Solutions to Ordinary Differential Equations Using Methods of Symmetry

Zachary Martinot May 2014

Introduction

The object of this paper is to explore some applications of the symmetries inher- ent to ordinary differential equations (ODEs) following the treatment in [3] with some useful material from [2]. The basic idea is to use a coordinate transforma- tion that takes a equation with no obvious solution to one that is integrable. The standard techniques presented in an introductory differential equations course can be seen as cases of symmetry transformations instead of a series of lucky guesses. In particular we will see that methods involving an integrating factor, homogeneous coordinates, or a reduction of order of a differential equation, can be explained in terms of symmetry transformations. These transformations are part of the theory of Lie groups.

Symmetries of Ordinary Differential Equations

Generally speaking symmetry can be understood as a mapping between math- ematical objects that preserves some properties of those objects. The most ob- vious examples are geometric shapes. Consider a square. If we rotate a square by 90 degrees we obtain the exact same square again; that is, a square with the same orientation. The same is true for rotations of 180, 270, or any integer multiple of 90 degrees. However, if we rotate the square by, say, 60 degrees we obtain an object that is distinguishable from the original by its orientation. So the rotations of the square by integer multiples of 90 degrees form a (discrete) set of mappings from the square to itself i.e. a square is symmetric under certain rotations. Now consider a circle. We can also rotate a circle in its plane and obtain the same circle we started with, but now we may rotate by any amount we like and still obtain the same object we started with. Thus the rotations of the circle form a continuous set of symmetries of the circle. This concept of symmetry can be applied analogously to algebraic expres- sions. Algebraic equations such as y = x2 or y = sin x also have symmetries. The transformation x 7→ −x maps the graph of y = x2 back to itself and the same is true for y = sin x under the transformation x 7→ x + 2π.

1 Similarly, the solutions to differential equations exhibit symmetries and this property can be exploited to find those solutions. The concept of (continuous) symmetry is formalized in the theory of Lie groups so we start with some general definitions from [1]. Definition 1. A group is a set G together with a group operation such that for any two elements g and h of G, the product g · h is again an element of G. Definition 2. An r-parameter Lie group is a group G which also caries the structure of an r-dimensional smooth manifold in such a way that both the group operation m : G × G 7→ G m(g, h) = g · h g, h ∈ G and the inversion i : G 7→ G i(g) = g−1, g ∈ G, are smooth maps between manifolds. In the study of ODE’s and their solutions, we are interested in particular 2 2 forms. Let At :(x, y) 7→ f((x, y), t) = (u, v) be a transformation from R to R depending on a parameter t ∈ R that has the following properties:

1. At is bijective

2. Aµ ◦ Aν = Aµ+ν

3. A0 = I, the identity. That is, f((x, y), 0) = (x, y).

4. For each µ there exists ν such that Aµ ◦ Aν = A0.

5. The function f is C∞ on D ⊂ R2,(x, y) ∈ D and analytic with respect to t.

Then the set G of transformations At is an additive transformation group. Further, G is a one-parameter Lie group. Definition 3. A symmetry of a differential equation is an invertible transfor- mation that maps solutions to solutions For the purposes of this paper we are interested in one-parameter Lie groups that are the symmetries of a first order ODE. These are transformations of the form At(x, y) = (u(x, y), v(x, y)) with a nonzero Jacobian uxvy − uyvx 6= 0. If the solutions of the ODE are defined on D ∈ R2, then the transformation A : D 7→ D maps solutions to solutions. Example 1. Consider the equation dy αy = , dx x where α is a positive integer. The solutions are curves y(x) = cxα. For any t t ∈ R, the transformation At :(x, y) 7→ (u(t), v(t)) = (x, e y) is a symmetry t α since At maps the solution y(x) to the solution v(u) = ce u .

