Download - Lecture 1: From symmetries to solutions · Sophus Lie’s symmetry methods answer these questions. Lecture 1: From symmetries to solutions Introduction to symmetries De nition A symmetry

Lecture 1: From symmetries to solutions

Lecture 1: From symmetries to solutionsSymmetry Methods for Differential and Difference Equations

Peter Hydon

University of Kent


Outline

1 First-order ODEs: solution by canonical coordinates

2 Introduction to symmetries

3 Canonical coordinates from Lie symmetries

4 Canonical coordinates for first-order O∆Es

5 Summary: the main results in Lecture 1


First-order ODEs: solution by canonical coordinates

Notation for scalar ordinary differential equations (ODEs)

Independent variable: x ∈ R; dependent variable: y ∈ R.

First-order ODEs (solved for highest derivative):

y ′ = ω(x , y), where y = y(x), y ′ =dy(x)

dx.

ODEs of order p (solved for highest derivative):

y (p) = ω(x , y , y ′, . . . , y (p−1)), where y (k) =dky(x)

dxk.

Assumption: ω is locally smooth in each argument.

The general solution has p arbitrary constants.



Some elementary solution methods for first-order ODEs

Linear ODEs: To solve the linear ODE

y ′ + a(x) y = b(x),

use an integrating factor. Equivalently, introduce new coordinates

r = x , s = y exp{∫

a(x) dx}−→ ds

dr= b(r) exp

{∫a(r)dr

}.

Homogeneous ODEs: To solve the homogeneous ODE

y ′ = F (y/x), F (z) 6= z ,

introduce new local coordinates

r = y/x , s = ln |x | −→ ds

dr=

1

F (r)− r.



Common aspects of these methods

In each case, local coordinates (r , s) put the ODE in the form

ds

dr= f (r) −→ s =

∫f (r)dr + c;

any local coordinates that do this are called canonical.Locally, s parametrizes the solutions.

solutioncurves

curves of constant r

s



Some questions

The solution curves of a first-order ODE foliate regions of the(x , y) plane. Does a set of canonical coordinates always exist?

If so, how can one find them?

How can one solve an ODE of unfamiliar type?

Example : y ′ =1− y2

xy+ 1.

Are canonical coordinates useful for higher-order ODEs?

Sophus Lie’s symmetry methods answer these questions.


Introduction to symmetries

Definition A symmetry of a geometrical object is an invertibletransformation that maps the object to itself. Individual points ofan object may be mapped to different points, but the object as awhole is unchanged by any symmetry.

Example Some symmetries of a square:

Γ1 Γ2

If the object has some associated structure, every symmetry mustpreserve this structure. (Otherwise, the object would change.)Examples include rigidity and smoothness.



The set of symmetries of an object is a group under composition oftransformations, which is an associative operation.

The identity (id) maps each point of the object to itself.

The group may be finite or infinite.

Γε

ε

Symmetries of the circle:

All rotations about centre

Γ

Reflection in each diagonal



A closer look at rotations of the circle

Γε

εx

x(x; ε) In cartesian coordinates:

x = (x , y) = (cos θ, sin θ),

x =(

cos(θ + ε), sin(θ + ε)).

Note: Γ0 = id,

Γδ Γε = Γδ+ε.

For sufficiently small |ε|, the Taylor expansion about ε = 0 gives

x =(x cos ε−y sin ε, x sin ε+y cos ε

)= (x , y) +ε (−y , x) +O(ε2).

The set of rotations is a one-parameter local Lie group.



Definition A parametrized set of transformations,

Γε : x 7→ x(x; ε), ε ∈ (ε0, ε1),

where ε0 < 0 < ε1, is a one-parameter local Lie group if:

1. Γ0 is the identity map, so that x = x when ε = 0.

2. Γδ Γε = Γδ+ε for every δ, ε sufficiently close to zero.

3. Each xα can be represented as a Taylor series in ε (in aneighbourhood of ε = 0 that is determined by x), and so

xα(x; ε) = xα + ε ζα(x) + O(ε2), α = 1, . . . ,N.

1 and 2 imply that Γ−1ε = Γ−ε when |ε| is sufficiently small.

Note: A local Lie group may not be a group; it need only satisfythe group axioms for sufficiently small parameter values.



Orbits of a one-parameter local Lie group acting on the plane

(x , y)

ε

(x , y)

x = x + εξ(x , y) + O(ε2)

y = y + εη(x , y) + O(ε2)

(ξ(x ,y)η(x ,y)

)(ξ(x ,y)η(x ,y)

) dxdε = ξ(x , y)

dydε = η(x , y)

The black curve is part of the orbit through (x , y).

Tangent vectors to the orbit are shown in red.

A point (x , y) is invariant if and only if ξ(x , y) = η(x , y) = 0.



