The relations are fairly easy to deduce (either by multiplication by matrices or geometrically), as one has. , R θ1 S θ2 = S θ1+θ 2

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1 10 PART 1 : SOLITON EQUATIONS 4. Symmetries of the KdV equation The idea behind symmetries is that we start with the idea of symmetries of general systems of algebraic equations. We progress to the idea that the full group of symmetries of a system of algebraic equations are generated by infiniteseminal generators. We then explore this idea in the context of partial differential equations 4.1. Symmetries for algebraic equations. Let us start with a somewhat simplified situation by considering the symmetry group of a system of algebraic equations (4.1) F ν (x) =0, for ν =1,...,n. where F ν are functions on a manifold. We will seek to generalize this to differential polynomials, but for the moment, let us concerntrate on the case in which each F ν is a polynomial. Definition 4.1. A group, G, is a symmetry group for this system of equations, (4.1), if for every g G and every x M solving (4.1), then g x also solves (4.1). Let us demonstrate this in the case of solutions of the equation (4.2) F (x, y) =x 2 + y 2 1=0. A rotation of the plane preserves the solutions, as do reflections. Our full group is G = S θ,r θ ; θ R, where ( ) ( )( ) x cos θ sin θ x (4.3) R θ = y sin θ cos θ y ( ) ( )( ) x cos θ sin θ x (4.4) S θ =. y sin θ cos θ y The relations are fairly easy to deduce (either by multiplication by matrices or geometrically), as one has which define the group. where R θ1 R θ2 = T θ1+θ 2, R θ1 S θ2 = S θ1+θ 2 = S θ2 R θ1, S θ1 S θ2 = R θ1 θ 2, Theorem 4.2. The group G is the symmetry group for (4.2). Proof. We suppose (x, y) R 2 solves (4.2), then R θ (x, y) =(x,y )=(x cos θ y sin θ, x sin θ + y cos θ) x 2 + y 2 =x 2 cos 2 θ 2xy cos θ sin θ + y 2 sin 2 θ + x 2 sin 2 θ +2xy cos θ sin θ + y 2 cos 2 θ = x 2 + y 2. This similarly holds for S θ.

2 PART 1 : SOLITON EQUATIONS 11 A second way in which we can consider a symmetry is to allow a point, (x, y), to vary continuously with respect to some variable s. For a circle, it is easy to see that we may associate a differential equation dy dx (4.5) = x, ds ds = y. By eliminating x from this set of equations we have d 2 y ds 2 = y which has the solution y =sins, x =coss. This evolution equation moves along the solutions of the algebraic equation in time. Lastly, let us now consider the case in which we allow the point (x, y) tomove by an infinitesemal value of θ. That is, in what direction is a point moving if we vary θ continuously around 0. We calculate this by taking the derivative with respect to θ at θ =0,sothat ( ) ( ( ) d x(θ) sin θ cos θ x(θ) = dθ y(θ) cos θ sin θ) y(θ) θ=0 ( )( ) 0 1 x(θ) =. 1 0 y(θ) What this means is that the rotation is completely determined by the initial conditions and the above differential equation. In fact, these just determines the vector field shown in Figure 3. We may express this result as being associated with the vector field v = x y y x whose effect on the circle is seen to be v(f )=2xy 2xy =0, hence, this vector field encapsulates the rotational part of the symmetries of a circle. Figure 3. The vector field associated with the infiniteseminal generator.

3 12 PART 1 : SOLITON EQUATIONS In a sense, we note that what is acting on the points is the exponential of the field. That is to say, to first order, what we have is that exp(ɛv) =I + vɛ + O(ɛ 2 ) and to that first order, we are moving along a solution of the system of equations Symmetries of the KdV equation. Therearethreewaysinwhichwe may consider the symmetries, the first is to consider these as evolution equations, where one may change u with respect to an auxilrary variable say s in a manner that the solution is preserved. The second is by utilizing the invariance of the equation under certain transformations. The third is an infinitesimal approach Evolution equations and KdV. One of the many common ways to write the KdV equation is to write (KdV) in the form of an evolution equation. Definition 4.3. An evolution equation is any partial differential equation of the form (4.6) u t = K(u), where K(u) is a differential polynomial (polynomial involving derivatives). In the case of (KdV) we may allow (through scaling of time and space) the KdV equation to take the form K(u) =u u x + 3 u x 3. More generally, we seek to specify ways to computing the symmetry groups of partial differential equations. Definition 4.4. A symmetry of (4.7) is an evolution equation in an auxilirary variable, say s, written as (4.7) u s = ˆK(u), where ˆK(u) is also a differential polynomial satisfying (4.8) on all solutions of (4.7). = ˆK(u), t For this section, we take the KdV equation to be u t = uu x + u xxx, however, it is easy to recover the 6 through scaling. Let us trial an expression like ˆK(u) =u, in which case we get = u s u x + uu sx + u xxxs, = uu x + uu x + u xxx. = u t = uu x + u xxx. This obviously fails. Not every choice of function works.

