POLONICI MATHEMATICI LXVII.1 (1997)
Borel resummation of formal solutions to nonlinear Laplace equations in 2 variables
by M. E. Pli´ s (Krak´ow) and B. Ziemian (Warszawa)
Abstract. We consider a nonlinear Laplace equation ∆u = f (x, u) in two variables.
Following the methods of B. Braaksma [Br] and J. Ecalle used for some nonlinear ordinary differential equations we construct first a formal power series solution and then we prove the convergence of the series in the same class as the function f in x.
0. Introduction. We consider a nonlinear Laplace equation of the form
(1) ∆u =
∂
2∂x
21+ ∂
2∂x
22u = f (x, u)
where x = (x
1, x
2) ∈ R
2. First we are going to construct a formal power series solution of (1) and then prove that every such solution is of the same class as the function f in x. Similar results for some nonlinear ordinary differential equations were proved by Braaksma [Br], following the ideas of J. Ecalle.
We denote by L the image of the positive quadrant R
2+= R
+×R
+under the unitary matrix
√12 1−i 1+i1+i 1−i.
Definition 1 ([Zie1]). A function F of the variable z = (z
1, z
2) ∈ C
2is said to be Laplace holomorphic on L if F is holomorphic on some polydisk centered at (0, 0) ∈ C
2, can be holomorphically continued to some sectorial neighbourhood S = S
1× S
2of L with vertex (0, 0), and is of exponential growth on S, i.e. for every closed subsector S
′= S
1′× S
2′⊂ S there exist constants c = (c
1, c
2) and C such that for z ∈ S
′,
(2) |F (z
1, z
2)| ≤ Ce
c1|z1|+c2|z2|.
Definition 2. A function f of the variable x = (x
1, x
2) ∈ R
2is said to be a 1-sum of a formal power series
1991 Mathematics Subject Classification: 35C20, 44A10.
Key words and phrases : Borel resummation, formal solutions, Laplace equation, Supported by KBN grant 2-PO3A-006-08.
[31]
f (x) = b X
∞ k,l=0g
kl1
(x
1+ ix
2)
k+1(ix
1+ x
2)
l+1if there exists a Laplace holomorphic function F on L such that
f (x) =
\
L
e
−xzF (z) dz and
F (z) = X
∞ k,l=0g
kli2
k+l+1k!l! (z
1− iz
2)
k(z
2− iz
1)
lnear zero. In that case we say that f is 1-resummable.
In this paper we assume that f (x, u) on the right hand side of (1) is the 1-sum in x of a formal power series
f (x, u) = b X
∞ k,l=0g
kl(u) 1
(x
1+ ix
2)
k+1(ix
1+ x
2)
l+1with coefficients g
kl(u) holomorphic for every (k, l) ∈ N
20, on some fixed neighbourhood U of zero in C, and g
kl(0) = 0.
Therefore, if we write f (x, u) = b
X
∞ k,l=0X
∞j=1
c
jklu
j1
(x
1+ ix
2)
k+1(ix
1+ x
2)
l+1= X
∞ j=1X
∞ k,l=0c
jkl1
(x
1+ ix
2)
k+1(ix
1+ x
2)
l+1u
jthen we can write f (x, u) = P
∞j=1
c
j(x)u
jwhere c
j(x) is the 1-sum of the formal series
X
∞ k,l=0c
jk,l1
(x
1+ ix
2)
k+1(ix
1+ x
2)
l+1, and f is holomorphic in u on U . Hence, we have c
j(x) =
T
L
e
−xzT
j(z) dz for some Laplace holomorphic functions T
j. Moreover, T
jare holomorphic on the same sector S for all j, and the constants c and C in (2) are independent of j.
Theorem. If T
j(0) 6= 0, then there exists a family of 1-resummable solutions of equation (1) of the form
u(x) = X
∞ ν=0d
ν1
(x
1+ ix
2)
ν1+1· 1
(ix
1+ x
2)
ν2+1.
This means that
(3) u(x) =
\
L
e
−xzT (z) dz
with T being a Laplace holomorphic function on L. Moreover , every formal solution b u of (1) of the above form is 1-resummable.
The proof will be divided into three parts.
