AMS 212B Perturbation Metho Lecture 18 Copyright by Hongyun Wang, UCSC Method of steepest descent Consider the integral I = f exp( λ g )d, g C =u + iv, λ We consider the situation where ƒ() and g() = u() + i v() are analytic. Outlines of Method of steepest descent 1. Find the roots of g They are saddle points of u(x, y) in the complex plane, not necessarily on C. 2. Move the integral path so that the global maximum of u(x, y) along the path is attained at a saddle point, 0 = () At the saddle point, we have dv Furthermore, we change the direction of the path at the saddle point to achieve d 2 v 2 3. Approximate the integral near the saddle point. + I ~ f ( 0 )exp λ g( 0 )+ 1 s s 2 2 0 2 Now we describe the method of steepest descent in details. Step 1: Recall that g() = u() + i v(). Suppose 0 = x0 + i y0 is a root of g. ==> u x 0 and u y 0-1 -
AMS 212B Perturbation Metho ==> (x0, y0) must be a saddle point of u(x, y) Question: Why is (x0, y0) not a maximum or minimum of u(x, y)? Answer: g() is analytic. ==> u(x, y) is harmonic. ==> A harmonic function has no internal maximum or minimum. ==> (x0, y0) must be a saddle point of u(x, y). Step 2: At saddle point, 0 = (), we have dg = g d!" # $# d =0 Recall g() = u() + i v(). Thus, we achieved dv. Next, we need to achieve d2 v 2, which is equivalent to imag d2 g 2. (It turns out that / 2 is analytically much more convenient to work with). We use the chain rule to calculate the derivatives. dg = g ( ) d = d ( 2 g ( )) d + g ( d )!"# d2 2 0 d = g ( 0 ) We write the two factors in the polar form 2 =0-2 -
AMS 212B Perturbation Metho = g ( 0 ) e iθ 0 g 0 d = e iθ 1 (Recall that s is the arc length) ==> = g 2 0 e i θ 0 +2θ 1 where θ0 = the phase angle of g (0), which is determined by the problem; θ1 = angle of the path at saddle point 0, which we can tune. We change the direction of the path at saddle point 0 to make θ 0 +2θ 1 = π ==> = g 2 0 Thus, we achieved d2 v = imag d2 g 2 2. Meaning of θ1 satisfying θ0 + 2θ1 = π Consider function u((s)) along the path near saddle point 0 = () Taylor expansion gives us u u ( + Δs) ( )+ 1 2 real + 1 2 g 0 = u ( ) 2 Δs 2 cos( θ 0 +2θ 1 )( Δs) 2 When cos(θ0 + 2θ1) < 0, function u(x, y) decreases as it leaves the saddle point along the path. The path that produces the steepest descent of u(x, y) is the one with θ0 + 2θ1 = π. Thus, θ1 satisfying θ0 + 2θ1 = π is the direction of steepest descent of u(x, y) at the saddle point, which gives the name of the method. Step 3: Approximate the integral near the saddle point. - 3 -
AMS 212B Perturbation Metho Near saddle point 0, along direction θ1 satisfying θ0 + 2θ1 = π, we have = g 2 0, ( ) = e iθ 1 + I ~ f ( 0 )exp λ g( 0 )+ 1 s s 2 2 0 change of variable: t = s 2 = f ( 0 )exp( λ g( 0 ))e iθ + 1 exp λ 2 g 0 = f ( 0 )exp( λ g( 0 ))e iθ Γ 1/2 1 λ g 0 Therefore, we obtain I ~ f ( 0 )exp( λ g( 0 ))e iθ 2π 1 g 0 /2 1λ t 2 dt Remarks: A. The goal of moving the integral path is to make sure the global maximum of u(x, y) along the path occurs at a saddle point. B. A saddle point is the minimum of a ridge. To determine whether or not the modified path nee to go through a particular saddle point, we look at whether or not the original path goes over the ridge, that is, whether or not the original path is trapped by the saddle point. C. To determine whether or not the original path is trapped by a saddle point 0, we look at the direction of the ridge and the location of the saddle point. Note that the direction of the ridge is perpendicular to the direction of steepest descent. The direction of steepest descent θ1 satisfies θ0 + 2θ1 = π. The direction of the ridge is θridge = (θ1 + π/2) = (π θ0/2) where θ0 = arg(g (0)). D. When there are more than one saddle points, we need to go through one or more saddle points. Of the saddle points we go through, only those saddle points with the largest value of u(x, y) will contribute to the leading term. - 4 -
AMS 212B Perturbation Metho Summary: Step 0: Step 1: Step 2: Steps in method of steepest descent Check the behavior of g() and ƒ() along path C at infinity ( ) a 1 real g f exp +a 2 Find roots of g () (they are saddle points of u(x, y)) Move the integral path so that the global maximum of u(x, y) along the path is attained at a saddle point. The modified integral path should go through the saddle point along the direction of steepest descent. This step will achieve dv, d 2 v 2 Step 3: Approximate the integral near the saddle point. Example (a simple example for warm-up) I = exp( λ 2 )d where the path C is C C : s s = se i π 6 +1 We identify function g() = 2. Step 0: g check the behavior of g() along path C at infinity = se i π 6 +1 2 = s 2 e iπ 3 +2se iπ 6 +1 = real g u 1 = s 2 2 + s 3 +1 1 4 s2 for large s - 5 -
AMS 212B Perturbation Metho 1 4 for large along path C Step 1: find roots of g () g () = 2 g (0) ==> 0 Step 2: find the direction of steepest descent at 0. g = 2= 2 iθ, θ 0 = π ==> θ1 =(π/2 θ0/2) ==> the integral path should go through 0 along the direction θ1. (Draw the new path to illustrate) Step 3: approximate the integral along the path near 0 Near 0, along direction θ1, we have g(0), g (0) = 2 g( ( s) ) g 0 ( ) = e iθ 1 = 1 1 2 g 0 ( s ) 2 = ( s ) 2 + I = exp( λ 2 )d ~ exp λ( s ) 2 = π λ C The simple example above is just for warm-up. The next example is about the asymptotics of Airy function Ai( u) = 1 π y3 cosu y + 3 dy, u ± 0 which is a much more serious example. - 6 -