Introduction

We begin this paper with the geometry of distributions. The main idea here is the various notions of symmetry and their use in solving a given differential equation. In the ‘Tangent and cotangent distribution’ section, we introduce the basic notions and definitions.

In the ‘Integral manifolds and maximal integral manifolds’ section, we describe the relation between differential equations and distributions. In the ‘Symmetries’ section, we present the geometry of distributions by their symmetries and find out the symmetries of the F-Gordon equation by this machinery. In the ‘A proof of the Frobenius theorem’ section, we introduce a simplified proof of the Frobenius theorem and some related corollaries. In the ‘Symmetries and solutions’ section, we describe the relations between symmetries and solutions of a distribution.

In all steps, we study the F-Gordon equation as an application and also a partial differential equation which appears in differential geometry and relativistic field theory. It is a generalized form of the Klein-Gordon equation u tt u xx + u = 0 as well as a relativistic version of the Schrodinger equation, which is used to describe spinless particles. It was named after Walter Gordon and Oskar Klein[1, 2].

Tangent and cotangent distribution

Throughout this paper, M denotes an (m + n)-dimensional smooth manifold.

Definition 2.1

A map D:MTM is called an m-dimensional tangent distribution on M, or briefly Tanm-distribution, if

D x : = D ( x ) T x M ( x M )

is an m-dimensional subspace of T x M. The smoothness of D means that for each xM, there exists an open neighborhood U of x and smooth vector fields X1,⋯,X m such that

D y = X 1 ( y ) , , X m ( y ) : = s pan R { X 1 ( y ) , , X m ( y ) } ( y U )

Definition 2.2

A map D : MTM is called an n-dimension cotangent distribution on M, or briefly Cotn-distribution, if

D x : = D ( x ) T x M ( x M )

is an n-dimensional subspace of T x M. The smoothness of D means that for each xM, there exists an open neighborhood U of x and smooth 1-forms ω1,⋯ωnsuch that

D y = ω 1 ( y ) , , ω n ( y ) : = s pan R { ω 1 ( y ) , , ω n ( y ) } ( y U )

In the sequel, without loss of generality, we can assume that these definitions are globally satisfied.

There is a correspondence between these two types of distributions. For Tanm-distribution D, there exist nowhere zero smooth vector fields X1,⋯,X m on M such that D = 〈X1,⋯,X m 〉, and similarly, for Cotn-distribution D, there exist global smooth 1-forms ω1,⋯,ωnon M such that D = 〈ω1,⋯,ωn〉.

Example 2.3

(Cartan distribution) Let M = Rk + 1. Denote the coordinates in M by x,p0,p1,..,p k , and given a function f(x,p0,⋯,pk−1), consider the following differential 1-forms

ω 0 = d p 0 p 1 dx , ω 2 = d p 1 p 2 dx , ω k 2 = d p k 2 p k 1 dx , ω k 1 = d p k 1 f ( x , p 0 , , p k 1 ) dx ,

and the distribution D = 〈ω0,⋯,ωk−1〉. This is the 1-dimensional distribution, called the Cartan distribution. This distribution can also be described by a single vector field X, D = 〈X〉, where

X = x + p 1 p 0 + p 2 p 1 + + p k 1 p k 2 + f ( x , p 0 , , p k 1 ) p k 1 .

Example 2.4

(F -Gordon equation) Let F : R5R be a differentiable function. The corresponding F-Gordon PDE is u xy = F(x,y,u,u x ,u y ). We construct 7-dimensional sub-manifold M defined by s = F(x,y,u,p,q), of

J 2 ( R 2 , R ) = { x , y , u , p = u x , q = u y , r = u xx , s = u xy , t = u yy } .

Consider the 1-forms

ω 1 = du p dx q dy , ω 2 = dp r dx F dy , ω 3 = dq F dx t dy.

This distribution can also be described by the following vector fields:

X 1 = x + p u + r p + F q , X 2 = y + q u + F p + t q , X 3 = r , X 4 = t .

Definition 2.5

Let D : MTM be a Tanm-distribution and set

A nn D x : = { ω x T x M ω x D x = 0 } .

It is clear that dimAnn D x = n. A 1-form ω ∈ Ω1(M)annihilates D on a subset NM, if and only if ω x ∈ Ann D x for all xM.

