Math 4381 / 6378 Symmetry Analysis

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1 Math 438 / 6378 Smmetr Analsis Elementar ODE Review First Order Equations Ordinar differential equations of the form = F(x, ( are called first order ordinar differential equations. There are a variet of techniques to solve these tpe of equations and main methods are: (i (ii (iii (iv (v (vi (vii (viii separable linear Bernoulli Ricatti homogeneous linear fractional exact Legendre transformations. Separable Equations A separable first order differential equation has the form: = f (xg( (2 The general solution is found b separating the differential equation and integrating including a single constant of integration, i.e. g( = For example, to solve f (x + c. = x + x ,

2 it is necessar to rewrite it as = (x + 2( +. Separating variables and writing in integral form gives + = x + 2, and integrating ields ln + = x x + c. Letting c = ln(k and solving for gives = k e x2 /2+2x. As a second example, consider Separating variables gives and integrating ields Letting c = 2 ln(k and solving for x gives Imposing the initial condition gives which gives on solving for k which gives the solution dt = x(2 x, x(0 = x 0. x(2 x = dt, (ln x ln 2 x = t + c. 2 x = x 0 = 2ke2t + ke 2t. 2k + k, k = x 0 2 x 0, x 0 = 2, x(t = 2x 0 e 2t 2 x 0 + x 0 e 2t. In the case where x 0 = 2, then the solution is x(t = 2 for all t. 2

3 .2 Linear Equations Equations of this tpe are in the form To solve this, we introduce the integrating factor + p(x = q(x. (3 µ = e p(x. (4 This is created so that when both sides of (3 are multiplied b µ, the left side (3 is a derivative of a product, that is, it becomes ( µ + p = d (µ, and then (3 can be integrated. For example, if then p(x = 2 x, so that µ On multipling (5 b µ gives 2 x = 2x3, (5 = e 2 x = e 2 ln x = x 2. x 2 2 x 3 = 2x x 2, which simplifies to Integrating gives and solving for gives ( d x 2 = 2x x 2. x 2 = x2 + x + c, = x 4 + x + cx 2..3 Bernoulli Equations of the form + p(x = q(xn (n = 0, (6 3

4 are called Bernoulli equations. Dividing both sides of (6 b n gives Let v =, then dv n into (7 gives n + p(x = q(x (7 n = ( n n or dv ( n = n. Upon making this substitution dv + p(xv = q(x n which is linear. So Bernoulli equations can be reduced to linear equations. Example Consider 2x = 3. This is an example of a Bernoulli equation where n = 3. Putting this into standard form gives Letting v = 2, then dv = x 2 = (8, and (8 is transformed to 2 dv 2x v =, or dv + v x = 2. As this is linear, then p(x = x, and the integrating factor for this is x, so that and thus or So that or x dv + v = d (xv = 2x, xv = x 2 + c, v = x + c x. 2 = x + c x, = ± x + c x. 4

5 .4 Ricatti Equations Ricatti equations have the form: = a(x2 + b(x + c(x. (9 To find a general solution to this requires having one solution first. Given this solution, it is possible to change equation (9 to a linear equation. If we let = 0 + u, where 0 is a solution to (9, then (9 is transformed to the linear equation u = (2a(x 0 + b(x u a(x, which is linear. To illustrate, we consider the following example = 2 x (0 x Since 0 = x is a solution to this equation, let = x + u, and (0 becomes u u 2 = (x 2 x xu + u 2 + ( x + +. x u Simplifing gives u u 2 = xu x 2 u 2, and multipling b u 2 and rearranging gives rise to the linear equation u x u = x 2. ( Here p(x = x so this has the integrating factor µ = e x and upon integration gives or Since = x + u, this gives as = x + u x u x 2 = d ( u = x x 3, u x = 2x 2 + c, u = cx 2x. cx 2x. = x, so ( becomes 5

6 .5 Homogeneous Equations Equations of the form ( = F x are called homogeneous equations. Substituting = xu will ield the equation (2 which separates to Consider the previous example, x du + u = F (u. du F(u u = x. If we let = xu, then (3 becomes or = 2 x (3 x d(xu = u 2 + u +, x du + u = u2 + u + and simplifing and separating gives Integrating gives or Since = xu this gives or solving for leads to the solution du u 2 = x. 2 ln u + u = ln x + ln c, 2 u + u = cx2. x + x = cx2, + x x = cx2, = cx3 + x cx 2. 6

