LECTURE5—09/30/2020THEMETHODOFSTATIONARYPHASE1.THE METHOD OF STATIONARY PHASE: SIMPLE MODELS Asymptotic series.When talking about the asymptotic behavioras h→o',we will use the notations“big O" and “"small o" as in mathematical analysis, namely for functions f, g whichare depending on h,. f = O(g) means “3 constant C > 0 such that Ifl ≤ Cgl for all small t"..f=o(g)means“ash→0,the quotient f/g→0"We will use the following conceptions/notations from asymptotic analysis:Definition 1.1. Let f=f(h)(1) We say f ~=oa,h if for each non-negative integer N,f-a=O(hN+1),h→0.(2) We say f = O(h) if f ~ 0, ie. f = O(hN) for all N.Remark. When we write f ~ E=o axhk, we don't require the series E=o axhk toconverge! Moreover, even if the series converges at a point, it need not converge tothe value of f at that point. For example, for any smooth function f = f(h), itsTaylor series is an asymptotic series,((0)f(h) ~k!However, the series converges for all small h to the value f(h) only if f is analyticat 0.EmpleHereisamoreconcree exmple Conside the function (h) -/,h<0.which is widely used in building cut-off functions.Then f is smooth everywherebut it is not analytic at 0. Moreover, by induction one can easily prove fk(O) = 0for all k. It follows that f ~ O, i.e. f = O(h).IIn this course, when we say as h -→ 0", we always means "as h -→ O+".1
LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE 1. The method of stationary phase: simple models ¶ Asymptotic series. When talking about the asymptotic behavior as ~ → 0 1 , we will use the notations “big O” and “small o” as in mathematical analysis, namely for functions f, g which are depending on ~, • f = O(g) means “∃ constant C > 0 such that |f| ≤ C|g| for all small ~”. • f = o(g) means “as ~ → 0, the quotient f/g → 0”. We will use the following conceptions/notations from asymptotic analysis: Definition 1.1. Let f = f(~). (1) We say f ∼ P∞ k=0 ak~ k if for each non-negative integer N, f − ❳ N k=0 ak~ k = O(~ N+1), ~ → 0. (2) We say f = O(~ ∞) if f ∼ 0, i.e. f = O(~ N ) for all N. Remark. When we write f ∼ P∞ k=0 ak~ k , we don’t require the series P∞ k=0 ak~ k to converge! Moreover, even if the series converges at a point, it need not converge to the value of f at that point. For example, for any smooth function f = f(~), its Taylor series is an asymptotic series, f(~) ∼ ❳∞ k=0 f (k) (0) k! ~ k . However, the series converges for all small ~ to the value f(~) only if f is analytic at 0. Example. Here is a more concrete example: Consider the function f(~) = ➝ e −1/~ , ~ > 0 0, ~ ≤ 0. which is widely used in building cut-off functions. Then f is smooth everywhere, but it is not analytic at 0. Moreover, by induction one can easily prove f k (0) = 0 for all k. It follows that f ∼ 0, i.e. f = O(~ ∞). 1 In this course, when we say “as ~ → 0”, we always means “as ~ → 0+”. 1
