On Complex Singularity Analysis of Holomorphic Solutions of Linear Partial Differential Equations

Transcription

1 Advances in Dynamical Systems and Applications ISSN , Volume 6, Number, pp ) On Complex Singularity Analysis of Holomorphic Solutions of Linear Partial Differential Equations S. Malek and C. Stenger University of Lille 1, Laboratoire Paul Painlevé Villeneuve d Ascq cedex, France University of La Rochelle, Avenue Michel Crépeau 1704 La Rochelle cedex France Stephane.Malek@math.univ-lille1.fr catherine.stenger@univ-lr.fr Abstract We construct formal series solutions of linear partial differential equations as linear combinations of powers of solutions of first order nonlinear differential equations, following the classical tanh method. We give sufficient conditions under which the constructed formal series define holomorphic functions on some punctured polydiscs of C. Moreover, we study the rate of growth of these solutions near their singular points. AMS Subject Classifications: 35C10, 35C0. Keywords: Linear Cauchy problems, Banach spaces of entire functions, formal power series, ordinary differential equations. 1 Introduction In the literature, the tanh method, which has been initiated by W. Malfliet in [11], is an effective algebraic method for finding explicit exact solutions of certain nonlinear partial differential equations with constant coefficients of the form Hut, x), t ut, x), x ut, x), xut, x),...) = 0 where H is a polynomial. The method consists in looking for solutions of the form ut, x) = j J u j ϕκx wt))) j Received March 16, 011; Accepted March 4, 011 Communicated by Martin Bohner

2 10 S. Malek and C. Stenger where J is a finite subset of Z, u j, κ, w are constants and ϕ is a solution of a Ricatti equation ϕ ξ) = a + bϕξ) + cϕξ)), where a, b, c are well chosen real numbers. Notice that this approach has been generalized in several directions and encounters a growing success these recent years, see for instance [3], [4], [13]. In this paper, we will consider special solutions of linear partial differential equations of the form S z Xt, z) = k=k 0,k 1 ) J a k t, z) k 0 t k 1 z Xt, z) with holomorphic coefficients a k t, z) on some domain D C with respect to t and near the origin in C with respect to z. Inspired by the tanh method, we are looking for solutions in the form of infinite sums Xt, z) = j 0 X j t, z)φt)) j 1.1) where X j t, z) are holomorphic functions on D with respect to t and near the origin in z and φt) is a solution of a general algebraic first order differential equation φ t) = P t, φt)) 1.) where P t, X) is a polynomial in X of degree larger than, with holomorphic coefficients on D. These kind of series are also called transasymptotic expansions in the terminology of J. Écalle, see [8]. In the framework of differential equations, similar series have been used to study the formation of complex singularities and local behaviour of the holomorphic solutions near the singular points along Stokes directions for systems of non linear differential equations with irregular singularity at 0 of the form t y t) = F t, yt)), where F is a holomorphic function from R R n into R n, for n 1. The constructed holomorphic solutions are of the form yt) = l 0 x l t)ϕt)) l, where x l t) are the k sums of formal series ˆx l t) C n [[t]] in some direction d, see [6], and where ϕt) is a solution of the differential equation t ϕ t) = ϕt). More recently, a similar study has been investigated for q-difference-differential equations by the first author, see [10]. From the result of P. Painlevé, see [9], we know that the only movable singularities in C of the solutions φt) of 1.) are poles and/or algebraic branch points. In this work, we will show that the special solutions 1.1) define holomorphic functions with respect to t near a movable singularity t 0 D of φ and near the origin with respect to

3 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 11 z. Moreover, we will analyse the local behaviour of the constructed solutions as t tends to t 0 in D. In Section, we consider a Cauchy problem in C 3 with holomorphic solutions V t, z, T ), which extends both the classical Cauchy Kowalevskii problem in holomorphic functions spaces near the origin in C with respect to t, z and a global version in entire functions spaces with exponential growth with respect to T given by J. Dubinskii, see [7]. The difficulty to combine these two results comes from the fact that the growth rate of the solution V t, z, T ) in the local variable t and the global variable T both depend on the variable z near the origin as observed in [1], [7]. In Section 3, we first construct the special solutions 1.1) in a formal sense. Then, in Section 4, using a majorizing series method, we reduce the problem of the description of their convergence domain to the construction of a solution of an auxiliary Cauchy problem in C 3 to which the results of Section can be applied. Finally, in the theorem 4., we show that the solutions 1.1) have at most an exponential growth rate as t approaches t 0 in D. A Cauchy Problem in a Class of Entire Functions with Exponential Growth and Finite Order.1 Weighted Banach Spaces of Holomorphic Functions in C 3 Definition.1. Let m, σ, δ 1, δ, q and b be positive real numbers such that q 1 and b > 1. We define a vector space G q δ 1, δ ; σ) which is a subspace of the vector space AC){t, z} of entire functions V t, z, T ) with respect to T which are holomorphic on a neighbourhood of the origin in C with respect to t, z). A function belongs to G q δ 1, δ ; σ), if the series converges, where V t, z, T ) = n,β 0 n,β 0 v n,β T ) tn z β n! β! v n,β T ) β;σ δ n 1 δ β n + β)! v n,β T ) β;σ = sup v n,β T ) 1 + T ) m exp σr b β) T q ) T C with r b β) = β n=0 1 n + 1) b.

4 1 S. Malek and C. Stenger We also define a weighted L 1 norm on G q δ 1, δ ; σ) as One can easily show that V t, z, T ) δ1,δ ;σ = n,β 0 v n,β T ) β;σ δ n 1 δ β n + β)!. ) G q δ 1, δ ; σ),. δ1,δ ;σ is a Banach space. Remark.. We have the following continuous inclusions G q δ 1, δ ; σ ) G q δ 1, δ ; σ) if δ 1 δ 1, δ δ, σ σ and r b β) is such that Riemann zeta function. lim r bβ) = ζb) where ζ is the β + In the next proposition, we study the rate of growth of the entire functions belonging to the constructed Banach spaces. Proposition.3. Let V t, z, T ) G q δ 1, δ ; σ). Then, there exists a positive constant C 1 such that T C, t C with t < δ 1 and z C with z < δ, we have V t, z, T ) C 1 1 t ) 1 1 z ) T ) m exp σζb) T q ). δ 1 δ Proof. We expand V t, z, T ) in powers of t and z V t, z, T ) = β 0 n 0 v n,β T ) tn z β n! β!. As V t, z, T ) G q δ 1, δ ; σ), there exists a positive constant c 1 such that v n,β T ) β;σ δ n 1 δ β n + β)! c 1 for all n, β 0. So that for all T C, we have v n,β T ) n!β! c T ) m exp σr b β) T q 1 ) δ1 n δ β n + β)! n!β!.1) for all n, β 0. Lemma.4. We have n + β)!/n!β!) n+β for all n, β 0.