1. Convolution equation. Applying ∆ to u in the form (3) we arrive at the complex symbol of ∆ as the complex polynomial
P (z
1, z
2) = z
12+ z
22=
1 + i
√ 2 z
1+ 1 − i
√ 2 z
21 − i
√ 2 z
1+ 1 + i
√ 2 z
2. In the new variables
ζ
1= 1 + i
√ 2 z
1+ 1 − i
√ 2 z
2, ζ
2= 1 − i
√ 2 z
1+ 1 + i
√ 2 z
2,
P becomes the polynomial e P (ζ
1, ζ
2) = ζ
1· ζ
2, and after changing variables on the left hand side of (1) we get
∆u(x
1, x
2) = (P (z
1, z
2)T )[e
−x1z1−x2z2]
= ( e P (ζ
1, ζ
2) e T )[e
−x1(1−i2√2ζ1+21+i√2ζ2)−x2(21+i√2ζ1+21−i√2ζ2)]
= ( e P (ζ
1, ζ
2) e T )[e
−(21−i√2x1+21+i√2x2)ζ1−(21+i√2x1+21−i√2x2)ζ2].
So we are looking for a solution u(y e
1, y
2) = e T [e
−y1ζ1−y2ζ2] = u
1 − i 2 √
2 x
1+ 1 + i 2 √
2 x
2, 1 + i 2 √
2 x
1+ 1 − i 2 √
2 x
2of the convolution equation
(4) ζ
1ζ
2T = f e
∗T e
where f
∗T = e P
∞j=1
T e
j∗ e T
∗jwith e T
∗jdenoting the jth convolution power of e T , i.e. T
∗j= T ∗ ... ∗ T (j times). From now on we write T instead of e T . We can assume that T
1(0) = 1, for otherwise we modify slightly the change of variables after dividing equation (1) by T
1(0).
Since our existence proof for the solution of (4) essentially follows that of Braaksma [Br], we shall consider T having the formal expansion
(5) T =
X
∞ k,l=0d
klζ e
1kζ e
2lwith e ζ
p= ζ
p/Γ (p + 1). Then due to the convolution formula
ζ e
l∗ e ζ
k= e ζ
l+k+1we find f
∗T =
X
∞ j=1X
∞ m1,m2=0c
jm1m2ζ e
1m1ζ e
2m2∗ X
∞k,l=0
d
klζ e
1kζ e
2l∗j= X
∞ j=1X
∞ m1,m2=0c
jm1m2ζ e
1m1ζ e
2m2∗
X
∞ ν1+...+νj=0d
ν1. . . d
νje ζ
ν1+...+νj+j−1= X
∞ j=1X
∞ m+ν1+...+νj=0c
jmd
ν1. . . d
νjζ e
m+ν1+...+νj+j= X
∞ k=0 k+1X
j=1
X
m+ν1+...+νj=k+1−j
c
jmd
ν1. . . d
νjζ e
k+1for k, m, ν
j∈ N
20, j = (j, j), 1 = (1, 1), k = min{k
1, k
2}.
Inserting this in (4) we find
(6) d
k(k + 1) =
k+1
X
j=1
X
m+ν1+...+νj=k+1−j
c
jmd
ν1. . . d
νj, since
ζ · e ζ
p= (p + 1)e ζ
p+1.
In particular, we can take d
00arbitrarily (since c
100= 1), d
10= c
101d
00, d
01= c
101d
00,
2d
20= c
120d
00+ c
110d
10, 2d
02= c
102d
00+ c
101d
01, 3d
11= c
111d
00+ c
110d
01+ c
101d
10+ c
200d
00, . . .
We are going to prove that T defined formally by (5) with coefficients d
νsatisfying the recurrence (6) is a holomorphic function of exponential growth in some sector S.
Before starting the resummation proof for the expansion (5), we consider the resummation problem with respect to one variable. Therefore, let us write (5) in the form
T = X
∞ k=0T
1k(ζ
1)e ζ
2kwhere T
1k(ζ
1) = P
∞l=0
d
lkζ e
1l.
In a way similar to that of deriving (6), we find that T
1ksatisfy the
convolution equation
ζ
1(k + 1)T
1k=
k+1
X
j=1
X
m+ν1+...+νj=k+1−j
T
mj∗ T
1ν1∗ . . . ∗ T
1νjfor k, m, ν
j∈ N
0, where T
mj= P
∞p=0
c
jpmζ e
1p. For k = 0 this gives (7) ζ
1T
10= T
01∗ T
10,
which is equivalent to the equation
(7
′) d
dt u
0= c
10(t)u
0, in the variable t =
21−i√2x
1+
21+i√2x
2.
For k = 1 we get
2ζ
1T
11= T
01∗ T
11+ T
11∗ T
10+ T
02∗ T
10∗2or equivalently 2 d
dt u
1= c
10(t)u
1+ c
11(t)u
0+ c
20(t)(u
0)
2.
