Coupling of Two Partial Differential Equations and its Application

Transcription

1 Publ. RIMS, Kyoto Univ , Coupling of Two Partial Differential Equations and its Application By Hidetoshi Tahara Abstract The paper considers the following two partial differential equations A u = F t, x, u, u and B w = G t, x, w, w in the complex domain, and gives an answer to the following question: when can we say that the two equations A and B are equivalent? or when can we transform A into B or B into A? The discussion is done by considering the coupling of two equations A and B, and by solving their coupling equation. The most important fact is that the coupling equation has infinitely many variables and so the meaning of the solution is not so trivial. The result is applied to the problem of analytic continuation of the solution. 1. Introduction In this paper, I will present a new approach to the study of nonlinear partial differential equations in the complex domain. Since the research is still in the first stage, as a model study I will discuss only the following two partial differential equations u A = F t, x, u, u where t, x C 2 are variables and u = ut, x is the unknown function and Communicated by T. Kawai. Received May 29, Mathematics Subject Classifications: Primary 35A22; Secondary 35A10, 35A20. Key words: Coupling equation, equivalence of two PDEs, analytic continuation. This research was partially supported by the Grant-in-Aid for Scientific Research No of Japan Society for the Promotion of Science. Department of Mathematics, Sophia University, Kioicho, Chiyoda-ku, Tokyo , Japan. h-tahara@hoffman.cc.sophia.ac.jp c 2007 Research Institute for Mathematical Sciences, Kyoto University. All rights reserved.

2 536 Hidetoshi Tahara B w = G t, x, w, w where t, x C 2 are variables and w = wt, x is the unknown function. For simplicity we suppose that F t, x, u 0,u 1 resp. Gt, x, w 0,w 1 is a holomorphic function defined in a neighborhood of the origin of C t C x C u0 C u1 resp. C t C x C w0 C w1. My basic question is Question. When can we say that the two equations A and B are equivalent? or when can we transform A into B or B into A? One way to treat this question is to consider the coupling of A and B, and to solve their coupling equation. The coupling of two partial differential equations A and B means that we consider the following partial differential equation with infinitely many variables t, x, u 0,u 1,... Φ φ + D m [F ]t, x, u 0,...,u m+1 φ = G t, x, φ, D[φ] u m m 0 where φ = φt, x, u 0,u 1,... is the unknown function, or the following partial differential equation with infinitely many variables t, x, w 0,w 1,... Ψ ψ + D m [G]t, x, w 0,...,w m+1 ψ = F t, x, ψ, D[ψ] w m m 0 where ψ = ψt, x, w 0,w 1,... is the unknown function. In the equation Φ resp. Ψ, the notation D means the following vector field with infinite many variables D = + u i+1 resp. D = u i + w i+1, w i and D m [F ]m =0, 1, 2,... are defined by D 0 [F ]=F, D[F ]=DF, D 2 [F ]= DD[F ], D 3 [F ]=DD 2 [F ] and so on. These two equations Φ and Ψ are called the coupling equations of A and B. In Sections 3 and 4, I will explain the meaning of the coupling of A and B, solve their coupling equations, and establish the equivalence of A and B under suitable conditions. Way of solving Φ and Ψ is as follows: first we

3 Coupling of Two PDEs 537 get a formal solution and then we prove the convergence of the formal solution. Since the solution has infinitely many variables, the meaning of the convergence is not so trivial. In the last Section 5, I will give an application to the problem of analytic continuation of the solution. The result is just the same as in Kobayashi [4] and Lope-Tahara [5]: this shows the effectiveness of our new approach in this paper. In the case of ordinary differential equations, the theory of the coupling of two differential equations was discussed in Section 4.1 of Gérard-Tahara [1]; it will be surveyed in the next Section 2. The results in this paper were already announced in Tahara [9]. 2. Coupling of Two Ordinary Differential Equations Before the discussion in partial differential equations, let us give a brief survey on the coupling of two ordinary differential equations in [Section 4.1 of Gérard-Tahara [1]]. First, let us consider the following two ordinary differential equations: du a = ft, u, dt dw b = gt, w, dt where ft, uresp. gt, w is a holomorphic function defined in a neighborhood of the origin of C t C u resp. C t C w. Definition 2.1. The coupling of a and b means that we consider the following partial differential equation 2.1 or 2.2: 2.1 φ + ft, u φ = gt, φ u where t, u are variables and φ = φt, u is the unknown function, or 2.2 ψ ψ + gt, w = ft, ψ w where t, w are variables and ψ = ψt, w is the unknown function. We call 2.1 or 2.2 the coupling equation of a and b. The convenience of considering the coupling equation lies in the following proposition.

4 538 Hidetoshi Tahara Proposition Let φt, u be a solution of 2.1. If ut is a solution of a then wt =φt, ut is a solution of b. 2 Let ψt, w be a solution of 2.2. If wt is a solution of b then ut =ψt, wt is a solution of a. Proof. We will prove only 1. Let φt, u be a solution of 2.1 and let ut be a solution of a. Set wt =φt, ut. Then we have dwt dt = φ = d dt φt, ut = φ φ t, ut + t, utdut u dt t, ut + ft, ut φt, ut = gt, φt, ut = gt, wt. u This shows that wt is a solution of b. Next, let us give a relation between two coupling equations 2.1 and 2.2. We have Proposition Let φt, u be a solution of 2.1 and suppose that the relation w = φt, u is equivalent to u = ψt, w for some function ψt, w; then ψt, w is a solution of Let ψt, w be a solution of 2.2 and suppose that the relation u = ψt, w is equivalent to w = φt, u for some function φt, u; then φt, u is a solution of 2.1. Proof. We will show only the part 1. Since w = φt, u isequivalentto u = ψt, w, we get u ψt, φt, u. By derivating this with respect to t and u we get 0 ψ 1 ψ w By using these relations we have ψ ψ + gt, w w = ψ w t, φt, u + ψ w t, φt, u φt, u. u w=φt,u = ψ t, φt, u φ t, φt, u φt, u, ψ t, φt, u + gt, φt, u t, φt, u w ψ t, u+gt, φt, u t, φt, u w = ψ w t, φt, u φ t, u+gt, φt, u = ψ t, φt, uft, u φt, u =ft, u. w u

5 Coupling of Two PDEs 539 Since w = φt, u isequivalenttou = ψt, w, we obtain ψ this proves the result 1. ψ + gt, w = ft, ψt, w; w Now, let us discuss the equivalence of two differential equations. Let ft, z and gt, z be holomorphic functions of t, z in a neighborhood of 0, 0 C t C z as before, and let us consider the following two equations: [a] du = ft, u, dt ut 0 as t 0, [b] dw = gt, w, dt wt 0 as t 0. Denote by S a resp. S b the set of all the holomorphic solutions of [a] resp. [b] in a suitable sectorial neighborhood of t = 0. Letφt, u be a holomorphic solution of 2.1 satisfying φ0, 0 = 0: if ut S a we have φt, ut φ0, 0 = 0 as t 0, and therefore the mapping Φ : S a ut φt, ut S b is well defined. Hence, if Φ : S a S b is bijective, solving [a] is equivalent to solving [b]. Thus: Definition 2.4. If Φ : S a S b is well defined and bijective, we say that the two equations [a] and [b] are equivalent. The following result gives a sufficient condition for Φ to be bijective. Theorem 2.5. If the coupling equation 2.1 has a holomorphic solution φt, u in a neighborhood of 0, 0 C t C u satisfying φ0, 0 = 0 and φ/ u0, 0 0, then the mapping Φ is bijective and so the two equations [a] and [b] are equivalent. Proof. Let us show that the mapping Φ : S a S b is bijective. Since φ/ u0, 0 0, by the implicit function theorem we can solve the equation w = φt, u with respect to u and obtain u = ψt, w for some holomorphic function ψt, w satisfying ψ0, 0 = 0. Moreover we know that this ψt, w is a solution of 2.2. Therefore, by 2 of Proposition 2.2 we can define the mapping Ψ : S b wt ψt, wt S a.

