Received 20 July 2017, Accepted 31 July 2017, Available Online 6 January 2021.
1. Introduction
When considering hyperbolic type equations
uxy=g(x,y,u,ux,uy)(1.1)
one finds an important special subclass, so called Darboux integrable equations, that is described in terms of
x-and
y-integrals. Recall that a function
W(
x,
y,
u,
ux,
ux x, …) is called a
y–integral of equation
(1.1) if
DyW(
x,
y,
u,
ux, …)|
(1.1) = 0, where
Dy represents the total derivative with respect to
y( see [
2] and [
8]). An
x-integral
W¯=W¯(x,y,u,uy,uyy,…) for equation
(1.1) is defined in a similar way. Equation
(1.1) is said to be Darboux integrable if it admits a nontrivial
x-integral and a nontrivial
y–integral.
The classification problem for Darboux integrable equations was considered by Goursat, Zhiber and Sokolov (see [2] and [8]). In his paper Goursat obtained a supposedly complete list of Darboux integrable equations of the form (1.1). A detailed discussion of the subject and corresponding references can be found in the survey [9].
Later Laine [7] published two Darboux integrable hyperbolic equations, which were absent in Goursat's list. The first equation found by Laine is
uxy=ux(uy+uyu−y+uyu−x).(1.2)
It has a second order
y-integral
W1=uxxux−12ux(1u−y+3u−x)+1u−x(1.3)
and a third order
x-integral
W¯=(uyyy−uyy22uy−uyy1+5uy12+4uyu−y)(uyy−2uy+2uy32+uy2u−y)−1 −(2uy+2uy32−6uy2−10uy52−4uy3(u−y)2)(uyy−2uy+2uy32+uy2u−y)−1.(1.4)
The second equation found by Laine is
uxy=2((u+X)2+ux+(u+X)(u+X)2+ux)(uy+uyu−y−uy(u+X)2+ux).(1.5)
It has a second order
y-integral
W2=uxx2ux(1−u+X(u+X)2+ux)+u+(u+X)2+2ux(u+X)2+ux −(u+X)2+ux+(u+X)(u+X)2+uxu−y(1.6)
and a third order
x-integral
(1.4). For the second equation Laine assumed
X to be an arbitrary function of
x. However Kaptsov (see [
6]) has shown that
X must be a constant function if equation
(1.5) admits the integrals
(1.6) and
(1.4). Thus it can be assumed, without loss of generality, that
X = 0.
One can also consider a semi-discrete analogue of Darboux integrable equations (see [1]). The notion of Darboux integrability for semi-discrete equations was developed by Habibullin (see [3]). For a function t = t(n, x) of the continuous variable x and discrete variable n we introduce notations
tk=t(n+k,x), k∈ℤ, t[m]=dmdxmt(n,x), m∈ℕ.
Then a hyperbolic type semi-discrete equation can be written as
t1x=f(x,n,t,t1,tx).(1.7)
A function
F of variables
x,
n, and
t,
t1,…,
tk is called an
x-integral of equation
(1.7) if
Dx F|
(1.7) = 0. A function
I of variables
x,
n,
t,
t[1],…,
t[m] is called an
n-integral of equation
(1.7) if
DI|
(1.7) =
I, where
D is a shift operator. Equation
(1.7) is said to be Darboux integrable if it admits a nontrivial
n-integral and a nontrivial
x-integral. In what follows we consider the equalities
Dx = 0 and
DI =
I, which define
x-and
n-integrals
F and
I, only on solutions of the corresponding equations. For more information on semi-discrete Darboux integrable equations see [
3], [
4] and [
5].
The interest in the continuous and discrete Darboux integrable models is stimulated by exponential type systems. Such systems are connected with semi-simple and affine Lie algebras which have applications in Liouville and conformal field theories.
The discretization of equations from Goursat’s list was considered by Habibullin and Zheltukhina in [5]. In the present paper we find semi-discrete versions of Laine equations (1.2) and (1.5). In particular we find semi-discrete equations that admit functions (1.3) or (1.6) as n-integrals, and show that these equations are Darboux integrable. This is the main result of our paper given in Theorem 1.1 and Theorem 1.2 below.