2LECTURE5-09/30/2020 THEMETHODOF STATIONARYPHASEWe can perform standard operations on asymptotic series. For example, iff(h)~a,hl,g(h)~b,hjthen wewill havef(h)±g(h) ~E(a,±b,)nand f(h)g(h) ~c,f),where c, = Ei-o abj-1. Similarly one can calculate the quotient of two asymptoticseries: If bo 0, thenf(h)/g(h)~d;hi,where d,'s are defined iteratively via do = ao/bo and d, = b-'(aj - Ei-o dibj-1).↑ Oscillatory integrals.Very often in semiclassical analysis we will need to evaluate the asymptoticbehavior of the oscillatory integrals of theform( eia(r)dr,(1)Ih=where Co(R", R) is called the phase, and a E C(R", C) is called the amplitude.The method of stationary phase is the correct tool for this purpose.?To illustrate, let's start with two extremal cases:: Suppose p(r) = c is a constant. ThenIn=eic/hAwhich is fast oscillating as h → O, where A = Jrna(r)dr is a constantindependent of h..Supposen =1 and (r)=.Thenby definition In=F(a)(-)Since ais compactly supported, F(a) is Schwartz. It follows I=O(hN) for any Ni.e.In = O(h).Intuitively,in the second case theexponential exp(i)oscillates fast in anyintervalof for small h, so that many cancellations takeplace and thus weget a functionrapidlydecreasing inh.Thisis infactthecaseatanypointrwhichisnotacriticalpoint of g. On the other hand, near a critical point of r the phase function doesn'tchange much, i.e.“looks like"a constant, so that we are in case l.According tothese intuitive observation, we expect that the main contributions to In should arisefrom the critical points of the phase function p.2Here we assume the phase function p is real-valued. In the case the phase function is complex,there is a similar way to evaluate the asymptotic of the integral: the method of steepest descent
2 LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE We can perform standard operations on asymptotic series. For example, if f(~) ∼ ❳aj~ j , g(~) ∼ ❳bj~ j , then we will have f(~) ± g(~) ∼ ❳(aj ± bj )~ j and f(~)g(~) ∼ ❳cj~ j , where cj = Pj l=0 albj−l . Similarly one can calculate the quotient of two asymptotic series: If b0 6= 0, then f(~)/g(~) ∼ ❳dj~ j , where dj ’s are defined iteratively via d0 = a0/b0 and dj = b −1 0 (aj − Pj−1 l=0 dlbj−l). ¶ Oscillatory integrals. Very often in semiclassical analysis we will need to evaluate the asymptotic behavior of the oscillatory integrals of the form (1) I~ = ❩ Rn e i ϕ(x) ~ a(x)dx, where ϕ ∈ C ∞(R n , R) is called the phase, and a ∈ C ∞ c (R n , C) is called the amplitude. The method of stationary phase is the correct tool for this purpose.2 To illustrate, let’s start with two extremal cases: • Suppose ϕ(x) = c is a constant. Then I~ = e ic/~A which is fast oscillating as ~ → 0, where A = ❘ Rn a(x)dx is a constant independent of ~. • Suppose n = 1 and ϕ(x) = x. Then by definition I~ = F(a)(− 1 ~ ). Since a is compactly supported, F(a) is Schwartz. It follows I~ = O(~ N ) for any N, i.e. I~ = O(~ ∞). Intuitively, in the second case the exponential exp(i x ~ ) oscillates fast in any interval of x for small ~, so that many cancellations take place and thus we get a function rapidly decreasing in ~. This is in fact the case at any point x which is not a critical point of ϕ. On the other hand, near a critical point of x the phase function ϕ doesn’t change much, i.e. “looks like” a constant, so that we are in case 1. According to these intuitive observation, we expect that the main contributions to I~ should arise from the critical points of the phase function ϕ. 2Here we assume the phase function ϕ is real-valued. In the case the phase function is complex, there is a similar way to evaluate the asymptotic of the integral: the method of steepest descent
