Monge equation of arbitrary degree in 1 + 1 space

0.14 MB·English

by  A. N. Leznov

Monge equation of arbitrary degree in 1 + 1
3
space
1
0
2

n A.N.Leznov and R.Torres-Cordoba
∗† ‡
a
J

1
3
Abstract

] Solution of Monge equation of arbitrary degree (∂nU = W(∂nU) is
h ∂xn ∂zn
p connected with solution of functional equation for 4 functions with 4
- different arguments. Some number solutions of this equation is repre-
h
t sented in explicit form.
a
m
[
1 Introduction

1
v
6 So called Monge-Amper equation degree n+1 in 1+1 dimensional x,z space
3 looks as [1]
0 ∂ ∂nU ∂ ∂nU ∂ ∂nU ∂ ∂nU
0 ( ) ( ) = ( ) ( )
. ∂x ∂xn ∂z ∂zn ∂y ∂xn ∂x ∂yn
2
0 The last equation may be considered as unity to zero Jacobian between
3
(∂nU), ∂nU, which means their function dependence or
1 ∂zn ∂xn
:
v ∂nU ∂nU
i = W( )
X ∂xn ∂zn
r
a The last equation we will call as Monge equation of n degree with notation
M . For initial Monge Ampher equation we use term (M A) . Each
n n+1
−
solution of M is simultaneous solution of (M A) . General solution of
n n+1
−
M depend on n functions of one argument and if it will be possible to find
n
it for arbitrary W function then it will be general solution for (M A)
n+1
−
depending on n+1 arbitrary functions of one argument.
∗e-mail: [email protected]
†Universidad Autonoma del Estado de Morelos, CIICAp, Cuernavaca, Mexico
‡Universidad Autonoma de Cd. Juarez, Chihuahua and Universidad Autonoma del
Estado de Morelos, CIICAp, Cuernavaca, Mexico
1
2 Trivial partial solution of (M A)
n+1
−
If in M W(x) = x then solution is obvious
n
n
U = f (x+λ z), λn = 1
k k k
kX+1
From this example it is clear that existence of the solution of M will deter-
n
mine function W for which solution is possible.
3 M
1
U = W(U )
x z
Solution is well known connected with classical Monge equation λ = λλ
z x
which can rewrite in the form of zero Jacobian (x+λz) λ (x+λz) λ = 0
x z z x
−
.The last condition means functional dependence x+λx+λz = F(λ) This
is exactly general solution in implicit form of classical Monge equation.
˜
4 M
2
The results of this section will be used below in sections M ,M . Equation
3 n
M˜ looks as
2
U = W(U ,U )
z,z x,x z,x
Reallyrelationaboveisnottheequationbecause3arbitraryfunctionsin1+1
are always functionally dependent. The talk is about some parametrization
of functions involved in this relation.
In notations a = U = W(b,c), b = U , c = U equation M˜)
z,z x,z xx 2
looks as system of equations a = W b +W c = b , b = c .
x b x c x z z x
This is typical hydrodynamic system. ([2]. By transformation x =
X(b,c),z = Z(b,c) we calculate derivatives
Z Z X X
c b c b
b = , c = , b = , c = , D = Jacob(X,Z)
x x z z
D −D − D D
Linear system of equation above is resolved in terms of one function R,X =
R ,Z = R
c b
R = W R +W R
cc b cb c bb
−
2
There are no known methods for solution of this equation under arbitrary
functionW. Belowweproposesometrick forfinding solutionofthisequation
together with function W. Let us try represent second order differential
equation in factorize form
∂2 ∂2 ∂2 ∂ ∂ ∂ ∂
+W W = ( +ν )( +ν ) (1)
∂c2 b∂b∂c − c∂b2 ∂c 1∂b ∂c 2∂b
Comparison left and right sides lead to conclusion
ν +ν = W , ν ν = W , ν2 +ν1ν2 = 0, ν1 +ν2ν1 = 0 (2)
1 2 b 1 2 − c c b c b
Inwhatfollowsalwaysν ν2, ν ν1. Thereareobviousthreepossibility
2 1
≡ ≡
in resolving the last system of equations, which will be considered on three
subsections below.
4.0.1 ν ,ν = Constants
2 1
In this case R = R (b ν c)+R (b ν c), W = (ν +ν )b ν ν c+w(R ).
