### 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

m## u

^{m}

## with c

1## 6= 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

_{1}

## y.

## 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

1## u =

## X

∞ m=2## c

m## u

^{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

^{−z}

^{1}

^{(k)x}

^{1}

^{−...−z}

^{n}

^{(k)x}

^{n}

## Φ(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 =

_{c}

^{1}

1

## P (D)u. We denote by H

m## for m ≥ 2 the special m-linear form H

m## [u

1## , . . . , u

m## ] = c

m## c

1## u

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

^{x}1

## , . . . , u

^{x}

_{m}

## ], Q being an m-linear form.

## Thus (1) and (2) can be written as

## (4) u = Lu −

## X

∞ m=2## H

m## (u, . . . , u) and

## (5) y = Ly.

## We are looking for a solution u represented by a formal series

## (6) u = y +

## X

∞ j=2## R

j## (y, . . . , y)

## with some j-linear form R

j## . From (6) we get y = u −

## X

∞ j=2## R

j## (y, . . . , y) (7)

## = u − R

2## (u, u) + X

∞ m=3## X

m j=2## X

|p|=m, pk≥1

## ( −1)

^{s}

^{j}

^{+1}

## R

j## (R

p1## , . . . , R

pj## ),

## with the convention R

1## (u) = u, |p| = p

1## + . . . + p

j## and s

j## = # {k : p

k## > 1 }.

## Thus

## Ly = Lu − LR

^{2}

## (u, u) (8)

## − X

∞ m=3## X

m j=2## X

|p|=m, pk≥1

## ( −1)

^{s}

^{j}

^{+1}

## LR

j## (R

p1## , . . . , R

pj## ).

## From (6) and (5) we get u = Ly

## + X

∞ j=2## 1

## 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=3## X

m j=2## X

|p|=m

## ( −1)

^{s}

^{j}

^{+1}

## LR

j## (R

p1## , . . . , R

pj## )

## + X

∞ j=2## 1

## 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=2## X

m j=2## X

|p|=m

## ( −1)

^{s}

^{j}

^{+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=2## X

|p|=m

## ( −1)

^{s}

^{j}

^{+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

m## we use the Laplace transformation. To this end set e

q## (x) = e

^{−qx}

## (so, according to our previous notation, e

^{0}

_{q}

## (τ ) = e

^{−qτ}

## ), and

## λ(q) = L[e

^{0}

_{q}

## ] (= L(e

q## )(0)), h

m## (q

1## , . . . , q

m## ) = H

m## [e

^{0}

_{q}

_{1}

## , . . . , e

^{0}

_{q}

_{m}

## ] = c

m## c

1## , r

m## (q

1## , . . . , q

m## ) = R

m## [e

^{0}

_{q}

_{1}

## , . . . , e

^{0}

_{q}

_{m}

## ].

## Here obviously q, q

j## , x, τ ∈ C

^{n}

## for j = 1, . . . , m, therefore qx means the

## scalar product.

## Applying (9) to the system [e

^{0}

_{q}

_{1}

## , . . . , e

^{0}

_{q}

_{m}

## ] we get, for m = 2, h

2## (q

1## , q

2## ) = r

2## (q

1## , q

2## )

## λ(q

1## + q

2## ) −

^{1}2

## (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=2## X

|p|=m

## ( −1)

^{s}

^{j}

^{+1}

## r

j## (q

1## + . . . + q

p1## , . . . , q

p^{′}

j−1+1

## + . . . + q

m## )

## × r

^{p}1

## (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## ) −

^{1}j

## [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)

^{s}

^{j}

^{+1}

## r

j## (q

1## + . . . + q

p1## , . . . , q

p^{′}

j−1+1

## + . . . + q

m## )