LECTURE5—09/30/2020THEMETHODOFSTATIONARYPHASE3 Non-stationary phase.Proposition 1.2. Suppose the phase function p has no critical point in a neighbor.hood of the support of a. Then(2)In = O(h~).Proof. Let x be a smooth cut-off function such that: x is identically one on the support of a.has nocritical point onthesupportof xThn)()() dmohmtioThe trick (histandard in semiclassical microlocal analysis)is to introduce an operator L given byLf(a)= Z)(0,(a)/V0()27Then it is easy to checkiveireX00L(ei)() =ite/V012九2It follows [, L(ei)a(r)da= ei(L*a)(r)drIn=Rwhere L* is the adjoint of L, explicitly given byrro-- a(Mp/a)Repeating this N times, we getL e(L*)Na(r)da =O(hN).In=口Thestationaryphaseformulafor quadraticphase.Nowwe consider phasefunctions o with very simple critical points,in which caseone can imagine that the main contribution to the oscillatory integral comes fromthe critical point. We start with a model, i.e. (r) = rTQr for some non-singularreal symmetric n × n matrix Q.Theorem 1.3 (Stationary phase for non-singular quadratic phase). Let Q be a real,symmetric, non-singular n × n matrir. For any a E Co(Rn), one has, eQra(r)d ~ (2h)/2 eg(0)Idet Q((-Po-(D) a(0),(3)where for a matrir A= (ai), we denote pA(D)=auD,D
LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE 3 ¶ Non-stationary phase. Proposition 1.2. Suppose the phase function ϕ has no critical point in a neighborhood of the support of a. Then (2) I~ = O(~ ∞). Proof. Let χ be a smooth cut-off function such that • χ is identically one on the support of a • ϕ has no critical point on the support of χ. Then a(x) = χ(x)a(x), and χ(x) |∇ϕ(x)| 2 is a smooth function. The trick (which is standard in semiclassical microlocal analysis) is to introduce an operator L given by Lf(x) = ❳ j χ(x)∂jϕ(x) |∇ϕ(x)| 2 ∂jf(x). Then it is easy to check L(e i ϕ(x) ~ )(x) = χ |∇ϕ| 2 ❳ j ∂jϕ i∂jϕ ~ e i ϕ(x) ~ = i ~ χei ϕ(x) ~ . It follows I~ = ~ i ❩ Rn L(e i ϕ(x) ~ )a(x)dx = ~ i ❩ Rn e i ϕ(x) ~ (L ∗ a)(x)dx, where L ∗ is the adjoint of L, explicitly given by L ∗ f(x) = − ❳ j ∂j ❶ χ(x)∂jϕ |∇ϕ(x)| 2 f(x) ➀ . Repeating this N times, we get I~ = ❶ ~ i ➀N ❩ Rn e i ϕ(x) ~ (L ∗ ) N a(x)dx = O(~ N ). ¶ The stationary phase formula for quadratic phase. Now we consider phase functions ϕ with very simple critical points, in which case one can imagine that the main contribution to the oscillatory integral comes from the critical point. We start with a model, i.e. ϕ(x) = 1 2 x TQx for some non-singular real symmetric n × n matrix Q. Theorem 1.3 (Stationary phase for non-singular quadratic phase). Let Q be a real, symmetric, non-singular n × n matrix. For any a ∈ C ∞ c (R n ), one has (3) ❩ Rn e i 2~ x T Qxa(x)dx ∼ (2π~) n/2 e i π 4 sgn(Q) | det Q| 1/2 ❳ k ~ k 1 k! ✒ − i 2 pQ−1 (D) ✓k a(0), where for a matrix A = (aij ), we denote pA(D) = P aklDkDl .
4LECTURE5-09/30/2020 THEMETHODOFSTATIONARYPHASEProof. Recall Theorem 2.3 in Lecture 4: If (r) = er Qr, then(2元)n/2eisgn(Q)Fp(s) = Idet(0)/2-e-0-sNow we apply the multiplication formula/F()(s)d= /(r)Fb(r)dato getFa(S)es(-h-1)d= (2m)n/2eign(--)eaTQra(r)drIdet(-hQ-1)1/2i.e.hn/2eiisgn(Q)erTQa(r)dre-Q-'Fa($)ds(2)n/2 /|det Q|1/2Using the Taylor's expansion formula for the exponential function, we see that forany non-negative integer N, the differenceZ(e-TQ-"Fa(E)de-)()(TQ-)Fa(E)dk!is bounded by (a multiple of)1hN+1I($TQ-1e)*Fa($)dE2N+1(N + 1)! JSotheconclusionfollowsfromLemma 1.4. For any a E and any polynomial pF(p(D)a)() = p($)Fa($)Proof. This is just a consequence of the fact F(Da) = saFa and口the linearity of F.which implies(sTQ-1s)*Fa($) = F(pQ-1(D)*a)($),togetherwith the Fourier inversion formula, which impliesPo-(D)a(0)=2) /.