Theorem 1.1.
The semi-discrete chain (1.7), which admits a minimal order n-integral
I1=txxtx−12tx(1t−ε(n)+3t−x)+1t−x,(1.8)
where ε(
n)
is an arbitrary function of n,
is
t1x=tx(t1−x)(t−x)B(n,t,t1),(1.9)
where B is a function of n,
t,
t1,
satisfying the following equation
(t1−ε)(t1−ε1)−2(t−ε)(t1−ε1)B+(t−ε)(t−ε1)B2=0.(1.10)
Moreover, chain (1.9) admits an x-integral of minimal order 3.
Theorem 1.2.
The semi-discrete chain (1.7), which admits a minimal order n-integral
I2=txx2tx(1−tt2+tx)+t+t2+2txt2+tx−t2+tx+tt2+txt−ε(n),(1.11)
where ε(
n)
is an arbitrary function of n,
is
t1x=2A(tA−t1)t2+tx+A2tx+2tA(tA−t1),(1.12)
where A is a function of n,
t,
t1,
satisfying the following system of equations
{At=−2t1(t1−ε1)A+(−ε+2t)(t1−ε1)A2−ε1(t−2ε)A32(t1−ε1)(t−ε)(t1−tA),At1=ε(t1−ε1)+(t−ε)(2t1−ε1)A−2t(t−2ε)A22(t1−ε1)(t−ε)(t1−tA).(1.13)
Moreover, chain (1.12) admits an x-integral of minimal order 2.
The paper is organized as follows. In Sections 2 and 3 we give proofs of Theorems 1.1 and 1.2 respectively. In Section 4 we show that function (1.4) can not be a minimal order n-integral for any equation (1.7).
2. Proof of Theorem 1.1
Discretization by n-integral: Let us find f(x, n, t, t1, tx) such that DI1 = I1, where I1 is defined by (1.8). Equality D I1 = I1 implies
fx+fttx+ft1f+ftxtxxf−f2(1t1−ε1+3t1−x)+1t1−x=txxtx−tx2(1t−ε+3t−x)+1t−x,(2.1)
where
ε =
ε(
n) and
ε1 =
ε(
n + 1).
By comparing the coefficients before txx in (2.1), we get ftxf=1tx, which implies that f = A(x, n, t, t1)tx. We substitute this expression for f in (2.1) and get
Ax+Attx+At1AtxA−Atx2(1t1−ε1+3t1−x)+1t1−x=−tx2(1t−ε+3t−x)+1t−x.(2.2)
The above equation is equivalent to a system of two equations
{AxA+1t1−x=1t−x,AtA+At1−A2(1t1−ε1+3t1−x)=−12(1t−ε+3t−x).(2.3)
The first equation of system
(2.3) can be written as
∂∂x(ln|A|−ln|t1−x|+ln|t−x|)=0 which implies that
A(x,n,t,t1)=t1−xt−xB(n,t,t1)(2.4)
for some function
B of variables
n,
t,
t1. Substituting expression
(2.4) for
A into the second equation of system
(2.3), we get
−1t−x+BtB+Bt−x+Bt1(t1−x)t−x−B(t1−x)2(t−x)(1t1−ε1+3t1−x)=−12(1t−ε+3t−x).(2.5)
Thus
(t−x)BtB+(t1−x)Bt1−B2(1+t1−xt1−ε1)=−12(1+t−xt−ε).(2.6)
We compare the coefficients before
x and
x0 in
(2.6) and obtain
{−BtB−Bt1+B2(t1−ε1)=12(t−ε),tBtB+t1Bt1−B2−t1B2(t1−ε1)=−12−t2(t−ε),(2.7)
which is equivalent to
{Bt=B(ε−2t+t1−εB+tB)2(t−ε)(t−t1),Bt1=−ε1+t1+ε1B+tB−2t1B2(t1−ε1)(t−t1).(2.8)
The last system is compatible, that is
Bt t1=Bt1t, if and only if equality
(1.10) is satisfied.