The set of all differential 1-forms on M which annihilates D, is called annihilator of D and denoted by Ann D.

Therefore, for each Tanm-distribution,

D : M TM , D : x D x ,

we can construct a Cotn-distribution

D : M T M , D : x D x = A nn D x

and vice versa. In the other words, for each Tanm-distribution D = 〈X1,⋯,X m 〉, we can construct a Cotn-distribution D = Ann D =〈ω1,⋯,ωn〉, and vice versa.

Theorem 2.6

  1. (a)

    D and its annihilator are modules over C (M).

  2. (b)

    Let X be a smooth vector field on M and ω ∈ Ann D, then

    L X ω ω L X mod D

Proof

  1. (a)

    is clear, and for (b), if Y belongs to D, then ω(Y) = 0 and

    ( L X ω ) Y = X . ( ω ( Y ) ) ω [ X . Y ] = ω [ X . Y ] = ( ω L X ) Y .

Integral manifolds and maximal integral manifolds

Definition 3.1

Let D be a distribution. A bijective immersed sub-manifold NM is called an integral manifold of D if one of the following equivalence conditions is satisfied:

  1. (1)

    T x ND x , for all xN.

  2. (2)
    N i = 1 n ker ω i

    .

Moreover, NM is called maximal integral manifold if for each xN, there exists an open neighborhood U of x such that there is no integral manifold N containing NU.

It is clear that the dimension of maximal integral manifold does not exceed the dimension of the distribution.

Definition 3.2

D is called a completely integrable distribution, or briefly CID, if for all maximal integral manifold N, one of the following equivalence conditions is satisfied:

(1) dim N = dim D.

(2) T x N = D x for all xN

(3)N i = 1 n ker ω i , and if Nbe an integral manifold with NN , then NN.

In the sequel, the set of all maximal integral manifolds is denoted by N.

Theorem 3.3

N= i = 1 n ker ω i

; that is ωi| N = 0 for i = 1,⋯,n.

Example 3.4

(Continuation of Example 2.3) If N is an integral curve of the distribution, then x can be chosen as a coordinate on N, and therefore,

N = { ( x , h 0 ( x ) , h 1 ( x ) , , h k 1 ( x ) ) | x R } .

Conditions ω0| N = 0,⋯,ωk−1| N = 0 imply that h 1 = h 0 , h 2 = h 1 ,, h k 1 = h k 1 , or that

N = J k 1 h = { ( x , h ( x ) , h ( x ) , , h ( k 1 ) ( x ) | x R }

for some function h : RR.

The last equation ωk−1| N = 0 gives us an ordinary differential equation h(k)(x) = f(x,h(x),h(x), ⋯,h(k−1)(x)).

The existence theorem shows us once more that the integral curves do exist, and therefore, the Cartan distribution is a CID.

Example 3.5

(Continuation of Example 2.4) This distribution in not a CID because there is no 4-dimensional integral manifold, and dim D = 4. For, if N be a 4-dimensinal integral manifold of the distribution, then (x,y,u,p) can be chosen as coordinates on N, and therefore,

N : q = h ( x , y , u , p ) , r = l ( x , y , u , p ) , t = m ( x , y , u , p ) , s = F ( x , y , u , p , h ) .

Condition ω1| N = 0 implies that −p dxh(x,y,u,p)dy + du = 0, which is impossible.

By the same reason, we conclude that there is no 3-dimensional integral manifold.

Now, if N be a 2-dimensinal integral manifold of the distribution, then (x,y) can be chosen as coordinates on N, and therefore,

N : u = h ( x , y ) , p = l ( x , y ) , q = m ( x , y ) , r = n ( x , y ) , t = o ( x , y ) , s = F ( x , y , u , p , q ) .

Conditions ω1| N = 0 and ω2| N = 0 imply that l = h x , m = h y , n = l x = h xx and o = m y = h yy .

The last equation ω3| N = 0 implies that h xy = F(x,y,h,h x ,h y ). This distribution is not a CID.

Symmetries

In this section, we consider a distribution D = 〈X1,⋯, X m 〉 = 〈ω1,⋯,ωn〉 on manifold Mn + m.