(Q-1),Fa(2)d,口This completes the proof.Remark.By calculating more carefully, one can prove that for any N, the remaindern e+Qra(r)dr -(2h)/2 e(@)E（-含(-2Po-(D)a(0)RN :=detQ|1/2is controlled byan explicitupper bound10°al.R≤Csup[a|≤2N+n+
4 LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE Proof. Recall Theorem 2.3 in Lecture 4: If ϕ(x) = e i 2 x T Qx, then Fϕ(ξ) = (2π) n/2 e i π 4 sgn(Q) | det(Q)| 1/2 e − i 2 ξ T Q−1 ξ . Now we apply the multiplication formula ❩ Fϕ(ξ)ψ(ξ)dξ = ❩ ϕ(x)Fψ(x)dx to get ❩ Rn Fa(ξ)e i 2 ξ T (−~Q−1 )ξ dξ = (2π) n/2 e i π 4 sgn(−~Q−1 ) | det(−~Q−1 )| 1/2 ❩ Rn e i 2~ x T Qxa(x)dx, i.e. ❩ Rn e i 2~ x T Qxa(x)dx = ~ n/2 (2π) n/2 e i π 4 sgn(Q) | det Q| 1/2 ❩ Rn e − i~ 2 ξ T Q−1 ξFa(ξ)dξ. Using the Taylor’s expansion formula for the exponential function, we see that for any non-negative integer N, the difference ❩ Rn e − i~ 2 ξ T Q−1 ξFa(ξ)dξ − ❳ N k=0 1 k! (− i~ 2 ) k ❩ Rn (ξ TQ −1 ξ) kFa(ξ)dξ is bounded by (a multiple of) ~ N+1 1 2 N+1(N + 1)! ❩ Rn |(ξ TQ −1 ξ) kFa(ξ)|dξ. So the conclusion follows from Lemma 1.4. For any a ∈ S and any polynomial p, F(p(D)a)(ξ) = p(ξ)Fa(ξ). Proof. This is just a consequence of the fact F(Dαa) = ξ αFa and the linearity of F. which implies (ξ TQ −1 ξ) kFa(ξ) = F(pQ−1 (D) k a)(ξ), together with the Fourier inversion formula, which implies pQ−1 (D) k a(0) = 1 (2π) n ❩ Rn (ξ TQ −1 ξ) kFa(ξ)dξ. This completes the proof. Remark. By calculating more carefully, one can prove that for any N, the remainder RN := ❩ Rn e i 2~ x T Qxa(x)dx − (2π~) n/2 e i π 4 sgn(Q) | det Q| 1/2 N ❳−1 k=0 ~ k 1 k! ✒ − i 2 pQ−1 (D) ✓k a(0) is controlled by an explicit upper bound RN ≤ CN sup |α|≤2N+n+1 |∂ α a|.
LECTURE5—09/30/2020THEMETHODOFSTATIONARYPHASE52.THEMETHODOFSTATIONARYPHASE:GENERALCASE Morse lemma.We first introduce some standard conceptions in global analysissDefinition 2.1. Let be a smooth function.(1) A point p is called a critical point of ifVp(p) = 0.(2) A critical point p of is called non-degenerate if the Hessian matrix d(p)isnon-degenerate,i.e.a0±0det d'(p) = detOron(3) A smooth function is called a Morse function if all of its critical points arenon-degenerate.3To study the stationary phase expansion for more general phase functions, onefirst convert the general (non-degenerate)phase function to a quadratic one by usingthe Morse lemma:Theorem 2.2 (Morse Lemma, version 1).Let E C(Rn).Suppose p is a non-degenerate critical point of p. Then there erists a neighborhood U of O, a neighbor-hood V of p and a diffeomorphism p:U-→V so that p(O) =p and(4)ps(a) = (p) + (a + + -+1-..