Existence of an x-integral: Let us show that equation (1.9) where function B satisfies (1.10) has a finite dimensional x-ring. We have,
t1x=t1−xt−xBtx, t2x=t2−xt−xBB1tx, and t3x=t3−xt−xBB1B2tx,(2.9)
where
B =
B(
n,
t,
t1),
B1 =
B(
n + 1,
t1,
t2) and
B2 =
B(
n + 2,
t2,
t3). We are looking for a function
F(
x,
n,
t,
t1,
t2,
t3) such that
DxF = 0, that is
Fx+Fttx+Ft1t1x+Ft2t2x+Ft3t3x=0.(2.10)
Thus
Fx+Fttx+Ft1t1−xt−xBtx+Ft2t2−xt−xBB1tx+Ft3t3−xt−xBB1B2tx=0,(2.11)
which is equivalent to
{Fx=0(t−x)Ft+(t1−x)BFt1+(t2−x)BB1Ft2+(t3−x)BB1B2Ft3=0.(2.12)
By comparing the coefficients of
x0 and
x in the last equality we get the following system
{tFt+t1BFt1+t2BB1Ft2+t3BB1B2Ft3=0,−Ft−BFt1−BB1Ft2−BB1B2Ft3=0.(2.13)
After diagonalization this system becomes
{Ft +BB1(t2−t1)t−t1Ft2+BB1B2(t3−t1)t−t1Ft3=0,Ft1+B1(t−t2)t−t1Ft2+B1B2(t−t3)t−t1Ft3=0.(2.14)
We introduce vector fields
V1=∂∂t+BB1(t2−t1)t−t1∂∂t2+BB1B2(t3−t1)t−t1∂∂t3,V2=∂∂t1+B1(t−t2)t−t1∂∂t2+B1B2(t−t3)t−t1∂∂t3.(2.15)
and
V = [
V1,
V2]. Then, we have
2(t−t1)2B1V=(t1−t2+B(t2−t+(t−t1)B1)∂∂t2+B2(t1−t3+B(t3−t+(t−t1)B1B2))∂∂t3.
Direct calculation show that
[V1,V]=3ε−4t+t12(ε−t)(t−t1)V and [V2,V]=3ε1+t−4t12(ε1−t1)(t1−t)V.(2.16)
Hence vector fields
V1,
V2 and
V form a finite-dimensional ring. By the Jacobi Theorem the system of three equations
V1(
F) = 0,
V2(
F) = 0,
V(
F) = 0 has a nonzero solution
F(
t,
t1,
t2,
t3). The function
F(
t,
t1,
t2,
t3) is an
x-integral of equation
(1.9).
3. Proof of Theorem 1.2
Discretization by n-integral: Let us find a function f(x, n, t, t1, tx) such that DI2 = I2, where I2 is given by (1.11). The equality DI2 = I2 implies that
fx+fttx+ft1f+ftxtxx2f(1−t1t12+f)−t12+f+t1t12+ft1−ε1+t1+t12+2ft12+f=txx2tx(1−tt2+tx)−t2+tx+tt2+txt−ε+t+t2+2txt2+tx,(3.1)
where
ε =
ε(
n) and
ε1 =
ε(
n + 1). Comparing the coefficients before
txx in equality
(3.1), we get
ftxf(1−t1t12+f)=1tx(1−tt2+tx).(3.2)
This can be written as
∂∂txln(ff+t12+t1f+t12−t1)=∂∂txln(txtx+t2+ttx+t2−t).(3.3)
Thus
f+t12+t1=(tx+t2+t)A(x,n,t,t1),(3.4)
where
A is some function of variables
x,
n,
t and
t1. The last equality is equivalent to
f=(2A2t−2At1)tx+t2+A2tx+t(2A2t−2At1).(3.5)
We substitute
f given by
(3.5) into equality
(3.1), use
(3.4) and equality
f+t12−t1=f(tx+t2−t)Atx
to get
1tx+t2f+t12(Λ1tx2+Λ2txtx+t2+Λ3tx+Λ4tx+t2++Λ5t2)=0,(3.6)
where
Λi=αi1Ax+αi2At+αi3At1+αi4, 1≤i≤5(3.7)
and
α11=0,α12=1,α13=A2,α14=At−ε−A3t1−ε1,α21=0,α22=t−t1A,α23=−3t1A+3tA2,α24=−t1+2tAt−ε+2t1A2−3tA3t1−ε1+A2−A,
α31=1,α32=t2,α33=2t12+5t2A2−6t1tA,α34=−t1t+t(t+2ε)At−ε+−5t2A3+4t1tA2−t12At1−ε1+t1+2tA2−t1A,
α41=t−t1A,α42=0,α43=4t3A2−6t1t2A+2t12t,α44=2εt2A+εtt1t−ε+−4t3A3+4t1t2A2−t12tAt1−ε1+2t2A2−t1tA,
We can solve the overdetermined system of linear equations Λ
i = 0,
i = 1,2 … 5, with respect to
Ax At,
At1 and obtain
{Ax=0,At=−At−ε+A22(t1−tA)(Aε1t1−ε1−εt−ε),At1=At1−ε1−12(t1−tA)(Aε1t1−ε1−εt−ε).(3.8)
By direct calculations one can check that
At t_{1} =
At_{1 t}, so the above system has a solution.