5 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 13 The lemma is a consequence of the classical binomial formula. From the estimates.1) we deduce that v n,β T ) n!β! ) n ) β c T ) m exp σr b β) T q ).) δ 1 δ for all T C, and for all n, β 0. From.), we get that T C, t C with t < δ 1 and z C with z < δ V t, z, T ) β 0 c T ) m 1 1 t δ 1 n 0 v n,β T ) t n z β n!β! ) β z exp σr b β) T q ). β 0 Now, as r b β) ζb) for all β N, we get that there exists a positive constant c 1, such that T C, t C with t < δ 1 and z C with z < δ V t, z, T ) c T ) m 1 exp σζb) T q ) 1 t δ 1 This proves the proposition. δ we have 1. 1 z δ. Linear Operators Defined on the Weighted Banach Spaces We define the integration operator 1 z and recall the classical estimates 1 z V ) t, z, T ) := : AC){t, z} AC){t, z} as Γx + γ) Γx) z 0 V t, ξ, T )dξ x γ.3) for all γ 0, as x tends to +. In this subsection, we study the continuity property of some linear operators defined on the spaces G q δ 1, δ ; σ). In the next proposition, we analyse some integro-differential operator. Proposition.5. Let s, ν, κ and S be non negative integers satisfying b s + κq 1)) q + ν < S.

6 14 S. Malek and C. Stenger Then the operator T s t ν ) T κ z S is a bounded linear operator from the Banach space G q δ 1, δ ; σ),. δ1,δ ;σ into itself. Moreover, there exists a constant C 5 > 0 such that T s t ν T κ z S V ) t, z, T ) C δ1,δ ;σ 5δ1 ν δ S V t, z, T ) δ1,δ ;σ for all V t, z, T ) G q δ 1, δ ; σ). Proof. For we have V t, z, T ) = β 0 n 0 T s t ν T κ z S V ) t, z, T ) = β S v n,β T ) tn z β n! β! G q δ 1, δ ; σ), n 0 T s T κ v n+ν,β S ) T ) tn z β n! β!. Therefore, by definition, we obtain that T s t ν T κ z S V ) t, z, T ) = δ1,δ T s κ δ1 n δ β ;σ T v n+ν,β S ) T ) β;σ n + β)! β S n 0 where T s T κ v n+ν,β S ) T ) β;σ = sup T s T κ v n+ν,β S ) T ) 1 + T ) m exp σr b β) T q ). T C From Cauchy s theorem, for T C, we have T s T κ v n+ν,β S ) T ) = κ! iπ ξ T =a T s v n+ν,β S ξ) ξ T ) κ+1 dξ where a > 0 can be chosen arbitrarily. We replace ξ by T +a expiθ), where θ [0, π[ and obtain T s κ T v n+ν,β S ) T ) π κ! T s πa κ Using the triangle inequality, we get 0 v n+ν,β ST ) β S;σ 1 + T + a expiθ) ) m exp σr b β S) T + a expiθ) q ) dθ. T s κ T v n+ν,β S ) T ) κ! a κ T s v n+ν,β S T ) β S;σ 1 + T + a) m exp σr b β S) T + a) q ).

7 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 15 We denote by I a,β T ) a part of the right-hand side of inequality below I a,β T ) := κ! a κ T s 1 + T + a) m exp σr b β S) T + a) q )..4) Now, our aim is to estimate T s T κ v n+ν,β S ) T ) β;σ ) T s T κ v n+ν,β S ) T ) β;σ sup I a,β T )1 + T ) m exp σr b β) T q ) T C So we have to prove the following lemma. Lemma.6. There exists a constant C 6 > 0 such that for all β S. v n+ν,β S T ) β S;σ. sup I a,β T )1 + T ) m exp σr b β) T q ) C 6 β bs+κq 1)) q T C Proof. We denote by J a,β the supremum to estimate J a,β := sup I a,β T )1 + T ) m exp σr b β) T q ). T C By definition of I a,β T ), we have ) T s m 1 + T + a J a,β = κ! sup exp σ {r T C a κ b β S) T + a) q r b β) T q }). 1 + T As T [0, + [ and a, m 0, 1 + T + a 1 + T we have ) m = 1 + ) m a 1 + a) m, 1 + T T s J a,β κ! sup T C a κ 1+a)m exp σ {r b β S) r b β)} T + a) q ) exp σr b β) { T + a) q T q }), using the equality γ 1 γ γ 3 γ 4 = γ 1 γ γ 3 ) + γ 3 γ 1 γ 4 ) in the exponential. After, we shall also need the estimates r b β) r b β S) = β n=β S+1 1 n + 1) b S β + 1) b.

8 16 S. Malek and C. Stenger Now, remember that the radius a in the Cauchy s theorem is still arbitrarily. As suggested by Y. Dubinskiĭ in [7], let us choose a > 0 such that T + a) q = T q + 1, thus a = T q + 1) 1/q T. Then, we obtain that J a,β κ! exp σr b β)) exp σs ) β + 1) b ) m 1 + T q + 1) 1/q T sup T s ) κ exp T C T q + 1) 1/q T σs T q β + 1) b To estimate the supremum on the right-hand side, we distinguish two cases. For the first case, using Taylor s formula at the origin ). 1 + x) 1/q = 1 + x q + xεx), x < 1 while εx) in the remainder term satisfies lim x 0 εx) = 0, there exists a constant 0 ς < 1 such that εx) < 1 q for x < ς. Therefore, for T > ς 1/q, there exists a constant c 1 > 0, independent of β such that ) m 1 + T q + 1) 1/q T ) κ T q + 1) 1/q T = T κq 1) + 1 ε q T q 1 T q q + ε )) m 1 T q T q )) κ c 1 T κq 1). Let A, B positive numbers, a study of the function fx) = x A exp Bx) shows that its supremum yields at x = A\B, thus sup T s+κq 1) exp σs ) T q T C, T q >ς 1 β + 1) b ) s+κq 1) s + κq 1) β + 1) b q qσs exp ) s + κq 1). q

9 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 17 It turns out that there exists a constant c > 0, independent of β, such that ) m 1 + T q + 1) 1/q T sup T s ) κ exp σs ) T T C, T q >ς 1 T q + 1) 1/q q c T β + 1) b β bs+κq 1)) q. For the second case T ς 1/q, it is easy to see that there exists a constant c 3 > 0, independent of β, such that ) m 1 + T q + 1) 1/q T sup T s ) κ exp σs ) T T C, T q ς 1 T q + 1) 1/q q c T β + 1) b 3 for all β S. Hence, we have two constants c 4, c 5 > 0, independent of β, such that J a,β c 4 κ! exp σζb)) exp σs ) β bs+κq 1))/q c β + 1) b 5 β bs+κq 1)) q for all β S. This proves the lemma.6. Now, we return to the proof of proposition.5. Using Lemma.6, we obtain that there exists a constant c 1 > 0, independent of β, such that T s T κ v n+ν,β S ) T ) β;σ c 1 β bs+κq 1)) q for all n 0 and for all β S, and hence T s t ν T κ z S β S n 0 V ) t, z, T ) δ1,δ ;σ c 1δ ν 1 δ S β bs+κq 1)) q n + ν + β S)! n + β)! Using the estimates.3) and our hypotheses that a constant c > 0, independent of β, such that v n+ν,β S T ) β S;σ v n+ν,β S T ) β S;σ δ1 n+ν δ β S n + ν + β S)!. b s + κq 1)) +ν < S, there exists q β bs+κq 1)) q n + ν + β S)! n + β)! β bs+κq 1)) q c c β S ν for all β S. To conclude the proof of Proposition.5, there exists a constant c 3 > 0 such that T s ν t κ T S z for all V t, z, T ) G q δ 1, δ ; σ). V ) t, z, T ) δ1,δ ;σ c 3δ ν 1 δ S V t, z, T ) δ1,δ ;σ

