Fall 2004/05 Solutions to Assignment 5: The Stationary Phase Method Provided by Mustafa Sabri Kilic. I(x) = e ixt e it5 /5 dt (1) Z J(λ) =

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1 8.35 Fall 24/5 Solution to Aignment 5: The Stationay Phae Method Povided by Mutafa Sabi Kilic. Find the leading tem fo each of the integal below fo λ >>. (a) R eiλt3 dt (b) R e iλt2 dt (c) R eiλ co t dt 2. Find the leading tem fo each of the integal below λ >>. I(x) = Conide both cae in which x> o x<. Solution:. (a) I(λ) = R eiλt3 dt Can wite I(λ) = Z + e ixt e it5 /5 dt () Z We obeve that the econd integal i the complex conjugate of the fit, hence I(λ) =2Re Z Moeove, we ee that the point t =i the only tationay phae point, which give the main contibution to the integal, theefoe we can make the appoximation whee I(λ) 2Re J(λ) = =2ReJ(λ) Note that we do not need to calculate the end point contibution, which would come out to be of mallle ode than the contibution of the tationay phae point-we ae only aked the leading tem. The integand of J i ocillatoy. We would like to change the contou into one on which i exponentially deceaing, fo then we we would be able to expe it a a gamma function. Thu we put z 3 = it 3 whee t ieal.thimeanthepathovewhichtheintegalitakenwouldbe z = i /3 t

2 Thee ae thee oot fo i /3 : e i/6,e 5i/6,e 3i/2. An analyi of thoe path bing that we can only change the domain of integal, which i oiginally the eal axi, into the taightline making an angle /6, which i the only cae whee we can cloe ou contou with a zeo contibution ac. Thu J = e i/6 e λt3 dt Makingonemoevaiablechange we aive at J = ei/6 3λ /3 Theefoe, the anwe i given by. (b) R e iλx2 dx τ = λt 3 e τ τ 2/3 dτ = ei/6 Γ(/3) /3 3λ I(λ) Γ(/3) /3 3λ Thee i no tationay phae point in the domain of the integal, hence thi i an integal in the fom (8.36) of the book, i.e of the fom I(λ) = Z b a e iλu(x) h(x)dx (2) fo which the end point contibution ae impotant. Hence the leading fom can be given by (8.39) in the book, which i e iλu(b) h(b) iλu (b) eiλu(a) h(a) iλu (a) In ou cae, we have only one end point, hence the leading tem i eiλu() h() iλu () = eiλ 2iλ. (c) R eiλ co x dx The tationay phae point ae olution to in x =, which ae x =,. Since tationay phae point contibute moe than the end point, we do not conide end point fo the pupoe of calculating the leading tem. In thi example, the end point ae tationay phae point, o they will aleady be taken cae of when calculating the tationay phae point, o in ome ene, we can ay that thee i no end point contibution. We ue the fomula (8.45) and (8.46) fom the book which ae e i/4 2 λu (x ) eiλu(x ) h(x ) and e i/4 2 λ u (x ) eiλu(x ) h(x ) (3) 2

3 depending on the ign of u (x ). Since the tationay phae point ae end point, they contibute only half of the quantity they would if they wee inteio point. Thu we finally find 2 2 e i/4 λ eiλ ei/4 λ e iλ = λ co(λ 4 ) a the leading tem. 2. The integal () i in the fom (2), with u(t) =t, h(t) =e it5 /5. The integal ha neithe point of tationay phae no finite end point. We fit teat the imple cae x<. We tat with caling the vaiable of integation by the tanfomation t =( x) /4 z, and the integal () become I(x) =( x) /4 e iλf(z) dz with f(z) = z + 5 z5 and Λ =( x) 5/4. Thi lat integand ha two tationay point z =,, which ae found by olving f (z) = +z 4 =. Theefoe the integal can be appoximated by making ue of (3), which give e i/4 2Λ e4iλ/5 + e i/4 2 e 4iΛ/5 theefoe 2 I(x) = ( x) /4 Λ co(4 5 Λ 4 ) = ( x) /4 2 ( x) co(4 5/4 5 ( x)5/4 4 )= 2( x) 3/8 co( 4 5 ( x)5/4 4 ) Fo the cae x>, we cale the oiginal integal with t = x /4 z, and the integal () become I(x) =x /4 e iλf(z) dz (4) with f(z) =z + 5 z5 and Λ = x 5/4. Thi integal till doe not have any tationay phae point, hence we look at the citical point of f by olving f (z) =+z 4 = which give z = e i/4,e 3i/4,e 5i/4,e 7i/4. Putting z = e iθ in f(z) =z + 5 z5 = (co θ + i in θ)+ 5 5 (co 5θ + i in 5θ). So the integand of (4) blow up fo lage in the egion whee in 5θ > and become exponentially mall in the egion whee in 5θ >. Thee ae hown in the figue below along with the citical point of f. The egion in which the integand become exponentially lage ae haded. We obeve that we can defom the domain of ou integal a hown in the figue, to upwad, o that the path of the integal now contain two of the citical point of f. We cannot defom the path of ou integal 3

4 Figue 4

5 downwad, a the black egion, ove which the integand become exponentially lagepeventufomdoingo. We calculate the elevant value which mean we have the expanion f(e i/4 ) = 4 5 ei/4,f (e i/4 )=4e 3i/4 f(e 3i/4 ) = 4 5 e3i/4,f (e 3i/4 )=4e i/4 iλf(z) iλ 4 5 ei/4 2Λe i/4 (z e i/4 ) 2 iλf(z) iλ 4 5 e3i/4 2Λe i/4 (z e 3i/4 ) 2 So the contibution fom thoe citical point ae calculated to be x /4 2Λe exp[iλ4 i/4 5 ei/4 ]+x /4 2Λe exp[iλ4 i/4 5 e3i/4 ] which i 2x 3/8 2 exp( x5/4 )co( x5/4 8 ) 5