6 540 Hidetoshi Tahara Since w = φt, uisequivalenttou = ψt, w, wt =φt, ut is also equivalent to ut =ψt, wt; this implies that wt =Φut is equivalent to ut = Ψwt. Thus we can conclude that Φ : S a S b is bijective and Ψ is the inverse mapping of Φ. In the application, we will regard 2.3 as an original equation and 2.4 as a reduced equation. Then, what we must do is to find a good reduced form 2.4 so that the following 1 and 2 are satisfied: is equivalent to 2.4, and 2 we can solve 2.4 directly. In a sense, 2.4 must be the best partner, and our coupling equation is a tool to find its best partner. The naming coupling comes from such a meaning. In Gérard-Tahara [1], one can see many concrete applications of this theory to the reduction problem of Briot-Bouquet s differential equation. 3. Coupling of Two Partial Differential Equations Now, let us generalize the theory in Section 2 to partial differential equations. In this section we will give only a formal theory, and in the next section we will give a substantial meaning to the formal theory. Let us consider the following two nonlinear partial differential equations: A u = F t, x, u, u where t, x C 2 are variables and u = ut, x is the unknown function and B w = G t, x, w, w where t, x C 2 are variables and w = wt, x is the unknown function. For simplicity we suppose that F t, x, u 0,u 1 resp. Gt, x, w 0,w 1 is a holomorphic function defined in a neighborhood of the origin of C t C x C u0 C u1 resp. C t C x C w0 C w1. For a function φ = φt, x, u 0,u 1,... with respect to the infinitely many variables t, x, u 0,u 1,... we define D[φ]t, x, u 0,u 1,...by D[φ] = φ t, x, u 0,u 1,...+ φ u i+1 t, x, u 0,u 1,... u i For m 2 we define D m [φ] as follows: D 2 [φ] =D [D[φ]], D 3 [φ] =D [ D 2 [φ] ], and so on. If φ is a function with p +3-variables t, x, u 0,...,u p thend m [φ]

7 Coupling of Two PDEs 541 is a function with p + m +3-variablest, x, u 0,...,u p+m. Of course, if ψ = ψt, x, w 0,w 1,... is a function with respect to the variables t, x, w 0,w 1,... the notation D[ψ]t, x, w 0,w 1,...isreadas D[ψ] = ψ t, x, w 0,w 1,...+ ψ w i+1 t, x, w 0,w 1,... w i We can regard D as a vector field with infinitely many variables x, u 0,u 1,... resp. x, w 0,w 1,...: D = + u i+1 resp. D = u i + w i+1. w i This operator D comes from the following formula: if Kt, x, u 0,u 1,... is a function with infinitely many variables t, x, u 0,u 1,...andifux isa holomorphic function, then under the relation u i = / i u i =0, 1, 2,... we have K t, x, u, u,... = K + K u 1 + K u 2 + = D[K]. u 0 u 1 Therefore, for any m N we have D m [K] t, x, u, u m [,... = K t, x, u, u ],.... Definition The coupling of two partial differential equations A and B means that we consider the following partial differential equation with infinitely many variables t, x, u 0,u 1,... Φ φ + D m [F ]t, x, u 0,...,u m+1 φ = G t, x, φ, D[φ] u m m 0 where φ = φt, x, u 0,u 1,... is the unknown function, or the following partial differential equation with infinitely many variables t, x, w 0,w 1,... Ψ ψ + D m [G]t, x, w 0,...,w m+1 ψ = F t, x, ψ, D[ψ] w m m 0 where ψ = ψt, x, w 0,w 1,... is the unknown function. We call Φ or Ψ the coupling equation of A and B.

8 542 Hidetoshi Tahara 3.1. The formal meaning of the coupling equation In this section we will explain the meaning of the coupling equations Φ and Ψ in the formal sense. Here, in the formal sense means that the result is true if the formal calculation makes sense. For simplicity we write φt,x,u, u/,...=φ t, x, u, u, 2 u u 2,..., n n,.... The convenience of considering the coupling equation lies in the following proposition. Proposition If φt, x, u 0,u 1,... is a solution of Φ and if ut, x is a solution of A, then the function wt, x =φt,x,u, u/,... is asolutionofb. 2 If ψt, x, w 0,w 1,... is a solution of Ψ and if wt, x is a solution of B, then the function ut, x =ψt,x,w, w/,... is a solution of A. Proof. We will show only 1. Let φt, x, u 0,u 1,... be a solution Φ and let ut, x be a solution of A. Set u i t, x = / i ut, x i =0, 1, 2,...: we have wt, x =φt,x,u, u/,...=φt, x, u 0,u 1,...and w/ = D[φ]t, x, u 0,u 1,... Therefore we have w = φ + = φ + = φ + φ u i u i = φ + φ [ i F u i φ u i ] t, x, u, u φ u i D i [F ]t, x, u 0,...,u i+1 = G t, x, φ, D[φ] = G This shows that wt, x is a solution of B. t, x, w, w. [ ] i u In order to state the relation between Φ and Ψ, let us introduce the notion of the reversibility of φt, x, u 0,u 1,... Definition Let φt, x, u 0,u 1,... be a function in t, x, u 0,u 1,... We say that the relation w = φt,x,u, u/,... is reversible with respect to u and w if there is a function ψt, x, w 0,w 1,...int, x, w 0,w 1,... such that

9 Coupling of Two PDEs 543 the relation is equivalent to w 0 = φt, x, u 0,u 1,u 2,..., w 1 = D[φ]t, x, u 0,u 1,u 2,..., w 2 = D 2 [φ]t, x, u 0,u 1,u 2,..., u 0 = ψt, x, w 0,w 1,w 2,..., u 1 = D[ψ]t, x, w 0,w 1,w 2,..., u 2 = D 2 [ψ]t, x, w 0,w 1,w 2,...,. In this case the function ψt, x, w 0,w 1,... is called the reverse function of φt, x, u 0,u 1,... By the definition, we see Lemma The reverse function ψt, x, w 0,w 1,... of φt, x, u 0,u 1,... is unique, if it exists. Proof. Let ψ 1 t, x, w 0,w 1,... be another reverse function of φt, x, u 0,u 1,... Then the system u 0 = ψ 1 t, x, w 0,w 1,w 2,..., u 1 = D[ψ 1 ]t, x, w 0,w 1,w 2,..., u 2 = D 2 [ψ 1 ]t, x, w 0,w 1,w 2,...,. is equivalent to 3.1.1, and so and are equivalent; this means that the equality ψt, x, w 0,w 1,w 2,...=ψ 1 t, x, w 0,w 1,w 2,... holds as functions with respect to t, x, w 0,w 1,w 2,... The following proposition gives the relation between two coupling equations Φ and Ψ: we can say that the equation Ψ is the reverse of Φ.