7 .6 Linear Fractional Equations that have the form are called linear fractional. Under a change of variables, = ax + b + e cx + + f, (4 x = x + α, = ȳ + β, we can change equation (4 to one that is either homogeneous (if ad bc = 0 or to one that is separable (if ad bc = 0. The following examples illustrate. Consider If we let = 2x x (5 x = x + α, = ȳ + β, then (5 becomes Choosing dȳ d x = 2 x 3ȳ + 2α 3β x 2ȳ + 3α 2β α 3β + 8 = 0, 3α 2β + 7 = 0, (6 leads to dȳ d x = 2 x 3ȳ 3 x 2ȳ, (7 a homogeneous equation. The natural question is, does (6 have a solution? In this case, it does and we can find that the solution is α = and β = 2. The solution of (7 is x 2 3 xȳ + ȳ 2 = c (8 and from our change of variables (x = x, = ȳ + 2 we find the solution to (5 is As a second example, consider (x + 2 3(x + ( 2 + ( 2 2 = c. (9 = 4x x + 7. (20 If we let x = x + α, = ȳ + β, 7

8 then We choose dȳ d x = 4 x 2ȳ + 4α 2β x ȳ + 2α β α 2β + 8 = 0, 2α β + 7 = 0, but this has no solution! However, if we let u = 2x, then (20 becomes which is separable! Its solution is given b du = 6 u + 7, u 2 + 4u = 2x + c, and upon back substitution, we obtain the solution of (20 as (2x 2 + 4(2x = 2x + c,.7 Exact Equations An ordinar differential equation of the form = F(x,, (2 has the alternate form M(x, + N(x, = 0. (22 If M and N have continuous partial derivatives of first order in some region R and M = N x, then the ODE (22 is said to be exact and can be integrated b setting φ x = M, and φ = N. For example, consider the differential equation = 2x x 2 + 2, (23 which can be written as 2x + (x = 0. (24 8

9 If we identif that M and N are so so (24 is exact. Setting and integrating the first gives M = 2x and N = x 2 + 2, φ x M = 2x = N x, = 2x, and φ = x2 + 2, φ = x 2 + g(, taking the partial of this with respect to gives Comparing this to φ = x2 + 2 gives that φ = x2 + g (. g ( = 2, so so that From Cal III we know that but in this case this is g( = c, φ = x c. dφ = φ x + φ, dφ = 2x + (x = 0 so φ is a constant. Thus we have as solutions to 2x + (x = 0 φ = k or x c = k, and absorbing the constant k into c gives as the set of possible solutions to (23. x c = 0 9

10 .7. Legendre Transformations Sometimes it is necessar to solve more general equations of the form F(x,, = 0, (25 sa, for example 2 x + 3 = 0. (26 One possibilit is to introduce a contact transformation that enables one to solve a given equation. Contact transformations, in general, are of the form with the contact condition that x = F(X, Y, Y, = G(X, Y, Y, = H(X, Y, Y, (27 G X + G Y + G Y Y F X + F Y + F Y Y = H. One such contact transformation is called a Legendre transformation and is given b One can verif that x = dy dx, = Substitution of (28 in (26 gives a linear ODE! Solving gives ( d dx Substituting (30 back into (28 gives Solving the first of (3 for X 2 gives d dx = X dy dx Y, = X. (28 X dy dx Y = X d2y dx 2 ( dy dx = X. d 2 Y dx 2 2X dy dx 3Y + X2 = 0 (29 Y = CX 3 2 X 2. (30 x = 3 2 cx 2 2X, = 2 cx 3 2 X 2. (3 X 3c ± 9c x 2 =, (32 8 and from the second of (3 gives ( = c 3c ± 3 ( 9c x 3c ± 4 9c x, the exact solution of (26. 0

