Georgia Southern University Digita Commons@Georgia Southern Mathematica Sciences Facuty Pubications Department of Mathematica Sciences 2009 Estabishment of Weak Conditions for Darboux- Goursat-Beudon Theorem Mekki Terbeche University of Oran Es-senia Broderick O. Ouyede Georgia Southern University, bouyede@georgiasouthern.edu Foow this and additiona works at: https://digitacommons.georgiasouthern.edu/math-sci-facpubs Part of the Mathematics Commons Recommended Citation Terbeche, Mekki, Broderick O. Ouyede. 2009. "Estabishment of Weak Conditions for Darboux-Goursat-Beudon Theorem." Internationa Journa of Mathematica Anaysis, 3 25: 1229-1237. source: http://www.m-hikari.com/ijma/ijma-password-2009/ijmapassword25-28-2009/ouyedeijma25-28-2009-2.pdf https://digitacommons.georgiasouthern.edu/math-sci-facpubs/266 This artice is brought to you for free and open access by the Department of Mathematica Sciences at Digita Commons@Georgia Southern. It has been accepted for incusion in Mathematica Sciences Facuty Pubications by an authorized administrator of Digita Commons@Georgia Southern. For more information, pease contact digitacommons@georgiasouthern.edu.
Int. Journa of Math. Anaysis, Vo. 3, 2009, no. 25, 1229-1237 Estabishment of Weak Conditions for Darboux-Goursat-Beudon Theorem Mekki Terbeche Department of Mathematics University of Oran Es-senia, Oran, Ageria terbeche2000@yahoo.fr Broderick O. Ouyede Department of Mathematica Sciences Georgia Southern University, Statesboro, GA 30460, USA Bouyede@GeorgiaSouthern.edu Abstract Abstract. This paper is devoted to the estabishment of weak conditions for Darboux-Goursat-Beudon DGB theorem in order to improve anaogous resuts in [1, 7]. By adapting a technique proposed by [7] in another setting [8] via majorant method, we obtain the generaization of DGB theorem. Mathematics Subject Cassifications: 35A10, 58A99 Keywords: Cauchy-Kovaevskaya theorem, connected open neighborhood, support of function 1 Introduction It is we known that the cassica Cauchy and Goursat theorems [1, 2, 5, 6, 7, 8] pay an important roe in the theory of differentia equations and their soutions. As a resut, the study of Darboux-Goursat-Beudon probem is attaining more prominence. In particuar, during the ast two decades many usefu and interesting contributions have been made in the investigation of existence and uniqueness of the soution. The extended DGB probem has been initiated in [1] and [7] and the theory of anaytic function of severa variabes [3] and [4] has been appied to ensure the existence and the uniqueness of the generaized DGB probem. In this paper, we sha continue this study and investigate the
1230 M. Terbeche and B. O. Ouyede existence and uniqueness of the soution of DGB probem with weak hypotheses in this framework. 2 Utiity notions and basic resuts We start by presenting some basic notations. Let R n1 be the n 1- dimensiona Eucidean space, R the set of rea numbers 0, R n1 be the set of a r =r 0,r 1,..., r n with r j R and C n1 be the n 1-dimensiona compex space with variabes z =z 0,z 1,..., z n and Ω be an open subset of C n1 containing the origin. We use the standard muti-index notation. More precisey, et Z be the set of integers, > 0or 0, and Z be the set of integers 0. Then Z n1 is the set of a α =α 0,α 1,..., α n with α j Z for each j =0, 1,..., n. The ength of α Z n1 is α = α 0 α 1... α n ; α β means α j β j for every j =0, 1,..., n; and α<βmeans α β and α β. If α Z n1 and β Z n1, we define the operation by α β =α 0 β 0,α 1 β 1,..., α n β n Z n1. Moreover, we et α! =α 0!α 1!...α n! and if α β β β0 β1 βn β! =... = α α 0 α 1 α n α!β α!, 1 α0 α1 αn D α =..., z 0 z 1 z n and use the notations D j = z j and D α = D α 0 0 Dα 1 1...Dαn n 2 respectivey. Let u be a continuous function in Ω; by the support of u, denoted by sup pu, we mean the cosure in Ω of {z : z Ω,uz 0}. Let C k Ω, k Z,0 k, denote the set of a functions u defined in Ω, whose derivatives D α uz exist and continuous for α k. Using the muti-index notation, we may write the Leibnitz formua D β uv = α β β! α!β α! Dβ α ud α v, 3 where we assume u, v C α Ω. If u C Ω, we may consider the Tayor expansion at the origin uz = α Z n1 D α u0 z α. 4 α!
