β
+
means αj ≤ βj for every j = 0, 1, ..., n; and α < β means α ≤ β and α = β. If
and β ∈ Zn+1
α ∈ Zn+1
+
+ , we define the operation + by
α + β = (α0 + β0 , α1 + β1 , ..., αn + βn ) ∈ Zn+1
+ .
Moreover, we let α! = α0 !α1 !...αn ! and if α ≤ β
β1
βn
β
β0
β!
,
...
=
=
α
α0
α1
αn
α! (β − α)!
α0 α1 αn
∂
∂
∂
α
...
,
D =
∂z0
∂z1
∂zn
and use the notations Dj =
∂
∂zj
(1)
and
D α = D0α0 D1α1 ...Dnαn
(2)
respectively. Let u be a continuous function in Ω; by the support of u, denoted
by sup pu, we mean the closure in Ω of {z : z ∈ Ω, u (z) = 0}. Let C k (Ω),
k ∈ Z+ , 0 ≤ k ≤ ∞, denote the set of all functions u defined in Ω, whose
derivatives D α u(z) exist and continuous for |α| ≤ k. Using the multi-index
notation, we may write the Leibnitz formula
D β (uv) =
α≤β
β!
D β−α uD α v,
α!(β − α)!
(3)
where we assume u, v ∈ C |α| (Ω). If u ∈ C ∞ (Ω), we may consider the Taylor
expansion at the origin
u(z) =
D α u(0)
zα.
α!
n+1
α∈
+
(4)
1231
Weak conditions
Let H(Ω) denote the set of all holomorphic functions in Ω, that is, functions
u(z) ∈ C ∞ (Ω) given by their Taylor expansion in some neighborhood of the
origin in Ω. A linear partial differential operator P (z; D) is defined by
aα (z) D α ,
(5)
P (z; D) =
|α|≤m
where the coefficients aα (z) are in H(Ω). If for some α of length m the the
coefficient aα (z) does not vanish identically in Ω, m is called the order of
P (z; D). We set Iβ = {(j, k) : j = 0, 1, ..., n, and k = 0, 1, ..., βj − 1}.
3
Weak conditions and main results
Consider Darboux-Goursat-Beudon problem:
⎧
Pu =
aα D α u = 0
⎪
⎪
⎨
|α|≤m
(P ) D0k (u − ϕ)
|z0 =0 = 0, 0 ≤ k < m − 1
⎪
⎪
⎩ (u − ϕ)
|z =0 = 0.
0
In fact, (m, 0, ..., 0) = β, aβ (0) = 0, with ϕ(z) = z0N , N ∈ Z+ ,
am,0,...,0 (0, z ) = 0,
(6)
D0 am,0,...,0 (0, z ) = 0,
(7)
am−1,1,0,...,0 (0, z ) = 0.
(8)
and
It is easy to observe that these conditions are weaker than those given in [1, 7].
Moreover, the problem mentioned above admits a unique analytic solution
uN = u. We shall prove that (ϕ = 0),
D0k u(0, z ) = 0, ∀k = 0, 1, ..., N − 1.
(9)
We have three cases to consider: N < m, N = m and N > m.
First case N < m: By hypothesis u = um satisfies the equality
D0k (u − ϕ)|z0=0 = 0,
0 ≤ k < m − 1.
(10)
0 ≤ k ≤ N − 1,
(11)
As N < m, we have
D0k (u − ϕ)|z0 =0 = 0,
1232
M. Terbeche and B. O. Oluyede
and
D0k u(0, z ) = 0,
0 ≤ k ≤ N − 1.
(12)
Second case N = m: Now, u = um satisfies the equation P u = 0, that is,
for every z ∈ Ω
aα (z)D α u(z)
|α|≤m
=
am,0,...,0 (z)D0m u(z)
+
n
am−1,0,...,0,
i=1
+ am−1,0,...,0 (z)D0m−1 u(z) +
1
ith place
m−1
u(z)
,0,...,0 (z)Di D0
aα (z)D α u(z)
|α|≤m
α0 <m−1
= 0.
(13)
Hence for z = (0, z ), we have
n
am−1,0,...,0,
i=1
1
ith place
,0,...,0 (0, z
)Di D0m−1 u(0, z )
+ am−1,0,...,0 (0, z )D0m−1 u(0, z )
= 0.
Set
D0m−1 u(0, z ) = U(z ),
am−1,0,...,0,
1
(14)
,0,...,0 (0, z
ith place
) = Ai (z ),
1 ≤ i ≤ n and am−1,0,...,0 (0, z ) = A0 (z ). By the hypothesis, we have
u(z0 , 0, z”) = ϕ(z0 , 0, z”) = z0N ,
(15)
for 0 ≤ k < m − 1 = N − 1, hence m = N > 1, consequently,
u(0, 0, z”) = ϕ(0, 0, z”) = 0,
(16)
therefore U(0, z”) = 0.
