Notes on Perturbation methods

Notes relevant to perturbation methods. The following is not intended to be a rigorous set of notes, rather, a summary and collection of bullet points that summarizes the ideas discussed at the course M8402, as taught by Dr. Mori, Yoichiro. 1 Multiple scale Method Consider the equation: y′′ + εy′ + y = 0 (1.1) where y(0) = 0 and y′(0) = 1. The “standard” power series expansion, namely: y(t) ∼ y0(t) + εy1(t) + · · · yields the solution: y(t) ∼ sin(t)− 1 2 εt sin(t). The problem is that this solution grows unbounded for large t. Hence, we should not expect that it agrees with the exact solution. In fact, the analytical solution is: y(t) = 1 √ 1− ε2/4 e−εt/2 sin ( t √ 1− ε2/4 ) . Without much motivation, we assume that our system depends on at least two time scales, u = t and v = εt. Then, for the function y(t) = y(u, v) we have d dt 7→ ∂ ∂u du dt + ∂ ∂v dv dt = ∂ ∂u + ε ∂ ∂v , and analogously, d dt2 7→ ∂ 2 ∂u2 + 2ε ∂ ∂u∂v + ε ∂ ∂v2 . With the above construction, we assume that y(u, v) has asymptotic expansion: y(u, v) ∼ y0(u, v) + εy1(u, v) + εy2(u, v) + · · · . (1.2) To avoid clutter, we omit the argument notation in what follows. Note that substitution of (1.2) into (1.1) yields: ( ∂ ∂u2 + 2ε ∂ ∂u∂v + ε ∂ ∂v2 ) (y0 + εy1 + · · · ) + ε ( ∂ ∂u + ε ∂ ∂v ) (y0 + εy1 + · · · ) + y0 + εy1 + · · · = 0 We balance the equations with α = 1 as to avoid terms that grow unbounded for large values of t. Now, we proceed to collect terms in terms of ε. We get the following: • O(1) : ( ∂ ∂u2 + 1 ) y0 = 0, with y0(0, 0) = 0 and ∂ ∂uy0(0, 0) = 1. The general solution of the above system is: y0 = A sin(u) +B cos(u) where A and B are both functions of v such that A(0) = 1 and B(0) = 0. • O(ε) : ( ∂ ∂u2 + 1 ) y1 = −2 ∂ 2 ∂u∂vy0 − ∂ ∂uy0, with y1(u, v) = 0, and ∂ ∂uy1 = − ∂ ∂vy0, when (u, v) = (0, 0). Using the solution found for y0 above we find the solution for y1; namely: y1 =C sin(u) +D cos(u) − 1 2 (2B′ +B)u cos(u)− 1 2 (2A′ +A)u sin(u) where C and D are functions of v, C(0) = A′(0), and D(0) = 0. Note that the solution of y1 contains unbounded terms; namely, those that have a factor of u. However, here we can avoid them by letting their respective coefficients be identically 0. In other words, we let: 2B′ +B = 0 and 2A′ +A = 0. Then, using some of the initial value conditions, we have that: A = e,