Weak conditions 1231 Let HΩ denote the set of a hoomorphic functions in Ω, that is, functions uz C Ω given by their Tayor expansion in some neighborhood of the origin in Ω. A inear partia differentia operator P z; D is defined by P z; D = a α z D α, 5 where the coefficients a α z are in HΩ. If for some α of ength m the the coefficient a α z does not vanish identicay in Ω, m is caed the order of P z; D. We set I β = {j, k :j =0, 1,..., n, and k =0, 1,..., β j 1}. 3 Weak conditions and main resuts Consider Darboux-Goursat-Beudon probem: Pu= a α D α u =0 P D0 k u ϕ z0 =0, 0 k<m 1 =0 u ϕ =0. z0 =0 In fact, m, 0,..., 0 = β,a β 0 = 0, with ϕz =z N 0, N Z, a m,0,...,0 0,z =0, 6 and D 0 a m,0,...,0 0,z =0, 7 a m 1,1,0,...,0 0,z 0. 8 It is easy to observe that these conditions are weaker than those given in [1, 7]. Moreover, the probem mentioned above admits a unique anaytic soution u N = u. We sha prove that ϕ 0, D k 0 u0,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 = u m satisfies the equaity D k 0 u ϕ z0 =0 =0, 0 k<m 1. 10 As N<m, we have D0 k u ϕ z0 =0, 0 k N 1, 11 =0
1232 M. Terbeche and B. O. Ouyede and D k 0u0,z =0, 0 k N 1. 12 Second case N = m: Now, u = u m satisfies the equation Pu=0, that is, for every z Ω a α zd α uz n = a m,0,...,0 zd0 m uz a m 1,0,...,0, 1,0,...,0zD i D0 m 1 uz i th pace a m 1,0,...,0 zd0 m 1 uz a α zd α uz = 0. Hence for z =0,z, we have 13 n a m 1,0,...,0, 1,0,...,00,z D i D0 m 1 u0,z i th pace a m 1,0,...,0 0,z D0 m 1 u0,z = 0. 14 Set D0 m 1 u0,z =Uz, a m 1,0,...,0, 1,0,...,00,z =A i z, i th pace 1 i n and a m 1,0,...,0 0,z =A 0 z. By the hypothesis, we have uz 0, 0,z = ϕz 0, 0,z = z N 0, 15 for 0 k<m 1=N 1, hence m = N>1, consequenty, u0, 0,z = ϕ0, 0,z = 0, 16 therefore U0,z = 0. Now, consider the foowing probem: n P A i z D i Uz A 0 z Uz =0, U =0, z1 =0 Note that A 1 0 0, and Uz = 0 is soution of probem P. It is unique by Cauchy-Kovaevskaya Theorem, that is, D m 1 0 u0,z = 0. Hence D k 0u0,z =0, 17
Weak conditions 1233 for every k Z such that 0 k m 1=N 1. Third case N>m: D m 1 0 u0,z =0, for 0 j k 1 <N 1 and m k. 18 Differentiate a α zd α uz =0,k m 1 times with respect to z 0 by using the Leibnitz formua, then z Ω, 0 = D k m 1 0 a α zd α uz = D k m 1 0 [a m,0,...,0 zd0 m uz n a m 1,0,...,0, 1,0,...,0zD i D0 m 1 uz i th pace a m 1,0,...,0 zd m 1 0 uz = k m 1 =1 n k m 1 k m 1 =1 D k m 1 0 We have k m 1 =0 k m 1 =0 k m 1 k m 1 a α z D α uz] D 0a m,0,...,0 zd k 1 0 uz D0 a m 1,0,...,0, 1 i th pace,0,...,0zd i D0 k uz a α z D α uz D 0a m 1,0,...,0 zd k 0 uz. 19 k m 1 k m 1 1 k m 1 = a m,0,...,0 zd k1 0 uz k m 1 =2 D 0a m,0,...,0 zd k 1 0 uz D 0 a m,0,...,0 zd k 0uz D 0 a m,0,...,0zd k 1 0 uz. 20 For z =0,z, the expression 20 vanishes according to 6 7 and 18.