1 1 2 2 1 2 1 2 b
x = R = R˙ +R′, z−= R = ν−R˙ ν R′. Resolving tw−o last equations
b ν cb = L1(x +2ν z), b c ν −c =1 L1 −(x +2 ν2z) or finally c = L2−L1, c =
− 2 1 1 − 1 2 2 ν2−ν1
ν2L2−ν1L1. InthesametermsW = ν22L2−ν12L1+w(x). Letusintroducenotation
ν2−ν1 ν2−ν1
lk = ν2sLk2−ν1sLk1. Lk k-derivatives of L function by their own argument. It is
s ν2−ν1 1[2
obvious (lk) = lk+1, (lk) = lk+1.
s x s s z s+1
4.0.2 ν = Constant,ν = ν (b ν c)
1 2 2 1
−
ν = ν ((b ν c)), R = R ((b ν c))+R ( (b−ν1cds 1 b))
2 2 − 1 1 − 1 2 ν1−ν2(s) −
R
ν ν ν
x = R = R′+R˙ 1 , z = R = ν R′+R˙ 2 1 , x+ν z = ν R˙ ,
b 1 2 ν ν c − 1 1 2 ν ν 1 − 1 2
2 1 2 1
− − − −
′
x+ν z = ∆R Θ(ν ) c = C(ν ) L(x+ν z), b = dν ν C ν L(x+ν z),
2 ≡ 2 2 − 1 Z 2 2 ν2− 1 1
a = dν ν2C ν2L(x+ν z)+θ(z),
Z 2 2 ν2 − 1 1
3
4.0.3 ν2 +ν1ν2 = 0, ν1 +ν2ν1 = 0
c b c b
In the system of equations in the title of this subsection let us perform trans-
formation b = B(ν1ν2), c = C(ν1ν2). Absolutely by the same technique as
above we pass to the system of equations with obvious solution
Bν2 ν2Cν2 = 0, Bν1 ν1Cν1 = 0
− −
c = C = C1(ν1)+C2(ν2), b = B = dν1ν1C1 + dν2ν2C2
Z ν1 Z ν2
W = db(ν +ν ) dcν ν +w(x) = dν1(ν1)2C1 + dν2(ν2)2C2 +w(x)
Z 1 2 −Z 1 2 Z ν1 Z ν2
From equation
∂R ∂R ∂R ∂R
( +ν ) = 0, ( +ν ) = 0
1 2
∂c ∂b ∂c ∂b
(both differential operators are commutative) it follows R = R1(ν )+R2(ν )
1 2
(∂R +ν ∂R) = (∂R (ν2)c∂R) = 0 from which follow result above. Further
∂c 1 ∂b ∂c − (ν2)b ∂b
z = R = R1 (ν ) +R2 (ν ) , x = R = R1 (ν ) +R2 (ν )
b ν1 1 b ν2 2 b c ν1 1 c ν2 2 c
or
x+ν z = R1 ((ν ) +ν (ν ) ), x+ν z = R2 ((ν ) +ν (ν ) )
1 ν1 1 c 1 1 b 2 ν2 2 c 2 2 b
After differentiation expressions above for b,c functions via ν ones we obtain
system of equations for determining derivatives of ν functions
C1 ν1 +C2 ν2 1 C1 ν1 +C2 ν2 0
ν1 c ν2 c = , ν1 b ν2 b =
(cid:18)ν1C1 ν1 +ν2C2 ν2(cid:19) (cid:18)0(cid:19) (cid:18)ν1C1 ν1 +ν2C2 ν2(cid:19) (cid:18)1(cid:19)
ν1 c ν2 c ν1 b ν2 b
Result of solution
ν2 ν1 1 1
C1 ν1 = , C2 ν2 = C1 ν1 = , C2 ν2 =
ν1 c ν2 ν1 ν2 c −ν2 ν1 ν1 b −ν2 ν1 ν2 b ν2 ν1
− − − −
Substituting these expressions in relations connected x,z variables with (ν)
one we obtain finally
R1 ((ν ) R2 ((ν )
x+ν z = ν1 1 c Θ1(ν1), x+ν z = ν2 2 c
1 C1 ≡ 2 C2 Θ2(ν2)
ν1 ν2 ≡
4
5 M
2
Thiscasecoincide withtheprevious oneunder conditionW = 0. Thesecond
b
order differential equation looks as
R = W R
cc c bb
Solution of this equation it is possible to find in a form R = dkekbw(k,c).