11 2 Linear Sstems A linear sstem of equations dt = ax + b, dt = cx +, (33 can be can be written as a matrix ODE where x = ( x ( a b and Ā = c d d x = A x (34 dt. If we consider solutions of the form then after substitution into (34 we obtain x = ce λt, λ c e λt = A c e λt from which we deduce (λi A c = 0. (35 In order to have nontrivial solutions c, we require that λi A = 0. (36 This is the eigenvalue-eigenvector problem. If ( a b A = c d then (36 becomes λ 2 (a + dλ + a d b c = 0, from which we have three cases: (i two distinct eigenvalues (ii two repeated eigenvalues, (iii two complex eigenvalues. Here we consider an example of the first, two distinct eigenvalues. If

12 ( d x dt = 2 0 x (37 then the characteristic equation is λ 2 λ = λ2 λ 2 = (λ + (λ 2 = 0, from which we obtain the eigenvalues λ = and λ = 2. Case : λ = From (35 we have ( 2 2 ( c c 2 = ( 0 0 from which we obtain upon expanding 2c + c 2 = 0 and we deduce the eigenvector Case 2: λ = 2 From (35 we have ( 2 2 c = ( 2 ( c c 2. = ( 0 0 from which we obtain upon expanding c c 2 = 0 and we deduce the eigenvector c = (.,, The general solution to (37 is then given b x = c ( 2 e t + c 2 ( e 2t. Alternate Form Sometimes a sstem of ODEs can be written as dt P(t, x, = Q(t, x, = This is similar to the alternate form for a single ODE = F(x, or M(x, + N(x, = 0. R(t, x,. (38 2

13 One could write (38 in terms of the usual sstem dt = Q P, dt = R P, and determine whether its linear or nonlinear and proceed as above but sometimes its not possible nor desirable. Consider the following x = 2 = du 3u. Here, it is easier to pick them in pairs, sa for example x = 2, x = du 3u. Each are easil solved giving rise to x 2 = c, u x 3 = c 2. Example 2 Consider u x = 2x = du u x. (39 Here we need to be somewhat choos in how we pick our pairs as not all pairs will work (i.e. a pair with onl two variables. The choice here is the first and third u x = du u x, as this simplifies to = du, which integrate to u = x + c. With this we substitute into the original sstem and obtain = c 2x =. c noting that we now in fact have onl a single pair = c 2x. Upon integration, we obtain x 2 = c + c 2, 3

14 and using c obtained previousl, we get x 2 = (u x + c 2. Example 3 Consider x = x + = Again, choose wisel. Here we choose the first pair x = dz x + + z. (40 x +, or = x + x which we find as its solution = x ln x + c x. Eliminating in the first and third pairing in (40 gives or x = which is linear in z. Integrating gives and eliminating c 2 gives Example 4 Consider dz x + x ln x + c x + z. dz = x + x ln x + c x + z. x z x = ln ln x + (c + ln x + c 2, z x = ln (ln x + ( x ln x + ln x + c 2. + z = = dz x. (4 Here it is impossible to choose a pair that onl involves 2 variables so we need to be ver clever. Consider the first and second as a pair and the second and third terms as a pair and re-write as = + z, dz = x (42 4

15 now here s the clever part, add and subtract the two ODEs in (42 d(x + z = x + z, d(x z = 2 x + z. (43 If we let u = x + z and v = x z, then (43 becomes du = u, dv = 2 v, from which we find the solution or, in term of the original variables u = c, v = 2 + c 2, x + z = c, (x z 2 = c 2. Example 5 Consider x = = du = dp 2p = dq 2q. (44 Here, there a 5 independent variables x,, u, p, and q. Again, we pick in pairs. First we pick onl the first two in (44 and obtain the solution = c x. Using this in (44 gives x =, (45 x = c c x = du = dp 2p = dq 2q, (46 noting the first two terms in (46 are identical (after cancellation and thus we reall onl have Now we pick another pair first and second in (47 so x = du = dp 2p = dq 2q, (47 x = du, so u ln x = c 2. 5

16 The first and third in (47 integrates to and the first and forth in (47 integrates to Thus, the solution to the sstem (44 is p x 2 = c 3, q x 2 = c 4. x = c, u ln x = c 2, p x 2 = c 3, q x 2 = c 4. (48 6

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