Now, consider the following problem:
⎧ n
⎨ A (z ) D U(z ) + A (z )U(z ) = 0,
i
i
0
(P ) i=1
⎩ U
= 0,
|z1 =0
Note that A1 (0) = 0, and U(z ) = 0 is solution of problem (P ). It is unique
by Cauchy-Kovalevskaya Theorem, that is, D0m−1 u(0, z ) = 0. Hence
D0k u(0, z ) = 0,
(17)
1233
Weak conditions
for every k ∈ Z+ such that 0 ≤ k ≤ m − 1 = N − 1.
Third case N > m :
for 0 ≤ j ≤ k − 1 < N − 1 and m ≤ k.
(18)
D0m−1 u(0, z ) = 0,
aα (z)D α u(z) = 0, k − (m − 1) times with respect to z0 by
Differentiate
|α|≤m
using the Leibnitz formula, then ∀z ∈ Ω,
⎞
⎛
k−(m−1) ⎝
0 = D0
aα (z)D α u(z)⎠
|α|≤m
k−(m−1)
[am,0,...,0 (z)D0m u(z)
D0
n
=
am−1,0,...,0,
+
i=1
1
m−1
u(z)
,0,...,0 (z)Di D0
ith place
+ am−1,0,...,0 (z)D0m−1 u(z) +
k−(m−1) =
l=1
|α|≤m
α0 <m−1
k − (m − 1)
l
n k−(m−1)
+
i=1
l=1
k−(m−1)
+ D0
k−(m−1) +
l=0
aα (z) D α u(z)]
⎛
D0l am,0,...,0 (z)D0k−l+1 u(z)
k − (m − 1)
l
D0l am−1,0,...,0,
⎞
1
ith place
k−l
,0,...,0 (z)Di D0 u(z)
⎟
⎜ α
⎟
⎜
a
(z)
D
u(z)
α
⎠
⎝
|α|≤m
α0 <m−1
k − (m − 1)
l
D0l am−1,0,...,0 (z)D0k−l u(z).
(19)
We have
k−(m−1) l=0
k − (m − 1)
l
D0l am,0,...,0 (z)D0k−l+1 u(z)
k − (m − 1)
+
D0 am,0,...,0 (z)D0k u(z)
=
1
k−(m−1) k − (m − 1)
(20)
+
D0l am,0,...,0 (z)D0k−l+1 u(z).
l
am,0,...,0 (z)D0k+1 u(z)
l=2
For z = (0, z ), the expression (20) vanishes according to (6) − (7)) and (18).
1234
M. Terbeche and B. O. Oluyede
The expression
n k−(m−1)
k − (m − 1) D0l am−1,0,...,0, 1 ,0,...,0 (z)Di D0k−l u(z)
l
ith place
i=1
(21)
l=1
equals
n k − (m − 1)
am−1,0,..,0, 1 ,0,..,0 (z)Di D0k−l u(z)
l
ith place
(22)
i=1
for z = (0, z ) and for the same reason as before. The expression
⎞
⎛
k−(m−1) ⎜
⎜
D0
⎝
|α|≤m
α0 <m−1
=
k−(m−1) |α|≤m
α0 <m−1
l=1
⎟
aα (z) D α u(z)⎟
⎠
k − (m − 1)
l
D0l aα0 ,α (z0 , z )
× Dzα D0k−m+1−l+α0 u(z0 , z ),
(23)
for z = (0, z ). This expression vanishes because α0 < m − 1 and 0 ≤ l ≤
k − (m − 1), hence k − m + 1 − l + α0 ≤ k − 1 and the result follows from (18).
The expression
k−(m−1) l=0
k − (m − 1)
l
D0l am−1,0,...,0 (z)D0k−l u(z)
= am−1,0,...,0 (z)D0k u(z)
k−(m−1) k − (m − 1)
D0l am−1,0,...,0 (z)D0k−l u(z),
+
l
(24)
l=1
for z = (0, z ), this expression becomes am−1,0,...,0 (0, z )D0k u(0, z ) (due to (18)).
Finally, we have
n
i=1
am−1,0,...,0,
1
ith place
,0,...,0 (0, z
)Di D0k u(0, z ) + am−1,0,...,0 (0, z )D0k u(0, z ) = 0.
Set
am−1,0,...,0,
1
ith place
D0k u(0, z ) = V (z ),
,0,...,0 (0, z ) = Bi (z ),
(25)
(26)
1235
Weak conditions
for 1 ≤ i ≤ n and
am−1,0,...,0 (0, z ) = B0 (z ).
(27)
We have B1 (0) = am−1,1,0,...,0 (0, 0) = 0, and
V (0, z”) = D0k u(0, 0, z”)
= D0k ϕ(0, 0, z”)
= 0,
(28)
due to the fact that ϕ(z) = z0N and k < N, therefore the following problem
⎧ n
⎨ B (z )D V (z ) + B (z )V (z ) = 0
i
i
0
i=1
⎩ V
=0
|z1 =0
admits the unique solution V = 0 (by Cauchy-Kovalevskaya Theorem for an
operator of order 1), in other words
D0k u (0, z ) = 0.