-),where r is the number of positive eigenvalues of the Hessian matrir d'p(p). The stationary phase formula for general phase.As we have seen, only the critical points of the phase function give an essentialcontribution to the oscillatory integraleira(r)daIh:In whatfollows we will assume that ECo(Rn)admits only non-degenerate criticalpoints in a neighborhood of the support of a. Since non-degenerate critical pointsmust be discrete, has only finitely many critical points in the support of a. Thusone can find apartition of unity[U;/1≤ i≤ N+1] of the support of a sothateach Ui, 1≤i≤N, contains exactly one critical point p of ,and Un+i contains3Note that non-degenerate critical points must be discrete. In more subtle examples where thecritical points are not necessary discrete, but still nice enough, say, form smooth manifolds, onehas an analogous conception, namely the Morse-Bott function. Many results for Morse functionscan begeneralized to Morse-Bott functions
LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE 5 2. The method of stationary phase: general case ¶ Morse lemma. We first introduce some standard conceptions in global analysis: Definition 2.1. Let ϕ be a smooth function. (1) A point p is called a critical point of ϕ if ∇ϕ(p) = 0. (2) A critical point p of ϕ is called non-degenerate if the Hessian matrix d 2ϕ(p) is non-degenerate, i.e. det d 2ϕ(p) = det ➊ ∂ 2ϕ ∂xi∂xj (p) ➍ 6= 0. (3) A smooth function is called a Morse function if all of its critical points are non-degenerate.3 To study the stationary phase expansion for more general phase functions, one first convert the general (non-degenerate) phase function to a quadratic one by using the Morse lemma: Theorem 2.2 (Morse Lemma, version 1). Let ϕ ∈ C ∞(R n ). Suppose p is a nondegenerate critical point of ϕ. Then there exists a neighborhood U of 0, a neighborhood V of p and a diffeomorphism ρ : U → V so that ρ(0) = p and (4) ρ ∗ϕ(x) = ϕ(p) + 1 2 (x 2 1 + · · · + x 2 r − x 2 r+1 − · · · − x 2 n ), where r is the number of positive eigenvalues of the Hessian matrix d 2ϕ(p). ¶ The stationary phase formula for general phase. As we have seen, only the critical points of the phase function ϕ give an essential contribution to the oscillatory integral I~ = ❩ Rn e i ϕ(x) ~ a(x)dx. In what follows we will assume that ϕ ∈ C ∞(R n ) admits only non-degenerate critical points in a neighborhood of the support of a. Since non-degenerate critical points must be discrete, ϕ has only finitely many critical points in the support of a. Thus one can find a partition of unity {Ui | 1 ≤ i ≤ N + 1} of the support of a so that each Ui , 1 ≤ i ≤ N, contains exactly one critical point pi of ϕ, and UN+1 contains 3Note that non-degenerate critical points must be discrete. In more subtle examples where the critical points are not necessary discrete, but still nice enough, say, form smooth manifolds, one has an analogous conception, namely the Morse-Bott function. Many results for Morse functions can be generalized to Morse-Bott functions
6LECTURE5-09/30/2020 THEMETHODOFSTATIONARYPHASEno critical points of .A partition of unity arguments converts the asymptoticbehaviorof Intothe sum of theasymptotic behavior ofR = /n exi(a)a(r)dr,where X;is a function compactly supported in U, and equals 1 identically near pi.According to the Morse lemma, for each U; (one can always shrink U, in the aboveintegral if necessary) one can find a diffeomorphism p :V→ U, so that p(o) =piandp*p(r) =p(pi) +++(+.+-+1-) =: p (),where r; is the number of positive eigenvalues of the matrix d(pi). It follows fromthe change of variable formula thatet"n,a(p(r)xi(p(r)| det dp(r)dr.:IPi=Moreover, since p:(r) has a unique non-degenerate critical point at O, moduloO(h)onehasiapi+a(p(a)| det dp(z)]dr + O(h).