On the summability of divergent power series solutions for certain first-order linear PDEs Masaki HIBINO (Meijo University) 1
1 Introduction (E) {1+x 2 +β(x,y)}y u x (x,y)+{x+b(x,y)}y2 u y (x,y) +u(x,y)=f(x,y) x,y C β,b,f: holomorphic at (x,y) = (,) C 2 (1.1)! û(x,y) = u n (x)y n : n= Problem When is û(x, y) summable? β(x,) b(x,) formal solution of (E) divergent 2
2 Definition and fundamental result Definition 2.1 (1) R>, B(R) ={x C; x R} O[R]: ring of holomorphic functions on B(R) (2) O[R][[y]] = û(x,y) = u n (x)y n ; u n (x) O[R] (3) O[R][[y]] 2 = û(x,y) = n= u n (x)y n O[R][[y]]; C, K > s.t. n= max u n(x) CK n n! (n =,1,2,...) x R (4) θ R, κ>, <ρ + S(θ,κ,ρ) = y; arg(y) θ < κ 2, < y <ρ 3
(5) û(x,y) = ρ<+ ρe iθ κ 2 ρ = + θ θ u n (x)y n O[R][[y]] 2 isborelsummable inθ n= S(θ,κ,ρ) (κ>π) u(x,y): holomorphoic onb(r) S(θ,κ,ρ) (< r R) s.t. N 1 u(x,y) u n (x)y n C K N N! y N max x r n= y S(θ,κ,ρ), N = 1,2,... û: Borel summable inθ u: unique Borelsumofûinθ κ 2 4
Definition2.2 û(x,y) = n= u n (x)y n O[R][[y]] 2 ˆB[û](x, ) = u n (x) n : formalboreltransform ofû n! n= ˆB[û](x,): holomorphic at (x,) = (,) Theorem2.1 û(x,y) O[R][[y]] 2 û(x, y): Borel summable in θ (BS) ˆB[û](x,): continued analytically tob(r ) S(θ,κ,+) ( r, κ ) C, δ> s.t. max ˆB[û](x,) Ce δ, S(θ,κ,+) x r Borel sum ofûinθ: u(x,)= 1 y e iθ e /y ˆB[û](x,)d 5
3 Motivation (3.1) A(x,y) u x (x,y)+b(x,y) u y (x,y)+u(x,y)=f(x,y) A,B,f: holomorphic at (x,y) = (,) C 2 (3.2) (3.3) (3.4) A(x,) A y (,)= B(x,) B y (x,) Remark 3.1 (3.2), (3.3), (3.4) (A,B) (x, y) = (x,y)=(,) A y (,) (nilpotent type) 6
Theorem 3.1(1999) (3.2), (3.3), (3.4)! û(x,y) = u n (x)y n O[R][[y]] 2 ( R): formal solution of (3.1) Problem θ: given n= Whenisû(x,y)Borelsummable inθ? (When does ˆB[û](x, ) satisfy(bs)?) Remark 3.2 û(x, y): Borel summable Borel sum u(x, y): holomorphic solution 7
(3.1) is rewritten (3.5) {α(x)+β(x,y)}y u x (x,y)+{a(x)+b(x,y)}y2 u y (x,y) +u(x,y)=f(x,y) (3.6) (3.7) α,β,a,b,f: holomorphic at the origin α()= β(x,) b(x,) 26 α(x): general, a(x) a(constant) 28 α(x) α(constant), a(x): general 21 α(x)=α +α 1 x, a(x): general In this talk we consider the case (3.8) α(x)=1+x 2, a(x)=x 8
4 Mainresult (E) {1+x 2 +β(x,y)}y u x (x,y)+{x+b(x,y)}y2 u y (x,y) +u(x,y)=f(x,y) Assumptions We define the regionω θ,κ by Ω θ,κ = tan[arcsin(τ)]; τ S(θ,κ,+) In order to assure the well-definedness ofω θ,κ we assume (A1) θ=,π (A2) f(x,y): C, δ> s.t. x Ω θ,κ (4.1) continued analytically toω θ,κ {y C; y c} ( c>) max f(x,y) Cexp δ sin(arctanx) y c 9