(29)
D0j u(0, z ) = 0,
(30)
Hence
for 0 ≤ j ≤ k < N −1 and m ≤ k. Now use the iterative method on k. Suppose
D0j u (0, z ) = 0,
(31)
for 0 ≤ j ≤ k ≤ N − 1, and m ≤ k, one can show this property remains
true for the rank k = N − 1. We use the same process as before instead of
differentiation k − (m − 1) times with respect to z0 with m > 1,
⎞
⎛
aα (z) D α u(z)⎠ = 0
(32)
D0N −m ⎝
|α|≤m
N
−m N −m
D0l am,0,...,0 (z) D0N −l u (z)
=
l
l=0
n N −l N −l
+
D0l am−1,0,...,0, 1 ,0,...,0 (z) Di D0N −l−1 u (z)
l
ith place
i=1 l=0
⎛
⎞
⎜ ⎟
α
⎟
+D0N −l ⎜
a
(z)
D
u
(z)
α
⎝
⎠
|α|≤m
α0 <m−1
N −l N −l
D0l am−1,0,...,0 (z)D0N −l−1 u(z).
+
l
l=1
(33)
1236
M. Terbeche and B. O. Oluyede
The conditions (6) −(7), (30) and the previous computations allow us to write:
n
am−1,0,...,0,
i=1
1
ith place
,0,...,0 (0, z
)Di D0N −1 u(0, z )
+ am−1,0,...,0 (0, z )D0N −1 u(0, z )
= 0.
(34)
Set
D0N −1 u (0, z ) = W (z ),
am−1,0,...,0,
1
,0,...,0 (0, z
ith place
(35)
) = Ci (z ),
(36)
for 1 ≤ i ≤ n and
am−1,0,...,0 (0, z ) = C0 (z ).
(37)
C1 (0) = am−1,1,0,...,0 (0, z ) = 0,
(38)
W (0, z”) = D0N −1 u(0, 0, z”)
= D0N −1 ϕ(0, 0, z”)
= 0,
(39)
We have
since D0N −1 (z0N )|z0 =0 = 0. This leads to the solution of the following problem
⎧ n
⎨ C (z ) D W (z ) + C (z ) W (z ) = 0
i
i
0
(P ) i=1
⎩ W
|z1 =0 = 0.
Note that W (z ) = 0 is the solution of the problem P and by CauchyKovalevskaya Theorem, it is unique, hence D0N −1 u(0, z ) = 0, consequently
we have showed that
D0j u (0, z ) = 0,
(40)
for 0 ≤ j ≤ k ≤ N and m ≤ k. We have 0 ∈ sup p u because u(z0 , 0, z”) = z0N .
Now we state the main result which improves the results in [2] and [7].
Theorem 3.1 Let Ω be an open set of C n+1 containing the origin and
(aα )|α|≤m ∈ C ∞ (Ω) . If am,0,...,0 (0, z ) = 0, D0 am,0,...,0 (0, z ) = 0 and also
am−1,1,0,...,0 (0, z ) = 0, then ∀N > 0, N ∈ N, ∃Ω ⊂ Ω connected
open neigh
aα D α u = 0 and
borhood of the origin, and uN = u, u ∈ C ∞ (Ω) such that
D0k u(0, z )
= 0 for every k, 0 ≤ k ≤ N − 1, 0 ∈ sup p u.
|α|≤m
Weak conditions
1237
Remark: We can extend the coefficients to complex domain, solve the
problem by the method discussed in this paper and obtain the solution by
restriction.
ACKNOWLEDGEMENT: The authors are very grateful to the referees
for their suggestions, comments and hints. They also would like to thank the
University of Oran Es-senia and Georgia Southern University for supporting
these research topics under CNEPRU B01820060148 and B3101/01/05.
References
[1] Hormander, L., Linear partial differential equations, Springer Verlag,
1963.
[2] Mati, J.A., Nuiro, S.P., Vamasrin, V.S., A non linear Goursat problem
with irregular data, Integral Transforms Spec. Funct. 6 (1-4) (1998), 229246, MR 991.350250Zol. 091123043.
[3] Rudin, W., Real and complex analysis, McGraw-Hill, New York 1974.
[4] Shih, W.H., Sur les solutions analytiques de quelques équations aux
dérivées partielles en mécanique des fluides, Hermann éditeur, Paris 1992.
[5] Terbeche, M., Necessary and sufficient condition for existence and uniqueness of the solution of Cauchy fuchsian operators, JIPAM, Vol. 2, Issue 2,
Art. 24, 2001, pp. 1-14.
[6] Terbeche, M., Problème de Cauchy pour des opé rateurs holomorphes de
type de Fuchs. Theses. Université des Sciences et Technologies de Lille-I,
France (1980), N◦ 818.
[7] Terbeche, M., and Oluyede, B. O., On some differential equations, International Journal of Mathematical Analysis, Vol. 3, No. 11, 2009, pp
505-526.
[8] Wagschal, C., Une généralisation du probl ème de Goursat pour des
systèmes d’équations intégro-diff érentielles holomorphes ou partiellement
holomorphes, J. math. pures et appl., 53, 1974, pp. 99-132.
Received: November, 2008