(5)IPi=The asymptotic of this integral is basically given by Theorem 1.3.In particular, to get the leading term in the asymptotic expansion above, oneonly need to evaluate |det dp(O)l, which follows fromLemma 2.3. Let , be smooth functions defined on V and U respectively, andsuppose O is a non-degenerate critical point of b, and p a non-degenerate criticalpoint of p. If p: V → U a diffeomorphism so that p(O) = p and p*p = b. Then(6)dpT (O)d(p)dp(0) = d-(O),Proof. We start with the equation o p = b. Taking derivatives of both sides onegetsa0pk-(p(r)OykariariTaking the second order derivatives at O of both sides, and using the conditionsp(0) = p and Vp(p) = 0, one gets020() (0 P(0) (0),ayroytorioriororjin other wordsdpT (0)d-p(p)dp(0) = d-(0),口
6 LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE no critical points of ϕ. A partition of unity arguments converts the asymptotic behavior of I~ to the sum of the asymptotic behavior of I pi ~ = ❩ Ui e i ϕ(x) ~ χi(x)a(x)dx, where χi is a function compactly supported in Ui and equals 1 identically near pi . According to the Morse lemma, for each Ui (one can always shrink Ui in the above integral if necessary) one can find a diffeomorphism ρ : Vi → Ui so that ρ(0) = pi and ρ ∗ϕ(x) = ϕ(pi) + 1 2 (x 2 1 + · · · + x 2 ri − x 2 ri+1 − · · · − x 2 n ) =: ψpi (x), where ri is the number of positive eigenvalues of the matrix d 2ϕ(pi). It follows from the change of variable formula that I pi ~ = ❩ Vi e i ψpi (x) ~ a(ρ(x))χi(ρ(x))| det dρ(x)|dx. Moreover, since ψpi (x) has a unique non-degenerate critical point at 0, modulo O(~ ∞) one has (5) I pi ~ = ❩ Rn e i ψpi (x) ~ a(ρ(x))| det dρ(x)|dx + O(~ ∞). The asymptotic of this integral is basically given by Theorem 1.3. In particular, to get the leading term in the asymptotic expansion above, one only need to evaluate | det dρ(0)|, which follows from Lemma 2.3. Let ψ, ϕ be smooth functions defined on V and U respectively, and suppose 0 is a non-degenerate critical point of ψ, and p a non-degenerate critical point of ϕ. If ρ : V → U a diffeomorphism so that ρ(0) = p and ρ ∗ϕ = ψ. Then (6) dρT (0)d 2ϕ(p)dρ(0) = d 2ψ(0). Proof. We start with the equation ϕ ◦ ρ = ψ. Taking derivatives of both sides one gets ∂ϕ ∂yk (ρ(x))∂ρk ∂xj = ∂ψ ∂xj . Taking the second order derivatives at 0 of both sides, and using the conditions ρ(0) = p and ∇ϕ(p) = 0, one gets ∂ 2ϕ ∂yk∂yl (p) ∂ρl ∂xi (0)∂ρk ∂xj (0) = ∂ 2ψ ∂xi∂xj (0), in other words dρT (0)d 2ϕ(p)dρ(0) = d 2ψ(0).
LECTURE5—09/30/2020THEMETHOD OFSTATIONARYPHASE1It follows that in our case,Idet dp(0)] = |det dp(pi)/-1/2(7)As a result, we concludeTheorem 2.4. As h 0, the oscillatory integral has the asymptotic erpansion(8)I～Z,dp(p:)=0wherea(pi)IPt~ (2h)n/2ei(psgn(d(p)(9)[det doo)WL,(a)(e),j>0where L = L,(c, D) is a differential operator (of order 2j which depends on thephase function ) with Lo =1. In particular, the leading term isa(pi)In =(2h)n/2 eiesgn(de(p:)I det d-o(p:)1/2 + (h+1).(10)dp(p:)=0Remark.We haveJc.f.L.Hormander,The Analysis of Linear Partial DifferentialOperators Vol. 1, section 7.7.]