We easily see that the jth equation is linear in u
jwith u
0, . . . , u
j−1regarded as coefficients. Since the solutions of linear equations with re- summable coefficients are resummable themselves (cf. [Br], [Zie1]), we see that all T
1kare Laplace holomorphic functions. The same is also true for
T
l2(ζ
2) = X
∞ j=0d
ljζ e
2j.
Now we pass to the proof of the convergence of the formal series (5) with d
νsatisfying (6). Since the series (5) satisfies (4), for a fixed N ∈ N
0the series
T
N= X
∞ l,j=N +1d
ljζ e
1lζ e
2j= T − S
Nsatisfies the equation
ζ
1ζ
2T
N= G
N(ζ, T
N) = X
∞ j=1X
j k=0j k
T
j∗ T
N∗k∗ S
N∗(j−k)− ζ
1ζ
2S
N= X
∞ k=0X
∞ j=k j≥1j k
T
j∗ S
N∗(j−k)∗ T
N∗k− ζ
1ζ
2S
N.
We write
(8) G
N(ζ, ψ) =
X
∞ k=0g
k(ζ) ∗ ψ
∗kwhere g
0= P
∞j=1
T
j∗ S
N∗j− ζ
1ζ
2S
N, and g
k= P
∞j=k j k
T
j∗ S
N∗(j−k)for k > 0. The series g
kare convergent near (0, 0) due to the remarks about the resummation problem with respect to one variable and the fact that the series P
∞j=k j k
T
j(ζ)u
j−kis convergent near 0. Moreover, we can see that for every subsector S
′⊂ S there exist K and c = (c
1, c
2) such that for ζ ∈ S
′, (9)
|g
0(ζ)| ≤ K|ζ
1|
N +1|ζ
2|
N +1e
c1|ζ1|+c2|ζ2|,
|g
k(ζ)| ≤ Ke
c1|ζ1|+c2|ζ2|for k ≥ 1.
For p = (p
1, p
2), p
i> 0, s = (s
1, s
2) ∈ R
2, we denote by W
s(p) the space of functions ψ holomorphic in the polydisc {|ζ
1| ≤ p
1, |ζ
2| ≤ p
2} and such that
kψk
s,p= sup
|ζi|≤pi
|ζ
−sψ(ζ)| < ∞.
Observe that for ζ ∈ {|ζ
i| ≤ p
i} and s
1> −1, s
2> −1,
|ψ
∗m(ζ)| ≤ kψk
ms,pΓ (s
1+ 1)
mΓ (s
2+ 1)
mΓ (m(s
1+ 1))Γ (m(s
2+ 1)) (10)
× |ζ
1|
m(s1+1)−1|ζ
2|
m(s2+1)−1.
Therefore by the properties of the Γ -function the function (8) makes sense for ψ ∈ W
s(p) if s is large enough.
Consider the operator (11) Rψ(ζ) = 1
ζ g
0(ζ) + 1
ζ (g
1∗ ψ)(ζ) + X
∞ m=21
ζ (g
m∗ ψ
∗m)(ζ).
Denoting the summands by R
0ψ, R
linψ and Qψ respectively, for ψ ∈ W
N −1(p) and ζ ∈ {|ζ
i| ≤ p
i, i = 1, 2} we get the estimates
(12)
|R
0ψ(ζ)| ≤ K|ζ
1|
N|ζ
2|
N,
|R
linψ(ζ)| ≤ Kkψk
N −1,p|ζ
1|
N −1|ζ
2|
N −1Γ (N ) Γ (N + 1)
2= K
N
2kψk
N −1,p|ζ
1ζ
2|
N −1,
|Qψ(ζ)| ≤ K
X
∞ m=2kψk
mN −1,pΓ (N )
2mΓ (mN )
2|ζ
1ζ
2|
(m−1)N −1|ζ
1ζ
2|
N. Set M =
1ζ
g
0(ζ)
N −1,p
. If kψk
N −1,p≤ 2M then by choosing p small and N large we may have (by (12))
kQψk
N −1,p≤
13M and kR
linψk ≤
13M.
Therefore the operator R acts in the space
B
N −1,p= {ψ ∈ W
N −1(p) : kψk
N −1,p≤ 2M}.