10 544 Hidetoshi Tahara Proposition If φt, x, u 0,u 1,... is a solution of Φ and if the relation w = φt,x,u, u/,... is reversible with respect to u and w, then the reverse function ψt, x, w 0,w 1,... is a solution of Ψ. To prove this we need the following lemma: Lemma For any functions φt, x, u 0,u 1,... and Ht, x, u 0, u 1,... we have [ ] φ D k D i [H] = D k [φ] D i [H], k =0, 1, 2,... u i u i 2 For a function Kt, x, w 0,w 1,... with respect to the variables t, x, w 0,w 1,... we write K φ = Kt, x, φ, D[φ],... which is a function with respect to the variables t, x, u 0,u 1,... We have D k[ ] K φ = k Dw [K], k =0, 1, 2,..., φ where D w isthesameoperatorasd with u 0,u 1,... replaced by w 0,w 1,... 3 If φt, x, u 0,u 1,... is a solution of the coupling equation Φ, we have D k [φ] + D i [F ] Dk [φ] = D k w [G], k =0, 1, 2,... u φ i as a function with respect to the variable t, x, u 0,u 1,... Proof. Let us show 1. By the definition of the operator D we have D[φ] = φ u i u i + u φ φ u i + u 0 u i 1 [ ] φ = D + φ for i =0, 1, 2,... u i u i 1 where φ/ u 1 = 0; therefore we obtain [ ] φ D D i [H] = [ φ ] D D i [H]+ φ D i+1 [H] u i u i u i = D[φ] φ D i [H]+ φ D i+1 [H] u i u i 1 u i = = D[φ] u i D[φ] u i D i [H] i 1 D i [H]. φ D i [H]+ φ D i+1 [H] u i 1 u i

11 Coupling of Two PDEs 545 This proves the equality in the case k = 1. The general case k 2can be proved by induction on k. Next let us show 2. The case k =1isverifiedby D [ K φ ] = K φ = = = = + K + K φ j 0 u i+1 K φ u i w j φ K + K φ j 0 w j φ K + K φ w j φ Dj+1 [φ] j 0 K + j 0 K w j+1 w j D j [φ] + K u i+1 j 0 Dj [φ] + D j [φ] u i+1 u i φ =D w[k] φ. w j φ D j [φ] u i The general case k 2 can be proved by induction on k. Lastly let us show 3. Since φt, x, u 0,u 1,... satisfies the equation Φ, by applying D k to Φ and by using the relation D k [G φ ]=D k w [G] φ we have D k [φ] + [ D k D i [F ] φ ] u i = D w k [G] φ. Therefore, to prove the equality we have only to notice the following equality: [ D k D i [F ] φ ] = D i [F ] Dk [φ], k =0, 1, 2,... u i u i which is already proved in the part 1. Proof of Proposition Since and are equivalent, we have the equality u 0 = ψ t, x, φ, D[φ],D 2 [φ],... as a function with respect to the variables t, x, u 0,u 1,u 2,... Therefore, by applying / u i to we have ψ j 0 w j φ D j [φ] u i = { 1, if i =0, 0, if i 1.

12 546 Hidetoshi Tahara Also, by applying / to and then by using and we have 0= = = = ψ + ψ φ j 0 D w j φ j [φ] w j φ D i [F ] Dj [φ] +D j w [G] φ u i ψ + ψ φ w D w j [G] φ D i [F ] ψ D j φ w j φ j [φ] u i j 0 j 0 ψ + ψ φ j 0 ψ + ψ φ w D w j [G] φ F. j φ j 0 Hence, as a function with respect to the variables t, x, u 0,u 1,u 1,...wehave the equality ψ + ψ φ w D w j [G] φ = F t, x, u 0,u 1. j φ j 0 Since and are equivalent, we can regard this equality as a function with respect to the variables t, x, w 0,w 1,...andobtain ψ + ψ D j [G] =F t, x, ψ, D[ψ]. w j j 0 This proves that ψt, x, w 0,w 1,... is a solution of the equation Ψ Equivalence of A and B Let F and G be function spaces in which we can consider the following two partial differential equations: [A] u = F t, x, u, u in F, [B] w = G t, x, w, w in G. Set S A = the set of all solutions of [A] in F, S B = the set of all solutions of [B] in G.

13 Coupling of Two PDEs 547 Then, if the coupling equation Φ has a solution φt, x, u 0,u 1...andifφt, x, u, u/,... G is well defined for any u S A, we can define the mapping Φ : S A ut, x wt, x =φt,x,u, u/,... S B. If the relation w = φt,x,u, u/,... is reversible with respect to u and w, and if the reverse function ψt, x, w 0,w 1,...satisfiesψt,x,w, w/,... F for any w S B, we can also define the mapping Ψ : S B wt, x ut, x =ψt,x,w, w/,... S A. Set wt, x =φt,x,u, u/,...; then by the definition of D we have w = φt,x,u, u/,..., w/ = D[φ]t,x,u, u/,..., w/ 2 = D 2 [φ]t,x,u, u/,...,. Similarly, if we set ut, x = ψt,x,w, w/,... wehave u = φt,x,w, w/,..., u/ = D[φ]t,x,w, w/,..., u/ 2 = D 2 [φ]t,x,w, w/,...,. Since is equivalent to we have the equivalence between and 3.2.4; therefore we have Ψ Φ=identity in S A and Φ Ψ=identity in S B.Thus,weobtain Theorem Suppose that the coupling equation Φ has a solution φt, x, u 0,u 1... and that the relation w = φt,x,u, u/,... is reversible with respect to u and w. If both mappings and are well defined, we can conclude that the both mappings are bijective and that one is the inverse of the other. By this theorem, we may say: Definition If the coupling equation Φ resp. Ψ has a solution φt, x, u 0,u 1... resp. ψt, x, w 0,w 1... and if the relation w = φt,x,u, u/,...resp. u = ψt,x,w, w/,... is reversible with respect to u and w or w and u, then we say that the two equations A and B are formally equivalent.

14 548 Hidetoshi Tahara 2 In addition, if both mappings and are well defined, we say that the two equations [A] and [B] are equivalent. In the application, we will regard [A] as an original equation and [B] as a reduced equation. Then, what we must do is to find a good reduced form [B] so that the following 1 and 2 are satisfied: 1 [A] is equivalent to [B], and 2 we can solve [B] concretely A Sufficient condition for the reversibility As is seen above, the condition of the reversibility of φt, x, u 0,u 1,...is very important. In this section we will give a sufficient condition for the relation w = φt,x,u, u/,... to be reversible with respect to u and w. Let us introduce the notations: D R = {x C ; x R}, O R denotes the ring of holomorphic functions in a neighborhood of D R,andO R [[u 0,...,u p ]] denotes the ring of formal power series in u 0,...,u p with coefficients in O R. Proposition If φt, x, u 0,u 1,... is of the form φ = u 0 + φ k x, u 0,...,u k t k k 0 O R [[u 0,...,u k ]] t k, the relation w = φt,x,u, u/,... is reversible with respect to u and w, and the reverse function ψt, x, w 0,w 1,... is also of the form ψ = w 0 + ψ k x, w 0,...,w k t k k 0 O R [[w 0,...,w k ]] t k. To prove this, let us consider the following equation with respect to the unknown function ψ = ψt, x, w 0,w 1,...: w 0 = φ t, x, ψ, D[ψ],D 2 [ψ],... in k 0 O R [[w 0,...,w k ]]t k. We have Lemma Let φt, x, u 0,u 1,... be of the form Then, the equation has a unique solution ψt, x, w 0,w 1,... of the form and it satisfies also u 0 = ψ t, x, φ, D[φ],D 2 [φ],... in k 0 O R [[u 0,...,u k ]]t k.