10 18 S. Malek and C. Stenger In the next proposition, we study the continuity of the operator of multiplication by a holomorphic function. Proposition.7. Let at, z) = β 0 a holomorphic function near the origin in C. We define n 0 a t, z) = β 0 t n z β a n,β n! β! n 0 Then there exists a constant C 7 > 0, such that a n,β tn z β n! β!. at, z)v t, z, T ) δ1,δ ;σ C 7 a δ 1, δ ) V t, z, T ) δ1,δ ;σ for all V t, z, T ) G q δ 1, δ ; σ). Proof. For we have V t, z, T ) = β 0 n 0 v n,β T ) tn z β n! β! G q δ 1, δ ; σ), t n 0 z β 0 n 0! β 0! V t, z, T ) = n!β! n β β 0 n n 0!n n 0 )!β 0!β β 0 )! v n n 0,β β 0 T ) tn z β n! β!. 0 Therefore, by definition, we obtain that t n 0 z β 0 V t, z, T ) n 0! β 0! δ1,δ ;σ = n!β! n 0!n n 0 )!β 0!β β 0 )! v n n 0,β β 0 T ) δ1 n δ β β β 0 β;σ n + β)!, n n 0 where n!β! n 0!n n 0 )!β 0!β β 0 )! v n n 0,β β 0 T ) and = β;σ n!β! n 0!n n 0 )!β 0!β β 0 )! v n n 0,β β 0 T ) β;σ v n n0,β β 0 T ) β;σ = sup v n n0,β β 0 T ) 1 + T ) m exp σr b β) T q ) T C for all n n 0 and for all β β 0. Observe that v n n0,β β 0 T ) β;σ sup v n n0,β β 0 T ) 1 + T ) m exp σr b β β 0 ) T q ) T C

11 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 19 since r b β) r b β β 0 ) for all β β 0. Therefore, we have v n n0,β β 0 T ) β;σ v n n0,β β 0 T ) β β0 ;σ for all β β 0. Using the upper mentioned inequalities, we obtain t n 0 z β 0 V t, z, T ) n 0! β 0! δn 0 1 δ β 0 n!β! n n 0 + β β 0 )! n 0! β 0! n n 0 )!β β 0 )! n + β)! δ1,δ ;σ β β 0 n n 0 δ n n 0 1 δ β β 0 v n n0,β β 0 T ) β β0 ;σ n n 0 + β β 0 )!. On the other hand, using the estimates.3), there exists a constant c 1 > 0 independent on n and β), such that n!β! n n 0 + β β 0 )! n n 0 )!β β 0 )! n + β)! for all n n 0 and for all β β 0, therefore t n 0 z β 0 V t, z, T ) n 0! β 0! δ1,δ ;σ δ n 0 1 c 1 n 0! c 1 δ β 0 β 0! V t, z, T ) δ 1,δ ;σ. So that, there exists a constant c > 0 such that at, z)v t, z, T ) δ1,δ ;σ a t n 0 z β 0 n 0,β 0 V t, z, T ) n β 0 0 n 0 0 0! β 0! δ1 δ ;σ a n0,β 0 t n 0 z β 0 V t, z, T ) n β 0 0 n 0 0 0! β 0! c a δ 1, δ ) V t, z, T ) δ1,δ ;σ. δ1,δ ;σ This proves proposition.7. In the next proposition, we analyse the action of some differential operator on functions of G q δ 1, δ ; σ) which are polynomial in the variable z. Proposition.8. Let s, ν, κ and θ non negative integers and let W t, z, T ) in the space G q δ 1,0, δ ; σ 0 ) with W t, z, T ) polynomial in the variable z. Then there exist δ 1 > 0, small enough δ 1 δ 1,0 ) and σ, big enough σ > σ 0 ) with σ 0 r b d) + 1 σ, such that the series T s t ν T κ zw ) θ t, z, T ) belongs to G q δ 1, δ ; σ). Moreover, there exists a constant C 8 > 0 such that T s t ν T κ zw ) θ t, z, T ) C δ1,δ ;σ 8δ1 ν δ θ W t, z, T ) δ1,0,δ ;σ 0 for all W t, z, T ) G q δ 1,0, δ ; σ 0 ) such that W t, z, T ) is a polynomial in the variable z.

12 0 S. Malek and C. Stenger Proof. For W t, z, T ) G q δ 1,0, δ ; σ 0 ) and W t, z, T ) = polynomial in the variable z, we have d β=0 n 0 T s t ν T κ zw ) d θ θ t, z, T ) = Therefore, by definition, we obtain that T s t ν T κ zw ) θ t, z, T ) = d θ δ1,δ ;σ where β=0 n 0 β=0 n 0 w n,β T ) tn z β n! β! T s T κ w n+ν,β+θ ) T ) tn z β n! β!. T s κ T w n+ν,β+θ ) T ) β;σ δ n 1 δ β n + β)!.5) T s T κ w n+ν,β+θ ) T ) β;σ = sup T s T κ w n+ν,β+θ ) T ) 1 + T ) m exp σr b β) T q ). T C As in the proof of Proposition.5, using Cauchy s theorem, we have, for all T C T s κ T w n+ν,β+θ ) T ) π 0 κ! T s πa κ w n+ν,β+θt ) β+θ;σ0 1 + T + a expiτ) ) m exp σ 0 r b β + θ) T + a expiτ) q ) dτ where a > 0 can be chosen arbitrarily and σ 0 will be chosen later. Using the triangular inequality, we get T s κ T w n+ν,β+θ ) T ) κ! a κ T s w n+ν,β+θ T ) β+θ;σ0 As in.4), we define 1 + T + a) m exp σ 0 r b β + θ) T + a) q ). I a,β;σ0 T ) := κ! a κ T s 1 + T + a) m exp σ 0 r b β + θ) T + a) q ). Now, our aim is to estimate T s κ T w n+ν,β+θ ) T ) β;σ. In fact, we have ) T s T κ w n+ν,β+θ ) T ) β;σ sup I a,β;σ0 T )1 + T ) m exp σr b β) T q ) T C w n+ν,β+θ T ) β+θ;σ0..6) So we have to prove the following lemma.

