Backlund Talk 1

Backlund transformations:Part one

Rob Thompson

University of MinnesotaMathematical Physics Seminar

March 2012

Introduction

A Backlund transformation relates solutions of PDE to each other.

Usefulness:

given one solution of a PDE, generate more. relate solutions of a difficult PDE to those of an easier PDE deeper applications in integrable equations

Trivial example

The CauchyRiemann equations

ux = vy uy = vxare a Backlund transformation from the Laplace equation to itself

uxx + uyy = 0 vxx + vyy = 0.

Introduction

Trivial example continued

Take u(x, y) = xy. The CR equations become

vy = y vx = x

The solution is

v(x, y) =1

2(y2 x2) + C

which also solves Laplace.

Since a solution of a PDE is taken to another of the same PDE,this is called an auto-Backlund transformation.

Classical Framework

For our purposes, a Backlund transformation is a relation

vx = B1(x, y, u, ux, uy, v) or ux = B1(x, y, v, vx, vy, u)vy = B2(x, y, u, ux, uy, v) uy = B2(x, y, v, vx, vy, u)

The integrability conditions follow:

B1y

=B2x

orB1y

=B2x

.

The integrability conditions = PDE we wish to study.

These PDE will then be related by the Backlund transformation.

Classical Framework

More generally we have differential equations of interest

(x, y, u, ux, uy, . . .) = 0 and (x, y, v, vx, vy, . . .) = 0

and relations

Bi(u, v, ux, uy, ux, vx, vy, . . . ;x, y) = 0 i = 1, 2

These relations are a Backlund transformation if

when = 0, the Bi are integrable for v and v solves = 0 when = 0, the Bi are integrable for u and u solves = 0

Liouvilles equation

A real example

Liouvilles equation: uxy = eu

Wave equation: vxy = 0

These equations are related by the Backlund transformation

vx ux =

2 e(u+v)/2

vy + uy =

2 e(uv)/2

Liouvilles equation is implied by:

(vy)x = (vx)y

Wave equation is implied by:

(uy)x = (ux)y

Liouvilles equation

Start with the relations:

vx ux =

2 e(u+v)/2

vy + uy =

2 e(uv)/2

Cross differentiate:

vxy uxy = 12

(uy + vy)e(u+v)/2

vyx + uyx = 12

(ux vx)e(uv)/2

Use the relations on the above:

vxy uxy = euvyx + uyx = e

u

Add them: vxy = 0 Subtract them: uxy = eu

Solving Liouvilles equation

We cant solve uxy = eu yet, but we can easily solve vxy = 0!

Plug v(x, y) = f(x) + g(y) into:

{vx ux =

2 e(u+v)/2

vy + uy =

2 e(uv)/2

ux f =

2 e(u+f+g)/2

uy + g =

2 e(ufg)/2

Multiply the top equation by e(u+f)/2 and bottom by e(ug)/2:

x

(e(u+f)/2

)=

12e(2f+g)/2

y

(e(ug)/2

)=

12e(f2g)/2

I probably made a mistake, but the idea is what counts....

History

Albert Victor Backlund (1845-1922)

Luigi Bianchi (1856-1928)

History

Two ideas seeded the invention of Backlund transformations

Lies work on contact transformations Geometric study of pseudospherical surfacesBour (1862): wrote down Sine-Gordon equation as a compatibilitycondition for Gauss equations for pseudospherical surfaces

Bianchi (1872): interpreted analytically a purely geometricconstruction for pseudospherical surfaces

Backlund (1882): introduced interative procedure for constructionof pseudospherical surfaces, related to Bianchis

Bianchi (1882): demonstrated the relation of Backlundsconstruction with the Sine-Gordon equation

Bianchi (1892): published permutability theorem, giving non-linearsuperposition formula for Sine-Gordon equation

Examples of pseudospherical surfaces

Images from R. Palais Math Museum

Pseudo-spherical surfaces and the SineGordon equation

S a surface with parametrization r(u, v).

rvr(u,v)

ru

First fundamental form: Edu2 + 2Fdudv+Gdv2

E = ru, ru F = ru, rv G = rv, rvSecond fundamental form: edu2 + 2fdudv+ gdv2

e = ruu,N f = ruv,N g = rvv,N

Pseudo-spherical surfaces and the SineGordon equation

IfE

v=G

u= 0 there is a parametrization of S with

E = 1 F = cos G = 1.

