Symmetry Methods for Differential and Difference Equations. Peter Hydon

Lecture 2: How to find Lie symmetries Symmetry Methods for Differential and Difference Equations Peter Hydon University of Kent

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

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 s (p) = Ω ( r, ṡ, s,..., s (p 1)), s (k) = dk s dr k. (1) The variable s is missing, because the ODE has the Lie symmetries (ˆr, ŝ) = (r, s + ε), which give ŝ (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) = Ω ( r, v, v,..., v (p 2)), where v = ṡ. (2) If (2) can be solved, the solution of (1) is s = v(r; c 1,... c p 1 ) dr + c p.

Reduction of order for ODEs and O Es Note Sometimes, it is more convenient to choose v to be a different function of r and ṡ. Any choice will do, provided that v depends locally diffeomorphically on ṡ. This is to ensure that (r, ṡ) 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) = Ω ( r, v, v,..., v (p 2)).

Reduction of order for ODEs and O Es Example The ODE y = 1 xy 2 (3) has Lie symmetries with ξ(x, y) = x 2, η(x, y) = xy. The canonical coordinates ( y (r, s) = x, 1 ) ( 1 give (ṡ, s) = x xy y, x 3 y ) (xy y) 3. With v = ṡ, the ODE reduces to v = v 3 r 2, which happens to be separable. Therefore, the general solution of (3) can be obtained by integration.

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 s p s p 1 = Ω ( n, s 1 s, s 2 s 1,..., s p 1 s p 2 ). (4) The variable s is missing, because the ODE has the Lie symmetries (ˆn, ŝ) = (n, s + ε), so ŝ k+1 ŝ k = s k+1 s k 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 v p 1 = Ω ( n, v, v 1,..., v p 2 ), where v = s1 s. (5) If (5) can be solved, the solution of (4) is s = v(n; c 1,... c p 1 ) + c p.

Reduction of order for ODEs and O Es Example The O E u 2 = u 1(uu 1 + u 1 u) (u 1 ) 2 + u 1 u has Lie symmetries whose characteristic is Q(n, u) = u 2. Thus s 2 = (s 1) 2 ss 1 + s s 1 s + 1, where s = 1 u. With the substitution v = s 1 s, this reduces to v 1 = v v + 1, whose general solution is v = 2 { 1 + c 1 ( 1) n} / { (c 1 ) 2 1 }.Then s = v(n) + c 2 = 2n + c { 1 1 ( 1) n } (c 1 ) 2 + c 2, 1 which is easily inverted to obtain u.

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 (ˆx, ŷ) and expand in powers of ε: F (ˆx, ŷ) = F (x, y) + ε { ξ(x, y)f,x + η(x, y)f,y } + O(ε 2 ), where the subscripts indicate the partial derivative with respect to each variable that follows the comma. Equivalently, F (ˆx, ŷ) = F (x, y) + εx {F (x, y)} + O(ε 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.)

The infinitesimal generator The infinitesimal generator turns up in the formal solution of the equations that define the tangent vector to the orbit at (ˆx, ŷ): dˆx dε = ξ(ˆx, ŷ), dŷ The formal solution is dε = η(ˆx, ŷ), (ˆx, ŷ) ε=0 = (x, y). ˆx = exp{εx }x, ŷ = exp{εx }y, where exp{εx } = k=0 ε k k! X k. The infinitesimal generator determines (ˆx, ŷ) locally, to all orders in ε. Indeed, wherever the above series converges, F (ˆx, ŷ) = exp{εx }F (x, y).

The infinitesimal generator Theorem F (x, y) is invariant if and only if X {F (x, y)} = 0. More generally, if a 1-parameter local Lie group acts on any space with coordinates x α, such that the infinitesimal generator is ˆx α = x α + ε ζ α (x) + O(ε 2 ), X = ζ α x α (summed over α). A locally-smooth function F (x) is invariant if and only if X {F (x)} = 0.

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 x α, X = (Xx α ) x α.