. One can write down an explicit formula for all these L,'s.. There are more complicated version of the stationary phase expansion wherethe phase function is allowed to have degenerate critical points.APPENDIX:APROOFOFMORSELEMMAWe shall prove thefollowing form of the Morse lemma which is obviously equiv-alent to the version stated above:Theorem 2.5 (Morse Lemma, version 2).Let Po and 1 be smooth functions onRn such that. Po(0) = P1(0) = 0,. VPo(0) = V1(0) = 0. d'po(O) = dpi(O) is non-degenerate.Then there erist neighborhoods Uo and Ui of O and a diffeomorphism p : Uo - Uisuch that p(O)=0 and ppi=po.Proof.There are many different proofs of Morse Lemma.Here weprovide a proof via the so-called Moser's trick, which is not verycommonly appeared in literature.Theidea is thefollowing:To con-struct a diffeomorphism p :Uo-→ Ui such that f*1 = po, we willconstruct smooth family of functions pt connecting po and i, and
LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE 7 It follows that in our case, (7) | det dρ(0)| = | det d 2ϕ(pi)| −1/2 . As a result, we conclude Theorem 2.4. As ~ → 0, the oscillatory integral has the asymptotic expansion (8) I~ ∼ ❳ dϕ(pi)=0 I pi ~ , where (9) I pi ~ ∼ (2π~) n/2 e i ϕ(pi ) ~ e iπ 4 sgn(d 2ϕ(pi)) a(pi) | det d 2ϕ(pi)| 1/2 ❳ j≥0 ~ jLj (a)(pi), where Lj = Lj (x, D) is a differential operator (of order 2j which depends on the phase function ϕ) with L0 = 1. In particular, the leading term is (10) I~ = (2π~) n/2 ❳ dϕ(pi)=0 e i ϕ(pi ) ~ e iπ 4 sgn(d 2ϕ(pi)) a(pi) | det d 2ϕ(pi)| 1/2 + O(~ n 2 +1). Remark. We have [c.f. L. H¨ormander, The Analysis of Linear Partial Differential Operators Vol. 1, section 7.7.] • One can write down an explicit formula for all these Lj ’s. • There are more complicated version of the stationary phase expansion where the phase function is allowed to have degenerate critical points. Appendix: A proof of Morse Lemma We shall prove the following form of the Morse lemma which is obviously equivalent to the version stated above: Theorem 2.5 (Morse Lemma, version 2). Let ϕ0 and ϕ1 be smooth functions on R n such that • ϕ0(0) = ϕ1(0) = 0, • ∇ϕ0(0) = ∇ϕ1(0) = 0. • d 2ϕ0(0) = d 2ϕ1(0) is non-degenerate. Then there exist neighborhoods U0 and U1 of 0 and a diffeomorphism ρ : U0 → U1 such that ρ(0) = 0 and ρ ∗ϕ1 = ϕ0. Proof. There are many different proofs of Morse Lemma. Here we provide a proof via the so-called Moser’s trick, which is not very commonly appeared in literature. The idea is the following: To construct a diffeomorphism ρ : U0 → U1 such that f ∗ϕ1 = ϕ0, we will construct smooth family of functions ϕt connecting ϕ0 and ϕ1, and
8LECTURE5-09/30/2020 THEMETHOD OF STATIONARYPHASEconstruct a time-dependent vectorfield Et so that the flow pt satisfiesthe stronger relation(11)PtPt=Po.Of courseifthis isdone,thenthetime-1fowmap p=piiswhatwearelooking for.Notethat the equation (11)is equivalent tod0=Piot=pi(o+Ct).SotoconstructsuchavectorfieldEt,itisenoughtosolvetheequation L=,Pt =-pt.Note that to guarantee the existence of time-1flow(at leastlocallynearO),wewill requireEt(O)=0.We apply Moser's trick as described above. To do so we let Pt = (1-t)po +tp1Then it is enough to find a time-dependent vector field E, such that Et(o) = 0 andL=+Pt=0o-01.Let 三t =ZA,(r,t),.Then our problem becomes: find functions A,(r,t) withA,(0,t) = 0, such thatZ A4(g,% - 0- 1(12)O (0) is non-degenerate.According to the second and the third conditions, [ar.It follows that the system of functions[0 = ,. .norform a system of coordinates near 0 with (0) = 0. So according to the nextlemma, one can find functions Bi,(r,t)so thatP0 -01 =EBg(2, 1)00:00.OrorObviously if we take A;(r,t) = E, Bi(r,t), then A,(0,t) = 0 and satisfies the口equation (12).This proves the existence of the diffeomorphism p.Lemma 2.6. Let be a smooth function with (0) = 0 and p(0) = 0 for alllal<k.Then there erists smooth functions Pa so that(13)0(r)= rcpa(r).lal=kProof.For k=1, one just take j,,,)dt.