(A3) β(x,y),b(x,y): continued analytically toω θ,κ {y C; y c} K, L>, p>1 s.t. x Ω θ,κ, m = 1, 2,... (4.2) (4.3) (4.4) 1 m β 1+x 2 y m(x,) KLm m! cos(arctanx) m x m β 1+x 2 y m(x,) m! cos(arctanx) m+1 KLm {1+ sin(arctanx) } p m b y m(x,) m! cos(arctanx) m+1 KLm {1+ sin(arctanx) } p Main theorem (A1), (A2), (A3) The formal solutionû(x,y) of (E) is Borel summable inθ 1
5 Proofof MainTheorem (easycase: f(x,y)=f(x),β(x,y) b(x,y) ) (E) (y+x 2 y) u x (x,y)+xy2 u y (x,y)+u(x,y)=f(x) û(x,y): formal solution of (E) ˆB[û](x,): formal Borel transform ofû ˆB[û](x,) =f tan arcsin sin(arctanx)+cos(arctanx) (A1)θ=,π; (A2)f(x) can be continued analytically to Ω θ,κ ={ tan[arcsin(τ)]; τ S(θ,κ,+)} with the estimate f(x) C exp δ sin(arctan x),x Ω θ,κ (BS) ˆB[û](x,) can be continued analytically to B(r ) S(θ,κ,+) ( r, κ ) and satisfies max ˆB[û](x,) C e δ, S(θ,κ,+) ( C, δ ) x r 11
6 Proofof MainTheorem For simplicity, we consider the caseβ(x,y) : (E) {1+x 2 }yd x u(x,y)+{x+b(x,y)}y 2 D y u(x,y) +u(x,y)=f(x,y) D x = x, D y= y û(x, y): formal solution of (E) v(x,) = ˆB[û](x,): formal Borel transform ofû It is sufficient to prove thatv(x,) satisfies (BS): (BS)v(x,) can be continued analytically tob(r ) S(θ,κ,+) ( r, κ ) and satisfies max v(x,) C e δ, S(θ,κ,+) ( C, δ ) x r 12
Step 1. Formal Borel transform of equation (E) formal Borel transform? A(x,y)yB(x,y) yd y C(x,y) D ˆB[C](x,) v(x, ) satisfies {1+x 2 } (6.1) + ˆB[A](x, t) ˆB[B](x,t)dt v x (x,t)dt+x t v (x,t)dt ˆB[b](x, t) t v (x,t)dt+v(x,) = ˆB[f](x,) (E) {1+x 2 }yd x u(x,y)+{x+b(x,y)}y 2 D y u(x,y) +u(x,y)=f(x,y) integration by parts (6.1) (6.2) 13
(6.2) {1+x 2 } + = ˆB[f](x,) v x (x,t)dt+x t v (x,t)dt ˆB[b] (x, t) t v(x,t)dt ˆB[b](x, t)v(x,t)dt+v(x,) (6.1) {1+x 2 } + v x (x,t)dt+x t v (x,t)dt ˆB[b](x, t) t v (x,t)dt+v(x,) = ˆB[f](x,) (6.2) D (6.3): 14
(6.3) {(1+x 2 )D x +(1+x)D }v(x,) = ˆB[b] (x,) v(x,) ˆB[b] (x, t) t v(x,t)dt + ˆB[b] (x, t)v(x,t)dt+ ˆB[f] (x,) v(x, ) = f(x, ) (integro-differential equation) (6.2) {1+x 2 } + = ˆB[f](x,) v x (x,t)dt+x t v (x,t)dt ˆB[b] (x, t) t v(x,t)dt ˆB[b](x, t)v(x,t)dt+v(x,) 15
(6.3) v(x,) = ˆB[b] (x,) v(x,) ˆB[b] (x, t) t v(x,t)dt + ˆB[b] (x, t)v(x,t)dt+g(x,) v(x, ) = f(x, ) (integro-differential equation) (6.4) = (1+x 2 )D x +(1+x)D g(x,) = ˆB[f] (x,) Does v(x, ) satisfy(bs)? 16
Step 2. Integro-differential equation Integral equation w(x,) =k(x,) (6.5) w(x,) = l tan (x,) + k w(x,) =l(x) tan (x, z), (x, z) z (x,) = arcsin sin(arctanx)+cos(arctanx) (x,) = cos(arctanx) cos( (x,)) (6.5) integration by substitution (6.3) (6.6) (x, z)dz 17