Prolongation formulae for scalar ODEs The symmetry condition for a p th -order ODE, is y (p) = ω(x, y, y,..., y (p 1) ), (6) ŷ (p) = ω (ˆx, ŷ, ŷ (1),..., ŷ (p 1)) when (6) holds. (Remember: solutions must be mapped to solutions.) This condition involves not only ˆx and ŷ, but also the derivatives ŷ (k) for k from 1 to p. Consequently, we must prolong the infinitesimal generator to act on the p th jet space, with local coordinates (x, y, y,..., y (p) ). To deal with all p at once, write X = ξ x + η y + η (1) y + η (2) y +. We will use (6) to evaluate the output on solutions of the ODE.

Prolongation formulae for scalar ODEs The symmetry condition gives, to first-order in ε, ( ) X y (p) ω(x, y,..., y (p 1) ) = 0 when y (p) = ω(x, y,..., y (p 1) ). Explicitly, this linearized symmetry condition (LSC) amounts to η (p) ξ ω,x η ω,y η (1) ω,y η (p 1) ω,y (p 1) = 0, (7) when (6) holds. [Here ω is shorthand for ω(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 = ω(x, y), the LSC is η (1) ξ ω,x η ω,y = 0 when y = ω(x, y); this holds as an identity in x and y (because the ODE eliminates derivatives of y).

Prolongation formulae for scalar ODEs The coefficients η (k) are defined by the expansion ŷ (k) = y (k) + εη (k) + O(ε 2 ). To calculate η (1), use the total derivative operator, to expand ŷ (1) as follows: D x = x + y y + y y + y y, ŷ (1) = D xŷ D x ˆx = y + εd x η + O(ε 2 ) 1 + εd x ξ + O(ε 2 ) = y +ε ( D x η y D x ξ ) +O(ε 2 ). Similarly, each η (k) is calculated recursively from ŷ (k) = y (k) + εd x η (k 1) + O(ε 2 ) 1 + εd x ξ + O(ε 2 ) = y (k) +ε ( D x η (k 1) y (k) D x ξ ) +O(ε 2 ).

Prolongation formulae for scalar ODEs Examples Assuming that ξ and η depend only on x and y, the prolongation formulae yield: η (1) = η,x + ( ) η,y ξ,x y ξ,y y 2, η (2) = η,xx + (2η,xy ξ,xx )y + (η,yy 2ξ,xy )y 2 ξ,yy y 3 + {η,y 2ξ,x 3ξ,y y }y, η (3) = η,xxx + (3η,xxy ξ,xxx )y + 3(η,xyy ξ,xxy )y 2 + (η,yyy 3ξ,xyy )y 3 ξ,yyy y 4 + 3{η,xy ξ,xx + (η,yy 3ξ,xy )y 2ξ,yy y 2 }y 3ξ,y y 2 + {η,y 3ξ,x 4ξ,y y }y.

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,..., y (p 1). The easiest way to understand how this works in practice is to study some examples.

How to solve the LSC Example The LSC for the simplest second-order ODE, y =0, is η (2) = 0 when y = 0, which amounts to η,xx + (2η,xy ξ,xx )y + (η,yy 2ξ,xy )y 2 ξ,yy y 3 = 0. As ξ and η are independent of y, the LSC splits into η xx = 0, 2η xy ξ xx = 0, η yy 2ξ xy = 0, ξ yy = 0. (8) The general solution of the last two of (8) is ξ(x, y) = A(x)y + B(x), η(x, y) = A (x)y 2 + C(x)y + D(x), where A, B, C and D are arbitrary locally-smooth functions.