Observe that for ψ, ψ + χ ∈ B
N −1,pwe have
|(ψ + χ)
∗m(ζ) − ψ
∗m(ζ)| =
X
m l=1m l
(ψ
∗(m−l)∗ χ
∗l)(ζ)
≤ X
ml=1
m l
kψk
m−lN −1,pkχk
lN −1,pΓ (N )
2mΓ (mN )
2|ζ
1ζ
2|
mN −1≤ Γ (N )
2mΓ (mN )
2|ζ
1ζ
2|
mN −1X
m l=1m l
kψk
m−lN −1,pkχk
lN −1,p. We have
X
m l=1m l
kψk
m−lN −1,pkχk
lN −1,p≤ kχk
N −1,pX
m l=1m l
(2M )
m−l(4M )
l−1≤ kχk
N −1,p4M (2M + 4M )
m= (6M )
m4M kχk
N −1,psince kχk ≤ kψ + χk + kψk ≤ 4M. Hence
|(ψ + χ)
∗m(ζ) − ψ
∗m(ζ)| ≤ (6M Γ (N )
2)
m4M Γ (mN )
2kχk
N −1,p(|ζ
1ζ
2|)
mN −1, and
1
ζ (g
m∗ ((ψ + χ)
∗m− ψ
∗m))(ζ)
≤ K
4M
(6M Γ (N )
2)
mΓ (mN )
2|ζ
1ζ
2|
(m−1)N −1|ζ
1ζ
2|
Nkχk
N −1,p. From this and from (12), we derive that
kR(ψ + χ) − Rψk
N −1,p≤ kR
linψk
N −1,p+ kQ(ψ + χ) − Qψk
≤
13kψk
N −1,p+ K
′kχk
N −1,p≤
23kχk
N −1,pprovided p is small enough. Therefore, for p small and N large, the operator R is a contraction on B
N −1,p. Hence we get a unique function ψ
Nsolving the nonlinear convolution equation
(13) ζ
1ζ
2ψ
N= G
N(ζ, ψ
N), ψ
N∈ B
N −1,p.
From the construction of G
Nit follows that for every N (sufficiently large) the function ψ
N+ S
Nsatisfies the equation (4), hence the kth Taylor coefficient of ψ
N(at 0) must satisfy (6) (for k
i≥ N + 1), so T defined formally by (5) and (6) converges on {|ζ
i| ≤ p
i}.
2. Analytic continuation of solutions. Define S(r) = {ζ ∈ C : |ζ| ≤ r}
∩ S
1(see Introduction) and let p be such that the solution ψ
Nof (13) is
holomorphic in the interior of S
2(p) = S(p) × S(p). We shall extend this
solution to a unique solution on some complex neighbourhood of R
2+.
Choose δ, p
1∈ R
+, δ < p
1< p. Define S
0= S(p
1) × S(p),
S
+= {ζ ∈ C
2: (ζ
1− p
1, ζ
2) ∈ S(δ) × S(p) or ζ
1= p
1}, S
1= S
0∪ S
+.
Then S
0∩ S
+= {p
1} × S(p).
Let W
0denote the space of functions on S
1which are continuous on S
1\ (S
0∩ S
+) and analytic in its interior. Next define e ψ ∈ W
0by setting ψ = ψ e
Non S
0and e ψ ≡ 0 on S
+. Introduce the space
V
N −1(δ) = {φ ∈ C
0(S
+) ∩ O(intS
+) : sup
ζ∈S+
|ζ
2−N +1φ(ζ)| < ∞}.
For φ ∈ V
N −1(δ) define φ
0∈ W
0by extending φ by zero on S
0. Then (φ
0∗ φ
0)(ζ) =
\
C(ζ)
φ
0(ζ − γ)φ
0(γ) dγ ≡ 0
where C(ζ) = C(ζ
1)×C(ζ
2), C(ζ
i) is a path from 0 to ζ
i. Hence also φ
∗m0≡ 0 for m ≥ 2. Clearly, e ψ
∗m= ψ
N∗mon S
0for all m. Therefore ( e ψ + φ)
∗m= ψ b
∗m+ m b ψ
∗(m−1)∗ φ
0.
Consequently, for G(ζ, ψ) = G
N(ζ, ψ) given by (8) we have G(ζ, e ψ + φ
0) = G(ζ, e ψ) + (B ∗ φ
0)(ζ) where
B(ζ) = g
1(ζ) + X
∞ m=2m(g
m∗ e ψ
∗(m−1))(ζ).
Thus the equation
ζ( e ψ + φ
0) = G(ζ, e ψ + φ
0) gives rise to a linear convolution equation
(14) φ
0= χ + 1
ζ (B ∗ φ
0)(ζ) for φ
0∈ V
N −1(δ), with χ(ζ) =
1ζG(ζ, e ψ) − e ψ.