Definition 4.1

A diffeomorphism F : MM is called a symmetry of D if FD x = DF(x)for all xM.

Therefore,we have the following theorem.

Theorem 4.2

The following conditions are equivalent:

(1) F is a symmetry of D;

(2) Fωis determine the same distribution D; that is D = 〈Fω1,⋯,Fωn〉;

(3) Fωi∧⋯∧ωn= 0 for i = 1,⋯,n;

(4) F ω i = j = 1 n a ij ω j , where a ij C(M);

(5) (FX i | x ) ∈ DF(x)for all xM and i = 1,⋯,n; and

(6) F X i = j = 1 n b ij X j , where b ij C(M).

Theorem 4.3

If F be a symmetry of D and N be an integral manifold, then F(N) is an integral manifold.

Proof

F is a diffeomorphism; therefore, F(N) is a sub-manifold of M. From other hand, if xN, then ωi|F(x)= (Fωi)| x = 0 for all i = 1,⋯,n; therefore, F(N) = {F(x) | xN} is an integral manifold. □

Theorem 4.4

Let N be the set of all maximal integral manifolds and F : MM be a symmetry, then F(N) = N.

Proof

If xN, then ωi|F(x)= (Fωi)| x = 0 for all i = 1,⋯,n; therefore, F(x) ∈ N and F(N) ⊂ N. □

Now, if yN, then there exists xM such that F(x) = y, since F is a diffeomorphism. Therefore, (Fωi)| x = ωi|F(x)= ωi| y = 0 for all i = 1,⋯,n; thus, xN and NF(N).

Definition 4.5

A vector field X on M is called an infinitesimal symmetry of distribution D, or briefly a symmetry of D, if the flow F l t X of X be a symmetry of D for all t.

Theorem 4.6

A vector field XX(M) is a symmetry if and only if

L X ω i | D = 0 for all i = 1 , , n .

Proof

Let X be a symmetry. If Ω = ω1∧⋯∧ωn, then {(FlX)ωi}∧ Ω = 0, by condition (3) in Theorem 4.2. Moreover, by the definition L X ω i := d dt 0 ( F l t X ) ω i , one gets

( L X ω i ) Ω = lim t 0 1 t ( F l t X ) ω i ω i Ω = lim t 0 1 t { ( F l X ) ω i } Ω ω i Ω 1 = 0 .

Therefore L X ωi| D = 0.

In converse, let L X ωi| D = 0 or L X ω i = j = 1 n b ij ω j for i = 1,⋯,n and b ij C(M). Now, if γ i (t):={ ( F l t X ) ω i }Ω, then

γ i ( 0 ) = { ( F l 0 X ) ω i } Ω = 0 ,
(1)

and

γ i ( t ) = d dt { ( F l t X ) ω i } Ω = ( ( F l t X ) L X ω i ) Ω = ( F l t X ) b ij ω i Ω = B ij { ( F l t X ) ω j } Ω

where B ij = ( F l t X ) b ij = b ij F l t X and

γ i ( t ) = B ij γ i ( t ) , i = 1 , , n .
(2)

Therefore, γ = (γ1⋯,γ n ) is a solution of the linear homogeneous system of ODEs (2) with the initial conditions (1), and γ must be identically zero.

Theorem 4.7

X is symmetry if and only if for all YD, then [X,Y] ∈ D.

Proof

By the above theorem, X is a symmetry if and only if for all ω ∈ Ann D, then L X ω ∈Ann D.

The Theorem comes from the Theorem 2.6 (b): L X ω = −ωL X on D. In other words, (L X ω)Y = −ω[X,Y] for all YD. □

Denote by Sym D the set of all symmetries of a distribution D.

Example 4.8

(Continuation of Example 3.4) Let k = 2. A vector fieldY=a x +b p 0 +c p 1 is an infinitesimal symmetry of D if and only if L Y ωi≡ 0 mod D, for i = 1,2. These give two equations:

c = Xb p 1 Xa , Xc = f Xa + Yf .

Example 4.9

(Continuation of Example 3.5) We consider the point infinitesimal transformation:

Z = X ( x , y , u ) x + Y ( x , y , u ) y + U ( x , y , u ) u + P ( x , y , u , p , q , r , t ) p + Q ( x , y , u , p , q , r , t ) q + R ( x , y , u , p , q , r , t ) r + T ( x , y , u , p , q , r , t ) t .