How to solve the LSC The remaining equations in (8) amount to A (x)y 2 +C (x)y +D (x) = 0, 3A (x)y +2C (x) B (x) = 0, which are split (by comparing powers of y) into A (x) = 0, C (x) = 0, D (x) = 0, B (x) = 2C (x). All that remains is to solve this simple system of ODEs (in terms of some arbitrary constants, c i ) and reconstruct ξ and η: ξ(x, y) = c 1 + c 3 x + c 5 y + c 7 x 2 + c 8 xy, η(x, y) = c 2 + c 4 y + c 6 x + c 7 xy + c 8 y 2. Therefore, every symmetry generator X is a linear combination of X 1 = x, X 2 = y, X 3 = x x, X 4 = y y, X 5 = y x, X 6 = x y, X 7 = x 2 x + xy y, X 8 = xy x + y 2 y.

How to solve the LSC Example The ODE y = y 3 arises in models of thin-film flow. The LSC is which amounts to η (3) + 3y 4 η = 0 when y = y 3, 0 = η,xxx + (3η,xxy ξ,xxx )y + 3(η,xyy ξ,xxy )y 2 + (η,yyy 3ξ,xyy )y 3 ξ,yyy y 4 + 3{η,xy ξ,xx + (η,yy 3ξ,xy )y 2ξ,yy y 2 }y 3ξ,y y 2 + {η,y 3ξ,x 4ξ,y y }y 3 + 3y 4 η. Although this looks horrible, only the red term is quadratic in y. Consequently, ξ = A(x).

How to solve the LSC Taking this into account, the LSC reduces to 0 = η,xxx + ( 3η,xxy A (x) ) y + 3η,xyy y 2 + η,yyy y 3 +3{η,xy A (x) + η,yy y }y + {η,y 3a (x)}y 3 + 3y 4 η. Only the blue terms are linear in y, so (equating powers of y in such terms) η,xy A (x) = 0, η,yy = 0 η = ( A (x) + c 1 ) y + B(x). Substitute this into the remaining terms of the LSC and compare powers of y, then y, to obtain the general solution of the LSC: ξ(x, y) = 4c 1 x + c 2, η(x, y) = 3c 1 y. Thus every symmetry generator X is a linear combination of X 1 = 4x x 3y y, X 2 = x.

How to solve the LSC For a given first-order ODE, the LSC has infinitely many linearly independent solutions. However, the LSC cannot be split with respect to derivatives of y, so one cannot find its general solution. The best that can be done is to restrict ξ and η and hope that this yields some symmetries. Example The LSC for the ODE y = (1 y 2 )/(xy) + 1 is ( 1 y 2 0 =η,x + xy ( 1 y 2 + x 2 y ( (η,y ) 1 y 2 + 1) ξ,x xy ( 1 + y 2 ) ξ + xy 2 ) η. + 1) 2 ξ,y Let us see whether there are any symmetry generators for which ξ = α(x), η = β(x)y + γ(x). (This is a common restriction.)

How to solve the LSC The LSC amounts to ( 1 y 0 = β y+γ +(β α 2 ) xy ) + 1 +α ( 1 y 2 x 2 y ) +(βy +γ) ( 1 + y 2 xy 2 ). Now one can split the (restricted) LSC in powers of y, to get (after some simplification) γ = 0, β α = 0, α + α/x = 0. There are no further constraints, so α = c 1 x 1 and β = c 1 x 2. Consequently, every generator of the restricted form is a multiple of X 1 = x 1 x x 2 y y. Exercise Use this symmetry generator to construct canonical coordinates and solve the ODE.

Beyond point symmetries Beyond point symmetries A nice version of the prolongation formula (given ξ and η) is η (k) = D k x Q + y (k+1) ξ, where Q = η y ξ is the characteristic; this is proved recursively. It leads to a very useful splitting of the infinitesimal generator: X = ξd x + Q y + (D x Q) y + (D 2 x Q) y +. The red term generates trivial Lie symmetries (for arbitrary ξ), because its restriction to solutions maps points along solution curves. So X is equivalent (up to a trivial symmetry generator) to X = Q y + (D x Q) y + (D 2 x Q) y +.