13 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 1 Lemma.9. There exists a constant C 9 > 0 such that for all 0 β d θ. sup I a,β;σ0 T )1 + T ) m exp σr b β) T q ) C 9 T C Proof. As in the proof of Lemma.6, we denote by J a,β;σ0,σ the supremum to estimate J a,β;σ0,σ := sup I a,β;σ0 T )1 + T ) m exp σr b β) T q ). T C By definition of I a,β;σ0 T ), we have ) T s m 1 + T + a J a,β;σ0,σ = κ! sup exp σ T C a κ 0 r b β + θ) T + a) q σr b β) T q ). 1 + T As in the proof of Lemma.6, as T [0, + [ and a, m 0, ) m ) m 1 + T + a a = a) m, 1 + T 1 + T and we have T s J a,β;σ0,σ κ! sup T C a 1 + κ a)m exp {σ 0 r b β + θ) σr b β)} T + a) q ) exp σr b β) { T + a) q T q }) using the equality γ 1 γ γ 3 γ 4 = γ 1 γ γ 3 ) + γ 3 γ 1 γ 4 ) in the exponential. Now, we will fix σ. It is necessary that σ 0 σ. Therefore, we can show that for all 0 β d θ σ 0 r b β + θ) σr b β) σ 0 r b d) σr b 0) = σ 0 r b d) σ. Let σ σ σ 0 ) such that σ σ 0 r b d) + 1, then we have that σ 0 r b β + θ) σr b β) 1 for all 0 β d θ. It turns out that T s J a,β;σ0,σ κ! sup T C a 1 + κ a)m exp 1 ) T + a)q exp σr b β) { T + a) q T q }). With the same choice for the radius a as before a = T q + 1) 1/q T,

14 S. Malek and C. Stenger we obtain that J a,β;σ0,σ κ! exp σr b β)) exp 1 ) sup T s T C ) m 1 + T q + 1) 1/q T ) κ exp 1 ) T T q + 1) 1/q q T for all 0 β d θ. On the other hand, there exists a constant c 1 > 0, such that ) m 1 + T q + 1) 1/q T sup T s ) κ exp 1 ) T T C T q + 1) 1/q q c 1 T and therefore, there exists a constant c > 0, independent of β, such that J a,β;σ0,σ c for all 0 β d θ. This proves the lemma.9. Now, we return to the proof of proposition.8. Using Lemma.9, we obtain for.6) that there exists a constant c 1 > 0, independent of β, such that T s κ T w n+ν,β+θ ) T ) β;σ c 1 w n+ν,β+θ T ) β+θ;σ0 for all n 0 and for all 0 β d θ, and hence.5) becomes T s t ν T κ zw ) θ t, z, T ) c δ1,δ ;σ 1δ1 ν δ θ d θ β=0 n 0 Using the estimates.3) n + ν + β + θ)! n + β)! n + ν + β + θ)! n + β)! w n+ν,β+θ T ) β+θ;σ0 n + β) ν+θ δ1 n+ν δ β+θ n + ν + β + θ)!. as n tends to infinity and for 0 β d θ, there exist two constants c 1, c 3 > 0, independent of n and β, such that n + ν + β + θ)! n + β)! c 3 c n+ν for all n 0 and for all 0 β d θ. We get that there exists a constant c 4 > 0 such that T s t ν T κ zw ) θ t, z, T ) d c δ1,δ ;σ 4δ1 ν δ θ c δ 1 ) n δ β w n,β T ) β;σ0 n + β)!. β=0 n 0

15 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 3 Now, choosing δ 1 > 0 such that δ 1 δ 1,0 /c, we conclude that T s t ν T κ zw ) θ t, z, T ) c δ1,δ ;σ 4δ1 ν δ θ W t, z, T ) δ1,0,δ ;σ 0 for all W t, z, T ) G q δ 1,0, δ ; σ 0 ). This proves the proposition.8..3 A Cauchy Problem in the Weighted Banach Spaces In the next two definitions, we introduce some linear operators. Definition.10. Let M the linear operator from AC){t, z} into AC){t, z} defined as M Ut, z, T )) = S z Ut, z, T ) q=q 0,q 1,q ) Q where Q, S and B q t, z, T ) satisfy the following assumptions. B q t, z, T ) q 0 t q 1 T q z Ut, z, T ).7) A1) S is a positive integer, Q is a finite subset of N 3 and we denote by q = q 0, q 1, q ) Q. A) For all q Q, B q t, z, T ) is polynomial in T σq) B q t, z, T ) = b q j j t, z)t where σq) is the degree of B q t, z, T ) with respect to the variable T and the coefficients b q j t, z) are holomorphic functions near the origin in C. A3) For all q = q 0, q 1, q ) Q, j=0 b σq) + q 1 q 1)) q + q 0 + q < S where q 1 and b > 1 are positive numbers defined from our Banach space G q δ 1, δ ; σ). Definition.11. Let N be the linear operator from AC){t, z} into AC){t, z} defined as N Ut, z, T )) = B q t, z, T ) q 0 t q 1 T q S z Ut, z, T ).8) q Q where Q, S and B q t, z, T ) satisfy the assumptions A1), A) and A3).

16 4 S. Malek and C. Stenger Using the definitions.7) and.8), we obtain the following relation between the operators M and N M z S = Id N.9) where Id is the identity operator from AC){t, z} into itself. In the next proposition, we show that N is a linear bounded operator on the constructed Banach spaces with small norm. Proposition.1. Let N be the operator defined by.8) and which satisfies the assumptions A1), A) and A3). Then, there exists a real number δ > 0, small enough which depends on the coefficients B q t, z, T ) for q = q 0, ) q 1, q ) Q), such that N is a bounded linear operator from G q δ 1, δ ; σ),. δ1,δ ;σ into itself. Moreover, there exists a constant C 10, 0 < C 10 < 1, such that for all Ut, z, T ) G q δ 1, δ ; σ). N Ut, z, T )) δ1,δ ;σ C 10 Ut, z, T ) δ1,δ ;σ Proof. By definition.8) of the operator N and the assumption A), we get N Ut, z, T )) δ1,δ ;σ σq) b q j t, z)t j q 0 t q 1 T q S z Ut, z, T ) δ 1,δ ;σ q Q j=0 for all Ut, z, T ) G q δ 1, δ ; σ). From the fact that b q j t, z), for j = 0,..., σq), are holomorphic functions near the origin in C, we deduce from Proposition.7 that there exists a constant c 1 > 0, such that σq) N Ut, z, T )) δ1,δ ;σ c 1 b q δ1, δ ) T j q 0 q Q j=0 j t q 1 T q S z Ut, z, T ) δ1,δ ;σ for all Ut, z, T ) G q δ 1, δ ; σ). Under the assumption A3), we can apply proposition.3 and we obtain the existence of a constant c > 0, such that N Ut, z, T )) δ1,δ ;σ c σq) b q j δ 1, δ ) δ q 0 1 δ S q Ut, z, T ) δ1,δ ;σ q Q j=0 for all Ut, z, T ) G q δ 1, δ ; σ). Now, with δ > 0 small enough, we have σq) 0 < c b q δ1, δ ) δ q 0 =: C 6 < 1. q Q This proves the proposition. j=0 j 1 δ S q