If Gauss curvature is assumed to be negative constant K = 1

,

e = 0 f =sin

g = 0.

The Gauss equations become

ruu = u cot ru u csc rvruv =

1

sin N

rvv = v csc ru + v cot rv

Compatibility requires 2uv = sin .

Pseudo-spherical surfaces and the SineGordon equation

Bonnets theorem ensures that, given E,F,G, e, f, g, a surface isdetermined, unique up to Euclidean motions.

Given a solution to the SineGordon equation

2uv = sin

we can create a pseudospherical surface.

Backlunds discovery amounts to the fact that new solutions ofSineGordon can be found from the relation(

2

)u

=

sin

( +

2

)( +

2

)v

=1

sin

(

2

)where is a real parameter. Bianchis initial method lacked .

A primer on EDS: basic definitions

Definitions

An exterior differential system (EDS) on a manifold M is an idealof differential forms I (M) closed under exterior derivative d.I is called a Pfaffian system if it is generated by one forms:

I = {1, . . . , s, d1, . . . , ds}

An integral manifold of I is an immersed submanifold h : S Msuch that

h = 0 for all IExample M = R5 with coordinates (x, y, u, p, q). I = {du pdx qdy, dp dx+ dq dy}

Integral manifolds correspond to prolonged functions u(x, y):

(x, y) 7 (x, y, u, p = ux, q = uy)

A primer on EDS: encoding PDE

More examples

Same M F (x, y, u, p, q) some function I = {F, du pdx qdy, dF, dp dx+ dq dy}

Integral manifolds correspond to functions u(x, y):

(x, y) 7 (x, y, u, ux, uy)

where u solves the PDE F (x, y, u, ux, uy) = 0.

We could take instead

M = {F (x, y, u, p, q) = 0}

rather than keeping 0 order form F in I.

A primer on EDS: encoding PDE

Even more examples The first order equations

ux = F (x, y, u)

uy = G(x, y, u)

are encoded by

I = {du FdxGdy, dF dx+ dG dy}

The F -Gordon equation

uxy = F (x, y, u, ux, uy)

is encoded by

I = {du ux dx uydy, dux dx+ duy dy, (dux Fdy) dx}

A primer on EDS: encoding PDE

Okay, last example

Burgers equationut = uxx + 2uux

is encoded on

M = R4 with coordinates (x, t, u, ux)

by

I = { (du ux dx) dt , du dx+ (dux + 2uux dx) dt }

A primer on EDS: existence of integral manifolds

Not all EDS possess integral manifolds. There is a simple sufficientcriterion for existence of integral manifolds for Pfaffian systems.

Frobenius TheoremIf M, I is a Pfaffian EDS (of constant rank) which is algebraicallygenerated by 1-forms, then it has integral manifolds.

ExampleRecall the EDS

I = { = du F (x, y, u)dxG(x, y, u)dy}which encodes ux = F (x, y, u), uy = G(x, y, u).

This EDS is algebraically generated by 1 forms if d = 0.This condition is equivalent to

Fy Gx +GFu FGu = 0which also comes from the condition (ux)y = (uy)x.

Integrable extensions of EDS

Let

pi : B M be a submersion J an EDS on B, I an EDS on M

Definition

J is an integrable extension of I if there is a Pfaffian system J onB such that

1) J is transverse to the fibers of pi

2) J is algebraically generated by J and pi(I)

By 1), integral manifolds of J project to IBy 2), integral manifolds of I lift to J

Integrable extensions of EDS

1) J is transverse to the fibers of pi

This ensures that pi s : S M is an immersion when s : S Bis an integral manifold of J .

