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Perturbation Methods, Physics 2400 Physics 2400 Spring 2016 Perturbation methods Lecture notes by M. G. Rozman Last modified: April 27, 2016 1 Introduction Perturbation theory is a collection of methods for obtaining approximate solutions to problems involving a small parameter ". These methods are so powerful that sometimes it is actually advisable to introduce a parameter " temporarily into a difficult problem having no small parameter, and then finally to set " = 1 to recover the original problem. The thematic approach of perturbation theory is to decompose a tough problem into an (infinite) number of relatively easy ones. The perturbation theory is most useful when the first few steps reveal the important features of the solution and the remaining ones give small corrections. We classify perturbation solutions into two varieties. A basic feature of regular perturbation problems (which we use to identify such problems) is that the exact solution for small but nonzero " smoothly approaches the unperturbed or zeroth-order solution as " ! 0. On the technical side, regular perturbation series is a power series in " having a nonvanishing radius of convergence. We define a singular perturbation problem as one whose solution for " = 0 is fundamentally different in character from the ”neighboring” solutions obtained in the limit " ! 0. Singular perturbation solution either does not take the form of a power series or, if it does, the power series does not converge. 2 Regular perturbation theory Here is an elementary example to introduce the ideas of regular perturbation theory. Example 1. Roots of a quintic polynomial. Let us find approximations to the roots of the following equation 1. 1 There are three real roots and two complex conjugate ones. The numerical values of the roots are x1 = 0:0625001, x2 = −2:01533, x3 = 1:98406, x4;5 = −0:0156155 ± 2:0003i. Those can be determined using e.g. the command NSolve x5 − 16x + 1 = 0; x . f g Page 1 of9 20160427143700 Physics 2400 Perturbation methods Spring 2016 x5 − 16x + 1 = 0: (1) As it stands, this problem is not a perturbation problem because there is no small parameter ". It may not be easy to convert a particular problem into a tractable perturbation problem, but in the present case the necessary trick is to replace the constant term in Eq. (1) with a parameter: x5 − 16x + " = 0: (2) When " = 1:, the original Eq. (1) is reproduced. We consider the roots to be functions of ". We further assume a perturbation series in powers of ": 1 X n x(") = an" : (3) n=0 To obtain the first term in this series, we set " = 0 in Eq. (2) and factor it as following x5 − 16x = 0 ! x(x2 − 4)(x2 + 4) = 0 ! x(x − 2)(x + 2)(x − 2i)(x + 2i) = 0: (4) Thus, in the zeroth-order perturbation theory xi = ±2;0;±2i; i = 1;:::;5: (5) A second-order perturbation approximation to the first of these roots consists of writing 2 x1 = −2 + a1" + a1" ; (6) substituting this expression into Eq. (2), and neglecting powers of " beyond "2. 2 The result is 2 (1 + 64a1)" + (−80a1 + 64a2)" = 0: (7) It is at this step that we realize the power of generalizing the original problem to a family of problems with variable ". It is because " is variable that we can conclude that the coefficient of each power of " in Eq. (7) is separately equal to zero. This gives a sequence of equations for the expansion coefficients ai: 2 1 + 64a1 = 0; −80a1 + 64a2 = 0; (8) with the solutions 1 5 5 a = − ; a = a = : (9) 1 64 2 4 1 16384 Therefore, the perturbation expansion for the root x1 is 1 5 x = −2 − " + "2: (10) 1 64 16384 2Computer algebra systems are perfectly suited for the tasl like this. E.g. s = Series " + x5 − 16x /. x ! a1 " + a2 "2 − 2; f";0;2g ; CoefficientList[s;"] f g Page 2 of9 20160427143700 Physics 2400 Perturbation methods Spring 2016 −5 If we now set " = 1, we obtain x1 = −2:01532 accurate to better than 10 . The same procedure gives 1 5 x = 2 − " − "2 = 1:98407 (11) 2 64 16384 1 x = " = 0:06250 (12) 3 16 1 5i x = −2i − " − "2 = −:015625 − 2:00031i (13) 4 64 16384 1 5i x = 2i − " + "2 = −:015625 + 2:00031i (14) 5 64 16384 This example illustrates the three steps of perturbative analysis: 1. Convert the original problem into a perturbation problem by introducing the small parameter ". 2. Assume an expression for the answer in the form of a perturbation series and compute the coefficients of that series. 