Symmetries of a given ODE

An ODE (of any order) may be represented by the set of itssolutions. For ODEs with a locally-smooth structure, symmetriesare defined as follows.

Definition A symmetry of a given ODE is a locally-defineddiffeomorphism, Γ, that maps the set of all solutions to itself.(Consequently, every solution is mapped to a solution.)

If Γ maps a solution to itself, that solution is invariant.

If every solution is invariant, Γ is said to be trivial.

In effect, the solutions are ‘points’ of the ODE; trivial symmetriesact like the identity transformation.



Action of Lie transformations on solution curves

(x , y)

(ξ(x ,y)η(x ,y)

)ε

(x , y)

Non-invariant solution curves:

orbits cross these transversely(1

y ′(x)

)Solution curve coincides with orbit at (x , y) if

Q(x , y , y ′) ≡∣∣∣ 1 ξ(x ,y)y ′ η(x ,y)

∣∣∣ = η(x , y)− y ′ξ(x , y)

is zero on curve; then the solution is invariant.

Q = 0 on all solutions ⇔ all solutions invariant ⇔ trivial symmetries



Symmetries of y ′ = 0

x

y

Γ1

Γ2

Γ3

Γ1 : (x , y)→ (x + ε1, y) (Lie, trivial)

Γ2 : (x , y)→ (x , y + ε2) (Lie, nontrivial)

Γ3 : (x , y)→ (x , −y) (discrete, nontrivial)


Canonical coordinates from Lie symmetries

Which ODEs have vertical translations?

The Symmetry Condition (SC) for dydx = ω(x , y):

dy

dx= ω(x , y) when

dy

dx= ω(x , y)

A first-order ODE y ′ = ω(x , y) admits all vertical translations,

(x , y) = (x , y + ε), ε ∈ R,

iff ω is a function of x only.

Proof :dy

dx=

dy

dx; SC→ ω(x , y+ε) = ω(x , y) → ω(x , y) = f (x).

These ODEs are easily solved: y =∫f (x) dx + c . (c 7→ c + ε)



Canonical coordinates

Idea Introduce local coordinates (r , s) in which Lie symmetrieslook (locally) like vertical translations:

r(x , y) = r(x , y); s(x , y) = s(x , y) + ε.

Note Any ODE with these symmetries can be written in terms ofthe invariant functions r , s, s, . . . , where the derivative with respectto r is denoted by a dot.

Method Solve

dx

ξ(x , y)=

dy

η(x , y)= dε, (x , y)

∣∣ε=0

= (x , y).

Potential problems: (a) invariant points (orbit is zero-dimensional),(b) patched solutions, (c) system too hard to solve (rare).



Example: The ODE

dy

dx=

y + 1

x+

y2

x3

has Lie symmetries with ξ(x , y) = x2, η(x , y) = xy .

Canonical coordinates are obtained from

dx

x2=

dy

x y= dε, (x , y)

∣∣ε=0

= (x , y).

Simple solution: r(x , y) = y/x , s(x , y) = −1/x .

The ODE reduces to s =1

1 + r2. Solution: y = −x tan

(1

x+c).


Canonical coordinates for first-order O∆Es

Notation for scalar ordinary difference equations (O∆Es)

Independent variable: n ∈ Z; dependent variable: u ∈ R.

First-order O∆Es (forward form):

u1 = ω(n, u), where u = u(n), u1 = u(n + 1).

O∆Es of order p (in forward form):

up = ω(n, u, u1, . . . , up−1), where uk = u(n + k).

Assumptions: ω is locally smooth in each continuous argument;∂ω/∂u 6= 0. (The O∆E is exactly pth-order.)

The general solution has p arbitrary constants.



0

y

x(x , y) (x+ε2, y)

(x , y+ε1)

u

0-1-2-3-4 1 2 3 4n

(n, u)

(n, u+ε)

The simplest O∆E is u1 − u = 0; its general solution is u = c.

Unlike y ′ = 0, it has no trivial Lie symmetries, because theindependent variable is discrete. However, the vertical translation(n, u) = (n, u + ε) is a symmetry for each ε ∈ R.



More generally, every O∆E of the form

u1 − u = f (n), (1)

has the one-parameter Lie group of symmetries

(n, u) = (n, u + ε) ε ∈ R.

Proof: u1 − u = (u1 + ε)− (u + ε) = u1 − u = f (n) = f (n).

No other first-order O∆E u1 = ω(n, u) has these Lie symmetries.

Just as y ′ = f (x) is solved by integration, (1) is solved bysummation:

u =∑

f (n) + c .



The summation operatorFor convenience, we use the following shorthand for indefinitesums:

∑f (n) =

n−1∑k=n0

f (k), n > n0,

0, n = n0,

−n0−1∑k=n

f (k), n < n0,

where n0 is an arbitrary fixed integer in the domain.