## × r

^{p}1

## (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

m## h

^{∞}

^{\}

−∞

## e

^{−z(k}

^{1}

^{)(x+τ}

^{1}

^{)}

## Φ(k

_{1}

## )dk

_{1}

## , . . . ,

∞

\

−∞

## e

^{−z(k}

^{m}

^{)(x+τ}

^{m}

^{)}

## Φ(k

m## ) dk

m## i

## =

\

R^{m}

## e

^{−[z(k}

^{1}

^{)+...+z(k}

^{m}

^{)]x}

## Φ(k

1## ) . . . Φ(k

m## )R

m## (e

_{z(k)}

## (0)) dk

1## . . . dk

m## =

\

R^{m}

## r

m## (z(k))e

^{−[z(k}

^{1}

^{)+...+z(k}

^{m}

^{)]x}

## Φ(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

k## Z

^{′}

## , 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\

R^{m}

## r

m## (z(k))Φ(k

1## ) . . . Φ(k

m## )e

^{−[z(k}

^{1}

^{)+...+z(k}

^{m}

^{)]x}

## dk, with r

m## defined by (10) and Φ as in (3).

## 3. Example. Consider the equation

## (12) ∆u = ∂

^{2}

## u

## ∂x

^{2}

_{1}

## + ∂

^{2}

## u

## ∂x

^{2}

_{2}

## = X

∞ j=1## c

j## u

^{j}

## . We can write it as

## ∆u − au = bu + X

∞ j=2## c

j## u

^{j}

## ,

## with a+b = c

1## , a > 0. Then P (D) = ∆ −a and λ(q) = q

1^{2}

## +q

^{2}

_{2}

## −a. We choose Z

^{′}

## = {(ik, √

## a + k

^{2}

## ) : k ∈ R} and we can see that Z

^{′}

## + . . . + Z

^{′}

## = nZ

^{′}

## = {(ix, y) : y ≥ √

## n

^{2}

## a + 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

^{2}

_{1}

## + . . . + q

## a + k

_{l}

^{2}

## )

^{2}

## − a

## = (l − 1)a + X

1≤m<j≤l

## 2( p

## a + k

^{2}

_{m}

## q

## a + k

^{2}

_{j}

## − k

^{m}

## k

j## ) ≥ (l

^{2}

## − 1)a > 0.

## For t

l## ∈ p

^{l}

## Z

^{′}

## (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

p_{1}

## )) + . . . + 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

^{n}

## such 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

1## defined 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

1## on φ(z) = e

^{−zx}

## . Now we assign to S

1## a defining function Ψ

1## :

## Ψ

1## (z) = 1 (2πi)

^{n}

## S

1## e

^{−(z−w)}

^{2}

## (z − w)

^{1}

## for 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))

^{1}

## Φ(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

R^{n}_{+}

## 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

R^{n}+

∞

\

−∞

## e

−zx−(z+iσε−z(k))^{2}

## (z + iσε − z(k))

^{1}

## Φ(k) dk dz.

## For m ≥ 2, the mth term of the formal solution u is the value on φ(z) = e

^{−zx}

## of the Laplace distribution S

m## given by

## S

m## [φ] =

\

R^{m}

## φ(z(k

1## ) + . . . + z(k

m## ))Φ(k

1## ) . . . Φ(k

m## )r

m## (z(k)) dk

## with r

m## given by (10). By analogy with the case m = 1, we obtain the defining function for S

m## :

## Ψ

m## (z) = 1 (2πi)

^{n}

## S

m## e

^{−(z−w)}

^{2}

## (z − w)

^{1}

## = 1

## (2πi)

^{n}

\

R^{m}

## e

^{−(z−(z(k}

^{1}

^{)+...+z(k}

^{m}

^{)))}

^{2}

## (z − (z(k

^{1}

## ) + . . . + z(k

m## )))

^{1}

## Φ(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

R^{n}+

## e

^{−zx}

## Ψ

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=1## S

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

## S

m## [φ].

**References**

### [1] *V. I. A r n o l d, Additional Topics in the Theory of Ordinary Differential Equations,* Nauka, Moscow, 1978 (in Russian).

### [2] *A. B o b y l e v, Poincar´e theorem, Boltzmann equation and KdV-type equations, Dokl.*

### Akad. Nauk SSSR 256 (1981), 1341–1346 (in Russian).

### [3] *R. R. R o s a l e s, Exact solutions of some nonlinear evolution equations, Stud. Appl.*

### Math. 59 (1978), 117–151.

### [4] *Z. S z m y d t and B. Z i e m i a n, Laplace distributions and hyperfunctions on R*

^{n}_{+}