On completeness of the Moutard transformations

6 Pages·0.09 MB·English

by  E. I. Ganzha

On completeness of the Moutard transformations∗
7
9 Ganzha E.I.
9
Dept. Mathematics,
1

n Krasnoyarsk State Pedagogical University
a
J Lebedevoi, 89, 660049, Krasnoyarsk, RUSSIA,

0
e-mail: [email protected]
2

2 9 June 1996
v
1
0
0
The Moutard equation
6
0
6 u = M(x,y)u, u = u(x,y), (1)
9 xy
/
t appeared initially in the classical differential geometry in the second half of the XIX century
n
i (see [1, 2]). It has now numerous applications in the theory of (2+1)-dimensional integrable
-
v
systems of partial differential equations. The other its form u − u = M(x,y)u and the
l tt zz
o analogous elliptic equation
s
: u +u = M(x,y)u (2)
v xx yy
i
X (the 2-dimensional Schr¨odinger equation) show the role of (1) in mathematical physics. In the
r framework of the classical differential geometry (1) played the key role in the central problems
a
of that epoch — the theory of surface isometries, the theory of congruences and conjugate nets.
In the modern soliton theory (1) was used to obtain solutions for the Kadomtsev-Petviashvili,
Novikov-Veselov equations and others ([3, 5]). The main instrument in applications of (1) is
the Moutard transformation which, provided we are given two different solutions u = R(x,y)
and u = ϕ(x,y) of (1) with a given ”potential” M = M (x,y), gives us (via a quadrature) a
0
solutionϑ(x,y)of(1)withtransformedpotentialM (x,y) = M −2(lnR) . Thecorresponding
1 0 xy
transformation formulas
2R R 1
x y
M = M −2(lnR) = −M + = R (3)
1 0 xy 0 R2 R
(cid:18) (cid:19)xy
(Rϑ) = −R2 ϕ ,
x R x (4)
 (Rϑ) = R2 ϕ(cid:16) (cid:17),
 y R y
(cid:16) (cid:17)
(and their analogues for (2), see [2, 3]) establish a (multivalued) correspondence between the
solution of the Moutard equation (M ) (i.e. (1) with the potential M (x,y)) and (M ) (i.e. (1)
0 0 1
∗The researchdescribed in this contribution was supported in part by grant RFBR-DFG No 96-01-00050
1
with the potential M (x,y)). Note that R = 1 is also a solution of (M ). Let us suppose that
1 1 R 1
we can find the complete solution of (1) for some given potential M ; such a solution has the
0
initial data given for example by the Goursat problem (2 functions of 1 variable), from (4) one
may obtain (via a quadrature) the general solution of (M ). For example for the case M = 0
1 0
the general solution is given by u = ϕ(x)+ψ(y), the Moutard transformation yields
u = α(x)ψ(y)−β(y)ϕ(x)+ (ψ β −β ψ)dy +
y y
(cid:18) Z
1
+ (ϕα −αϕ )dx
x x
α(x)+β(y)
Z (cid:19)
for the potential M = −2[ln(α(x)+β(y))] = 2αxβy (i.e. R = α(x)+β(y)).
1 xy (α+β)2
If we will build the following chain of Moutard transformations (M ) → (M ) → (M ) →
0 1 2
... (M ) → ...wewillseethat(apriori)thek-thpotentialdependsonthechoiceof2k functions
k
of 1 variable — the initial data of the solutions R (x,y) of (M ), s = 0,1,...,k −1.
s s
In [3] a method was found that gives us the possibility to express the potential M and all
k
solutions of(M ) via 2k solutionsof theinitialequation(M )(the ”pfaffianformula”, analogous
k 0
to the ”wronskian formula” for the case of the Darboux transformations for (1+1)-dimensional
integrable equations [5, 6]).
Nevertheless the fundamental question how ”wide” is the obtained class of potentials M (in
k
the space of all smooth functions of two variable) was open. In the theory of (1+1)-dimensional
integrable equations the question of density of the finite gap solutions of the Korteweg-de Vries
equation in the class of all quasiperiodic functions was answered positively. In this paper we
show that the set of the potentials M obtainable from any fixed M is ”locally dense” in the
k 0
space of all smooth functions in a sense to be detailed below. Hence we prove that the found
in [3, 4] families of solutions of the corresponding (2+1)-dimensional integrable equations give
locally ”almost every” solution.
Theorem 1 Let an initial potential M (x,y) ∈ C∞ be given in a neighborhood of (0,0).
0
Then for any N = 0,1,2,... one may find some K such that for any given numbers P ,
x1... xk
0 ≤ k ≤ N, x ∈ {x,y}, the corresponding derivatives of M (in the chain of the Moutard
s K
transformations (M ) → (M ) → (M ) → ...(M ) → ...) at (0,0) coincide with P :
0 1 2 n x1... xk
∂
∂ ∂ ... ∂ M (0,0) = P , ∂ = , k ≤ N. (5)
x1 x2 xk K x1... xk z ∂z
Proof will be given by induction. For N = 0 we set K = 1, from (3)
R R
x y
M (0,0) = −M (0,0)+ 2
1 0 R2
(cid:12)(0,0)
(cid:12)
(cid:12)
where R is a solution of (M ). As one canprove (see for examp(cid:12)le [1]), the values ϕ(x) = R(x,0),
0
ψ(y) = R(0,y) (ϕ(0) = ψ(0)) may be chosen as the initial data for (1). So the quantities
R(0,0), R (0,0), R (0,0), ... , R (0,0), R (0,0), ... are independent. Set R(0,0) = 1,