For w we pass to ordinary differential equation of the secondR order w =
cc
k2W w. This is typical one dimensional Schrodinger equation with zero en-
c
ergy and potential function k2W (c). All cases when it is possible to find its
c
solution in explicit form are described in corresponding monographies.The
linear equation of second order have two fundamental solution w ,w and
1 2
general solution is their linear combination. Thus
W = dkekb(f (k)w (k)+f (k)w (k))
1 1 2 2
Z
Both fundamental solutions are depended from k because potential energy
k2W depends from this parameter. Thus (M A) have trivial solution
c 3
−
depending on two one dimensional functions and series solutions connected
with the cases of integrability of corresponding ordinary differential equation
also depending on two one dimensional functions f ,f .
1 2
5.1 Degenerate solution
In the main text of this section was assumed no one 2 functions from 3
ones a,b,c functional dependent. Let us consider opposite situation. c =
C(a) = W−1(a), b = B(a) (W−1 inverse with respect W function). From
linear system of equations a = b , b = c = C a we immediately obtain
z x z x a x
B2 = C and the first equation is usual one dimensional Monge equation
a a
1 1
a = C2a with solution in implicit form x + C2z = G(a). Thus we have
z a x a
solution in degenerate case M ,(M A) equations depending on 2 arbitrary
2 3
functions (W−1(a),G(a)). −
6 M
3
U = W(U )
xxx zzz
5
Thisequationasinsectionpossiblesystem forminnotationsa = U , b =
z,z,z
U , c = U d = U . d = c = W(a) , a = b , b = c . In
x,z,z x,x,z x,x,x z x z z x x z
two dimensional all threefunctions arefunctionallydepended. Thus it ispos-
sible represented a = f(b,c) The system of equations under such assumption
take form
f b +f c = b , b = c , c = W b +W c , W = W(f)
b x c x z x z x b z c z
After the same transformation as in section M˜ we pass to two equations in
2
partial derivatives which must be self consistent
f R f R = R , R = W R +W R , W = W(f), x = R , z = R
b b,c c b,b c,c b,b b c,c c c,b c b
− − − −
(3)
6.1 ν ,ν = Constants
2 1
In this case R = R (b ν c)+R (b ν c), W = (ν +ν )b ν ν c+w(R ).
1 1 2 2 1 2 1 2 b
x = R = R˙ +R′, z−= R = ν−R˙ ν R′. Resolving tw−o last equations
b ν cb = L1(x +2ν z), b c ν −c =1 L1 −(x +2 ν2z) or finally c = L2−L1, b =
− 2 1 1 − 1 2 2 ν2−ν1
ν2L2−ν1L1. In the same terms f = ν22L2−ν12L1 +σ(x) For determination W we
ν2−ν1 ν2−ν1
have additional equations
ν−1 +ν−1 = W , (ν ν )−1 = W , ν2 +ν1ν2 = 0, ν1 +ν2ν1 = 0 (4)
1 2 c 1 2 − b c b c b
ν−1L ν−1L
W = c(ν−1 +ν−1) b(ν ν )−1 +θ(z) = 2 2 − 1 1 +θ(z)
1 2 − 1 2 ν ν
2 1
−
Thus in notation of the section M˜ we have
2
c = l0, b = l0, a = f = l0 +θ(z),W = l0 +σ(x)
0 1 2 −1
But W = W(f). This fact is equivalent as equality to zero Jacobian be-
tween this functions Jacbian(f,W) = 0 with arbitrary two arguments of the
problem. Choosing x,z as such arguments have
f W = f W , (l1 +θ )(l1 x+σ ) = l1l1
x z z x 3 z −1 x 2 0
or
[]
σ θ )+σ l1 +θ l1 + L˙ L′ = 0 (5)
x z x 3 z −1 ν ν 2 1
1 2
6
In what follows following notations are used
ν +ν )ν ν
∆ = ν ν , [] = (ν2 +ν ν +ν2), ρ = 1 2 1 2
2 − 1 1 1 2 2 []
Equation above is functional equation connected 4 functions L˙ ,L′,σ ,θ
2 1 x z
with 4 different arguments x + ν z,x + ν z,x,z. This functional equation
2 1
may be rewritten in many different forms. We present one of them from
which we will be able to obtain some number of partial solutions
θ +ν2L′ θ +ν2L˙
z 1 1 = z 2 2 (6)
ν−1L′ ν σ ν−1L˙ ν σ
1 1 − 2 x 2 2 − 1 x
Both functional equations are invariant with respect to the following trans-
formation
[] [] 1 1
σ , θ , L′ , L˙
x → ν4ν4σ z → θ 1 → ν2L′ 2 → ν2L˙
1 2 x z 1 1 1 2
We would like to show that if one derivatives is constant functional equation
have some explicit partial solution.
