POLONICI MATHEMATICI LXIX.2 (1998)
Poincar´ e theorem and nonlinear PDE’s
by Maria E. Pli´ s (Krak´ow)
Abstract. A family of formal solutions of some type of nonlinear partial differen- tial equations is found. Terms of such solutions are Laplace transforms of some Laplace distributions. The series of these distributions are locally finite.
1. Introduction. In this paper we consider the nonlinear partial differ- ential equations of the form
(1) P (D)u = f (u),
with f analytic at zero, f (0) = 0, and f
′(0) 6= 0. This means that f(u) = P
∞m=1
c
mu
mwith c
16= 0. The operator P (D) is a linear differential opera- tor P (∂/∂x), where P (z) = P (z
1, . . . , z
n) is a complex polynomial without constant term.
Our aim is to find the behaviour of a solution of the equation (1) at infinity, writing this solution as some series of integrals involving a solution of the linear part of (1), i.e.
(2) P (D)y = c
1y.
The method of construction used here follows Bobylev [2] (see also Ros- ales [3]). They solve only evolution type nonlinear equations (KdV, sine- Gordon etc.) in this way. Applying essentially Bobylev’s idea to use the Poincar´e theorem on normal forms (see Arnold [1]) we can solve a wide enough class of equations (1). The same method can be used in the case of the right hand side of (1) being a function not only of u but also of the derivatives of u.
1991 Mathematics Subject Classification: 44A10, 46F20.
Key words and phrases : Laplace distributions, Laplace transforms, formal solutions.
Supported by KBN grant 2-P03A-006-08.
[99]
2. Construction of formal solutions. We can write (1) as P (D)u − c
1u =
X
∞ m=2c
mu
m.
Let Z = {(z
1, . . . , z
n) ∈ C
n: P (z) − c
1= 0 } and assume that Z contains a curve Z
′⊂ Z described by
Z
′= {z = (z
1, . . . , z
n) ∈ Z : z = z(k) = (z
1(k), . . . , z
n(k)), k ∈ R}, such that (Z
′+ Z
′) ∩ Z
′= ∅. Let y = y(x
1, . . . , x
n) be a solution of the linear equation (2) given by the integral over Z
′:
(3) y(x
1, . . . , x
n) =
∞
\
−∞
e
−z1(k)x1−...−zn(k)xnΦ(k) dk
with Φ being an arbitrary function such that the integral (3) makes sense for x = (x
1, . . . , x
n) large enough.
We use the following notations. Let Lu =
c11
P (D)u. We denote by H
mfor m ≥ 2 the special m-linear form H
m[u
1, . . . , u
m] = c
mc
1u
1(0) . . . u
m(0).
Let u
x(t) = u(x+t), x, t ∈ C
n, and let Q(u
1, . . . , u
m) denote the function x 7→ Q[u
x1, . . . , u
xm], Q being an m-linear form.
Thus (1) and (2) can be written as
(4) u = Lu −
X
∞ m=2H
m(u, . . . , u) and
(5) y = Ly.
We are looking for a solution u represented by a formal series
(6) u = y +
X
∞ j=2R
j(y, . . . , y)
with some j-linear form R
j. From (6) we get y = u −
X
∞ j=2R
j(y, . . . , y) (7)
= u − R
2(u, u) + X
∞ m=3X
m j=2X
|p|=m, pk≥1
( −1)
sj+1R
j(R
p1, . . . , R
pj),
with the convention R
1(u) = u, |p| = p
1+ . . . + p
jand s
j= # {k : p
k> 1 }.
Thus
Ly = Lu − LR
2(u, u) (8)
− X
∞ m=3X
m j=2X
|p|=m, pk≥1
( −1)
sj+1LR
j(R
p1, . . . , R
pj).
From (6) and (5) we get u = Ly
+ X
∞ j=21
j {R
j(Ly, y, . . . , y) + R
j(y, Ly, . . . , y) + . . . + R
j(y, . . . , y, Ly) }.
Now by (7) and (8) we have u = Lu − LR
2(u, u) −
X
∞ m=3X
m j=2X
|p|=m
( −1)
sj+1LR
j(R
p1, . . . , R
pj)
+ X
∞ j=21
j {R
j(Lu − LR
2(u, u) − . . . , . . . , u − R
2(u, u) − . . .) + . . . + R
j(u − R
2(u, u) . . . , . . . , Lu − LR
2(u, u) . . .) }
= Lu − X
∞ m=2X
m j=2X
|p|=m
( −1)
sj+1{LR
j(R
p1, . . . , R
pj)
− 1
j (R
j(LR
p1, . . . , R
pj) + . . . + R
j(R
p1, . . . , LR
pj)) }.