15 Coupling of Two PDEs 549 Proof. Since the equation is written in the form w 0 = ψ + φ k x, ψ, D[ψ],...,D k [ψ] t k, under the expression ψ = ψ it i we have w 0 = ψ i t i + φ k x, ψ i t i,..., D k [ψ i ]t i t k. By setting t =0wehaveψ 0 = w 0, and so this equation is reduced to the form ψ i t i = φ k x, ψ i t i,..., D k [ψ i ]t i t k. i 1 Let us look at the coefficients of t in the both sides of 3.3.5; we have ψ 1 = φ 1 x, ψ 0,D[ψ 0 ] = φ 1 x, w 0,D[w 0 ] = φ 1 x, w 0,w 1 in the second equality we used the fact ψ 0 = w 0. In general, for p 1 we consider the equation in the modulo class Ot p+1 ; then we have p p p k p k p ψ i t i φ k x, ψ i t i,..., D k [ψ i ]t i t k mod Ot p+1. i=1 k=1 i=0 Note that only the terms ψ 0 = w 0,ψ 1,...,ψ p 1 appear in the right hand side of p ; therefore, if ψ 0 = w 0,ψ 1,...,ψ p 1 are known, by looking at the coefficients of t p in the both sides of p we can uniquely determine ψ p. Moreover, if ψ i has the form ψ i x, w 0,...,w i fori =1,...,p 1, we have the result that ψ p also has the form ψ p x, w 0,...,w p. Thus, we have proved that the equation has a unique solution and it has the form Next, let us show that the above solution ψt, x, w 0,w 1,... satisfies the equation To do so, it is enough to prove that p u 0 i=0 k=0 k=0 i=0 p p i p i ψ i x, φ k t k,..., D i [φ k ]t k t i mod Ot p+1 with φ 0 = u 0 and ψ 0 = w 0 holdsforallp =0, 1, 2,... Let us show this by induction on p. Our conditions are: ψ 0 = w 0, ψ 1 = φ 1 x, w 0,w 1 andthe relations p for p 1. When p = 0, the relation is nothing but the equality u 0 = ψ 0 x, φ 0 = φ 0.Whenp = 1, the right hand side of is ψ 0 x, φ 0 + φ 1 t+ψ 1 x, φ 0,D[φ 0 ]t = φ 0 + φ 1 t + ψ 1 x, u 0,u 1 t = u 0 +φ 1 x, u 0,u 1 +ψ 1 x, u 0,u 1 t = u 0

16 550 Hidetoshi Tahara which is verified by φ 0 = u 0 and the definition of ψ 1 x, u 0,u 1 : this proves Let p 2 and suppose that p 1 is already proved: then we have p 1 u 0 i=0 p 1 i=0 p 1 i ψ i x, k=0 p 1 ψ i x, k=0 p 1 i φ k t k,..., p 1 φ k t k,..., k=0 k=0 D i [φ k ]t k t i mod Ot p D i [φ k ]t k t i mod Ot p. Since D j [u 0 ]=u j holds, by applying D j to the both sides of the above equality and by using 2 of Lemma we have p 1 u j i=0 D j [ψ i ] for j =0, 1, 2,... p 1 x, k=0 p 1 φ k t k,..., k=0 D i+j [φ k ]t k t i mod Ot p Since p is a known relation we have p ψ i x, w 0,...,w i t i i=1 p k=1 p 1 p 1 φ k x, ψ i t i,..., D k [ψ i ]t i t k mod Ot p+1 i=0 i=0 as functions with respect to the variables t, x, w 0,w 1,... By substituting p 1 w i = D i [φ k ]x, u 0,...,u k+i t k, i =0, 1, 2,... k=0 into the both sides of we have the relation p i=1 p 1 p 1 ψ i x, φ k t k,..., D i [φ k ]t k t i p k=1 k=0 mod Ot p+1 p 1 φ k x, i=0 p 1 k=0 p 1 ψ i x, k=0 p 1 D k [ψ i ] x, i=0 p 1 φ k t k,..., k=0 k=0 p 1 φ k t k,..., D i [φ k ]t k t i,..., k=0 D i+k [φ k ]t k t i t k

17 Coupling of Two PDEs 551 as functions with respect to the variables t, x, u 0,u 1,... Therefore, by applying to the right hand side of we obtain p i=1 p 1 p 1 ψ i x, φ k t k,..., D i [φ k ]t k t i k=0 k=0 p φ k x, u0 + Ot p,...,u k + Ot p t k mod Ot p+1 k=1 p φ k x, u0,...,u k t k k=1 which immediately leads us to p. Thus, Lemma is proved. mod Ot p+1 Proof of Proposition Let ψt, x, w 0,w 1,... be the solution of in Lemma Let us show the equivalence between the relations u j = D j [ψ]t, x, w 0,w 1,..., j =0, 1, 2,... and w j = D j [φ]t, x, u 0,u 1,..., j =0, 1, 2,... Let us apply D j to the both sides of the equality 3.3.3; by 2 of Lemma we have w j = D j [φ]t, x, ψ, D[ψ],D 2 [ψ],..., j =0, 1, 2,... and therefore under the relation we can get the relation Similarly, by applying D j to the both sides of the equality we have u j = D j [ψ]t, x, φ, D[φ],D 2 [φ],..., j =0, 1, 2,... and therefore under the relation we can get the relation This proves Proposition Composition of two couplings Let us consider three partial differential equations A, B and C z = H t, x, z, z

18 552 Hidetoshi Tahara where t, x C 2 are variables and z = zt, x is the unknown function. The coupling equation A and B is φ D i [F ]t, x, u 0,...,u i+1 φ = G t, x, φ, D[φ] u i and the coupling equation of B and C is η D i [G]t, x, w 0,...,w i+1 η = H t, x, η, D[η] w i where η = ηt, x, w 0,w 1,... is the unknown function. We have Proposition Let φt, x, u 0,u 1,u 2,... be a solution of 3.4.1, and let ηt, x, w 0,w 1,w 2,... be a solution of Then, the composition ζ = ηt, x, φ, D[φ],D 2 [φ],... is a solution of ζ D i [F ]t, x, u 0,...,u i+1 ζ = H t, x, ζ, D[ζ] u i which is the coupling equation of A and C. Proof. Set ζt, x, u 0,u 1,u 2,...=ηt, x, φ, D[φ],D 2 [φ],...=η φ. Then, by Lemma 3.1.5, and we have ζ + D i [F ] ζ u i η = + η φ w j φ j 0 η = + η φ w j φ j 0 = = = η + η φ w j φ Dj j 0 η + η φ w j φ Dj j 0 D j [φ] + D i [F ] η j 0 [ D j [φ] + ] D i [F ] Dj [φ] u i [ φ + D i [F ] φ ] u i [ ] Gt, x, φ, D[φ] w j φ η + η φ w j φ Dj [G]t, x, φ, D[φ],...,D 1+j [φ] j 0 D j [φ] u i

19 Coupling of Two PDEs 553 η = + η D j [G] w j φ = Ht, x, η, D[η] φ j 0 = Ht, x, η φ,d[η] φ =Ht, x, η φ,d[η φ ] = Ht, x, ζ, D[ζ]. This proves Proposition By combining this with Definition 3.2.2, we can easily see Proposition If A and B are formally equivalent resp. equivalent, andifb and C are formally equivalent resp. equivalent, thena and C are also formally equivalent resp. equivalent. As an application of Proposition 3.4.1, we will give another criterion of the reversibility of φt, x, u 0,u 1,... In the context of Section 3.1, let φt, x, u 0,u 1,... be a solution of Φ, and let ψt, x, w 0,w 1,... be a solution of Ψ. Then, by Proposition we see that the composition ξ = ψt, x, φ, D[φ],D 2 [φ],... is a solution of ξ D i [F ]t, x, u 0,...,u i+1 ξ = F t, x, ξ, D[ξ], u i and the composition η = φt, x, ψ, D[ψ],D 2 [ψ],... is a solution of η + D i [G]t, x, w 0,...,w i+1 η = G t, x, η, D[η]. w i It is easy to see that the equation has a solution ξ = u 0 ; therefore, if the solution of ξ D m [F ] ξ = F t, x, ξ, D[ξ], ξ = u 0 u m t=0 m 0 is unique and if ψt, x, φ, D[φ],... t=0 = u 0 holds, we have u 0 = ψt, x, φ, D[φ],... Similarly, if the uniqueness of the solution of η + D m [G] η = Gt, x, η, D[η], u m m 0 η = w 0 t=0 is valid and if φt, x, ψ, D[ψ],... t=0 = w 0 holds, we have w 0 = φt, x, ψ, D[ψ],... Thus, we obtain