An O∆E will be regarded as solved if all that remains is to carryout summations, whether or not we can evaluate the sums inclosed form.



The simplest Lie symmetries of are of the form

n = n, u = u + εQ(n, u) + O(ε2);

here Q(n, u) is the characteristic with respect to (n, u).

[Vertical translations, (n, u) = (n, u + ε), have Q(n, u) = 1.]

To see how these Lie symmetries transform the shifted variablesuk , simply replace the free variable n by n + k :

uk = uk + εQ(n + k , uk) + O(ε2).

This is the prolongation formula for O∆Es. It is much simpler thanthe corresponding formula for ODEs (see Part 2 of this course).



The change-of-variables formula

Now consider the effect of changing coordinates from (n, u) to(n, v), where v ′(n, u) 6= 0. (Here v ′ is shorthand for ∂v/∂u.)

Apply Taylor’s Theorem to obtain

v ≡ v(n, u) = v(n, u+εQ(n, u)+O(ε2)

)= v+εv ′(n, u)Q(n, u)+O(ε2).

Therefore the characteristic with respect to (n, v) is Q(n, v), where

Q(n, v(n, u)

)= v ′(n, u)Q(n, u).



A canonical coordinate for O∆Es

Just as for ODEs, we seek a local canonical coordinate, s, suchthat the symmetries amount to translations in s:

(n, s) = (n, s + ε).

The characteristic with respect to (n, s) is Q(n, s) = 1; so, by thechange-of-variables formula,

s(n, u) =

∫du

Q(n, u),

in any neighbourhood in which Q(n, u) 6= 0.



Example The O∆E

u1 =u

1 + nu, n ≥ 1,

has the Lie symmetries (n, u) =(n, u

1−εu

), whose characteristic is

Q(n, u) = u2. The canonical coordinate

s(n, u) =

∫u−2 du = −u−1

transforms the O∆E to

s1 − s = −(u1)−1 + u−1 = −n,

and hence

s = c1 −n−1∑k=1

k = c1 − n(n − 1)/2.



Consequently, the general solution of

u1 =u

1 + nu, n ≥ 1,

is

u =2

n(n − 1)− 2c1.

One cannot define a canonical coordinate s at u = 0, becauseQ(n, 0) = 0. Consequently, the points (n, 0) are invariant.

Clearly, u = 0 is a solution, although it is not part of the generalsolution. Any solution that consists entirely of invariant points iscalled an invariant solution.



Compatibility

Canonical coordinates for a given O∆E must satisfy an extracondition: locally, one must be able to write the whole O∆E interms of s. Unlike ODEs, O∆Es involve several different basepoints; s must be consistent across all of them for the O∆E to betransformed correctly.

Any canonical coordinate s that meets this condition will be calledcompatible with the O∆E.

Problem: It is not always possible to find a real-valued compatiblecanonical coordinate!

Solution: Use a complex-valued canonical coordinate instead.



Example The O∆E

u1 =u − n

nu − 1, n ≥ 2,

has a characteristic Q(n, u) = (−1)n(u2 − 1). It is easy to showthat |u1| is greater (less) than 1 whenever |u| is less (greater) than1 and u 6= 1/n.

No real-valued compatible s exists, so use

s(n, u) =(−1)n

2Log

(u − 1

u + 1

).

Here Log is the principal value of the complex logarithm:

Log(z) = ln(|z |) + iArg(z), Arg(z) ∈ (−π, π].



Whether |u| is greater or less than 1, the O∆E amounts to

s1 − s = 12 (−1)n+1

{ln((n − 1)/(n + 1)

)+ iπ

},

whose general solution is

s = 12 (−1)n

{ln((n − 1)/n

)+ iπ/2

}+ c1,

where c1 is a complex-valued constant that can be written in termsof u(2).A routine calculation yields the general solution of theoriginal O∆E; here c = u(2).

u =

(c + 1)n + 2(c − 1)(n − 1)

(c + 1)n − 2(c − 1)(n − 1), n even;

2(1− c)n + (c + 1)(n − 1)

2(1− c)n − (c + 1)(n − 1), n odd.


Summary: the main results in Lecture 1

Summary: the main results in Lecture 1

Most well-known methods for solving a given first-order ODEor O∆E use canonical coordinates to transform the equationinto a simple solvable form.

Symmetries of a given ODE or O∆E map the set of solutionsto itself (invertibly, preserving the locally-smooth dependenceon arbitrary constants).

Nontrivial one-parameter local Lie groups of symmetries yielduseful local canonical coordinates.

Download - Lecture 1: From symmetries to solutions · Sophus Lie’s symmetry methods answer these questions. Lecture 1: From symmetries to solutions Introduction to symmetries De nition A symmetry

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