1234 M. Terbeche and B. O. Ouyede The expression n equas k m 1 =1 k m 1 n k m 1 D0 a m 1,0,...,0, 1,0,...,0zD i D0 k uz 21 i th pace a m 1,0,..,0, 1,0,..,0zD i D0 k uz 22 i th pace for z =0,z and for the same reason as before. The expression = D k m 1 0 k m 1 =1 a α z D α uz k m 1 D 0a α0,α z 0,z D α z Dk m1 α 0 0 uz 0,z, 23 for z =0,z. This expression vanishes because α 0 <m 1 and 0 k m 1, hence k m 1 α 0 k 1 and the resut foows from 18. The expression k m 1 =0 k m 1 = a m 1,0,...,0 zd0 k uz k m 1 k m 1 =1 D 0 a m 1,0,...,0zD k 0 uz D 0 a m 1,0,...,0zD k 0 uz, 24 for z =0,z, this expression becomes a m 1,0,...,0 0,z D k 0 u0,z due to 18. Finay, we have Set n a m 1,0,...,0, 1,0,...,00,z D i D0 k u0,z a m 1,0,...,0 0,z D0 k u0,z =0. i th pace D k 0 u0,z = V z, 25 a m 1,0,...,0, 1,0,...,00,z = B i z, 26 i th pace
Weak conditions 1235 for 1 i n and We have B 1 0 = a m 1,1,0,...,0 0, 0 0, and a m 1,0,...,0 0,z =B 0 z. 27 V 0,z = D0 k u0, 0,z = D0 k ϕ0, 0,z = 0, 28 due to the fact that ϕz =z0 N and k<n,therefore the foowing probem n B i z D i V z B 0 z V z =0 V z1 =0 =0 admits the unique soution V = 0 by Cauchy-Kovaevskaya Theorem for an operator of order 1, in other words Hence D k 0u 0,z =0. 29 D j 0u0,z =0, 30 for 0 j k<n 1 and m k. Now use the iterative method on k. Suppose D j 0u 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 z 0 with m>1, D N m 0 a α z D α uz = 0 32 = N m =0 N m n N N =0 D N 0 N N =1 D 0 a m,0,...,0 z D N 0 u z D0 a m 1,0,...,0, 1,0,...,0 z D i D0 N 1 u z i th pace a α z D α u z D 0 a m 1,0,...,0zD N 1 0 uz. 33
1236 M. Terbeche and B. O. Ouyede The conditions 6 7, 30 and the previous computations aow us to write: n a m 1,0,...,0, 1,0,...,00,z D i D0 N 1 u0,z i th pace a m 1,0,...,0 0,z D0 N 1 u0,z = 0. 34 Set D N 1 0 u 0,z =W z, 35 for 1 i n and a m 1,0,...,0, 1,0,...,00,z =C i z, 36 i th pace a m 1,0,...,0 0,z =C 0 z. 37 We have C 1 0 = a m 1,1,0,...,0 0,z 0, 38 W 0,z = D0 N 1 u0, 0,z = D0 N 1 ϕ0, 0,z = 0, 39 since D0 N 1 z0 N z0 =0 =0. This eads to the soution of the foowing probem n P C i z D i W z C 0 z W z =0 W z1 =0 =0. Note that W z = 0 is the soution of the probem P and by Cauchy- Kovaevskaya Theorem, it is unique, hence D0 N 1 u0,z = 0, consequenty we have showed that D j 0 u 0,z =0, 40 for 0 j k N and m k. We have 0 sup pubecause uz 0, 0,z = z N 0. Now we state the main resut which improves the resuts in [2] and [7]. Theorem 3.1 Let Ω be an open set of C n1 containing the origin and a α C Ω. If a m,0,...,0 0,z = 0, D 0 a m,0,...,0 0,z = 0 and aso a m 1,1,0,...,0 0,z 0, then N >0, N N, Ω Ω connected open neighborhood of the origin, and u N = u, u C Ω such that a α D α u =0and D k 0u0,z =0for every k, 0 k N 1, 0 sup pu.
Weak conditions 1237 Remark: We can extend the coefficients to compex domain, sove the probem by the method discussed in this paper and obtain the soution by restriction. ACKNOWLEDGEMENT: The authors are very gratefu to the referees for their suggestions, comments and hints. They aso woud ike 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 partia differentia equations, Springer Verag, 1963. [2] Mati, J.A., Nuiro, S.P., Vamasrin, V.S., A non inear Goursat probem with irreguar data, Integra Transforms Spec. Funct. 6 1-4 1998, 229-246, MR 991.350250Zo. 091123043. [3] Rudin, W., Rea and compex anaysis, McGraw-Hi, New York 1974. [4] Shih, W.H., Sur es soutions anaytiques de queques équations aux dérivées partiees en mécanique des fuides, Hermann éditeur, Paris 1992. [5] Terbeche, M., Necessary and sufficient condition for existence and uniqueness of the soution of Cauchy fuchsian operators, JIPAM, Vo. 2, Issue 2, Art. 24, 2001, pp. 1-14. [6] Terbeche, M., Probème de Cauchy pour des opé rateurs hoomorphes de type de Fuchs. Theses. Université des Sciences et Technoogies de Lie-I, France 1980, N 818. [7] Terbeche, M., and Ouyede, B. O., On some differentia equations, Internationa Journa of Mathematica Anaysis, Vo. 3, No. 11, 2009, pp 505-526. [8] Wagscha, C., Une généraisation du prob ème de Goursat pour des systèmes d équations intégro-diff érentiees hoomorphes ou partieement hoomorphes, J. math. pures et app., 53, 1974, pp. 99-132. Received: November, 2008