20 554 Hidetoshi Tahara Proposition Let φt, x, u 0,u 1,... be a solution of Φ with φ t=0 = u 0, and let ψt, x, w 0,w 1,... be a solution of Ψ with ψ t=0 = w 0. If the uniqueness of the solution is valid in two Cauchy problems and 3.4.7, then we can conclude that the relation w = φt,x,u, u/,... is reversible with respect to u and w, andψt, x, w 0,w 1,... is the reverse function of φt, x, u 0,u 1, Equivalence of Two Partial Differential Equations In this section, we will give a concrete meaning to the formal theory in Section 3 and establish the equivalence of two partial differential equations. Let t, x C t C x be the variables, and let F t, x, z 1,z 2 be a holomorphic function defined in a neighborhood of the origin of C t C x C z1 C z2. Let us consider the following partial differential equation u = F t, x, u, u and let us seek for a reduction to a simple form. As is seen in the case of ordinary differential equations in [1], it will be reasonable to treat the equation w =0 as a candidate of the reduced form of In order to justify this assertion, it is enough to discuss the following coupling equation Φ φ + D m [F ]t, x, u 0,...,u m+1 φ =0, u m m 0 or Ψ ψ = F t, x, ψ, D[ψ]. In this section we denote by RC \{0} the universal covering space of C \{0}, and we write S θ = {t RC \{0}; arg t <θ} and S θ r ={t RC \ {0}; 0< t r, arg t <θ} for θ>0andr> Formal solutions of Φ and Ψ First, let us look for a formal solution of the coupling equation Φ in the form φ = u 0 + φ k x, u 0,...,u k t k O R [[u 0,...,u k ]] t k. k 0 We have

21 Coupling of Two PDEs 555 Proposition The coupling equation Φ has a unique formal solution of the form Moreover we have the following properties: i φ 1 x, u 0,u 1 = F0,x,u 0,u 1, ii φ k x, u 0,...,u k for k 1 is a holomorphic function in a neighborhood of {x, u 0,...,u k C C k+1 ; x R, u 0 ρ and u 1 ρ} for some R>0 and ρ>0 which are independent of k, and iii φ k x, u 0,...,u k for k 2 is a polynomial with respect to u 2,...,u k. Proof. Set φ 0 = u 0 and F t, x, z 1,z 2 = F i x, z 1,z 2 t i. By substituting into the equation Φ we have kφ k t k 1 + D m [F i ]x, u 0,...,u m+1 φ j t i+j =0. u m j 0 0 m j If we set t = 0 in the above equality we see φ 1 = F 0 x, u 0,u 1 φ 0 = F 0 x, u 0,u 1 = F0,x,u 0,u 1, u 0 and also by looking at the coefficients of t k with k 1 we have the following recurrent formulas: φ k+1 = 1 k +1 i+j=k 0 m j D m [F i ]x, u 0,...,u m+1 φ j u m. This proves that φ k x, u 0,...,u k k =2, 3,... are determined uniquely by the formula inductively on k. The other conditions follow from and the definition of D. Next, let us look for a formal solution of the coupling equation Ψ in the form ψ = w 0 + ψ k x, w 0,...,w k t k O R [[w 0,...,w k ]] t k. k 0 We have Proposition The coupling equation Ψ has a unique formal solution of the form Moreover we have the following properties: i ψ 1 x, w 0,w 1 =F0,x,w 0,w 1, ii ψ k x, w 0,...,w k for k 1 is a holomorphic function in a neighborhood of {x, w 0,...,w k C C k+1 ; x R, w 0 ρ and w 1 ρ} for some R>0 and ρ>0 which are independent of k, and iii ψ k x, w 0,...,w k for k 2 is a polynomial with respect to w 2,...,w k.

22 556 Hidetoshi Tahara Proof. Set ψ 0 = w 0 and F t, x, w 0 + Y,w 1 + Z = i,j,α a i,j,α x, w 0,w 1 t i Y j Z α. Then, by substituting into the equation Ψ and by comparing the coefficients of t k in the both sides of the equation Ψ we have the result that ψ k x, w 0,...,w k k =1, 2,... are determined uniquely by the following recurrent formulas: ψ 1 x, w 0,w 1 =a 0,0,0 x, w 0,w 1 =F 0,x,w 0,w 1 and ψ k+1 x, w 0,...,w k+1 = 1 k +1 l + m =k i a i,j,α x, w 0,w 1 1 i+j+α k ψ l1 ψ lj D[ψ m1 ] D[ψ mα ] for k 1, where l = l l j and m = m m α. This proves Proposition Let φt, x, u 0,u 1,... and ψt, x, w 0,w 1,... be the formal solution in Propositions and 4.1.2, respectively. Then, by Proposition we see that the relation w = φt,x,u, u/,... is reversible with respect to u and w and that ψt, x, w 0,w 1,... is the reverse function of φt, x, u 0,u 1,... By Definition we have Theorem equivalent. The two equations and are formally 4.2. Convergence of ψt, x, w 0,w 1,... In this section we will prove the convergence of the formal solution ψ = w 0 + ψ k x, w 0,...,w k t k of the equation Ψ in Proposition For R 1 > 0, ε>0andη>0weset W k R 1,ε,η= { x, w 0,...,w k C C k+1 ; x R 1, w 0 0!ε/η 0, w 1 1!ε/η,..., w k k!ε/η k}

23 Coupling of Two PDEs 557 k =1, 2,... By Proposition and by taking R 1 > 0, ε>0, η>0sothat R 1, ε and ε/η are sufficiently small we may suppose that each ψ k x, w 0,...,w k is a holomorphic function on W k = W k R 1,ε,ηandso We have ψ k Wk =max W k ψ k x, w 0,...,w k <, k =1, 2,... Proposition Let R 1 > 0 be sufficiently small. Then, for any η>0 we can find an ε>0 such that the series ψ k Wk z k with W k = W k R 1,ε,η is convergent in a neighborhood of z =0 C. To prove this result, we suppose: c-1 F t, x, z 1,z 2 is a holomorphic function in a neighborhood of K = {t, x, z 1,z 2 C 4 ; t r, x R 0, z 1 ρ 1, z 2 ρ 2 } for some r>0, R 0 > 0, ρ 1 > 0andρ 2 > 0, and c-2 F t, x, z 1,z 2 M on K. Take any 0 <ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2. Set K 0 = {x, w 0,w 1 C 3 ; x R 0, w 0 ρ 0 1, w 1 ρ 0 2}, andleta i,j,α x, w 0,w 1 i, j, α N 3 be the coefficients in 4.1.5: by Cauchy s inequality we have ai,j,α x, w 0,w 1 M r i ρ 1 ρ 0 1 j ρ 2 ρ 0 2 α on K 0 for any i, j, α N 3. Let wx be a holomorphic function on D R = {x C ; x R} for some R with 0 <R R 0 and suppose the condition w R ρ 0 1 and w ρ 0 2 R where we used the notation By we have w R =max x D R wx ai,j,α x, w, w/ M r i ρ 1 ρ 0 1 j ρ 2 ρ 0 on D R. 2 α The following is the key result on the convergence of ψt, x, w 0,w 1,...