x xx y yy
R (0,0) = 1/2. Changing R (0,0) one can make M (0,0) equal to the given number P which
y x 1
2
proves Theorem for the case N = 0. In addition if we will suppose that all the other nonmixed
derivatives ∂kR, ∂kR, k > 1, are equal to 0 then the higher derivatives ∂m∂nM are not
x y x y 1
arbitrary and are uniquely determined by P = M (0,0) (for the given M (x,y)). Therefore we
1 0
have proved for the case N = N = 0 the following main inductive proposition:
0
foranyN = N onecanfindK suchthatthecorrespondingderivatives∂m∂nM ofanappropri-
0 x y K
ately constructed MK at (0,0)coincide with the given numbers Px... xy... y for m+n ≤ N0,
m n
and the higher derivatives ∂m∂nM , m+ n > N , depend (for the fixed M (x,y)) only upon
x y K 0 | {z }| {z0}
Px... xy... y = ∂xp∂yqMK(0,0) for p+q ≤ N0, p ≤ m.
p q
The step of induction. Provided the main inductive proposition is proved for all deriva-
| {z }| {z }
tives of orders ≤ N = N we will prove it for N = N + 1. Let K be the corresponding to
0 0 0
N = N number of the potential M for which we have proved the inductive proposition.
0 K0
Performing another N +2 Moutard transformations we get M , P = K +N +2; let us check
0 P 0 0
the validity of the inductive proposition for this function. ¿From (3) we get
2R(0)R(0) 2R(1)R(1) 2R(N0+1)R(N0+1)
M = (−1)N0 M − x y + x y −···± x y ,
P K0 (R(0))2 (R(1))2 (R(N0+1))2 !
where R(s) are solutions of (M ), s = 0,... ,N +1. Let us call the derivative ∂s+1R(s) at
K0+s 0 x
the origin the principal derivative of R(s) and ∂N0+2−sR(s) its auxiliary derivative. We suppose
y
further that at (0,0) the values of all R(s) are equal to 1 and the values of their auxiliary
derivatives are equal to 1/2, the principal derivatives are (so far) undefined, all the other
nonmixed derivatives R(s) , ... , R(s) , ... of all orders are set to 0 at the origin. Consider
x... x y... y
now the derivative
∂N0+1M ∂N0+1M 2R(0)∂N0+2R(0)
P = (−1)N0 K0 − x y +
∂yN0+1 ∂yN0+1 (R(0))2
2R(N0+1)∂N0+2R(N0+1) (6)
x y
+···± +
(R(N0+1))2 !
+F(R(s),R(s),∂kR(s),M ,∂mM ),
x y K0+s y K0+s
(at (0,0)), where F comprises all the terms except the given in parentheses. The mixed deriva-
tives ∂m∂nR(s) are eliminated using (1). As one can easily see the only principal derivative R(0)
x y x
does not appear in F.
We will use an additional induction over k to prove that the values of M and their
K0+s
derivatives w.r.t y included in F (they have the orders ≤ N ) at the origin are determined
0
uniquely by the equations ∂ykMP = Py... y, k = 0,1,... ,N0. Indeed for k = k0 = 0
k
| {z }2R(0)R(0) 2R(N0+1)R(N0+1)
P = M | = (−1)N0 M − x y +···± x y ,
P (0,0) K0 (R(0))2 (R(N0+1))2 !(cid:12)
(cid:12)(0,0)
(cid:12)
(cid:12)
where all the derivatives of R(s) except R(0) are not principal i.e. fixed. Since t(cid:12)he coefficient of
x
the principal derivative R(0) is R(0) = 0, we can find M (0,0). The values M are found
x y K0 K0+s
3
from
2R(0)R(0) 2R(s−1)R(s−1)
M = (−1)s M − x y +···± x y . (7)
K0+s K0 (R(0))2 (R(s−1))2 !
For k = k +1 ≤ N (the step of the additional induction)
0 0
Py... y = ∂yk0+1MP = (−1)N0 ∂yk0+1MK0 − Ns=00+1(−1)s2R(xs()R∂yk(s0)+)22R(s)!+
k0+1 P
+F(R(q),R(q),∂kR(q),M ,∂mM ),
| {z } x y K0+q y K0+q
whereagaintheonlyprincipalderivativeR(0) haseverywherezerocoefficients(sincek +1 ≤ N )
x 0 0
and the derivatives ∂mM have the orders m ≤ k i.e. are already found. The derivatives
y K0+q 0
∂k0+1M are found differentiating (7). The additional induction is finished.
y K0+q
Since according to the main inductive proposition we can actually choose ∂mM , 0 ≤
y K0
m ≤ N , arbitrarily, we set them in (6) to be equal to the values found above. Among the
0
still indefinite quantities in (6) only the principal derivative R(0) is present. Its coefficient is
x
equal to 2∂yN0+2R(0) = 1. Choosing appropriately the value of R(0) we will obtain the desired
(R(0))2 x
equality Py... y = ∂yN0+1MP for arbitrary ∂yk0+1MK0. Since due to the inductive proposition
N0+1
∂N0+1M depends only upon ∂mM , m ≤ N , the chosen R(0) depends only on them and
y K0| {z } y K0 0 x
Py... y = ∂yN0+1MP. Consequently calculating all the higher derivatives ∂ymMP of the orders
N0+1
m|>{zN0}+1, we conclude that they depend only on the quantities Py... y, m ≤ N0 +1.
m
The possibility to obtain Px... xy... y = ∂xn∂yN0+1−nMP, n ≤ |N0{z+}1, will be proved by
n N0+1−n
an auxiliary induction over n. The case n = 0 has been just considered.
| {z }| {z }
Provided we have proved for all n ≤ n that
0
a) the principal derivatives R(0), R(1), ... , Rx(n)... x, are already chosen in such a way that
x xx
n+1
Px... xy... y = ∂xk∂yN0+1−kMP, k ≤ n, ho|ld{,z }
k N0+1−k
b) in| {z }| {z }
∂k∂N0+1−kM = (−1)N0 ∂k∂N0+1−kM
x y P x y K0

