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How to Find Lie Symmetries Lecture 2: How to find Lie symmetries Lecture 2: How to find Lie symmetries Symmetry Methods for Differential and Difference Equations Peter Hydon University of Kent Lecture 2: How to find Lie symmetries Outline 1 Reduction of order for ODEs and O∆Es 2 The infinitesimal generator 3 Prolongation formulae for scalar ODEs 4 How to solve the LSC 5 Beyond point symmetries 6 The LSC for a given O∆E 7 Inherited symmetries 8 Summary: the main results in Lecture 2 Lecture 2: How to find Lie symmetries Reduction of order for ODEs and O∆Es Reduction of order for ODEs A given ODE of order p may be written in canonical coordinates as dk s s(p) = Ωr; s_; ¨s;:::; s(p−1); s(k) = : (1) dr k The variable s is missing, because the ODE has the Lie symmetries (^r; ^s) = (r; s + "), which give ^s(k) = s(k) for each k ≥ 1. In other words, the ODE can be written entirely in terms of differential functions that are invariant under the Lie symmetries. Consequently, (1) reduces to v (p−1) = Ωr; v; v_ ;:::; v (p−2); where v =s _: (2) If (2) can be solved, the solution of (1) is Z s = v(r; c1;::: cp−1) dr + cp: Lecture 2: How to find Lie symmetries Reduction of order for ODEs and O∆Es Note Sometimes, it is more convenient to choose v to be a different function of r ands _. Any choice will do, provided that v depends locally diffeomorphically ons _. This is to ensure that (r; s_) can be obtained from (r; v) and vice versa, and that the local smooth structure is preserved. Whichever choice of v is made, the reduced ODE will always be of the form v (p−1) = Ωr; v; v_ ;:::; v (p−2): Lecture 2: How to find Lie symmetries Reduction of order for ODEs and O∆Es Example The ODE 1 y 00 = (3) xy 2 has Lie symmetries with ξ(x; y) = x2; η(x; y) = xy. The canonical coordinates y 1 1 x3y 00 (r; s) = ; − give (_s; ¨s) = ; : x x xy 0 − y (xy 0 − y)3 With v =s _, the ODE reduces to v 3 v_ = ; r 2 which happens to be separable. Therefore, the general solution of (3) can be obtained by integration. Lecture 2: How to find Lie symmetries Reduction of order for ODEs and O∆Es Reduction of order for O∆Es A given O∆E of order p may be written in canonical coordinates as sp − sp−1 = Ω n; s1 − s; s2 − s1;:::; sp−1 − sp−2 : (4) The variable s is missing, because the ODE has the Lie symmetries (^n; ^s) = (n; s + "), so ^sk+1 − ^sk = sk+1 − sk for each k ≥ 0. In other words, the O∆E can be written entirely in terms of functions that are invariant under the Lie symmetries. Consequently, (1) reduces to vp−1 = Ω n; v; v1;:::; vp−2 ; where v = s1 − s: (5) If (5) can be solved, the solution of (4) is X s = v(n; c1;::: cp−1) + cp: Lecture 2: How to find Lie symmetries Reduction of order for ODEs and O∆Es Example The O∆E u1(uu1 + u1 − u) u2 = 2 (u1) + u1 − u has Lie symmetries whose characteristic is Q(n; u) = u2. Thus 2 (s1) − ss1 + s 1 s2 = ; where s = − : s1 − s + 1 u With the substitution v = s1 − s, this reduces to v v = − ; 1 v + 1 n 2 whose general solution is v = 2 1 + c1(−1) = (c1) − 1 .Then n X 2n + c1 1 − (−1) s = v(n) + c2 = 2 + c2; (c1) − 1 which is easily inverted to obtain u. Lecture 2: How to find Lie symmetries The infinitesimal generator The infinitesimal generator Question How does an arbitrary locally-smooth function F (x; y) vary along the orbit of a 1-parameter local Lie group? Answer Replace (x; y) by (^x; y^) and expand in powers of ": 2 F (^x; y^) = F (x; y) + " ξ(x; y)F;x + η(x; y)F;y + O(" ); where the subscripts indicate the partial derivative with respect to each variable that follows the comma. Equivalently, F (^x; y^) = F (x; y) + "X fF (x; y)g + O("2); where X = ξ(x; y)@x + η(x; y)@y is the infinitesimal generator of the local Lie group. @ (We use the shorthand @a for @a here and from now on.) Lecture 2: How to find Lie symmetries The infinitesimal generator The infinitesimal generator turns up in the formal solution of the equations that define the tangent vector to the orbit at (^x; y^): dx^ dy^ = ξ(^x; y^); = η(^x; y^); (^x; y^) = (x; y): d" d" "=0 The formal solution is 1 X "k x^ = expf"X gx; y^ = expf"X gy; where expf"X g = X k : k! k=0 The infinitesimal generator determines (^x; y^) locally, to all orders in ". Indeed, wherever the above series converges, F (^x; y^) = expf"X gF (x; y): Lecture 2: How to find Lie symmetries The infinitesimal generator Theorem F (x; y) is invariant if and only if X fF (x; y)g = 0. More generally, if a 1-parameter local Lie group acts on any space with coordinates xα, such that x^α = xα + " ζα(x) + O("2); the infinitesimal generator is α X = ζ @xα (summed over α): A locally-smooth function F (x) is invariant if and only if X fF (x)g = 0: Lecture 2: How to find Lie symmetries The infinitesimal generator The change-of-variables formula is straightforward for infinitesimal generators. Given local coordinates (a; b) on the (x; y)-plane, X = (Xa) @a + (Xb) @b: In particular, canonical coordinates (r; s) satisfy Xr = 0; Xs = 1; so X = @s : [By the above theorem, r is invariant under the action of the 1-parameter Lie group.] More generally, in a space with coordinates xα, α X = (Xx ) @xα : Lecture 2: How to find Lie symmetries Prolongation formulae for scalar ODEs The symmetry condition for a pth-order ODE, y (p) = !(x; y; y 0;:::; y (p−1)); (6) is y^(p) = !x^; y^; y^(1);:::; y^(p−1) when (6) holds: (Remember: solutions must be mapped to solutions.) This condition involves not onlyx ^ andy ^, but also the derivatives y^(k) for k from 1 to p. Consequently, we must prolong the infinitesimal generator to act on the pth jet space, with local coordinates (x; y; y 0;:::; y (p)). To deal with all p at once, write (1) (2) X = ξ@x + η@y + η @y 0 + η @y 00 + ··· : We will use (6) to evaluate the output on solutions of the ODE. Lecture 2: How to find Lie symmetries Prolongation formulae for scalar ODEs The symmetry condition gives, to first-order in ", X y (p)−!(x; y;:::; y (p−1)) = 0 when y (p) = !(x; y;:::; y (p−1)): Explicitly, this linearized symmetry condition (LSC) amounts to (p) (1) (p−1) η − ξ !;x − η !;y − η !;y 0 − · · · − η !;y (p−1) = 0; (7) when (6) holds. [Here ! is shorthand for !(x; y;:::; y (p−1)).] The LSC holds as an identity in x; y;:::; y (p−2) and y (p−1). Example For a given first-order ODE, y 0 = !(x; y), the LSC is (1) 0 η − ξ !;x − η !;y = 0 when y = !(x; y); this holds as an identity in x and y (because the ODE eliminates derivatives of y). Lecture 2: How to find Lie symmetries Prolongation formulae for scalar ODEs The coefficients η(k) are defined by the expansion y^(k) = y (k) + εη(k) + O("2): To calculate η(1), use the total derivative operator, 0 00 000 Dx = @x + y @y + y @y 0 + y @y 00 ··· ; to expandy ^(1) as follows: 0 2 (1) Dx y^ y + "Dx η + O(" ) 0 0 2 y^ = = 2 = y +" Dx η − y Dx ξ +O(" ): Dx x^ 1 + "Dx ξ + O(" ) Similarly, each η(k) is calculated recursively from (k) (k−1) 2 (k) y + "Dx η + O(" ) (k) (k−1) (k) 2 y^ = 2 = y +" Dx η −y Dx ξ +O(" ): 1 + "Dx ξ + O(" ) Lecture 2: How to find Lie symmetries Prolongation formulae for scalar ODEs Examples Assuming that ξ and η depend only on x and y, the prolongation formulae yield: (1) 0 02 η = η;x + η;y − ξ;x y − ξ;y y ; (2) 0 02 03 η = η;xx + (2η;xy − ξ;xx )y + (η;yy − 2ξ;xy )y − ξ;yy y 0 00 + fη;y − 2ξ;x − 3ξ;y y gy ; (3) 0 02 η = η;xxx + (3η;xxy − ξ;xxx )y + 3(η;xyy − ξ;xxy )y 03 04 + (η;yyy − 3ξ;xyy )y − ξ;yyy y 0 02 00 + 3fη;xy − ξ;xx + (η;yy − 3ξ;xy )y − 2ξ;yy y gy 002 0 000 − 3ξ;y y + fη;y − 3ξ;x − 4ξ;y y gy : Lecture 2: How to find Lie symmetries How to solve the LSC How to solve the LSC We are now able to write out the LSC as a set of partial differential equations (PDEs) for ξ(x; y) and η(x; y). Usually, for ODEs of order p ≥ 2, one can solve these equations. This is because the LSC can be split according to the dependence of each term on y 0;:::; y (p−1). The easiest way to understand how this works in practice is to study some examples.
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