17 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 5 As a consequence of the proposition.1, we deduce the invertibility of the operator M S z. Proposition.13. Let M the operator defined by.7) and which satisfies the assumptions A1), A) and A3). Then, there exists a real number δ > 0, small enough which depends on the coefficients B q t, z, T ) for q = q 0, q 1, q ) Q), such that, M z S is a bounded invertible operator from Moreover, there exists a constant C 11 > 0, such that G q δ 1, δ ; σ),. δ1,δ ;σ ) M z S 1 V t, z, T )) C 11 V t, z, T ) δ1,δ δ1,δ;σ ;σ for V t, z, T ) G q δ 1, δ ; σ). Now, we are able to prove the main result of this section. Theorem.14. Consider a partial differential equation of the form ) into itself. S z Ut, z, T ) = q Q B q t, z, T ) q 0 t q 1 T q z Ut, z, T ).10) which satisfies the assumptions A1), A) and A3). We impose the following initial conditions. For all 0 j S 1, j z U ) t, 0, T ) = ψ j t, T ).11) where the ψ j t, T ) belong to G q δ 1,0, δ,0 ; σ 0 ), for given δ 1,0, δ,0 and σ 0 > 0. Then, there exists δ 1, δ and σ > 0, such that the problem.10) and.11) has a unique solution Ut, z, T ) G q δ 1, δ ; σ). Proof. A formal series Ut, z, T ) AC)[[t, z]] which satisfies the equation.11) can be written in the form where Ut, z, T ) = S z V t, z, T ) + W t, z, T ) S 1 W t, z, T ) := ψ j t, T ) zj j!. j=0 From the fact that the initial conditions.11) belong to G q δ 1,0, δ,0 ; σ 0 ) and Proposition.7, we get that W t, z, T ) belongs to G q δ 1,0, δ,0 ; σ 0 ). Moreover, Ut, z, T ) is a solution of.10) and.11) if and only if V t, z, T ) satisfies the equation ) M S z V t, z, T )) = M W t, z, T ))..1) From the fact that W t, z, T ) is a polynomial in the variable z, and that W t, z, T ) G q δ 1,0, δ,0 ; σ 0 ),

18 6 S. Malek and C. Stenger using Proposition.7 and Proposition.8, we get that there exist δ 1 and σ > 0 such that M W t, z, T )) G q δ 1, δ,0 ; σ). Therefore, by Proposition.13, for δ > 0 small enough, there exists a unique solution V t, z, T ) of.1) which belongs to G q δ 1, δ ; σ). To conclude, it remains to show that Ut, z, T ) G q δ 1, δ ; σ). Indeed, we have Ut, z, T ) = z S V t, z, T ) + W t, z, T ). Using Proposition, we get that z S V t, z, T ) G q δ 1, δ ; σ). Moreover, we have also W t, z, T ) G q δ 1, δ ; σ). In conclusion, Ut, z, T ) belongs to G q δ 1, δ ; σ). This proves our theorem. 3 Formal Solutions of a Linear Cauchy Problem Let us consider φt) a holomorphic solution of the ordinary differential equation φ t) = P t, φt)) 3.1) where P t, X) is a holomorphic function on some domain D C with respect to t and polynomial in the variable X of degree larger than. Consider the following linear partial differential equation S z Xt, z) = k=k 0,k 1 ) J where S is a positive integer and J is a finite set which satisfies a k t, z) k 0 t k 1 z Xt, z) 3.) J { k = k 0, k 1 ) N k 1 S 1 }. 3.3) The coefficients a k t, z) are holomorphic functions on the domain D with respect to t and near the origin in C with respect to z. Our aim is to look for transseries solution Xt, z) of 3.) in powers of φt) and z ˆXt, z) = β 0 l 0 X l,β t) φt)l z β β! 3.4) where X l,β t) are holomorphic functions on the domain D C. To fix the notations, for all integers n 1, for all vectors p = p 1,..., p n ) N n, we define p 1 := p 1 + p p n and p := p 1 + p np n. We define the set P n) as By convention, we put P 0) = {0}. P n) := {p = p 1,..., p n ) N n p = n}.

19 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 7 Remark 3.1. If p P n), then p 1 n. Indeed, n p 1 = p 1 +p +...+np n p 1 + p p n ) = p +p n 1)p n 0. Proposition 3.. Let X l,β, l 0, 0 β S 1, be given holomorphic functions on D. Then the partial differential equation 3.) has a formal solution ˆXt, z) of the form 3.4) for given initial conditions j z ˆX)t, 0) = l 0 X l,j t) φt)l, 0 j S ) Proof. Our goal is to construct a sequence of holomorphic functions X l,β t), l, β 0, such that the formal series ˆXt, z) is a solution of the problem 3.), 3.5). In a first step, we plug ˆXt, z) into the equation 3.) using the formal expansion 3.4) and we deduce a recursion for the coefficients X l,β t) by identification of the powers of φt) and z. For the left-hand side term of 3.), we obtain S z ˆXt, z) = β 0 l 0 X l,β+s t) φt)l z β β!. For the right-hand side term, it will be more complicated. A first calculus shows us that Using Leibniz formula for k 0 t k 1 z ˆXt, z) = β 0 l 0 D k l,β t, φt)) := k 0 t k 0 t } {X l,β+k1 t) φt)l z β β!. 3.6) } {X l,β+k1 t) φt)l, we have D k l,β t, φt)) = k 0 m=0 k 0! m!k 0 m)! k 0 m t X l,β+k1 t) ) t m φt) l ). To expand m t m t φt) l, we use Faà di Bruno formula φt) l = p P m) p 1 l m! φt) l p 1 p 1!... p m! l p 1 )! m ) φ k) pk t). 3.7) k=1 k!

20 8 S. Malek and C. Stenger We introduce the notation Ω k β t, φt)) := Dl,β k t, φt)) = l 0 l 0 Therefore, the equation 3.) yields β 0 l 0 X l,β+s t) φt)l k 0 m=0 p P m ) p 1 l z β β! = Using for a k t, z) the Taylor expansion β 0 k=k 0,k 1 ) J a k t, z) = β 0 l 0 k 0! k 0 m t X l,β+k1 t) ) m!k 0 m)! m! φt) l p 1 m φ k) t) p 1!... p m! l p 1 )! a k t, z) β 0 a k βt) zβ β!, k=1 k! ) pk. Ω k β t, φt)) zβ β!. 3.8) where the coefficients a k βt) are holomorphic functions on D, we obtain from 3.8) ) X l,β+s t) φt)l z β β! = β 0 k=k 0,k 1 ) J β =0 By identification of the coefficients of zβ, we get β! l 0 X l,β+s t) φt)l = β β! β!β β )! ak β β t)ω k β t, φt)) zβ β!. β k=k 0,k 1 ) J β =0 β! β!β β )! ak β β t)ω k β t, φt)) 3.9) for all β N. On the right-hand side, the term Ω k β t, φt)) has the form where ω k,m l t, φt)) = Ω k β t, φt)) = l 0 k 0! k 0 m)! k 0 m=0 p P m) p 1 l k 0 m t X l,β +k 1 t) ) ω k,m l t, φt)) 3.10) 1 φt) l p 1 p 1!... p m! l p 1 )! m ) φ k) pk t), k! k=1