24 558 Hidetoshi Tahara Proposition Suppose the conditions c-1 and c-2. Ifwx is a holomorphic function on D R for some R with 0 <R R 0 and satisfies the condition for some ρ 0 1 and ρ 0 2 with 0 <ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2,then the series ψ k x, w, w w,..., k k z k is convergent at least in the domain { Ω= x, z C 2 ; x <R, z R x < r R +4Mr R/ρ 1 ρ e/ρ 2 ρ 0 2 Moreover, we have the following estimates: ψ k x, w, w w,..., k z k k ρ 1 ρ on Ω, 2 D[ψ k ] x, w, w,..., k+1 w z k k+1 ρ 2 ρ on Ω. 2 Before the proof of this result, we recall Nagumo s lemma: }. Lemma Nagumo s lemma. on {x C ; x <R} and satisfies If fx is a holomorphic function C f s R s a for 0 < s <R for some C 0 and a 0, we have f a +1eC s R s a+1 for 0 < s <R. This was first proved by Nagumo [6] in a more general form; one can see a simple proof in the book [3] Lemma of [3]. In the proof of Proposition this lemma will play an important role. Proof of Proposition By Proposition 4.1.2, we may assume that all the terms ψ k x, w, w w,..., k k, k =1, 2,... are holomorphic functions on D R.

25 Coupling of Two PDEs 559 We will prove the convergence of by the method of majorant series. As an equation of majorant series, we adopt the following analytic functional equation with respect to Y : Y = z Y 0 = 0, i+j+α 0 M R i+jz i r i ρ 1 ρ 0 1 j ρ 2 ρ 0 Y j ey α, 2 α R s R s where s is a parameter with 0 <s<r. By the implicit function theorem we see that has a unique holomorphic solution Y = Y z in a neighborhood of z =0 C. If we expand this into the form Y = Y k z k we easily see that the coefficients Y k k =1, 2,... are determined uniquely by the following recurrent formulas: Y 1 = M and Y k+1 = 1 i+j+α k M R i+j r i ρ 1 ρ 0 1 j ρ 2 ρ 0 2 α R s eym1 eymα Y l1 Y lj R s R s l + m =k i for g 1, where l = l l j and m = m m α. Moreover, by induction on k we can see that Y k has the form Y k = C k, k =1, 2,... R s k 1 where C k 0k =1, 2,... are constants independent of the parameter s. We write Y k = Y k s when we emphasize the fact that Y k depends on the parameter s with 0 <s<r. The following lemma guarantees that Y z = Y ksz k is a majorant series of Lemma For any k =1, 2,... we have: ψ k x, w, w w s k,..., k k Y k s for 0 < s <R, D[ψ k ] x, w, w,..., k+1 w s k+1 ey ks k for 0 < s <R. R s

26 560 Hidetoshi Tahara Proof of Lemma We will prove this by induction on k. Sinceψ 1 is defined by ψ 1 x, w 0,w 1 =F0,x,w 0,w 1, we have ψ 1 x, w, w s F 0,x,w, w R M = Y 1 for 0 < s <R. By Lemma we have D[ψ 1 ] x, w, w, 2 w s 2 = ψ 1 x, w, w s em R s = ey 1 R s for 0 < s <R. This proves and Next, let k 1 and suppose that p and p are already proved for p =1,...,k. Then, by 4.1.6, and the induction hypothesis we have ψ k+1 x, w, w,..., k+1 w s k+1 1 k +1 1 i+j+α k l + m =k i M r i ρ 1 ρ 0 1 j ρ 2 ρ 0 2 α eym1 eymα Y l1 Y lj. R s R s Therefore, by comparing this with and by using R/R s > 1wehave ψ k+1 x, w, w,..., k+1 w s k+1 1 k +1 Y k+1 = 1 C k+1 k +1R s k for 0 < s <R: this yields k+1. Moreover, by applying Lemma to we have D[ψ k+1 ] x, w, w,..., k+2 w s k+2 = ψ k+1 x, w, w,..., k+1 w s k+1 1 k +1eC k+1 k +1 R s k+1 = ec k+1 R s k+1 = ey k+1 for 0 < s <R. R s This proves also k+1. Now, let us prove that the series is convergent in the domain Ω in By Lemma 4.2.4, for any fixed s with 0 <s<rwe have ψ k x, w, w w s,..., k k z k Y k s z k = Y z

27 Coupling of Two PDEs 561 where Y z is the unique solution of Therefore, to prove the convergence of the series on Ω it is sufficient to show that Y z isconvergent on Ω, or equivalently that Y z is holomorphic on Ω. Note that Y z 0andso M R i+j Y = z z i r i ρ 1 ρ 0 i+j+α 0 1 j ρ 2 ρ 0 Y j ey α 2 α R s R s R z i R Y j e Y α = Mz R s r R s ρ 1 ρ 0 1 R s ρ 2 ρ 0 2 i+j+α 0 1 R R s z r Mz 1 R/ρ 1 ρ 0 1 +e/ρ 2 ρ 0 2 R s, Y where i a iz i i b iz i means that a i b i holds for all i. Therefore, if we consider the equation Mz Z = R z 1 R/ρ 1 ρ 0 1+e/ρ 2 ρ 0 2, Z R s r R s Z0 = 0, we see that this equation has a unique holomorphic solution Zz in a neighborhood of z =0 C and we have Zz Y z. Moreover, by solving concretely we have Zz = βMz 2β 1 αz with α = R 1 and β = R/ρ 1 ρ 0 1+e/ρ 2 ρ 0 2. R s r R s Thus, we conclude that Zz is holomorphic in the domain { Ω s = z C ; 4βM z } 1 α z < 1 { } z = z C ; R s < r R +4MrR/ρ 1 ρ 0 1 +e/ρ 2 ρ 0 2. Since Zz Y z holds, we see that Y z is also holomorphic on Ω s and so the series ψ k x, w, w w s,..., k k z k is convergent in the domain Ω s.

28 562 Hidetoshi Tahara Since s can be taken arbitrarily in 0 <s<r, we obtain the result that the series is convergent in the domain } {x, z C 2 ; x s, z Ω s =Ω. 0<s<R This proves the former half of Proposition Let us show the estimates and Since Y z Zz, we have Y z Z z ; therefore, if x = s we have ψ k x, w, w w,..., k z k k Y z Z z = 1 1 2β 1 2β = R s 2R/ρ 1 ρ 0 1 +e/ρ 2 ρ 0 2 R 2R/ρ 1 ρ 0 1 = ρ 1 ρ on Ω; this proves Similarly, we have D [ ] ψ k x, w, w,..., k+1 w z k k+1 ψ k x, w, w 1 4βM z 1 α z w s,..., k k z k e R s Y z e R s Z z = e βM z R s 2β 1 α z e 1 R s 2β = e 2R/ρ 1 ρ 0 1 +e/ρ 2 ρ 0 2 e 2e/ρ 2 ρ 0 2 = ρ 2 ρ on Ω; this proves Thus, all the parts of Proposition are proved. For ax = a ix i we write a x = a i x i 0; for ax, y = i,j a i,jx i y j we write a x, y = i,j a i,j x i y j 0; and so on. Let F t, x, z 1,z 2 be as before, let F t, x, z 1,z 2 be defined as above, and let us consider Ψ + ψ + = F t, x, ψ +,D[ψ + ].

29 Coupling of Two PDEs 563 It is clear by Proposition that Ψ + has a unique formal solution ψ + t, x, w 0,w 1,...oftheform ψ + = w 0 + ψ + k x, w 0,...,w k t k O R [[w 0,...,w k ]] t k k 0 and that ψ + k x, w 0,...,w k 0holdsforallk =1, 2,... By Proposition we have Proposition Suppose that F t, x, z 1,z 2 satisfies the conditions c-1 and c-2 with F and M replaced by F and M +, respectively. If wx is a holomorphic function on D R for some R with 0 <R R 0 and satisfies the condition for some ρ 0 1 and ρ 0 2 with 0 <ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2,then the series ψ + k x, w, w w,..., k k z k is convergent at least in the domain { Ω + = x, z C 2 ; x <R, z R x < r R +4M + r R/ρ 1 ρ 0 1 +e/ρ 2 ρ 0 2 Moreover, we have the following estimates: ψ + k x, w, w w,..., k z k k ρ 1 ρ on Ω +, 2 D[ψ + k ] x, w, w,..., k+1 w z k k+1 ρ 2 ρ on Ω +. 2 By using this proposition, let us give a proof of Proposition Proof of Proposition Let r>0, R 0 > 0, ρ 1 > 0andρ 2 > 0beas in Proposition 4.2.5, and let 0 <R R 0,0<ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2.Take a>0andl>rsuch that a L R ρ0 1 and a L R 2 ρ0 2 }. hold. Set wx = a L x.