For ζ ∈ S
+and φ ∈ V
N −1(δ) we have
1
ζ (B ∗ φ
0)(ζ) =
1 ζ
ζ\1
p1
ζ\2
0
B(ζ
1− η
1, ζ
2− η
2)φ(η
1, η
2) dη
1dη
2= 1 ζ
ζ2
\
0
h
ζ1−p\10
B(ζ
1− p
1− γ
1, ζ
2− η
2)φ(γ
1+ p
1, η
2) dγ
1i dη
2≤ 1
|ζ| kφk
N −1ζ2
\
0
η
N −12h
ζ1−p\10
B(τ, ζ
2− η
2) dτ i dη
2with kφk
N −1= sup
ζ∈S+|ζ
2−N +1φ(ζ)|. Now, from the definition of B, we see that for τ ∈ S(δ) × S(p),
|B(τ)|≤ C
1 +
X
∞ m=2mkψ
Nk
m−1N −1,pΓ (N )
2(m−1)Γ ((m − 1)N)
2(|τ
1| · |τ
2|)
(m−1)N< M.
Thus for ζ ∈ S
+(and consequently for |ζ
1| ≥ p
1) we have
1
ζ (B ∗ φ
0)(ζ)
≤ Kkφk
N −1|ζ
2N −1| with K = M δ N p
1.
Hence if we take δ < N p
1/M , then the operator φ →
1ζB ∗φ
0is a contraction in the space V
N −1(δ). Thus there exists a unique solution φ ∈ V
N −1(δ) satisfying (14). Hence φ = ψ
Non the interior of S
2(p) ∩ S
+and it is clear that φ extends ψ
Nto S
+.
A repeated application of this procedure yields an extension of ψ
Nto some region U × S(p), where U is a sectorial neighbourhood of R
+in C.
By interchanging variables and proceeding by the same method we get an extension of ψ
Nto some region S(p) × V with V being a sectorial neigh- bourhood of R
+in C. Finally, in the same way we obtain an extension of ψ
Nto some sector U × V .
3. Exponential estimation. It follows from the results on analytic continuation of the solution of (13) that there exists a function ψ, holomor- phic in some sector S containing R
2+, satisfying (13) and such that ζ
−N +1ψ is locally bounded. We shall prove a global exponential estimate: for every closed subsector S
′⊂ S,
|ψ(ζ
1, ζ
2)| ≤ K|ζ
N −1|e
c1|ζ1|+c2|ζ2|for ζ ∈ S
′, with appropriate constants K and c
1, c
2. The proof is again a two-dimensional variant of the reasoning given in [Br].
For p > 0 define
M (p) = sup{|ζ
1−N +1ψ(ζ
1, ζ
2)| : 0 < |ζ
1| < 1, |ζ
2| = p, ζ ∈ S
′}.
It follows from the local estimates for ψ that M (p) makes sense for each fixed p > 0. Then for 0 < |ζ
1| < 1, |ζ
2| = p, ζ ∈ S
′,
|ψ(ζ
1, ζ
2)| ≤ M(p)|ζ
1N −1|, and as in (10),
|ψ
∗m(ζ
1, ζ
2)| ≤ M
∗m(p)
Γ (N )
mΓ (mN ) |ζ
1(m−1)N −1|
|ζ
1|
N. Then, by (9), we find that for any bc
2> c
2,
1
ζ (g
m∗ ψ
∗m)(ζ)
≤ Ke
ˆc2p∗ q
mM
∗m(p)|ζ
1|
Nwhere q is a sufficiently small constant such that
Γ (N )
mΓ (mN ) |ζ
1(m−1)N −1|
1/m≤ q for m ∈ N.
Therefore we have for 0 < |ζ
1| < 1, |ζ
2| = p, ζ ∈ S
′,
|ψ(ζ)| = |Rψ(ζ)| ≤ K|ζ
1|
Ne
cˆ2p+ K|ζ
1|
Ne
ˆc2p∗
X
∞ m=1q
mM
∗m(p) and for all p > 0,
(15) |ζ
1−N +1ψ(ζ)| ≤ e Ke
ˆc2p+ e K e
ˆc2p∗
X
∞ m=1q
mM
∗m(p) .
Denoting the right hand side of (15) by SM we get M (p) ≤ SM(p) for p > 0.
Consider the equation
(16) N (p) = SN (p).
Under the Laplace transformation v(s) = LN(s) =
∞\
0
e
−psN (p) dp equation (16) becomes
v(s) = K e
s − bc
2+ K e s − bc
2·
X
∞ m=1(qv(s))
m= K e
s − bc
2· 1 1 − qv(s) or equivalently
qv
2− v + K e
s − bc
2= 0.
This equation has a unique solution analytic in 1/s at infinity, of the form v(s) = K e
s + X
∞l=1
b
ls
l+1for s large enough with coefficients b
l∈ R. Hence
N (p) = e K + X
∞l=1