3. Recover the answer to the original problem by summing the perturbation series for the appropriate value of ". Step1 is sometimes ambiguous because there may be many ways to introduce a small parameter in the equation. However, it is preferable to introduce " in such a way that the zeroth- order solution (the leading term in the perturbation series) is obtainable as a closed-form analytic expression. Of course, step1 may be omitted when the original problem already has a small parameter if a perturbation series can be developed in powers of that parameter. Let’s apply the perturbation theory to a boundary value problem for an ordinary differential equation. Example 2. Approximate solution of an boundary-value problem Consider the nonlinear two-point boundary-value problem cos(x) π y00 + y = ; y(0) = y = 2: (15) 3 + y2 2 As a first step, we convert Eq. (15) into a perturbation problem by introducing " in the right side. Then we obtain a first-order approximation to the answer. Finally, we return to the original equation by assigning " = 1. cos(x) y00 + y = " : (16) 3 + y2 We assume a perturbation expansion for y(x) in the form 1 X n y(x) = " yn(x); (17) n=0 π π where y0(0) = y0 2 = 2, yn(0) = yn 2 = 0 for n ≥ 1. Page 3 of9 20160427143700 Physics 2400 Perturbation methods Spring 2016 The zeroth-order problem y00 + y = 0 is obtained by setting " = 0. The solution that satisfies the initial conditions is y0(x) = 2 cos(x) + 2 sin(x): (18) The first order problem is obtained by substituting Eq. (17) into Eq. (16) and setting the coefficient of " equal to 0: cos(x) cos(x) y00 + y = = : (19) 1 1 2 7 + sin(2x) 3 + y0 Eq. (19) can be solved by variation of parameters. We seek the solution of Eq. (19) in the form π y (x) = a(x) cos(x) + b(x) sin(x); a(0) = 0; b = 0; (20) 1 2 where a(x) and b(x) are subject to the constraint a0(x) cos(x) + b0(x) sin(x) = 0: (21) The solution is, ! ! x 7 1 + 7 tan(x) 7 1 a(x) = − + p arctan p − p arctan p ; (22) 2 8 3 4 3 8 3 4 3 ! 1 1 1 + 7 tan(x) 1 π b(x) = log(7 + sin(2x)) + p arctan p − log(7) − p : (23) 4 8 3 4 3 4 16 3 The graph of the first order solution y(x) = y0(x) + y1(x) (24) is presented in Fig.1 together with the numerical solution of Eq. (15). 2.8 2.6 Figure 1: The graphs of the solu- tion of the boundary value problem 2.4 Eq. (15). 2.2 0.5 1.0 1.5 Page 4 of9 20160427143700 Physics 2400 Perturbation methods Spring 2016 3 Singular perturbation theory 3.1 The Method of Dominant Balance Suppose you are given an equation in the form A + B + C + D = 0; (25) where A, B, C, D are different terms in the equation. These terms could represent elements of a polynomial equation (e.g. x5) or could represent terms in an ordinary differential equation or for that matter terms in a nonlinear partial differential equation. It is almost invariably the case that two of the terms are larger than all of the others. For example, it could be that A, D are larger than B, C. If this were the case we can then “approximate” the original equation by the neglecting B, C entirely and instead solving A+ D = 0. We can then check that this reduced equation is “consistent”, by taking the solution to the reduced equation, plugging it back into the original equation, and verifying that the neglected terms are indeed smaller than the two terms we have kept. If they are, our hypothesized solution is “consistent”; if not the solution is “inconsistent” and we must find another dominant balance. Let’s return to Eq. (1) and convert it to a perturbation problem as following. "x5 − x + 1 = 0: (26) We begin by setting " = 0 to obtain the unperturbed problem −x + 1 = 0, with the solution x = 1. Note that the unperturbed equation has only one root while the original equation has five roots. This abrupt change in the character of the solution, namely the disappearance of four roots when " = 0, implies that Eq. (26) is a singular perturbation problem. Part of the exact solution ceases to exist when " = 0. The explanation for this behavior is that the four missing roots tend to 1 as " ! 0. Thus, for those roots it is no longer valid to neglect "x5 compared with −x + 1 in the limit " ! 0. Of course, for the roots near 1 the term "x5 is indeed small as " ! 0 and we may assume a regular perturbation expansion for these roots of the form of Eq. (3), as we did in Example Example 1.. To track down the four missing roots we first estimate their orders of magnitude as " ! 0. We do this by considering all possible dominant balances between pairs of terms in Eq.
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