2 Definition 4. For a given point (x0, y0) and a continuous symmetry At of an 2 ODE, the set of points {At(x, y): a < t < b} ⊂ R is an orbit of At. 2 As t varies continuously At(x0, y0) traces a continuous curve in R that is transverse to the solution curves of the ODE. Along an orbit, a change in t maps a solution curve to another solution curve.

Application of Symmetry Methods

We would like to develop a way to find the symmetries of a particular equation. Lemma 1. A differential equation with a translational symmetry in the depen- dent variable is separable.

Proof. Let At(x, y) = (u(t), v(t)) = (x, y + t) be a symmetry of the ODE dy = w(x, y). dx Then dv d(y + t) dy w(x, y + t) = w(u(t), v(t)) = = = = w(x, y). du dx dx

dy Thus w is independent of y and dx = w(x), a separable equation that is integrable: y(x) = R w(x)dx.

If we have a way to convert any general symmetry into a translational sym- metry, then we can integrate to find the solution. To check whether a transformation is a symmetry, we consider two solutions y(x) and v(u) that solve the same first order ODE: dy = w(x, y) dx dv = w(u, v) du A symmetry of the ODE is a transformation that maps (x, y) 7→ (u(x, y), v(x, y)). If we expand the differentials, we find:

0 dv vxdx + vydy vxdx + vyy (x)dx vx + vyw(x, y) w(u, v) = = = 0 = (1) du uxdx + uydy uxdx + uyy (x)dx ux + uyw(x, y) In principle we might like to solve this nonlinear partial differential equation for u and v to explicitly obtain the symmetry (x, y) 7→ (u(x, y), v(x, y)), but this is generally impossible. However, we can find a linearization from the Taylor series expansion of (1). From this we can construct the symmetry. This amounts

3 to finding the vector field that is everywhere tangent to the coordinate curves in the new coordinate system (u,v). This is accomplished by expanding w(u, v), u, and v in Taylor series about t = 0:

dU dV w(u, v) =w(x, y) + t(w (x, y) | ) + t(w (x, y) | ) + O(t2) x dt t=0 y dt t=0 dU u(x, y) =x + t( | ) + O(t2) dt t=0 dV v(x, y) =y + t( | ) + O(t2) dt t=0

dU Since (u(t), v(t)) = At(x, y) = (U(x, y, t),V (x, y, t)), we have dt |t=0 = 0 dV 0 0 0 u (0) = ε(x, y) and dt |t=0 = v (0) = η(x, y). The vector fields (u (t), v (t)) are the slopes of the orbits of At and thus can be integrated to obtain the or- bits. If ε(x, y) = 0, then the orbits of At are vertical lines. Otherwise they are the solutions to dy η(x, y) = . dx ε(x, y)

At is not necessarily a translational symmetry of (1) but using At we can find one that is. This will be the a mapping to the canonical coordinates (r, s) Definition 5. The canonical coordinates (r(x, y), s(x, y)) of a differential equa- tion are the coordinates in which the equation becomes separable.

In the simplest case we look for coordinates that admits a symmetry St : (r, s) 7→ (r, s + t). Then the tangent vector at (r(At(x, y)), s(At(x, y)) = (r(x, y), s(x, y)) is   dr d(s + t) , = (0, 1) dt t=0 dt t=0 Taking derivatives with respect to t at t = 0, we get

d
dr dx dr dy dr dr r(x, y) = + = ε + η = 0 dt t=0 dx dt t=0 dy dt t=0 dx dy

d
ds dx ds dy ds ds s(x, y) = + = ε + η = 1 dt t=0 dx dt t=0 dy dt t=0 dx dy

or

sxε + syη = 1 (2)

rxε + ryη = 0 (3)

4 a system of first order linear partial differential equations that can be solved for r = r(x, y) and s = s(x, y). In the canonical coordinates equation (1) becomes the separable equation

0 ds sx + syy = 0 dr rx + ryy

which can be solved for s(r). In order to recover the solution in the original (x, y) coordinates the transformation (x, y) 7→ (r, s) must be invertible i.e rxsy − rysx 6= 0.

