Asymptotic Analysis and Singular Perturbation Theory

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(a) if a(x) > 0, then the boundary layer is at x = 0; (b) if a(x) < 0, then the boundary layer is at x = 1; (c) if a(x) changes sign and a ′ (x) > 0, then a boundary layer cannot occur at either endpoint (in this case a corner layer typically occurs in the interior); (d) if a(x) changes sign and a ′ (x) < 0, then a boundary layer can occur at both endpoints. The first two cases are treated by a straightforward modification of the expansion for constant coefficients. The other two cases are more difficult, and we illustrate them with some examples. Example 4.1 Consider the BVP εy ′′ + xy ′ − y = 0 − 1 < x < 1, y(−1) = 1, y(1) = 2. This ODE can be solved exactly. One solution is y(x) = x. A second linearly independent solution can be found by reduction of order, which gives y(x) = e −x2 /(2ε) + x ε e −x2 /(2ε) dx. We will use the MMAE to construct an asymptotic solution without relying on an exact solution. The inner solution grows exponentially into the interior at either end, so we cannot construct a boundary layer solution. We use instead left and right outer solutions where y(x, ε) = y0(x) + εy1(x) + O(ε 2 ), xy ′ 0 − y0 = 0. As we will see, matching implies that the left and right outer solutions are valid in the intervals (−1, 0) and (0, 1), respectively. Imposing the boundary conditions at the left and right, we therefore get + −x as ε → 0 with −1 ≤ x < 0 fixed, y(x, ε) ∼ 2x as ε → 0 + with 0 < x < 1 fixed. These outer solutions meet at x = 0, where the coefficient of y ′ in the ODE vanishes. The outer solution has a ‘corner’ at that point. We seek an inner solution inside a corner layer about x = 0. To find the appropriate scalings, we introduce the inner variables X = x , Y = ηY, δ 60
and use a dominant balance argument. The rescaled ODE is ε δ 2 Y ′′ + XY ′ − Y = 0. The dominant balance for the inner solution occurs when δ = ε1/2 , and all three terms are of the same order of magnitude. Matching the inner solution with the left and right outer solutions, we find that −δX as X → −∞, ηY (X, ε) ∼ 2δX as X → ∞. We therefore choose η = δ. The leading order inner solution is then given by y(x, ε) ∼ ε 1/2 x Y0 ε1/2 , where Y0(X) satisfies Y ′′ 0 + XY ′ 0 − Y0 = 0, Y0(X) ∼ −X as X → −∞, 2X as X → ∞. In this case, the ODE does not simplify at all; however, we obtain a canonical boundary value problem on R for matching the two outer solutions. The solution of this inner problem is Y0(X) = −X + 3 √ 2π e −X2 X /2 + X e −∞ −t2 /2 dt and this completes the construction of the leading order asymptotic solution. (Other problems may lead to ODEs that require the use of special functions.) Example 4.2 Consider the BVP εy ′′ − xy ′ + y = 0 − 1 < x < 1, y(−1) = 1, y(1) = 2. The coefficients of y and y ′ have the opposite sign to the previous example, and we can find an inner, boundary layer solution at both x = 0 and x = 1. The leading order outer solution y(x, ε) ∼ y0(x) satisfies with solution where C is a constant of integration. −xy ′ 0 + y0 = 0, y0(x) = Cx, 61 ,
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Asymptotic Analysis and Singular Pe
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4.1.1 Outer solution . . . . . . .
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Once we have constructed such an as
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For example, setting ε = 0 in (1.2
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Solving the first equation, we get
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and A ε : R n → R n is a linear
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Page 77 and 78: Chapter 5 Method of Multiple Scales
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Page 83 and 84: 5.3 The method of averaging Conside
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Page 87 and 88: 5.5 The WKB method for ODEs Suppose
Page 89: Thus, the amplitude of the oscillat
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