Then, Z is an infinitesimal symmetry of D if and only if L Z ωi≡ 0 mod D, for i = 1,2,3. These give ten equations:

P r = P t = Q r = Q t = 0 , p 2 X y + qp Y y + q U x + pQ = pq X x + qP + q 2 Y x + p U y , rp X y + F Y y + q P x + ( qr pF ) P p + ( qF pt ) P q + pX F x + pY F y + pU F u + pP F p + pQ F q = qr X x + qF Y x + p P y + qR , pF X y + pt Y y + ( qr pF ) Q p + ( qF pt ) Q q + q Q x + pT = qF X x + qt Y x + p Q y + qP F p + qY F y + qX F x + qQ F q + qU F u , Q y + q Q u + F Q p + t Q q = t Y y + tq Y u + F X y + qF X u + T , U y + q U u = p X y + pq X u + q Y y + q 2 Y u + Q , P y + q P u + F P p + t P q = r X y + qr X u + F Y y + qF Y u + ( X F x + Y F y + U F u + P F p + Q F q ) .

Complicated computations using Maple show that

P = p X x p 2 X u q Y x pq Y u + U x + p U u , Q = 1 p pq X x p 2 X y + q 2 Y x pq Y y q U x + p U y + qP , R = 1 q pq X x p 2 X y + q 2 Y x pq Y y q U x + p U y + qP . F q + ( qr pF ) . P p + ( qF pt ) . P q + q P x p P y + pX F x + pY F y + pU F u + pP F p + ( p Y y q Y x ) .F pr . ( q X x p X y ) , T = 1 p 3 ( p 2 t + q 2 r 2 pqF ) P + p 2 ( pt + q 2 F q ) X x + p 2 ( qr 2 pF pq F q ) X y + q ( q 2 ( r + p F q ) + 3 p ( pt qF ) ) Y x p 2 ( q 2 F q + 2 pt + qF ) Y y ( q 2 r + p q 2 F q + p 2 t + 2 pq ) U x + p 2 q F q U y p q 2 P x + p 2 q P y + pq ( pF qr ) P p + pq ( pt qF ) P q + pq ( pX F x + pY F y + pU F u + pP F p + qP F q ) p 2 q 2 X xx + 2 p 3 q X xy p 4 X yy p q 3 Y xx + 2 p 2 q 2 Y xy p 3 q Y yy + p q 2 U xx 2 p 2 q U xy + p 3 U yy ,

and X = X(x,uqy), Y = Y(y,upx), and U(x,y,u) must satisfy in PDE:

( p F p F ) X x + p ( p F p 2 F ) X u + ( q F q F ) Y y + q ( q F q 2 F ) Y u F p U x F q U y + ( F p F p q F q ) U u + U xy + q U xu + p U yu + pq U uu = X F x + Y F y + U F u .

A proof of the Frobenius theorem

Theorem 5.1

Let X ∈ Sym D D and N be maximal integral manifold. Then, X is tangent to N.

Proof

Let X(x) ∉ T x N. Then, there exists an open set U of x and sufficiently small ϵ such that N ̄ := ϵ < t < ϵ F l t X N U is a smooth sub-manifold of M.

Since XD, So N ̄ is an integral manifold.

Since X ∈ Sym D , so tangent to F l t X N U belongs to D, for all −ϵ < t < ϵ.

On the other hand, tangent spaces to N ̄ are sums of tangent spaces to F l t X (NU) and the 1-dimensional subspace generated by X, but both of them belong to D, and their means are N ̄ N. □

Theorem 5.2

If XD ∩ Sym D and N be a maximal integral manifold, then F l t X (N)=N for all t.

Theorem 5.3 (Frobenious)

A distribution D is completely integrable, if and only if it is closed under Lie bracket. In other words, [X,Y] ∈ D for each X,YD.

Proof

Let N be a maximal integral manifold with T x N = D x . Therefore, for all X,YD, X and Y are tangent to N, and so [X,Y] is also tangent to N.