Beyond point symmetries X X (ˆx, ŷ) y (x, y) x ξd x ε Non-invariant solution curves: orbits cross these transversely For first-order ODEs, the geometrical picture is shown above: the generator X splits into a trivial tangential component, ξd x, and a nontrivial vertical component, X. Here, vertical means that there is no component in the direction of the independent variable x. The same picture holds (in the appropriate jet space) for higher order ODEs. The LSCs with X and X give the same characteristics.

Beyond point symmetries So far, we have restricted attention to point symmetries, which act as local diffeomorphisms on the (x, y)-plane. More symmetries are available, which act as local diffeomorphisms on various jet spaces. Given an ODE of order p, the types of symmetries are as follows. Point symmetries have a characteristic that depends only on x, y and y, and is linear in y, so Q = η(x, y) ξ(x, y)y. Contact symmetries have a characteristic Q(x, y, y ), with ξ = Q,y, η = Q y Q,y, η (1) = Q,x + y Q,y. Dynamical symmetries have a characteristic that depends on some or all of x, y, y,... y (p 1). It is usual to set ξ = 0, so that ˆx = x.

Beyond point symmetries To find any of these characteristics, first pick a suitable restriction on Q. Then solve the LSC in the form ( ) X y (p) ω(x, y,..., y (p 1) ) = 0 on solutions, using y (p) = ω(x, y,..., y (p 1) ) to eliminate y (p+k) for all k 0. Example Every characteristic of Lie contact symmetries for is a linear combination of y = x(x 1)y 3 2xy 2 + y Q 1 = 1, Q 2 = x, (point symmetry characteristics) Q 3 = e y, Q 4 = (xy y x)e y, Q 5 = e y xy +y +x. These are obtained by splitting the LSC in powers of y.

Beyond point symmetries Just as for first-order ODEs, there are infinitely many dynamical symmetry characteristics Q(x, y,..., y (p 1) ). To find any of them by splitting the LSC, one must make a further restriction. The most common restriction is to look for characteristics that are independent of y (p 1). All nontrivial symmetries, dynamical or otherwise, act by changing the arbitrary constants in the general solution of the ODE. (One can regard them as local point transformations on the p-dimensional space of first integrals.) The same idea applies to O Es.

The LSC for a given O E The LSC for a given O E The LSC for a given O E is written using the forward shift operator, S n, which replaces n by n + 1 in all functions of n: S n {f (n)} = f (n + 1), S n u = u 1, S n u i = u i+1. The last two equations can be combined by using u 0 as another notation for u; we will use this alternative whenever it is helpful. For simplicity, we will assume that u is defined for all n Z, though what follows can be adapted to any domain. For later use, we also introduce the identity operator I, which maps every function of n to itself.

The LSC for a given O E For scalar O Es, Lie symmetries are of the form ˆn = n, û = u + εq + O(ε 2 ). In particular, Lie point symmetries have û (and thus, Q) dependent on n and u only. The prolongation formula is obtained by replacing n by n + k: û k = u k + ε S k nq + O(ε 2 ). Therefore, the infinitesimal generator (prolonged to all orders) is X = k Z ( S k nq ) u k.

The LSC for a given O E The symmetry condition for is u p = ω(n, u,..., u p 1 ) (9) û p = ω(ˆn, û,..., û p 1 ) when (9) holds. Just as for ODEs, the LSC for Lie symmetries is obtained by expanding the symmetry condition to first order in ε: X { u p ω(n, u,..., u p 1 ) } = 0 when (9) holds. Using ω as shorthand for ω(n, u,..., u p 1 ), the LSC amounts to S p nq ω,1 Q ω,2 S n Q ω,p S p 1 n Q = 0 when (9) holds, where the subscript, i denotes the partial derivative with respect to the i th continuous argument. For example, ω,p = ω/ u (p 1).

The LSC for a given O E For most nonlinear O Es of order p > 1, one can find all Lie point symmetries from the LSC, without having to restrict Q. To see how, let us examine the LSC for a given second-order O E, u 2 = ω(n, u, u 1 ), (10) which is a functional equation (not an ODE or PDE) for Q: Q(n + 2, ω) ω,2 Q(n + 1, u 1 ) ω,1 Q(n, u) = 0. (11) Although functional equations are generally hard to solve, Lie symmetries are diffeomorphisms. Consequently, Q is a locally smooth function of its continuous argument and so the LSC can be solved by the method of differential elimination. Before discussing this method in general, we begin with a simple example.