(6.6) v(x,) = f tan (x,), + + i=1 g tan (x, z), (x, z) z 3 I i v(x,) (integralequation) (x, z)dz I 1 v(x,) = (x, z) 2 z ˆB[b] tan (x, z), v tan (x, z), (x, z) z dz z I 2 v(x,) = (x, z) 3 ˆB[b] tan (x, z), (x, z) (z s) s v tan (x, z), (x, z) s dsdz z I 3 v(x,) = (x, z) 2 ˆB[b] tan (x, z), (x, z) (z s) v tan (x, z), (x, z) s dsdz Does asolutionv(x,) of(6.6) satisfy (BS)? 18
Step 3. Iteration v (x,) =f tan (x,), + g tan (x, z), (x, z) z (x, z)dz v n+1 (x,) =v (x,)+ 3 I i v n (x,) (n ) i=1 w (x,) =v (x,) w n (x,) =v n (x,) v n 1 (x,) (n 1) Lemma 6.1 w n (x,): continued analytically tob(r ) S(θ,κ,+) ( r, κ > ) C, δ, M > s.t. S(θ,κ,+) max w n (x,) C e δ M n 1+ 1 n n (6.7) x r p 1 n! 19
(A3) β(x,y),b(x,y): continued analytically toω θ,κ {y C; y c} K, L>, p>1 s.t. x Ω θ,κ, m = 1, 2,... (4.2) (4.3) (4.4) 1 m β 1+x 2 y m(x,) KLm m! cos(arctanx) m x m β 1+x 2 y m(x,) m! cos(arctanx) m+1 KLm {1+ sin(arctanx) } p m b y m(x,) m! cos(arctanx) m+1 KLm {1+ sin(arctanx) } p Main theorem (A1), (A2), (A3) The formal solutionû(x,y) of (E) is Borel summable inθ 1
Lemma 6.1 max x r n= w n (x,) C e δ n= M n 1+ 1 n n p 1 n! = C e δ1, S(θ,κ,+) δ 1 =δ +M 1+ 1 p 1 v(x,) = lim n v n(x,) = w n (x,): n= holomorphic solution of (6.6) onb(r ) S(θ,κ,+) max v(x,) C e δ1, S(θ,κ,+) x r 2
Proofof Lemma6.1 (A1)and(A2)= Lemma6.1forn= (A3)and analyticcontinuationpropertyforw n (x,) = analyticcontinuationpropertyfor w n+1 (x,) w n+1 (x,)=i 1 w n (x,)+i 2 w n (x,)+i 3 w n (x,) I 1 w n (x,) = (x, z) 2 z ˆB[b] tan (x, z), w n tan (x, z), (x, z) z dz z I 2 w n (x,) = (x, z) 3 ˆB[b] tan (x, z), (x, z) (z s) s w n tan (x, z), (x, z) s dsdz z I 3 w n (x,) = (x, z) 2 ˆB[b] tan (x, z), (x, z) (z s) w n tan (x, z), (x, z) s dsdz 21
On the estimate(6.7)? max w n (x,) C e δ M n 1+ 1 n n (6.7) x r p 1 n! A(x,y) = = m= A m (x)y m B[A](x, t)v(x,t)dt = m= A m (x) 1 m m= A m (x) m! ( t) m V(x,t)dt V(x, m+1 )d m+1 d 2 d 1 (6.8) w n+1 (x,)= 1 w n (x,)+ 2 w n (x,) 22
1w n (x,) =I 1 w n (x,)+i 2 w n (x,) = b m tan (x, 1 ) (x, 1 ) m+1 m=1 1 m 1 2w n (x,) =I 3 w n (x,) = b m tan m=1 1 m m w n tan (x, 1 ), (x, 1 ) m d m d 2 d 1 (x, 1 ) (x, 1 ) m+1 w n tan (x, 1 ), (x, 1 ) m+1 d m+1 d 2 d 1 b m (x)= 1 m b m! y m(x,) Estimate (6.7)forw n (x,), (4.4),(6.8) = Estimate (6.7)forw n+1 (x,) 23
(A3) β(x,y),b(x,y): continued analytically toω θ,κ {y C; y c} K, L>, p>1 s.t. x Ω θ,κ, m = 1, 2,... (4.2) (4.3) (4.4) 1 m β 1+x 2 y m(x,) KLm m! cos(arctanx) m x m β 1+x 2 y m(x,) m! cos(arctanx) m+1 KLm {1+ sin(arctanx) } p m b y m(x,) m! cos(arctanx) m+1 KLm {1+ sin(arctanx) } p Main theorem (A1), (A2), (A3) The formal solutionû(x,y) of (E) is Borel summable inθ 1