21 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 9 contains powers of φt) and powers of φ k) t) for k = 1,..., m. In the next two lemmas, we are going to specify the form of ω k,m l t, φt)). Lemma 3.3. Let us consider the ordinary differential equation 3.1) with P t, X) = d j=0 p j t) Xj j!, d where the coefficients p j t) are holomorphic on some domain D C and d is the degree of P t, X) with respect to the variable X in particular p d t) 0). Then, for all positive integer n, there exists a function Q n t, X) such that with φ n) t) = Q n t, φt)) 3.11) Q n t, X) = d n j=0 q n,j t) Xj j! where the coefficients q n,j t) are holomorphic on D. Moreover, d n := deg X Q n t, X) = nd 1) + 1 is the degree of Q n t, X) with respect to the variable X and q n,dn t) = p d t)) n n j=1 d j! d!d j d)!. Proof. We prove this lemma by induction on n. The assertions are true for n = 1, with Q 1 t, X) = P t, X), therefore d 1 = d = d 1) + 1 and q 1,d1 t) = p d t) = p d t) 1 j=1 d 1! d!d 1 d)!. We suppose, that the assertions are true until n 1. Differentiating 3.11) gives φ n+1) t) = d n j=0 q n,jt) φt)j j! d n 1 + φ t) q n,j+1 t) φt)j. 3.1) j! j=0 By 3.1), the highest power in 3.1) of φt) is in the second sum, it is p d t)q n,dn t) φt)d 1+dn d!d n 1)! = d 1 + d n)! p d t)q n,dn t) φt)d 1+dn d!d n 1)! d 1 + d n )!.

22 30 S. Malek and C. Stenger Therefore, the right-hand side of 3.1) can be written as Q n+1 t, φt)) := d n+1 j=0 q n+1,j t) φt)j j! where the coefficients q n+1,j t) are holomorphic on D which are sums and products of the p j t), q n,j t) and q n,jt). Moreover, d n+1 := deg X Q n+1 t, X) = d 1 + d n = n + 1)d 1) + 1 and q n+1,dn+1 t) = n+1 d n+1! d!d n+1 d)! p dt)q n,dn t) = p d t)) n+1 d j! d!d j d)!. j=1 This proves our lemma. Lemma 3.4. Let m N with m 1 and p P m). Under the assumptions of Lemma 3.3, there exist a positive integer D m and holomorphic coefficients Q m j t) on D, such that where and m ) φ k) pk t) = k! k=1 D m j=0 Q m j t) φt)j j! D m = md 1) + p 1 Q m D m t) = D m! p d t)) m m Proof. Using Lemma 3.3, we have m ) φ k) pk t) = k! k=1 k=1 m 1 k! k=1 k j=1 d k j=0 In the right-hand side, the highest power of φt) is d j! d!d j d)! k!d k! )p k. ) pk q k,j t) φt)j. j! D m := m d k p k = k=1 m m m kd 1) + 1) p k = d 1) kp k + p k = md 1)+ p 1 k=1 k=1 k=1 because p P m). Also, the coefficient Q m D m t) of φt)dm is given by D m! m ) pk Q m qk,dk t) D m t) = D m!. k!d k! k=1

23 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 31 By Lemma 3.3, we get This proves our lemma. Q m D m t) = D m! p d t)) m m k j=1 k=1 d j! d!d j d)! k!d k! )p k. We return to the proof of our proposition 3.. By Lemma 3.4, ω k,m l t, φt)) has the form ω k,m l t, φt)) = k 0! k 0 m)! p P m) p 1 l 1 p 1!... p m! D m j=0 Q m j t) φt)l+j p 1 l p 1 )!j!. For fixed l, ω k,m l t, φt)) is a polynomial in φt) with powers of φt) between l m and l + D m p 1 = l + md 1). To avoid confusion for the identification of the coefficients of φt)l form, we will replace l by L in ω k,m l t, φt)). Therefore 3.10) has the Ω k β t, φt)) = L 0 and we obtain for equation 3.9) l 0 X l,β+s t) φt)l = β k=k 0,k 1 ) J β =0 k 0! k 0 m)! p P m) p 1 L k 0 m=0 β! k 0 m t β!β β )! ak β β t) L 0 1 p 1!... p m! D m j=0 X L,β +k 1 t) ) ω k,m L k 0 m=0 t, φt)) k 0 m t X L,β +k 1 t) ) Q m j t) L + j p 1)! L p 1 )!j! φt) L+j p 1 L + j p 1 )! 3.13) for all β N. By identification of the coefficients of φt)l, we get the following recursion for the coefficients X l,β t), X l,β+s t) = β k=k 0,k 1 ) J β =0 β! β!β β )! ak β β t) k 0 m=0 k 0! k 0 m)! 3.14)

24 3 S. Malek and C. Stenger where L,p,j) l,m k 0 m t X L,β +k 1 t) ) 1 p 1!... p m! Qm j t) L p 1 )!j! l,m = {L, p, j) N N m N p P m), j = 0,..., D m, L + j p 1 = l} for all l, β) N. In a second step of the proof, we consider now given holomorphic functions X l,β t) on D, for all l 0, for β = 0,..., S 1. From 3.3), we deduce that there exists a unique sequence X l,β t)) l 0,β 0 of holomorphic functions on D which satisfies the recurrence 3.14) for the given initial holomorphic functions X l,β t), for all l 0, for β = 0,..., S 1. From the first part of the proof, we deduce that the formal transseries ˆXt, z) = l 0,β 0 X l,β t) φt)l is a solution of the Cauchy problem 3.), 3.5). This ends the proof of Proposition Holomorphic Solutions of the Linear Cauchy Problem In this section, we will give sufficient conditions under which the formal transasymptotic series solutions of the Cauchy problem 3.), 3.5) constructed in the previous section are convergent in some punctured polydiscs of C. 4.1 Classification of Singularities for First Order Ordinary Differential Equations Consider our first order non linear differential equation ) φ t) = P t, φt)), where P t, X) is polynomial in X of degree larger than, with holomorphic coefficients in D. As a consequence of a result of P. Painlevé see [9], Theorem 3.3.), we know that the only movable singularities in C of the solutions φt) of ) are poles and/or algebraic branch points. In the following, we define D θ t, r) as an open disc Dt, r) centered at t C with radius r > 0, minus the segment [ t, re iθ ), for θ R. Let φt) a solution of ) defined on D θ t 0, r 0 ) D, with r 0 > 0, and θ R, where φt) can be represented by a Puiseux series of the form φt) = f n t t 0 ) n/µ 4.1) n n 0 z β β!