6.1.1 σ = A = Constant
x
In this case let us introduce in the last equation notation u−1(x + ν z) =
1
ν−1L′ ν A, v−1(x+ν z) = ν−1L˙ nu A and determine θ in this terms.
1 1− 2 2 2 2− 1 z
Result is the following one
ν3 ν3 ν3ν Au ν ν3Av
θ = 2 − 1 + 2 1 − 2 1
z
u v u v
− −
Derivative with respect to x the last expression equal to zero and u =
x
u′,v = v˙ The final result
x
u′ v˙
= = k, ln(v+(A˜)−1) = k(x+ν z)+d kD , A˜ ρA
u+(A˜)−1 (v +(A˜)−1 2 2 ≡ 2 ≡
v−1 = (ekD2 (A˜)−1)−1 = ν−1L˙ ν A, u−1 = (ekD1 1)−1 = ν−1L′ ν A,
− 2 2− 1 − 1 1− 2
ν2ekD1 ν2eD2
θ = Aν ν 2 − 1
z 2 1 ekD1 ekD2
−
7
Trivial integration of the last expressions lead to
ν A˜ ν A˜
L = 2 ln(1 A˜e−kD2)+ν ν AD , L = 1 ln(1 A˜eD1)+ν ν AD
2 k − 2 1 2 1 k − 2 1 2
ν3 ν3
θ = A˜ 2 − 1 ln(e−kD2 e−kD1) Aν ν [(ν +ν )x+[]z] σ = Ax
− k∆ − − 2 1 2 1
ν D ν D A˜ 1 (A˜)−1e−kD2 A˜ 1 (A˜)−1e−kD2
W = l0 +σ = A 1 2 − 2 1+ ln − +Ax = ln −
−1 ∆ k 1 A˜e−kD1 k 1 (A˜)−1e−kD1
− −
ν2D ν2D ) A˜
f = a = l0 +θ = A 2 2 − 1 1 +ν3 ln(1 A˜e−kD2)
2 ∆ 2∆k − −
A˜ ν3 ν3
ν3 ln(1 A˜e−kD1) A˜ 2 − 1 ln(e−kD2 e−kD1) Aν ν [(ν +ν )x+[]z] =
1∆k − − k∆ − − 2 1 2 1
A˜ (1 A˜e−kD2) A˜ (1 A˜e−kD1
ν3 ln − ν3 ln − )
2∆k (e−kD2 e−kD1 − 1∆k (e−kD2 e−kD1
− −
or
∆k Wk
A˜ f = (ν23 −ν13)ln(eWAk˜ −1)+ν13 A˜ (7)
6.1.2 L′ = D = Constant
1
In this case let us rewrite the main equation in equivalent form
ν−1L˙ ν σ L′ θ +ν2L˙
2 2 − 1 x 1 = z 2 2 (8)
ν−1L′ ν σ θ +ν2L′
1 1 − 2 x z 1 1
and introduce notations u−1u−1(x) = ν−1D ν σ , v−1(z) = θ +ν2D. In
this notations the last equation resolved1 wit−h re2spxect to L˙ looksz as 1
2
1 ν1 ν−1u ν2v
L˙ = − ν2 +D 2 − 1
2 ν−1u ν2v ν−1u ν2v
2 − 2 2 − 2
After action on the this expression by operator H = ∂ ν ∂ keeping in
∂z − 2∂x
mind that HL = 0, Hu = ν u , Hv = v we pass to equation for u,v
2 2 x z
functions with obvious solutio−n (D˜ = D(ν ν ))
+ 2
u v
x +ν−1 z = 0, ν−1D˜−1+v = e−kν2z, ν2D˜−1+u = ekx
ν2D˜−1 +u 2 ν−1D˜−1 +v 2 2
2 2
8
Substituting these expressions ion equation for L˙ and using definitions of
2
u,v functions we obtain explicit expressions for all derivations
e−kD2 1
L˙ = D +D(ν2 ν2) , σ = ν−1ν−1D ν−1
2 2 − 1 ν−1 ν2e−kD2 x 1 2 − 2 ekx ν2D˜−1
2 − 2 − 2
1
L′ = D, θ = ν2D+
1 z − 1 e−ν2kz ν−1D˜−1
− 2
Trivial integration leads to explicit expressions of L ,L ,θ,σ functions