30 564 Hidetoshi Tahara Then, we see that wx is a holomorphic function on D R and k w k = k!a, k =0, 1, 2,... L x k+1 and so we have w R ρ 0 1 and w/ R ρ 0 2. Hence, by Proposition and its proof we see: for any fixed s with 0 <s<rthe series ψ + k x, w, w w s,..., k k z k Ω + s = is convergent in the domain { z C ; z R s < Since ψ + t, x, w 0,...,w k 0holds,wehave ψ + k x, w, w w s,..., k k = ψ + 0!a k x, = ψ + k s, } r R +4M + rr/ρ 1 ρ 0 1 +e/ρ 2 ρ 0 2. L x, 1!a L x 2,..., k!a s L x k+1 0!a L s, 1!a L s 2,..., k!a L s k+1 ; therefore, if we set ε = a/l s, η = L s and we have W k s, ε, η = { x, w 0,...,w k C C k+1 ; x s, w 0 0!ε/η 0, w 1 1!ε/η,..., w k k!ε/η k}, ψ k Wk ψ + k W k = ψ + k s, = ψ + k = s, 0!ε η 0, 1!ε η,...,k!ε η k 0!a L s, 1!a L s 2,..., k!a L s k+1 ψ + k x, w, w w s,..., k k. Thus, by the convergence of we obtain the result that the series ψ k Wk z k with W k = W k s, ε, η is convergent on Ω + s.

31 Coupling of Two PDEs 565 Now, let us justify the assertion of Proposition Take any 0 <R 1 < R 0 and fix it. For any η>0wetakeε so that <ε min { ρ 0 1/2, ρ 0 2η/4 }. Set L = R 1 + η, a = εl R 1 andr =min{r 0,R 1 + η/2}; thenwehave η = L R 1, ε = a/l R 1, L R L R 1 η/2 =η/2, a L R = εl R 1 L R εη η/2 =2ε ρ0 1, a L R 2 = εl R 1 L R 2 εη η/2 2 = 4ε η ρ0 2. Thus, by setting s = R 1 we can conclude that the series with s = R 1 is convergent in the domain Ω + R 1 with R =min{r 0,R 1 + η/2}. This proves Proposition By the proof we have Proposition Suppose that F t, x, z 1,z 2 satisfies the conditions c-1 and c-2 with F and M replaced by F and M +, respectively. Let 0 < ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2. Then, for any 0 <R 1 <R 0, η>0 and 0 <ε min{ρ 0 1/2, ρ 0 2η/4} the series ψ + k W k z k with W k = W k R 1,ε,η is convergent in the domain Ω + R 1 in with s = R 1 and R =min{r 0, R 1 + η/2}. Moreover, we have the following estimate: ψ + k W k z k ρ 1 ρ on Ω + R 1. The following corollary explains how to use this result. Corollary Let ψt, x, w 0,w 1,... be the unique solution of Ψ of the form 4.2.1, andletwt, x be a holomorphic function on S θ r D R for some θ>0, r>0 and R>0. If sup wt, x < min { ρ 1 /2, ρ 2 R/4 } S θ r D R holds, the function ψt,x,w, w/,... is convergent and defines a holomorphic function on S θ r 1 D R1 for some r 1 > 0 and R 1 > 0.

32 566 Hidetoshi Tahara Proof. By the assumption we can take ε>0, 0 <η<r,0<ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2 such that sup wt, x ε min { ρ 0 1/2, ρ 0 2η/4 }. S θ r D R Then, by Cauchy s inequality we have k w k t R η k!ε η k on S θr, k =0, 1, 2,... and therefore by setting R 1 = R η and W k = W k R 1,ε,ηk =1, 2,...we have ψ k x, w, w w,..., k R1 k t k ψ k Wk t k ψ + k W k t k <, if t S θ r Ω + R 1. This proves Corollary Remark Of course, this corollary can be verified directly by Proposition 4.3. Convergence of φt, x, u 0,u 1,... In this section we will prove the convergence of the formal solution φ = u 0 + φ k x, u 0,...,u k t k of the equation Φ in Proposition For R 1 > 0, ε>0andη>0weset U k R 1,ε,η= { x, u 0,...,u k C C k+1 ; x R 1, u 0 0!ε/η 0, u 1 1!ε/η,..., u k k!ε/η k} k =1, 2,... By Proposition and by taking R 1 > 0, ε>0, η>0sothat R 1, ε and ε/η are sufficiently small we may suppose that each φ k x, u 0,...,u k is a holomorphic function on U k = U k R 1,ε,ηandso We have φ k Uk =max U k φ k x, u 0,...,u k <, k =1, 2,...

33 Coupling of Two PDEs 567 Proposition Let R 1 > 0 be sufficiently small. Then, for any η>0 we can find an ε>0 such that the series φ k Uk z k with U k = U k R 1,ε,η is convergent in a neighborhood of z =0 C. Ψ + let To prove this, we consider ψ ψ + = w 0 + = F t, x, ψ +,D[ψ + ] ; ψ + k x, w 0,...,w k t k be the unique solution of this equation. Then, Proposition follows easily from Proposition 4.2.6, Lemma given below and Stirling s formula 2πx x+1/2 e x Γ1 + x 2πx x+1/2 e x+1 for x>0. Lemma For any k =0, 1, 2,... we have k φ k x, ψ +,...,D k [ψ + ] k + q k ψ + k+q k! x, w 0,...,w k+q t q q 0 with φ 0 = u 0 and ψ + 0 = w 0. In particular, by setting t =0we have k φ k x, w 0,...,w k kk k! ψ+ k x, w 0,...,w k. Proof. When k =0,wehaveφ 0 x, u 0 =u 0 and φ 0 x, ψ + =ψ + = q 0 ψ + q t q ; this proves Whenk =1,wehaveφ 1 x, u 0,u 1 = F 0,x,u 0,u 1 and so φ 1 x, ψ +,D[ψ + ] = F 0,x,ψ +,D[ψ + ] F t, x, ψ +,D[ψ + ] = ψ+ = ψ p + t p = 1 + qψ 1+q + tq ; p 0 q 0

34 568 Hidetoshi Tahara this proves Let k 1 and suppose that p is already proved for p =0, 1,...,k. Then, by we have and so φ k+1 1 k +1 i+j=k 0 m j D m [ F i ]x, u 0,...,u m+1 φ j u m φ k+1 x, ψ +,...,D k+1 [ψ + ] t k 1 D m [ F i ] x, ψ +,...,D m+1 [ψ + ] t i k +1 i+j=k 0 m j φ j u m x, ψ +,...,D j [ψ + ] t j. Since we have by 2 of Lemma D m [ F i ]x, ψ +,...,D m+1 [ψ + ] t i = D m[ F i x, ψ +,D[ψ + ] t i] D m[ F t, x, ψ +,D[ψ + ] ] = D m[ ψ + ] = Dm [ψ + ], by applying this to we have φ k+1 x, ψ +,...,D k+1 [ψ + ] t k 1 k +1 = 1 k +1 0 j k 0 m j 0 j k D m [ψ + ] φ j x, ψ +,...,D j [ψ + ] t j u m φ j x, ψ +,...,D j [ψ + ] t j. Thus, by the induction hypothesis we obtain φ k+1 x, ψ +,...,D k+1 [ψ + ] t k 1 j + q j ψ + j+q k +1 j! x, w 0,...,w j+q t q t j = 1 k +1 = 1 k +1 0 j k 0 j k q 1 0 j k p j q 0 j + q j qψ + j+q j! x, w 0,...,w j+q t q 1+j p +1 j p +1 j ψ p+1 + j! x, w 0,...,w p+1 t p.