N0+1 2∂k+1R(s)∂N0+2−kR(s)
− (−1)s x y (8)
(R(s))2
s=0
X
+F(R(q),∂mR(q),∂kR(q),M ,∂r∂mM ) ,
x y K0+q x y K0+q
!
for k ≤ n in the terms collected in F only the principal derivatives defined during the
previous steps as well as ∂r∂mM with r ≤ n, m+r ≤ N , K +q < P, are present,
x y K0+q 0 0
4
c) all the higher derivatives ∂m∂kM , m + k > N + 1, m ≤ n, and the already defined
x y P 0
principal derivatives ∂s+1R(s), s ≤ n, depend (according to our construct of M ) only on
x P
the choice of Px... xy... y, p+q ≤ N0 +1, p ≤ m.
p q
we will make the in|du{cztiv}e|st{ezp}n = n + 1. Then in the rest F only the already defined
0
principal derivatives of R(s) and ∂r∂mM with r ≤ n +1, m+r ≤ N , will appear alongside
x y K0+q 0 0
with the still unknown ∂n0+1∂mM .
x y K0+q
In order to determine these quantities we will make another imbedded induction over m
using the equations Px... xy... y = ∂xn0+1∂yN−n0−1MP, N ≤ N0. Indeed for m = 0 in
n0+1 N−n0−1
| {z }| {z }
N0+1 2∂n0+2R(s)R(s)
∂n0+1M = (−1)N0 ∂n0+1M − (−1)s x y +F (9)
x P  x K0 (R(s))2 
s=0
X
 