25 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 33 where µ 1 and n 0 1 are positive integers such that f n0 0. If µ is equal to 1, the point t 0 is called a pole of order n 0 and then the function φt) is holomorphic on Dt 0, r) \ {t 0 }, otherwise the point t 0 is called an algebraic branch point. 4. An Auxiliary Linear Cauchy Problem Let v n0,l,β := sup t D n 0 t X l,β t) for all n 0, l, β 0, where X l,β t)) l 0,β 0 is the sequence constructed in Proposition 3.. In this subsection, we will construct a majorizing sequence u n0,l,β of v n0,l,β such that the formal series Ut, z, T ) = β 0 l 0 n 0 0 t n 0 u n0,l,β satisfies a linear Cauchy problem. We recall the recursion formula 3.14) for the coefficients X l,β t) n 0! T l z β β! X l,β+s t) = β k=k 0,k 1 ) J β =0 β! β!β β )! k 0 m=0 k 0! k 0 m)! 4.) L,p,j) l,m 1 p 1!... p m! j!l j)! Ak,m j,β β t) k 0 m t X L,β +k 1 t) ) where A k,m j,β β t) := a k β β t)q m j t) with a k βt) the coefficient of z β /β! in the expansion of a k t, z) with respect to z and Q m j t) are defined in Lemma 3.4 and l,m = {L, p, j) N N m N j = 0,..., D m ; p P m); L + j p 1 = l} for all l, β 0. From the Leibniz formula, if we differentiate n 0 times equation 4.), we have n 0 t X l,β+s t) = j!l j)! β k 0 β! k=k 0,k 1 ) J β =0 m=0 L,p,j) l,m n 0 n 1 =0 n 0! n 1!n 0 n 1 )! n 0 n 1 t k 0! 1 β!β β )! k 0 m)! p 1!... p m! ) A k,m k j,β β t) 0 m+n 1 t X L,β +k 1 t) ) 4.3)

26 34 S. Malek and C. Stenger for all n 0, l, β 0. Therefore, we get the following inequalities for the v n0,l,β v n0,l,β+s = sup t D n 0 t X l,β+s t) β k 0 β! k=k 0,k 1 ) J β =0 m=0 L,p,j) l,m 1 p 1!... p m! j!l j)! n 0 n 1 =0 k 0! β!β β )! k 0 m)! n 0! n 1!n 0 n 1 )! sup n 0 n 1 t A k,m j,β β t) t D sup k 0 m+n 1 t X L,β +k 1 t) 4.4) t D for all n 0, l, β 0. As A k,m j,β t) is holomorphic on D, using Cauchy estimates, there exist two constants M j,β and r r depends only on D) such that sup t D n 0 n 1 t A k,m j,β β t) Therefore, we obtain from 4.4) M j,β β r) r n 0 n 1 n 0 n 1 )! =: A n0 n 1,j,β β. 4.5) v n0,l,β+s β k 0 n 0 k=k 0,k 1 ) J β =0 m=0 L,p,j) l,m n 1 =0 β! k 0! β!β β )! k 0 m)! 4.6) 1 n 0! p 1!... p m! j!l j)! n 1!n 0 n 1 )! A n 0 n 1,j,β β v k0 m+n 1,L,β +k 1 for all n 0, l, β 0. Now, let u n0,l,β) n0 0,l 0,β 0 the sequence satisfying u n0,l,β+s = β k 0 n 0 k=k 0,k 1 ) J β =0 m=0 L,p,j) l,m n 1 =0 β! k 0! 1 β!β β )! k 0 m)! p 1!... p m! n 0! j!l j)! n 1!n 0 n 1 )! A n 0 n 1,j,β β u k0 m+n 1,L,β +k 1 4.7) for all n 0, l, β 0. From 3.3), we have 0 k 1 S 1 for all k = k 0, k 1 ) J. We deduce that the sequence u n0,l,β) n0 0,l 0,β 0 is uniquely determined by 4.7) and the initial conditions u n0,l,β := v n0,l,β = sup n 0 t X l,β t) 4.8) t D for β = 0,..., S 1 and for all n 0, l 0. We deduce from 4.6), 4.7) and 4.8) that for all n 0, l, β 0. v n0,l,β u n0,l,β 4.9)

27 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 35 Now, we define the formal series Ut, z, T ) = β 0 l 0 n 0 0 t n 0 u n0,l,β n 0! T l z β β! 4.10) where the coefficients u n0,l,β satisfy 4.7) and 4.8) for all n 0, l, β 0. We have the following proposition. Proposition 4.1. The formal series 4.10) is the unique solution of the Cauchy problem z S Ut, z, T ) = B q t, z, T ) q 0 t q 1 T q z Ut, z, T ) 4.11) for given initial conditions where and q=q 0,q 1,q ) Q j zu)t, 0, T ) = l,n 0, 0 t n 0 v n0,l,j n 0! T l, 0 j S 1, 4.1) Q = { q = q 0, q 1, q ) N 3 k = k 0, k 1 ) J ; m = 0,..., k 0 ; p P m); B q t, z, T ) = k 0! q 0! 1 p 1!... p m! where the coefficients A n,j,β are defined by 4.5). q 0 = k 0 m; q 1 = p 1 ; q = k 1 } D m β 0 j=0 n 0 t n A n,j,β n! T j z β j! β! 4.13) Proof. We plug Ut, z, T ) into 4.11) using the expansion 4.10). For the left-hand side term of 4.11), we obtain z S Ut, z, T ) = t n 0 T l z β u n0,l,β+s n β 0 l 0 n 0 0 0! β!. 4.14) For the right-hand side term, a first calculus shows us that q 0 t q 1 T q z Ut, z, T ) = t n 0 u n0 +q 0,l+q 1,β+q n β 0 l 0 n 0 0 0! T l z β β!. Using for B q t, z, T ) the expansion 4.13), we obtain B q t, z, T ) q 0 t q 1 T q z Ut, z, T ) = β l n 0 β! β β 0 l 0 n 0 0 β =0 l 1 =0 n 1 =0!β β )! n 0! k 0! 1 l 1!l l 1 )! n 1!n 0 n 1 )! q 0! p 1!... p m! A n0 n 1,l l 1,β β u n1 +q 0,l 1 +q 1,β +q ) tn 0 n 0! T l z β β! 4.15)