Existence of an x-integral: We are looking for a function F(t, t1, t2) such that DxF = 0 that is
Fttx+Ft1t1x+Ft2t2x=0,(3.9)
where
t satisfies equation
(1.7) with function
f given by
(3.5). We use
t1x=A2(t,t1)tx+2A(t,t1)(tA(t,t1)−t1)(tx+t2+t)
and
f+t12=(tx+t2+t)A−t1,
to get
t2x=A2(t,t1)A2(t1,t2)tx+2(tx+t2+t)(tA(t,t1)−t1)A(t,t1)A2(t1,t2)+2(tx+t2+t)(t1A(t1,t2)−t2)A(t,t1)A(t1,t2).
By substituting these expressions for
t1x and
t2x into equality
(3.9) and comparing the coefficients of
tx+t2,tx and
tx0, we obtain the following system of equations
{Ft2A(t,t1)(tA(t,t1)−t1)Ft1+2A(t,t1)A(t1,t2)(tA(t,t1)A(t1,t2)−t2)Ft2=0,+A2(t,t1)Ft1+A2(t,t1)A2(t1,t2)Ft2=0,2tA(t,t1)(tA(t,t1)−t1)Ft1+2tA(t,t1)A(t1,t2)(tA(t,t1)A(t1,t2)−t2)Ft2=0.
To check for the existence of a solution we transform the above system to its row reduced form
{Ft +A2(t,t1)A(t1,t2)(t2−t1A(t1,t2))tA(t,t1)−t1Ft2=0,Ft1+A(t1,t2)(t2−tA(t,t1)A(t1,t2))−tA(t,t1)+t1Ft2=0.(3.10)
The corresponding vector fields
V1=∂∂t+A2(t,t1)A(t1,t2)(t2−t1A(t1,t2))tA(t,t1)−t1∂∂t2,V2=∂∂t1+A(t1,t2)(t2−tA(t,t1)A(t1,t2))−tA(t,t1)+t1∂∂t2
commute, that is [
V1,
V2] = 0, provided
A satisfies system
(3.8). Thus by the Jacobi theorem, system
(3.10) has a solution. To solve the system define a function
E(
t,
t1,
t2) by
Et=A2tA−t1,Et2=1A1(t1A1−t2),Et1=t2−tAA1(tA−t1)(t1A1−t2)+1t1−ε1E,
where
A =
A(
t,
t1) and
A1 =
A(
t1,
t2).
One can check that Ett1=Et1t and Et1t2=Et2t1, so such a function E exists. Function E is a first integral of the first equation of system (3.10). We write system (3.10) using new variables
t˜=t,t˜1=t1,t˜2=E(t,t1,t2)
and obtain
{Ft˜=0Ft˜1+t˜2t˜1−ε1Ft˜2=0.(3.11)
Therefore one of the
x-integrals is
F(
t,
t1,
t2) =
E(
t,
t1,
t2) /(
t1−
ε(
n + 1)) where function
E defined above.
4. Nonexistence of a chain (1.7) admitting the minimal order n-integral (1.4)
Let us find a function f(x, n, t, t1, tx) such that equation (1.7) has the n-integral
I=txxx−txx22tx−txx1+5tx+4txt−x−2tx+2txtx−6tx2−10tx2tx−4tx3(t−x)2txx−2tx+4txtx+2tx2t−x.