(n + 1 ≤ N + 1) only the known quantities appear except ∂n0+1M which we may choose
0 0 x K0
according to the main inductive proposition in such a way that Px... x = ∂xn0+1MP|(0,0) hold.
n0+1
Differentiating (7) we find ∂n0+1M | , K + q < P inductively over q. The step of the
x K0+q (0,0) 0 | {z }
imbedded induction we perform as earlier applying the operator ∂m to (9). Again according to
y
the main inductive proposition (valid for M ) ∂n0+1∂N0−n0M depends only upon ∂n∂mM ,
K0 x y K0 x y K0
m + n ≤ N , n ≤ n + 1, which due to (8) implies the same dependence of the principal
0 0
derivative ∂n0+2Rn0+1. The validity of a), b), c) is easily checked now for n = n +1.
x 0
The auxiliary induction over n proves that we can consecutively choose definite values of
R(0), R(1), ..., ∂N0+2R(N0+1) in such a way that (5) hold for N = N +1 and (5) holds also for
x xx x 0
N ≤ N after an appropriate choice of ∂m∂mM , m + r ≤ N , which is possible due to the
0 x y K0 0
inductive proposition. The higher derivatives ∂m∂mM , m+ r > N +1, depend only upon
x y K0 0
the lower order ones as indicated in the inductive proposition. Therefore the main inductive
proposition as well as Theorem are proved.
References
[1] Darboux G. Lec¸ons sur la th´eorie g´en´erale des surfaces et les applications g´eom´etriques du
calcul infinit´esimal, t. 1–4, Paris, (1887–1896).
[2] Bianchi L. Lezioni di geometria differenziale, 3-a ed., v. 1–4, Bologna, 1923–1927.
[3] Athorne C., Nimmo J.J.C. On the Moutard transformation for integrable partial differen-
tial equations, Inverse Problems, 1991, v. 7, p 809–826.
[4] Nimmo J.J.C. Darboux transformations in (2 + 1) dimensions, in: Proc. NATO ARW
”Applications of Analytic and Geometric Methods to Nonlinear Differential Equations”
(ed. P. Clarkson), NATO ASI Series, Kluwer, 1992, p. 183–192.
5
[5] Oevel W., Rogers C. Gauge transformations and reciprocal links in 2+1 dimensions, Rev.
Mod. Phys., 1993, v. 5, p. 299–330.
[6] Matveev V.B., Salle M.A. Darboux transformations and solitons. Springer-Verlag, 1991.
6