Now we compare the mth order terms and obtain a recurrence system of equations for R
m, m ≥ 2:
H
m= X
m j=2X
|p|=m
( −1)
sj+1{LR
j(R
p1, . . . , R
pj) (9)
− 1
j (R
j(LR
p1, . . . , R
pj) + . . . + R
j(R
p1, . . . , LR
pj)) }.
To find R
mwe use the Laplace transformation. To this end set e
q(x) = e
−qx(so, according to our previous notation, e
0q(τ ) = e
−qτ), and
λ(q) = L[e
0q] (= L(e
q)(0)), h
m(q
1, . . . , q
m) = H
m[e
0q1, . . . , e
0qm] = c
mc
1, r
m(q
1, . . . , q
m) = R
m[e
0q1, . . . , e
0qm].
Here obviously q, q
j, x, τ ∈ C
nfor j = 1, . . . , m, therefore qx means the
scalar product.
Applying (9) to the system [e
0q1, . . . , e
0qm] we get, for m = 2, h
2(q
1, q
2) = r
2(q
1, q
2)
λ(q
1+ q
2) −
12(λ(q
1) + λ(q
2)) , and for m > 2, defining p
′l= p
1+ . . . + p
l(l = 1, . . . , j),
h
m(q
1, . . . , q
m)
= X
m j=2X
|p|=m
( −1)
sj+1r
j(q
1+ . . . + q
p1, . . . , q
p′j−1+1
+ . . . + q
m)
× r
p1(q
1, . . . , q
p1) . . . r
pj(q
p′j−1+1, . . . , q
m)
×
λ(q
1+ . . . + q
m)
− 1
j (λ(q
1+ . . . + q
p1) + . . . + λ(q
p′j−1+1
+ . . . + q
m))
. Therefore if ∆
j(t
1, . . . , t
j) = λ(t
1+ . . . + t
j) −
1j[λ(t
1) + . . . + λ(t
j)] then we have the recurrence system of formulas for r
m, with q = (q
1, . . . , q
m):
(10)
r
2(q)∆
2(q) = c
2/c
1, r
m(q)∆
m(q) = c
m/c
1+ . . .
+
m−1
X
j=2
X
|p|=m
( −1)
sj+1r
j(q
1+ . . . + q
p1, . . . , q
p′j−1+1
+ . . . + q
m)
× r
p1(q
1, . . . , q
p1) . . . r
pj(q
p′j−1+1
, . . . , q
m)
× ∆
j(q
1+ . . . + q
p1, . . . , q
p′j−1+1
+ . . . + q
m).
Now, with y given by (3), we can write R
m(y, . . . , y)(x)
= R
mh
∞\−∞
e
−z(k1)(x+τ1)Φ(k
1)dk
1, . . . ,
∞
\
−∞
e
−z(km)(x+τm)Φ(k
m) dk
mi
=
\
Rm
e
−[z(k1)+...+z(km)]xΦ(k
1) . . . Φ(k
m)R
m(e
z(k)(0)) dk
1. . . dk
m=
\
Rm
r
m(z(k))e
−[z(k1)+...+z(km)]xΦ(k
1) . . . Φ(k
m) dk.
Thus we have proved
Theorem. Assume that |∆
j(t
1, . . . , t
j) | ≥ A > 0 on the set
W
j(p
1, . . . , p
j) = {(t
1, . . . , t
j) : t
k∈ p
kZ
′, k = 1, . . . , j }
for every j ∈ N, j ≥ 2 and every (p
1, . . . , p
j) ∈ N
j. Then we have a solution u of (1) given by a formal series
(11) u(x) = X
∞ m=1\
Rm
r
m(z(k))Φ(k
1) . . . Φ(k
m)e
−[z(k1)+...+z(km)]xdk, with r
mdefined by (10) and Φ as in (3).
3. Example. Consider the equation
(12) ∆u = ∂
2u
∂x
21+ ∂
2u
∂x
22= X
∞ j=1c
ju
j. We can write it as
∆u − au = bu + X
∞ j=2c
ju
j,
with a+b = c
1, a > 0. Then P (D) = ∆ −a and λ(q) = q
12+q
22−a. We choose Z
′= {(ik, √
a + k
2) : k ∈ R} and we can see that Z
′+ . . . + Z
′= nZ
′= {(ix, y) : y ≥ √
n
2a + x
2, x, y ∈ R} ⊂ C
2, 2Z
′∩Z
′= ∅ and nZ
′⊂ (n−1)Z
′. Moreover, S
∞n=2
(C
2\ nZ
′) = C
2. We have
λ(z(k
1) + . . . + z(k
l))
= (i(k
1+ . . . + k
l))
2+ ( q
a + k
21+ . . . + q
a + k
l2)
2− a
= (l − 1)a + X
1≤m<j≤l
2( p
a + k
2mq
a + k
2j− k
mk
j) ≥ (l
2− 1)a > 0.