2 1
ν2 ν2 ν2
L = 1Dx+ν Dz +D 2 − 1 ln(ν−1ekx ν2e−kν2z), L = D(x+ν z)
2 ν2 2 kν2 2 − 2 1 1
2 2
ν +ν D˜
θ = D[]z D 2 1Dln(e−kν2z ν−1D˜−1), σ = ν−3ν−1[]Dx ν−3 ln(ekx ν2D˜−1)
− − k − 2 2 1 − 2 k − 2
With help of these formulae and definitions above we have
D˜ e−kν2z ν−1D˜−1
W = l0 +σ = ν−3 ln(ν−1 ν2 − 2 ),
−1 2 k 2 − 2 ekx ν−1D˜−1
− 2
D˜ ekx ν−1D˜−1
a = f = l0 +θ = ln(ν−1 − 2 ν2)
2 k 2 e−kν2z ν−1D−1 − 2
− 2
or functions W,f are functionally dependent
ν2−1e−kνD˜23W −ν22e−D˜kf = 1 (9)
6.1.3 θ = E = Constant
z
After introduction new notation 1(x + ν z) = E + ν2L′, 1(x + ν z) =
u 1 1 1 v 2
E +ν2L˙ and resolving the main equation (?? with respect Toσ we have
2 2 x
σ (ν v ν u = ν−3 ν−3 +E(ν−3u ν−3v)
x 1 − 2 2 − 1 1 − 2
After calculation derivative with respect to z argument and trivial manipu-
lations as in previous sub subsections we pass to equations foe u,v functions
with obvious solution
ν1u′ = k = ν2v˙ u = eνk1D1 ν1 , v = eνk2D2 ν2
u+ ν1 v + ν2 − Eρ − Eρ
Eρ Eρ
9
From definitions of u,v functions above σ via these functions we obtain
x
ν2L˙ = E+ 1 , ν2L′ = E+ 1 , σ = Eν1−3eνk1D1 −ν1−3eνk2D2
2 2 − eνk2D2 − Eν2ρ 1 1 − eνk1D1 − Eν1ρ x ν1eνk2D2 −ν2eνk1D1
Result of integration
ρ ν ρ ν
ν22L2 = −ED2+ k ln(1− E2ρe−νk2D2), ν12L1 = −ED1+ k ln(1−E1ρe−νk1D1)
σ = −ν1−3ν2−1 − ν1ν3+ν3ν1D2 − νk1ν+2νν22(≡ kν[]3ρν3)ln(ν1e−ν1kD1 −ν2e−νk2D2)
2 1 2 1 2 1
ρ (1 ν2e−νk2D2) ρ ν ν
f = ln − Eρ , W = [ν−3ln(1 2 e−νk2D2) ν−3ln(1 1 e−νk1D1)
k∆ (1 ν1e−νk1D1) k∆ 2 −Eρ − 1 −Eρ −
− Eρ
(ν2−3 −ν1−3)ln(ν1e−νk1D1 −ν2e−νk2D2)].
From the last two relations above we obtain
ρ
W = ν−3ln(1 efkρ∆) ν−3ln(e−fkρ∆ 1)
−k∆ 2 − − 1 −
6.2 ν = Constant,ν = ν (b ν c)
1 2 2 1
−
ν = ν ((b ν c)), R = R ((b ν c))+R ( (b−ν1cds 1 b))
2 2 − 1 1 − 1 2 ν1−ν2(s) −
R
ν ν ν
x = R = R′+R˙ 1 , z = R = ν R′+R˙ 2 1 , x+ν z = ν R˙ ,
b 1 2 ν ν c − 1 1 2 ν ν 1 − 1 2
2 1 2 1
− − − −
′
x+ν z = ∆R Θ(ν ) c = C(ν ) L(x+ν z), b = dν ν C ν L(x+ν z),
2 ≡ 2 2 − 1 Z 2 2 ν2− 1 1
a = dν ν2C ν2L(x+ν z)+θ(z), W = dν ν−1C ν−1L(x+ν z)+σ(x)
Z 2 2 ν2− 1 1 Z 2 2 ν2 1 1
and only one equation remains
∆2[]
W = W(a), σ θ +θ (ν−1C (ν ) ν−1L′)+σ (ν3C (ν ) ν3L′) = (ν ) C L′.
x z z 2 ν2 2 x− 1 x 2 ν2 2 x− 1 ν ν 2 x ν2
1 2
also functional equation of the same kind as above and below ones.
10