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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