28 36 S. Malek and C. Stenger where A n0 n 1,l l 1,β β = 0 for l l 1 > D m. By identification of the coefficients of t n 0 T l z β in 4.11), using 4.14) and 4.15), we get n 0! β! u n0,l,β+s = β l n 0 q=q 0,q 1,q ) Q β =0 l 1 =0 n 1 =0 k 0! q 0! β! n 0! β!β β )! l 1!l l 1 )! n 1!n 0 n 1 )! 1 p 1!... p m! A n 0 n 1,l l 1,β β u n1 +q 0,l 1 +q 1,β +q for all n 0, l, β 0 and where A n0 n 1,l l 1,β β = 0 for l l 1 > D m. Using the definition of the set Q, we have u n0,l,β+s = k=k 0,k 1 ) J k 0 m=0 β l n 0 p P m) β =0 l 1 =0 n 1 =0 β! n 0! β!β β )! l 1!l l 1 )! n 1!n 0 n 1 )! k 0! 1 k 0 m)! p 1!... p m! A n 0 n 1,l l 1,β β u n1 +k 0 m,l 1 + p 1,β +k 1 for all n 0, l, β 0 and where A n0 n 1,l l 1,β β formulation, we obtain = 0 for l l 1 > D m. Or with an other u n0,l,β+s = k=k 0,k 1 ) J k 0 m=0 β D m n 0 p P m) β =0 l =0 n 1 =0 β! n 0! β!β β )! l!l l )! n 1!n 0 n 1 )! k 0! 1 k 0 m)! p 1!... p m! A n 0 n 1,l,β β u n1 +k 0 m,l l + p 1,β +k 1 for all n 0, l, β 0, which is exactly equation 4.7). This proves our proposition. 4.3 The Main Result We are now in position to state the main result of the paper. Theorem 4.. Consider an ordinary differential equation φ t) = P t, φt)) 4.16) where P t, X) is a holomorphic function on some domain D C with respect to t and polynomial in the variable X of degree d. Let φt) be a holomorphic solution

29 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 37 of 4.16) on some punctured disc D θ t 0, r 0 ) D, where an expansion 4.1) holds. Consider the partial differential equation S z Xt, z) = k=k 0,k 1 ) J where S is a positive integer and J is a finite set which satisfies J { k = k 0, k 1 ) N k 1 S 1 }. a k t, z) k 0 t k 1 z Xt, z) 4.17) The coefficients a k t, z) are holomorphic functions on the domain D with respect to t and near the origin in C with respect to z. We make the following assumptions. A4) There exist two real constants b > 1 and q 1 such that max max 0 m k 0 p 1,...,p m) P m) bmd 1) + qp p m )) q + k 0 m + k 1 < S for all k = k 0, k 1 ) J, where P m) = {p 1,..., p m ) N m /p 1 + p mp m = m}, for all m 1, and P 0) = {0} by convention. A5) Let X l,β t), l 0, 0 β S 1, be bounded holomorphic functions on D, satisfying the estimates for all β = 0,..., S 1 t n X l,β t) T l δ1,0 n < +. n! Then, the formal series 3.4) n 0 Xt, z) = β 0 l 0 l 0 sup t D X l,β t) φt)l constructed in Proposition 3., solution of 4.17), defines a holomorphic function on the open set D θ t 0, r 0 ) D 0, δδ ), for a given δ 0, 1) and δ > 0. Moreover, the function Xt, z) has the following rate of growth on the set D θ t 0, r 0 ) D 0, δδ ). If r 0 is small enough, there exist positive constants C 11 and C 1 such that sup z δδ Xt, z) C 11 1 δ) 1 t t 0 n0m/µ expc 1 t t 0 n0q/µ ) 4.18) for all t D θ t 0, r 0 ). z β β! 0;σ 0

30 38 S. Malek and C. Stenger Proof. From Proposition 4.1, the formal series 4.10) Ut, z, T ) = β 0 l 0 n 0 0 t n 0 u n0,l,β is solution of the auxiliary partial differential equation 4.11) z S Ut, z, T ) = B q t, z, T ) q 0 t q 1 T q z Ut, z, T ) where and q=q 0,q 1,q ) Q Q = { q = q 0, q 1, q ) N 3 k = k 0, k 1 ) J ; m = 0,..., k 0 ; p P m); B q t, z, T ) = k 0! q 0! 1 p 1!... p m! D m β 0 j=0 n 0 n 0! T l z β β! q 0 = k 0 m; q 1 = p 1 ; q = k 1 } t n A n,j,β n! The formal series Ut, z, T ) satisfies also the initial conditions β z Ut, 0, T ) = l 0 n 0 0 t n 0 u n0,l,β n 0! T l =: ψ β t, T ) T j z β j! β!. for all β = 0,..., S 1. From the assumptions A5), we get that the series ψ β t, T ) belong to the Banach space G q δ 1,0, δ,0 ; σ 0 ), for all β = 0,..., S 1, for some δ,0 > 0. Moreover, the equation 4.11) satisfies the assumptions A1) and A). From the assumption A4) one can easily check that the equation 4.11) also satisfies the condition A3). We get that all the assumptions of Theorem.14 are satisfied and we deduce the existence of δ 1, δ and σ > 0 such that the formal series Ut, z, T ) belongs to G q δ 1, δ ; σ) for b, q given in the assumption A4). Now, consider the formal series in powers of T and z W t, z, T ) = X l,β t) T l z β β! β 0 l 0 where X l,β t)) l 0,β 0 is the sequence constructed in Proposition 3.. We have from Proposition.3 and 4.9), that there exist a positive constant c 1 such that, for T C, t D and z C, z δδ, where δ ]0, 1[ is given, sup sup W t, z, T ) t D β 0 l 0 z δδ v 0,l,β T l 1 β! ) β δδ U 0, δδ ), T c 1 1 δ) T ) m expσζb) T q ) 4.19)

31 On Complex Singularity Analysis of Holomorphic Solutions of Linear PDEs 39 So that W t, z, T ) defines a holomorphic function on D with respect to the variable t, on D 0, δδ ) with respect to the variable z and an entire function with respect to the variable T with at most an exponential growth of order q. We deduce that the function Xt, z) = W t, z, φt)) solution of 4.17) defines a holomorphic function on D θ t 0, r 0 ) D 0, δδ ). On the other hand, from the expansion 4.1), there exist two constants m 1, m > 0 such that m 1 t t 0 n 0/µ φt) m t t 0 n 0/µ, 4.0) for all t D θ t 0, r 0 ). In conclusion, we deduce from 4.19) and 4.0) that the estimates 4.18) hold, if r 0 > 0 is small enough. References [1] W. Balser, From divergent power series to analytic functions. Theory and application of multisummable power series. Lecture Notes in Mathematics, 158. Springer-Verlag, Berlin, x+108 pp. [] W. Balser, Formal power series and linear systems of meromorphic ordinary differential equations, Springer-Verlag, New-York, 000. [3] Y. Chen, Q. Wang, A unified rational expansion method to construct a series of explicit exact solutions to nonlinear evolution equations. Appl. Math. Comput ), no. 1, [4] R. Conte, M. Musette, Link between solitary waves and projective Riccati equations. J. Phys. A 5 199), no. 1, [5] O. Costin, Asymptotics and Borel summability. Chapman & Hall/CRC Monographs and Surveys in Pure and Applied Mathematics, 141. CRC Press, Boca Raton, FL, 009. xiv+50 pp. [6] O. Costin, R. Costin, On the formation of singularities of solutions of nonlinear differential systems in antistokes directions. Invent. Math ), no. 3, [7] J. Dubinskii, Analytic pseudo-differential operators and their applications. Translated from the Russian. Mathematics and its Applications Soviet Series), 68. Kluwer Academic Publishers Group, Dordrecht, xii+5 pp.

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