4 PART 1 : SOLITON EQUATIONS 13 We could search for a number of terms, however, one clever observation, if we let u have weight 2, and derivatives in x increase the weights by one and derivatives in then we find that u t }{{} 5 = uu }{{} x 5 + u }{{} xxx = K(u), 5 is a polynomial whose terms are all of weight 5, hence, we can consider terms all of weight 3, which only includes ˆK(u) =u x which indeed turns out to be a symmetry as we find the left hand side is ˆK(u) = u xx u + u 2 x + u xxxx t while the right hand side is = u xs u + u x u s + u sxxxx = u xx u + u 2 x + u xxxx. with the left and right hand coinciding. If consider all terms of weight 5, then we naturally arise at a symmetry of the form ˆK(u) =c 1 u x u + c 2 u xxx We find the left hand side gives (4.9) =(c 1 u x u + c 2 u xxx ) which we solve for c 1, c 2. In particular, this gives = uu x + u xxx which coincides with the time derivate. If we consider those terms of degree 7, we have a general form ˆK(u) =c 1 u 2 u x + c 2 uu xxx + c 3 u x u xx + C 4 u 5x. These coefficients can be calculated in a similar way, the result being (assuming C 1 =1)that ˆK(u) =u 2 u x +2uu xxx +4u x u xx u 5x which is just another symmetry Transformations that leave the equation invariant. First of all, if you have a solution, u(x, t), then since (KdV) does not depend explicitly on t or x, then we have that v(x, t) =u(x a, t b) is also a solution. We also notice that we may scale variables u μu, t τt, x χx then μu t + μ2 uu x τ χ + μu xxx χ 3 =0 if we let μτ/χ = 1 and τ/χ 3 = 1. Relating these back to τ we find that if we scale x by λ, t by λ 3 and u by 1/λ 2, then we obtain another solution.

5 14 PART 1 : SOLITON EQUATIONS The last symmetry is known as a Galilean boost. Suppose we let x x λt by letting v(x, t) =u(x λt, t) then by differentiating v by t gives v t = u t λu x which means that by differentiating v by t, we introduce an additive factor of λu x. Since putting on an additive constant factor to v only adds an extra multiple of u x via the uu x term, we find if v(x, t) =u(x λt, t)+μ then v t = u t λu x = u xxx + uu x λu x = v xxx + v x (v μ λ) =0 then if μ = λ we have another solution given by v(x, t) =u(x λt, t) λ. This is actually the full group of transformations that we require, but to show this we need to consider the full set of infinitesimal generators Prolongation. The above symmetries were derived in a rather ad-hoc fashion. We wish to show that the infinitesemal symmetry group is generated. In fact, from the first subsection on evolution equations we have an interepretation of the term of degree 3 is that s = x and the interpretation of the term of degree 5 is that s = t. However, to find the infinitesimal generators, we turn to the idea of prolongation. Intuitively, the k-th prolongation is the space obtained by adjoining the k-th mixed derivatives. In the case of the first prolongation on a curve, adjoining the first derivative has the same consequence as blowing up a point on the curve. The method resolves singularities in the same way, however, there are some technical difficulties is dealing with exceptional divisors For algebraic curves. For example, let the set of independent variables be X =(x, y), then the graph of a function u = f : X U C q,is Γ=X U =(x, y, f(x, y)) the second prolongation of f is given by ( f (2) =pr (2) f(x, y) = f, f x, f y, 2 f x 2, 2 ) f x y, 2 f y 2 where we denote the image of f (2) by U (2) as it encodes higher derivatives. In this way, we extend the graph of the function to a higher space, so that the graph of the prolongation above is the set Γ (2) = X U (2) So, if you have two spaces defined by functions that agree at a point on Γ, having their k-th prolongations agree, so the points intersect on Γ (k), encodes a higher order contact point in the sense that their k-th order Taylor series will agree. So we may, for example consider the curve defined by x 2 + y 2 1=0 by letting y = y(x), in which case, the first prolongation is pr (1) y(x) =(y(x),y (x)).