35 Coupling of Two PDEs 569 Since the degree of each term of the left hand side of in t is greater than or equal to k, bylookingatthetermswhosedegreeint is greater than or equal to k and then by canceling t k from the both sides we obtain φ k+1 x, ψ +,...,D k+1 [ψ + ] 1 k +1 = 1 k +1 = 1 k +1 0 j k p k q 0 0 j k p +1 j p +1 j ψ p+1 + j! x, w 0,...,w p+1 t p k [ q 0 = 1 k +1 q 0 0 j k k +1+q j k +1+q j j! ψ + k+1+q x, w 0,...,w k+1+q t q ] k +1+q j+1 j k +1+qj j! j! ψ + k+1+q x, w 0,...,w k+1+q t q k +1+q k+1 ψ + k+1+q k! x, w 0,...,w k+1+q t q = k +1+q k+1 ψ + k+1+q k +1! x, w 0,...,w k+1+q t q ; q 0 this proves k+1. Since k k /k! e k / 2π holds for any k =1, 2,..., we have the following precise form of Proposition 4.3.1: Proposition Suppose that F t, x, z 1,z 2 satisfies the conditions c-1 and c-2 with F and M replaced by F and M +, respectively. Let 0 < ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2. Then, for any 0 <R 1 <R 0, η>0 and 0 <ε min{ρ 0 1 /2, ρ0 2η/4} the series φ k Uk z k with U k = U k R 1,ε,η is convergent in the domain { Ω R 1,η = z C ; e z R R 1 < with R =min{r 0,R 1 + η/2}. Moreover, we have the following estimate: φ k Uk z k ρ 1 ρ } r R +4M + rr/ρ 1 ρ 0 1 +e/ρ 2 ρ 0 2 on Ω R 1,η.

36 570 Hidetoshi Tahara Corollary Let φt, x, u 0,u 1,... betheuniquesolutionofφ of the form 4.3.1, andletut, x be a holomorphic function on S θ r D R for some θ>0, r>0 and R>0. If sup ut, x < min { ρ 1 /2, ρ 2 R/4 } S θ r D R holds, the function φt,x,u, u/,... is convergent and defines a holomorphic function on S θ r 1 D R1 for some r 1 > 0 and R 1 > Equivalence of and Now, let us establish the equivalence of two equations and Let r>0, R>0, ρ 1 > 0andρ 2 > 0. We suppose: H 1 F t, x, z 1,z 2 is a holomorphic function on { t, x, z 1,z 2 C 4 ; t <r, x <R, z 1 <ρ 1, z 2 <ρ 2 }. Definition We denote by X the set of all the functions wt, x satisfying the following properties: i wt, x is a holomorphic function on S θ r 1 D R1 for some θ>0, 0 <r 1 <rand 0 <R 1 <R,and ii wt, x ρ 0 1 and w/t, x ρ 0 2 hold on S θ r 1 D R1 for some 0 <ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2. 2WedenotebyH the set of all the functions wt, x satisfying the following properties: i wt, x is a holomorphic function on D r1 D R1 for some 0 <r 1 <r, and 0 <R 1 <R,and ii wt, x ρ 0 1 and w/t, x ρ 0 2 hold on D r1 D R1 for some 0 <ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2. Then we have Theorem Suppose the condition H 1. The following two equations are equivalent: u = F t, x, u, u in X resp. in H, w =0 in X resp. in H.

37 Coupling of Two PDEs 571 This follows from Theorem and Proposition Let φt, x, u 0,u 1,... be the solution of Φ in If u X resp. u H, then we have φt,x,u, u/,... X resp. φt,x,u, u/,... H. 2 Let ψt, x, w 0,w 1,... be the solution of Ψ in If w X resp. w H, then we have ψt,x,w, w/,... X resp. ψt, x, w, w/,... H. Let us show this proposition now. For a function wt, x = l 0 w ltx l we write w x t, x = l 0 w lt x l.ifwt, x X resp. wt, x H holds, the coefficients w l t are all holomorphic on S θ r 1 resp. ond r1 ; but w l t l 0 are only continuous on S θ r 1 resp. ond r1. In order to treat this function w x t, x, let us introduce the following function spaces CX and CH : Definition WedenotebyCX the set of all the functions wt, x satisfying the following properties: i wt, x is a continuous function on S θ r 1 D R1 for some θ>0, 0 <r 1 <rand 0 <R 1 <R, ii wt, x is holomorphic in x D R1 for any fixed t S θ r 1, and iii wt, x ρ 0 1 and w/t, x ρ 0 2 hold on S θ r 1 D R1 for some 0 <ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2. 2 We denote by CH the set of all the functions wt, x satisfying the following properties: i wt, x is a continuous function on D r1 D R1 for some 0 <r 1 <r, and 0 <R 1 <R, ii wt, x is holomorphic in x D R1 for any fixed t D r1,and iii wt, x ρ 0 1 and w/t, x ρ 0 2 hold on D r1 D R1 for some 0 <ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2. Then we have Lemma If wt, x X resp. wt, x H we have w x t, x CX resp. w x t, x CH. Proof. Suppose that wt, x = l 0 w ltx l X. Then, by the conditions wt, x ρ 0 1 and w/t, x ρ0 2 on S θr 1 D R1 with 0 <ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2 we have w l t ρ0 1 R 1 l l 0 and lw l t ρ0 2 R 1 l 1 l 1,

38 572 Hidetoshi Tahara and so we see: ρ 0 1 R 1 l xl = w x t, x = w l t x l l 0 l 0 w x t, x = l w l t x l 1 l 1 l 1 ρ x/r 1, ρ 0 2 R 1 l 1 xl 1 = ρ x/r 1. Therefore, if we take R 2 > 0 sufficiently small so that ρ 0 1/1 R 2 /R 1 <ρ 1 and ρ 0 2/1 R 2 /R 1 <ρ 2 we have the properties: i w x t, x is a continuous function on S θ r 1 D R2, ii w x t, x is holomorphic in x D R2, and iii w x t, x ρ 1 and w x/t, x ρ 2 on S θr 1 D R2 for ρ 1 = ρ0 1 /1 R 2 /R 1 <ρ 1 and ρ 2 = ρ 0 2/1 R 2 /R 1 <ρ 2. This proves the result: w x t, x CX. The case H canbeprovedinthesameway. Let ψ + = w 0 + ψ + k x, w 0,...,w k t k be the unique solution of ψ + = F t, x, ψ +,D[ψ + ]. If w = wt, x X resp. w = wt, x H, we have w x = w x t, x CX resp. w x = w x t, x CH and φ k x,w, w/,... φ k x, w x, w x /,... kk k! ψ+ k x, w x, w x /,..., k =1, 2,..., ψ k x,w, w/,... ψ k x, w x, w x /,... ψ + k x, w x, w x /,..., k =1, 2,... as formal power series in x with a parameter t. Therefore, to prove Proposition it is sufficient to show the following result. Lemma If w = wt, x CX resp. w = wt, x CH we have ψ + t,x,w, w/,... CX resp. ψ + t,x,w, w/,... CH. Proof. Let θ>0, 0 <r 1 <r,0<r 1 <R,0<ρ 0 1 <ρ 1 and 0 <ρ 0 2 <ρ 2. Let wt, x be a continuous function on S θ r 1 D R1 which is holomorphic in x