2) J is algebraically generated by J and pi(I)

This implies that, if S is an integral manifold of I, then J |pi1(S)is Frobenius.

An integral manifold down below lifts to a family of integralmanifolds above, parametrized by the points of the fiber.

Integrable extensions of EDS

Example: Hopf-Cole transformation

ut = (ux + u2)x{

M = R4 with coordinates (x, t, u, p)I = { (du p dt) dt , du dx+ (dp+ 2up dx) dt }

vx = u vt = ux + u2{

B = M R with new coordinate vJ = {dv udx (p+ u2)dt}

Backlund transformations

Definition

A Backlund transformation between EDS I1 and I2 is a commonintegrable extension J :

J , Bpi2

$$

pi1

zzI1,M1 I2,M2

Lift an integral manifold from I1 to J , then project to an integralmanifold of I2, or vice versa.

The trick is to be able to construct integrable extensions, and toconstruct them for equations that matter.

Symmetry reduction of EDS

Let G be a Lie group which acts regularly on M . Then thequotient is a nice submersion of smooth manifolds:

piG : M M/G

DefinitionThe reduction of the EDS I by G is the EDS

I/G = { a differential form on M/G | piG() I }

Motivation1) When G is a symmetry group of I, the reduction is relativelyeasy to compute.

2) Pairs of integrable extensions arise naturally from symmetryreduction.

Symmetry reduction of EDS

Definitions

G is a symmetry group of I if gI = I for all g G.G is transverse to I if one forms in I do not annihilate theinfinitesimal generators G of G.

AlgorithmSuppose G is a transverse symmetry group of I. Then I/G can becomputed using two ingredients:

A cross section : M/GM to the group action A special collection of forms in I, called G-semibasic forms:

Isb = { I|X y = 0, X G}

Then(basis for Isb) = basis for I/G

Symmetry reduction of EDS

Example

M = R5 with coordinates (x, y, u, v, ux, vy) I = {du ux dx, dv vy dy, dx dux, dy dvy} G generated by X = u + v M/G = R4, and a cross section is given by

(x, y, w,wx, wy) = (x, y, w, v = 0, wx,wy)

A basis of G-semibasic forms is{du ux dx, dx dv + vy dy, dx dux, dy dvy}

The reduced EDS isI/G = {dw wxdx wydy, dx dwx, dy dwy}

The main idea

Let I be a Pfaffian system on M , with symmetry groups G1, G2with a common subgroup H. Suppose all actions are regular andtransverse to I.Then the following commutative diagram consists of integrableextensions, and thus I/H defines a Backlund transformation forI/G1 and I/G2.

I

piH

piG2

piG1

I/H

p1{{p2 ##

I/G1 I/G2

Backlund transformation for Liouville revisited

I

Vy = eW ,Wx = eV

p1

vv

p2

))vxy = 0 uxy = e

u

p1(x, y, V,W ) = (x, y, v = V W )p2(x, y, V,W ) = (x, y, u = V +W + log(2))

Writing V,W in terms of u, v, the equations defined by theintegrable extension become:

vx ux =

2e(u+v)/2 vy + ux =

2e(uv)/2

Behind the curtain

I

I/H

p1

{{

p2

##I/G1 I/G2

M = J2 J2, with coordinates (x, y, z, w, zx, wy, zxx, wyy) I = product of contact systemsH = {z w, zz + ww}G1 = H + {z + w}G2 = H + {z2z w2w}

References

P. Olver, Equivalence, Invariants and Symmetry, Cambridge Press

I. Anderson and M. Fels, Exterior Differential Systems withSymmetry, Acta Appl. Math. (2005)

I. Anderson and M. Fels, Backlund Transformations for DarbouxIntegrable Differential Systems, Arxiv (2011)

C. Rogers and W.F. Shadwick, Backlund Transformations andTheir Applications, Academic Press

T.A. Ivey and J.M. Landsberg, Cartan For Beginners, AMS Volume61