(p;(c) =Or口For larger k, apply the above formula and induction
8 LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE construct a time-dependent vector field Ξt so that the flow ρt satisfies the stronger relation (11) ρ ∗ tϕt = ϕ0. Of course if this is done, then the time-1 flow map ρ = ρ1 is what we are looking for. Note that the equation (11) is equivalent to 0 = d dtρ ∗ tϕt = ρ ∗ t ( ˙ϕt + LΞtϕt). So to construct such a vector field Ξt , it is enough to solve the equation LΞtϕt = −ϕ˙t . Note that to guarantee the existence of time-1 flow (at least locally near 0), we will require Ξt(0) = 0. We apply Moser’s trick as described above. To do so we let ϕt = (1−t)ϕ0 +tϕ1. Then it is enough to find a time-dependent vector field Ξt such that Ξt(0) = 0 and LΞtϕt = ϕ0 − ϕ1. Let Ξt = P Aj (x, t) ∂ ∂xj . Then our problem becomes: find functions Aj (x, t) with Aj (0, t) = 0, such that (12) ❳Aj (x, t) ∂ϕt ∂xj = ϕ0 − ϕ1. According to the second and the third conditions, [ ∂ 2ϕt ∂xi∂xj (0)] is non-degenerate. It follows that the system of functions ➝ ∂ϕt ∂xi | i = 1, 2, · · · , n➠ form a system of coordinates near 0 with ∂ϕt ∂xi (0) = 0. So according to the next lemma, one can find functions Bij (x, t) so that ϕ0 − ϕ1 = ❳Bij (x, t) ∂ϕt ∂xi ∂ϕt ∂xj . Obviously if we take Aj (x, t) = P i Bij (x, t) ∂ϕt ∂xi , then Aj (0, t) = 0 and satisfies the equation (12). This proves the existence of the diffeomorphism ρ. Lemma 2.6. Let ϕ be a smooth function with ϕ(0) = 0 and ∂ αϕ(0) = 0 for all |α| < k. Then there exists smooth functions ϕα so that (13) ϕ(x) = ❳ |α|=k x αϕα(x). Proof. For k = 1, one just take ϕj (x) = ❩ 1 0 ∂ϕ ∂xj (x1, · · · , xj−1, txj , 0, · · · , 0)dt. For larger k, apply the above formula and induction.
9LECTURE5—09/30/2020THEMETHODOFSTATIONARYPHASEBy exactly the same method, one can prove the following Morse Lemma withparameters:Theorem 2.7 (Morse Lemma with parameters). Let i Co(R") be two familyof smooth functions, depending smoothly on the parameter s E Rk. Suppose: g(0) = pi(0) = 0,V(0) = Vi(0) = 0,·. dpo(O) = d-pi(o) are non-degenerate.Then there erist an e > 0, a neighborhood U of 0 and for all Isl < e an openembedding ps : U - Rn, depending smoothly on s, so that ps(O) = 0 and pspi = po
LECTURE 5 — 09/30/2020 THE METHOD OF STATIONARY PHASE 9 By exactly the same method, one can prove the following Morse Lemma with parameters: Theorem 2.7 (Morse Lemma with parameters). Let ϕ s i ∈ C ∞(R n ) be two family of smooth functions, depending smoothly on the parameter s ∈ R k . Suppose • ϕ s 0 (0) = ϕ s 1 (0) = 0, • ∇ϕ s 0 (0) = ∇ϕ s 1 (0) = 0, • d 2ϕ s 0 (0) = d 2ϕ s 1 (0) are non-degenerate. Then there exist an ε > 0, a neighborhood U of 0 and for all |s| < ε an open embedding ρs : U → R n , depending smoothly on s, so that ρs(0) = 0 and ρ ∗ sϕ s 1 = ϕ s 0