We have,
t1x=f(x,n,t,t1,tx),t1xx=fx+fttx+ft1f+ftxtxx,t1xxx=(fxx+fxttx+fxt1f+fxtxtxx)+tx(fxt+ftttx+ftt1f+fttxtxx)+fttxx+f(fxt1+ftt1tx+ft1t1f+ft1txtxx)+ft1(fx+fttx+ft1f+ftxtxx)+txx(fxtx+fttxtx+ft1txf+ftxtxtxx)+ftxtxxx.
Equality
DI =
I is equivalent to
J: =
L(
DL)(
DI −
I) = 0, where
L=2tx(t−x){txx(t−x)−2tx(tx+1)2}. We have
J=Λ1txxx+Λ2txx3+Λ3txx2+Λ4txx+Λ5,
where Λ
k, 1 ≤
k ≤ 5, are some functions of variables
x,
n,
t,
t1,
tx. In particular,
Λ12(t−x)(t1−x)txf=2(t−x)f(1+f)2−2(t1−x)txftx(1+tx)2−(t1−x)(t−x)(fx+fttx+ft1f),Λ2=(t−x)2(t1−x)2{fftx−txftx2+2txfftxtx},Λ3(t−x)(t1−x)=(t−x)f[4f3/2+2f2+(x−t1)fx+f(2+(x−t1)ft1)]+10(x−t1)tx3/2fftx+tx[10(t−x)f3/2ftx+2(t−x)(t1−x)ftxfx+4(t−x)f2(2ftx+(x−t1)ft1tx)]+txf(2(t−t1)ftx+(t−x)(x−t1)(3ft+4fxtx))−2(t1−x)tx2[2f(2ftx−ftxtx+(t−x)fttx)+ftx(ftx+(x−t)ft)]−4(ftx2−2fftxtx)(t1−x)tx5/2−2(ftx2−2fftxtx)(t1−x)tx3.
Equality Λ
2 = 0 implies that
fftx−txftx2+2txfftxtx=0, thus
f2ftx∂∂tx{txftx2f}=0.
Hence,
txftx2f=A2(x,n,t,t1) for some function
A depending on
x,
n,
t,
t1 only. Therefore,
ftxf=Atx and hence
∂∂tx{f−Atx=0}. We have,
f=Atx+B
where
A =
A(
x,
n,
t,
t1) and
B =
B(
x,
n,
t,
t1). We substitute
f=A2tx+2ABtx+B2 into
Λ1=0 and get
α1tx2+α2tx3/2+α3tx+α4tx+α5=0.
We solve the system of equations
αk = 0,1 ≤
k ≤ 5, and obtain
B = 0, that is
{Ax=B2ABt−32ABBt1+2(t1−x)B+A{2(t−t1)+6(t−x)B+3(t−x)B2}2(t−x)(x−t1),At=A2BBt+A32BBt1+A{2(t1−x)A+2(x−t1)B−(t−x)A2(2+B)}2(t−x)(x−t1)B,At1=−12ABBt−A2BBt1+2(x−t1)+(t−x)A(2+3B)2(t−x)(x−t1)B,Bx=−B2Bt1−B(1+B)2t1−x.(4.1)
We substitute
f=A2tx+2ABtx+B2 into Λ
3 = 0 and get
β1tx3+β2tx5/2+β3tx2+β4tx3/2+β5tx+β6tx+β7=0.