6 PART 1 : SOLITON EQUATIONS 15 where the relevant graph is Γ (1) =(x, y(x),y (x)). Now, given a point, (x, y) Γ, we can consider the action R θ (rotation by an angle θ), it is easy to see that there is an extension of the prolongation to the extended graph, then we need to know how R θ changes our derivative, that is d x 1 x2 = dx = x 1 x 2 y hence, R θ y (x) = x sin θ y x cos θ cos θ + y x sin θ. So when we pass to the prolongation graph we find ( R θ Γ (1) = x cos θ y sin θ, x sin θ + y cos θ, sin θ u ) x cos θ. cos θ + u x sin θ 4.7. For vector fields. We wish to pass this onto the vector field. Suppose we have a parameter space X =(x 1,...,x p )withf : X U C q be given by (f 1,...,f q ), then we have vector field of the form v = p ξ i (x, f) xi + i=1 q φ j fj. then the prolongation of this vector field adjoins higher derivative terms, such as the mixed derivatives, and this can be calculated by pr (n) v = d dɛ pr (n) [exp(ɛv)] (x, f (n) ) ɛ=0 Recall that the relevant vector field inducing the infinitesimal is given by j=1 v = x y y x. which includes the terms x and y,then pr (1) v = x y y x +(1+y 2 x) yx. We could continue with higher prolongations For KdV. The application to the KdV equation requires a lot of algebra. We intend to expand upon this section significantly, however, for the moment, we shall outline the approach. I works the same way as the case above. When we apply prolongation to the KdV equation we consider the vector field u t + u xxx + uu x =0 v = ξ x + τ t + φ u

7 16 PART 1 : SOLITON EQUATIONS We are required to consider the third prolongation, which is pr (1) v = v + φ x + φ t ux ut pr (2) v =pr (1) v + φ xx + φ xt + φ tt uxx uxt utt pr (3) v =pr (2) v + φ xxx + φ xxt + φ xtt + φ ttt. uxxx uxxt uxtt uttt Let us calculate what a few of these are in general; φ t and φ x are the coefficients of the first prolongations of t and x respectively, given by at the second level, we have φ x = D x (φ + ξu x + τu t )+ξu xx + τu xt = D x φ u x D x ξ u t D x τ = φ x +(φ u ξ x )u x τ x u t ξ u u 2 x τ u u x u t φ t = D t (φ + ξu x + τu t )+ξu xt + τu tt = D t φ + u x D t ξ u t D t τ = φ t ξ t u x +(φ u τ t )u t ξ u u x u t τ u u 2 t φ xx = D 2 x(φ ξu x τu t )+ξu xxx + τu xxt = Dxφ 2 u x Dxξ 2 u t Dxτ 2 2u xx D x ξ 2u xt D x τ = φ xx +(2φ xu ξ xx )u x τ xx u t +(φ uu 2ξ xu )u 2 x 2τ xu u x u t ξ uu u 3 x τ uu u 2 xu t +(φ u 2ξ x )u xx 2τ x u xt 3ξ u u x u xx τ u u t u xx 2τ u u x u xt. and so on, in terms of the mixed derivatives. For the third, we have φ xxx = D 3 x(φ ξu x τu t )+ξu xxxx + τu xxxt = D 3 xφ u x D 3 xξ u y D 3 xτ 3u xx D 2 xξ 3u xt D 2 xτ 3u xxx D x ξ 3u xxt D x τ which is a HUGE calculation. We need to apply pr (3) to the KdV equation, while we included the second term, it is not actually needed, because it is clear the third prolongation, applied to to the KdV equation is (4.10) φ t + φ xxx + uφ x + u x φ =0. Now substituting the KdV equation and the derivatives, i.e., we make the following replacements u t = u xxx uu x u xt = u xxx u 2 x uu xx u xxt = u xxxx 2u xx u x u x u xx uu xxx u xxxt = u xxxxx 2u xxx u x 3u 2 xx u x u xxx u x u xxx uu xxxx wherever they appear. This is some huge differential polynomial, however, a careful analysis shows D t τ = 0 indicating it is a function of x and u alone, whose coefficients

8 PART 1 : SOLITON EQUATIONS 17 are equated with 0 gives (4.11) (4.12) (4.13) τ x = τ u = ξ u = φ t = φ x =0 φ = ξ t 2/3uτ t φ x = 2 3 τ t giving the result τ = c 1 +4c 4 t, ξ = c 2 + c 3 t + c 4 x φ = c 3 2c 4 u. The coefficients of c 1,...,c 4 give the basis v 1 = x, v 2 = t, v 3 = x x +3t t 2u u, v 4 = u + t x. which then correspond to the invariances (v 1 ) v(x, t) =u(x a, t) (v 2 ) v(x, t) =u(x, t b) (v 3 ) v(x, t) = 1 λ 2 u(λx, tλ3 ) (v 4 ) v(x, t) =v(x λt, t) λ where the first two actually coincide with the evolution equations from the second subsection.