The LSC for a given O E Example The LSC for Lie point symmetries of u 2 = (u 1 ) 2 /u is Q ( n + 2, (u 1 ) 2 /u ) 2(u 1 /u) Q(n + 1, u 1 ) + (u 1 /u) 2 Q(n, u) = 0. In each term, Q depends on only one continuous variable (and one discrete variable). We could differentiate the LSC in order to eliminate that term. In particular, { u2 (u 1 ) 2 u u 2u 1 u 1 } Q ( n + 2, (u 1 ) 2 /u ) = 0. By applying the differential operator in braces to the LSC, we eliminate the first term in the LSC and obtain Q (n + 1, u 1 ) 1 Q(n + 1, u 1 ) Q (n, u) + 1 Q(n, u) = 0. (12) u 1 u This has more terms than the LSC does, but Q and Q only take two pairs of arguments, instead of three.

The LSC for a given O E Continue the process, by differentiating (12) with respect to u, in order to eliminate Q (n + 1, u 1 ). This gives [ Q (n, u) + 1 u Q(n, u) ] = 0. (13) Having found a (parametrized) differential equation for Q, we can start to solve the system. Integrate (13) once, to obtain Q (n, u) 1 Q(n, u) = α(n), (14) u where α(n) is to be determined. Now substitute (14) into the intermediate equation (12), which gives the O E α 1 α = 0. Consequently α = c 1, so the general solution of (12) is Q(n, u) = c 1 u ln u + β(n) u. (15)

The LSC for a given O E Finally, substitute (15) into the LSC to obtain (after cancellation) β 2 2β 1 + β = 0. The general solution of this linear O E is β(n) = c 2 n + c 3, and hence the general solution of the LSC is Q(n, u) = c 1 u ln u + c 2 nu + c 3 u. In other words, every characteristic is a linear combination of Q 1 = u ln u, Q 2 = nu, Q 3 = u.

The LSC for a given O E This example illustrates the general technique for obtaining Lie point of any difference equation of order p 2: 1 Write down the LSC. 2 Apply appropriate differential operators to reduce the number of unknown functions. 3 Having reached a differential equation, back-substitute and solve the resulting linear difference equations. 4 Iterate, if necessary. Exactly the same approach may be used to obtain dynamical symmetries, with Q dependent on n, u, u 1,..., u p 1, provided that Q is restricted is some way. For first-order O Es, it is necessary to assume that Q has a particular dependence on u. A common assumption is Q(n, u) = α(n) u 2 + β(n) u + γ(n). (16)

The LSC for a given O E Example The LSC for the O E u 1 = u/(1 + nu) is Q ( n + 1, u/(1 + nu) ) Q(n, u)/(1 + nu) 2 = 0. With the restriction (16), the LSC reduces to u 2 α 1 + u(1 + nu)β 1 + (1 + nu) 2 γ 1 u 2 α u β γ = 0, which we split into a system of O Es by comparing powers of u: u 2 terms : α 1 + n β 1 + n 2 γ 1 α = 0, u terms : β 1 + 2n γ 1 β = 0, other terms : γ 1 γ = 0.

The LSC for a given O E Systems in this triangular form are solved by starting at the bottom and working up. In this case, each of the scalar equations is easy to solve, so there is no need to restrict α, β or γ. We obtain γ = c 1, β = c 2 c 1 n(n 1), α = c 3 c 2 n(n 1)/2 + c 1 n 2 (n 1) 2 /4. Therefore every characteristic that is quadratic in u is of the form ( n(n 1) 2 ( ) n(n 1) Q(n, u) = c 1 1 u) + c 2 u 1 u + c 3 u 2. 2 2 Note For systems not in triangular form, one may need to combine the O Es to solve them. If necessary, restrict α, β and γ.