We solve the system of equations
βk = 0,1 ≤
k ≤ 7, and obtain
B = 0, or
{Ax=3B8ABt−2324ABBt1+21(t1−x)B+A{16(t−t1)+51(t−x)B+23(t−x)B2}24(t−x)(x−t1),At=3A8BBt+3A38BBt1+A{7(t1−x)A+8(x−t1)B−(t−x)A2(7+3B)}8(t−x)(x−t1)B,At1=−38ABBt−3A8BBt1+7(x−t1)+(t−x)A(7+11B)8(t−x)(x−t1)B,Bx=−B2Bt1−B(1+B)2t1−x.(4.2)
We equate expressions for
Ax and
At from
(4.1) and
(4.2) and find
{Bt=−A{2(t1−x)B+A((t−t1)+(t−x)B)}2(t−x)(x−t1)B,Bt1=t−t1+3(t−x)B+2(t−x)B22(t−x)(x−t−1)B.(4.3)
Then, it follows from
(4.1) that
{Ax=(t1−x+(t−x)A)B2(t−x)(x−t1),At=A((t1−x)A+(x−t)A2+2(x−t1)B)2(t−x)(x−t1)B,At1=x−t1+(t−x)A(1+2B)2(t−x)(x−t1)B,Bx=B(t+t1−2x+(t−x)B)2(t1−x)(x−t).(4.4)
Equality
Att1−At1t=0 becomes
(t1−x)2−(t−x)2A3(t−x)2(t1−x)2B=0, thus
A3=(t1−x)2(t−x)2.(4.5)
Equality
Axt1−At1x=0 becomes
−(t1−x)2+(t−x)2A(1+B)2(t−x)2(t1−x)2B=0, thus
A(1+B)2=(t1−x)2(t−x)2.(4.6)
Equality
Axt−
Atx = 0 becomes
(t1−x)2(A−B)2−(t−x)2A3(t−x)2(t1−x)2B=0. It implies that
A3(A−B)2=(t1−x)2(t−x)2,(4.7)
or
A =
B, that leads to
A =
B = 0 and
f = 0. It follows from
(4.5) and
(4.7) that
A −
B = 1 or
A −
B = −1. It follows from
(4.5) and
(4.6) that 1 +
B =
A or 1 +
B = −
A. This gives rise to four possibilities:
- 1)
- 2)
A − B = 1 and A + B = − 1 which gives A = 0, B = − 1 and therefore f = 1;
- 3)
A − B = −1 and A − B = 1 which is an inconsistent system;
- 4)
A − B = −1 and A + B = −1 which gives A = −1, B = 0 and therefore f = tx
We have to study case 1 ) only. In this case we get B = A − 1 and equation t1x=Atx+Bbecomes t1x+1=A(tx+1), that can be written as well as
(t1x+1)3=A3(tx+1)3.(4.8)
Due to
(4.5), our equation
(4.8) becomes
(t1x+1)3(t1−x)2=(tx+1)3(t−x)2.
The last equation admits an
n-integral
I=(tx+1)3(t−x)2 of order one.
Let us consider case B = 0. We write DI − I = 0 for the chain t1 x = C(x, n, t, t1) tx and get
Λ1txxx+Λ2txx2+Λ3txx+Λ4=0
where Λ
k = Λ
k(
x,
n,
t,
t1,
tx), 1 ≤
k ≤ 4. Equation Λ
1 = 0 implies
α1tx+α2tx+α3=0
where
αk =
αk(
x,
n,
t,
t1), 1 ≤
k ≤ 3. In particular,
α2=4C(−(t1−x)+(t−x)C). Since
α2 = 0 we have
C = (
t1−
x)
2(
t −
x)
− 2. The chain becomes
t1 x = (
t1−
x)
2(
t−
x)
− 2tx. It admits the
n-integral
I = (
t −
x)
−2 tx of order one.
Therefore, if equation (1.7) admits n-integral (1.4) then (1.4) is not a minimal order integral.
Acknowledgment
We are thankful to Prof. Habibullin for suggesting the Laine equations discretization problem and for his interest in our work.
References
[2]Goursat Goursat, Recherches sur quelques équations aux dérivés partielles du second ordre, Annales de la faculté des Sciences de l’Université de Toulouse 2e série, Vol. 1, No. 1, 1899, pp. 31-78. [7]Laine Laine, Sur l’a application de la method de Darboux aux equations s = f (x, y, z, p, q), Comptes rendus, Vol. V.182, 1926, pp. 1127-1128. [9]Zhiber Zhiber, Murtazina Murtazina, Habibullin Habibullin, and Shabat Shabat, Characteristic Lie rings and integrable models in mathematical physics, Ufa Math. J.,, Vol. 4, No. 3, 2012, pp. 17-85.