Standard Integration Techniques

We can now examine some standard techniques for solving differential equations and see that they are unified by methods of symmetry.

Integrating Factors For an inhomogeneous first-order linear equation dy + F (x)y = G(x) (4) dx the standard approach is to multiply both side by an integrating factor and integrate to obtain Z R x F dτ − R x F dτ y = e 0 e 0 G(x)dx.

We observe how this solution arises using symmetries. The homogeneous equation

0 yh + F (x)yh = 0 is separable and directly integrable:

R x F dτ yh = e 0

Since equation (4) is linear , if yp is a solution to (4) , y = yp + yh is also a solution. Hence the transformation At :(x, y) 7→ (x, y + tyh(x)) is a symmetry of (4). The orbits of At are vertical lines, so we can set r = x for one of the 0 canonical coordinates. We also have v (t) = yh(x) so from equation (?) we get s = y/yh. In the canonical coordinates (r, s), equation (4) becomes:

5 ds s dx + s dy = x y dr rxdx + rydy 0 yyh(x) 1 dy = − 2 + yh(x) yh(x) dx

yG(x)yh(x) 1 dy = 2 + yh(x) yh(x) dx F (x)y 1 = + G(x) − F (x)y
yh(x) yh(x) G(x) G(r) = = yh(x) yh(r) − R x F dτ = e 0 G(x).

Integrating obtains Z − R x F dτ s(r) = s(x) = e 0 G(x)dx

and inverting the transformation yields the solution in the original coordi- nates (x,y): Z − R x F dτ − R x F dτ y = s(x)yh = e 0 e 0 G(x)dx.

Homogeneous Coordinates A homogeneous equation has the form y y0 = F ( ) (5) x t t The transformations At :(x, y) 7→ (e x, e y) = (u, v) are symmetries of (5) 0 0 t and since (u (t), v (t)) = e (x, y) the orbits of At are straight lines through the origin plus the origin itself. Thus r = y/x is constant on the orbits. If we let s = ln x equation (2) is satisfied and the Jacobian of the transformation (x, y) 7→ (r, s) is non-zero. In these coordinates (5) is transformed into the separable equation

0 ds sx + syy = 0 dr rx + ryy 1/x 1 = 2 y = . −y/x + F ( x )/x F (r) − r

6 Reduction of Order We can use symmetry to reduce the order of a second order ODE. Consider the homogeneous second order linear equation

y00 + F (x)y0 + G(x)y = 0 (6)

A transformation of the form At :(x, y) 7→ (x, f(t)y) is a symmetry of (6) since f(t) can be factored out of each term:

d2 d (f(t)y) + F (x) (f(t)y) + G(x)(f(t)y) = f(t)(y00 + F (x)y0 + G(x)y) = 0 dx2 dx

The orbits of At are again vertical lines so we take r = x. Additionally, ε = 0 0 t 0 and (2) is 1 = syη = sy(f (0)y). Now let f(t) = e so that sy = 1/(f (0)y) = 1/y. This has s = ln y as a solution which is invertible by y = es. Then we have

0 s y = e sx 00 s 2 s s 2 y = e sx + e sxx = e (sxx + sx) so (6) becomes 2 sxx + sx + F (x)sx + G(x) = 0 (7) Since s does not appear explicitly in the transformed equation we can make the substitution z = sx so that (7) becomes

2 zx + z + F (x)z + G(x) = 0 which is a first order ODE.

7 References

[1] Olver, P. J. (2000). Applications of Lie groups to differential equations (Vol. 107). Springer. [2] Starrett, J. (2007). Solving differential equations by symmetry groups. The American Mathematical Monthly, 778-792.

[3] Yap, S. L. (2010). Differential Equations-Not Just a Bag of Tricks!. Mathe- matics Magazine, 83 (1), 3-14.

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