For t
l∈ p
lZ
′(l = 1, . . . , r) and m = p
1+ . . . + p
r(r ≥ 2) we also have
∆
r(t
1, . . . , t
r)
= ∆
r(z(k
1) + . . . + z(k
p1), . . . , z(k
p′r−1+1
) + . . . + z(k
m))
= λ(z(k
1) + . . . + z(k
m)) − 1
r [λ(z(k
1) + . . . + z(k
p1)) + . . . + λ(z(k
p′r−1+1
) + . . . + z(k
m))]
≥
1 − 1 r
m
2+ 2
1 − 1
r
a = r − 1
r (m
2+ 2)a > 0.
So the assumptions of the Theorem are satisfied.
4. Locally finite representation of formal solutions. Assume now that there exists a linear map A : C
n→C
nsuch that A(Z
′) ⊂ R
n+. We define b
R
n+= A
−1(R
n+). We consider the space L
′(ω)(R
n+) of Laplace distributions
(of type ω ∈ R
n, see [4] or [5]), supported by b R
n+, and denote by L
(ω)(b R
n+) the space of test functions φ = ψ ◦ A for some ψ ∈ L
(ω)(R
n+).
Consider a functional S
1defined for φ ∈ L
(ω)(b R
n+) by
(13) S
1[φ] =
∞\
−∞
φ(z(k))Φ(k) dk
with Φ as in (3). Obviously y in (3) is the value of S
1on φ(z) = e
−zx. Now we assign to S
1a defining function Ψ
1:
Ψ
1(z) = 1 (2πi)
nS
1e
−(z−w)2(z − w)
1for z ∈ C
n# b R
n+:= A
−1(C
n# R
n+), where by C
n# R
n+we understand (C \ R
+)
n. We can see by (13) that
Ψ
1(z) = 1 (2πi)
n∞\
−∞
e
−(z−z(k))2(z − z(k))
1Φ(k) dk.
Therefore, for φ ∈ L
(ω)(b R
n+), we can write (cf. [5]) S
1[φ] = X
σ∈{−1,1}n
sgn σ lim
ε→0+
\
b
Rn+
φ(u)Ψ
1(u + iσε) du.
Hence for the solution y of (2) we get the following formula:
y(x) = S
1[e
−zx]
= X
σ∈{−1,1}n
sgn σ lim
ε→0+
1 (2πi)
n\
b
Rn+
∞
\
−∞
e
−zx−(z+iσε−z(k))2(z + iσε − z(k))
1Φ(k) dk dz.
For m ≥ 2, the mth term of the formal solution u is the value on φ(z) = e
−zxof the Laplace distribution S
mgiven by
S
m[φ] =
\
Rm
φ(z(k
1) + . . . + z(k
m))Φ(k
1) . . . Φ(k
m)r
m(z(k)) dk
with r
mgiven by (10). By analogy with the case m = 1, we obtain the defining function for S
m:
Ψ
m(z) = 1 (2πi)
nS
me
−(z−w)2(z − w)
1= 1
(2πi)
n\
Rm
e
−(z−(z(k1)+...+z(km)))2(z − (z(k
1) + . . . + z(k
m)))
1Φ(k
1) . . . Φ(k
m)r
m(z(k)) dk.
Hence we get
S
m[e
−zx] = X
σ∈{−1,1}n
sgn σ lim
ε→0+
\
b
Rn+
e
−zxΨ
m(z + iσε) dz.
We see that for m ≥ 2, supp S
m= mZ
′, and R
m(y, . . . , y)(x) = S
m[e
−zx].
So if (m + 1)Z
′⊂ mZ
′, then we have u(x) =
X
∞ m=1S
m[e
−zx] = S[e
−zx]
and this formal series is locally finite, that is, for every φ ∈ L
(ω)(b R
n+) with supp φ bounded, S[φ] is the sum of a finite number of terms. More precisely, let φ(x) = 0 for |x| ≥ M, x ∈ b R
n+. Since we can find N ∈ N such that N Z
′∩ {x : |x| < M} = ∅ we have S
m[φ] = 0 for m ≥ N. So S[φ] = P
N−1m=1