On the other hand, let for all X,YD, their [X,Y] ∈ D. By the Theorem, all XD is a symmetry too, and so all XD is tangent to N, and this means T x N = D x , for all xN. □

Theorem 5.4

A distribution D is completely integrable if and only if D ⊂ Sym D .

Theorem 5.5

Let D = 〈ω1,⋯,ωn〉 be a completely integrable distribution and XD. Then, the differential 1-forms ( F l t X ) ω 1 ,, ( F l t X ) ω n vanish on D for all t.

Proof

If D is completely integrable, then X is a symmetry. Hence,

( F l t X ) ω i = j a ij ω j .

Symmetries and solutions

Definition 6.1

If an (infinitesimal) symmetry X belongs to the distribution D, then it is called a characteristic symmetry. Denote by Char(D) := S D D the set of all characteristic symmetries[3, 4].

It is shown that Char(D) is an ideal of the Lie algebra S D and is a module on C(M). Thus, we can define the quotient Lie algebra

Shuf ( D ) : = Sym D / Char ( D ) .

Definition 6.2

Elements of Shuf(D) are called shuffling symmetries of D.

Any symmetry X ∈ Sy m D generates a flow on N (the set of all maximal integral manifolds of D), and, in fact, the characteristic symmetries generate trivial flows. In other words, classes X mod Char(D) mix or ‘shuffle’ the set of all maximal manifolds.

Example 6.3

(Continuation of Example 4.8) Let k = 2. In this case,

x p 1 p 0 f p 1 mod C har ( D ) .

Therefore, Shuf(D) is spanned byZ=(ba p 1 ) p 0 +(caf) p 1 , where

c = Xb p 1 Xa , Xc = f Xa + Yf .

Example 6.4

(Continuation of Example 4.9) In this case, we have

x p u r p F q , r 0 , y q u F p t q , t 0 ,

in Shuf(D). Therefore, Shuf(D) is spanned by

W = ( U pX qY ) u + ( P rX FY ) p + ( Q FX tY ) q ,

where

P = p X x p 2 X u q Y x pq Y u + U x + p U u , Q = 1 p pq X x p 2 X y + q 2 Y x pq Y y q U x + p U y + qP ,

and X = X(x,uqy), Y = Y(y,upx), and U(x,y,u) must satisfy in PDE:

( p F p F ) X x + p ( p F p 2 F ) X u + ( q F q F ) Y y + q ( q F q 2 F ) Y u , F p U x F q U y + ( F p F p q F q ) U u + U xy + q U xu + p U yu + pq U uu = X F x + Y F y + U F u .
(3)

Example 6.5

(Quasilinear Klein-Gordon Equation) In this example, we find the shuffling symmetries of the quasilinear Klein-Gordon equation

u tt α 2 u xx + γ 2 u = β u 3

as an application of the previous example, where α, β, and γ are real constants. The equation can be transformed by definingξ= 1 2 (xαt) andη= 1 2 (x+αt). Then, by the chain rule, we obtain α2u ξη + γ2u = β u3. This equation reduces to

u xy = au + b u 3 ,
(4)

by t = y, a = −(γ/α)2, and b = β/α2.

By solving the PDE (3), we conclude that Shuf(D) is spanned by the three following vector fields:

X 1 = ( px qy ) u ( p + y u 2 ( a + bu ) rx ) p + ( q + x u 2 ( a + bu ) ty ) q , X 2 = q u + u 2 ( a + bu ) p + t q , X 3 = p u + r p + u 2 ( a + bu ) q .

For example, we have

F l s X 3 ( x , y , u , p , q , r , t ) = x , y , u + sp + s 2 2 r , p + sr , q + s . u 2 ( a + bu ) + s 2 40 . up ( 2 a + 3 bu ) + s 3 42 . ( 14 a p 2 + 42 bu p 2 + 14 aur + 21 b u 2 r ) + s 4 4 . p ( b p 2 + ar + 3 bur ) + s 5 20 . r ( 6 b p 2 + ar + 3 bur ) + s 6 8 . bp r 2 + s 7 56 . b r 3 , r , t ,

and if u = h(x,y) be a solution of (4), then F l s X 3 (x,y,h, h x , h y , h xx , h yy ) is also a new solution of (4), for sufficiently small sR.