The LSC for a given O E Potential problems with differential elimination 1. Generally, each elimination increases the number of terms, a phenomenon known as expression swell. This makes it difficult to find dynamical symmetries of high-order O Es. 2. In view of the above, it is wise to use computer algebra to keep track of the calculations. However, beware of computer algebra solvers for differential and difference equations. They may overlook some solutions! 3. It may be difficult (or even impossible) to solve some of the linear difference equations in Step 3. However, this is unusual for nonlinear O Es.

Inherited symmetries Inherited symmetries For ODEs and O Es, the set of Lie symmetry characteristics that satisfy a particular restriction is a vector space. However, for the set of all Lie point symmetries, there is an extra structure. Theorem If X i and X j generate Lie symmetries (of any type), so does [X i, X j ] = X i X j X j X i. Corollary The space L of all Lie point symmetry generators is a Lie algebra under the commutator [, ]. Note The space of all dynamical symmetry generators is also a Lie algebra, but it is easier to find the set of all Lie point symmetries (when p 2) and to construct their canonical coordinates!

Inherited symmetries An R-dimensional Lie algebra L is solvable if the derived series, L (0) = L, L (i+1) = [L (i), L (i) ], stops at {0}. If L is solvable, one can use the derived series to construct a chain of subalgebras, {0} = L 0 L 1 L R = L, such that each L k is a k-dimensional ideal of L k+1. Given such a chain, a canonical basis for L satisfies L k = Span(X 1,..., X k ), for k = 1,..., R. Theorem Given an ODE or O E of order p 2, suppose that L is solvable and that X 1,..., X R is a canonical basis. When the ODE or O E is reduced using canonical coordinates corresponding to X 1, the reduced equation inherits the symmetries generated by X 2,..., X R.

Inherited symmetries Example has a canonical basis The Lie algebra L of point symmetry generators for u 3 = u 2 + (u 2 u 1 ) 2 u 1 u X 1 = u, X 2 = u u. Reduce the order by using X 1, whose canonical coordinate is s = u: v 2 = v 2 1 /v v = u 1 u. The reduced O E inherits the symmetries generated by X 2, as X 2 v = u v u + u v 1 = u 1 u = v. u 1 Therefore X 2 reappears as the scaling symmetry generator X 2 = v v, which enables one to carry out a second reduction of order.

Inherited symmetries By contrast, let us use X 2 first, with s = ln u and ṽ = u 1 /u for simplicity. (For the same reason as for ODEs, we can take ṽ to be a locally invertible function of s 1 s.) This reduction of order works (as it is guaranteed to do), yielding ṽ 2 = 1 + ṽ (ṽ 1 1) 2 ṽ 1 (ṽ 1). (17) However, the reduced O E (17) has no Lie point symmetries. What has happened to the symmetries generated by X 1? As before, apply X 1 (prolonged) to ṽ, to get X 1 ṽ = ṽ u + ṽ = 1 u 1 u u 1 u 2 = 1 ṽ. u This is not a function of ṽ only, so it cannot be a generator of Lie point symmetries for the reduced O E (17).

Summary: the main results in Lecture 2 Summary: the main results in Lecture 2 Given an ODE or O E of order p 2, canonical coordinates reduce the order by one. If the reduced equation can be solved, the original equation is solved by one more integration or summation. The infinitesimal generator X determines the local behaviour of Lie symmetries. A function is invariant if and only if it is in the kernel of X. The prolonged infinitesimal generator yields the LSC. For an ODE, this is a differential equation; for an O E, the LSC is a functional equation. By restricting the characteristic, Q, one can seek all Lie symmetries that satisfy the restriction. Point symmetries act on the space of independent and dependent variables; dynamical symmetries act on the space of first integrals (or arbitrary constants).

Summary: the main results in Lecture 2 Conclusion: There are substantial differences between O E and ODE symmetry methods!