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LECTURE 5: THE METHOD OF STATIONARY PHASE 1. A crash course on Fourier transform ¶ Some notions. • For j = 1, · · · , n, – ∂j = ∂x∂ j . – Dj = 1i ∂j . • For any multi-index α = (α1 , · · · , αn ) ∈ Nn . – |α| = α1 + · · · + αn . – α! = α1 ! · · · αn !. – xα = xα1 1 · · · xαnn . – ∂ α = ∂xα11 · · · ∂xαnn . – Dα = D1α1 · · · Dnαn . • For any ϕ ∈ C ∞ (Rn ), – Gradient vector ∇ϕ = (∂1 ϕ, · · · , ∂n ϕ). – Hessian matrix d2 ϕ = [∂i ∂j ϕ]n×n . ¶ Schwartz functions. Definition 1.1. A function ϕ ∈ C ∞ (Rn ) is called a Schwartz function (or a rapidly decreasing function) if sup |xα ∂ β ϕ| < ∞ (1) Rn for all multi-indices α and β. We will denote the set of all Schwartz functions by S (Rn ), or by S for simplicity. Obviously S is an infinitely dimensional vector space. It is easy to see • C0∞ (Rn ) ⊂ S (Rn ) ⊂ Lp (Rn ) for any 1 ≤ p ≤ ∞. 2 • For any λ ∈ C with Re(λ) > 0, ϕ(x) = e−λ|x| ∈ S (Rn ). • If ϕ is Schwartz, so are xα Dβ ϕ, Dα xβ ϕ, where α, β are any multi-indices. Recall that a semi-norm on a vector space V is function | · | : V → R so that for all λ ∈ C and all u, v ∈ V , (1) (Absolute homogeneity) |λv| = |λ||v|. (2) (Triangle inequality) |u + v| ≤ |u| + |v|. On S one can define, for each pair of multi-indices α, β a semi-norm (2) |ϕ|α,β = sup |xα ∂ β ϕ|. Rn 1 2 LECTURE 5: THE METHOD OF STATIONARY PHASE We say that ϕj → ϕ in S if |ϕj − ϕ|α,β → 0 as j → ∞ for all α, β. The topology induced by these semi-norms turns S into a Fréchet space. Equivalently, one can define a metric d on S via ∞ X kϕ − ψkk (3) d(ϕ, ψ) = , 2−k 1 + kϕ − ψkk k=0 where for each k ≥ 0, kϕkk = max sup |xα Dβ ϕ| |α|+|β|≤k x∈Rn is a norm on S . Then we have a metric topology on S which coincides with the one described above. Exercise 1. Under the topology described above, C0∞ (Rn ) is dense in S . ¶ Fourier transform on S . Let ϕ ∈ S be a Schwartz function. By definition its Fourier transform Fϕ is Z e−ix·ξ ϕ(x)dx. (4) Fϕ(ξ) = ϕ̂(ξ) = Rn For example, one can prove F(e− (5) |x|2 2 n ) = (2π) 2 e− |ξ|2 2 . By defintion we have the obvious inequality kFϕkL∞ ≤ kϕkL1 . It is easy to check that for any ϕ ∈ S and any multi-index α, F(xα ϕ) = (−1)|α| Dξα (F(ϕ)) (6) and F(Dxα ϕ) = ξ α F(ϕ). (7) As a consequence, we see ϕ ∈ S =⇒ Fϕ ∈ S . From definition it is also easy to check Z Z (8) Fϕ(ξ)ψ(ξ)dξ = Rn ϕ(x)Fψ(x)dx. Rn Lemma 1.2. Suppose ϕ ∈ S . (1) For any λ ∈ R+ , if we let ϕλ = ϕ(λx), then ϕλ ∈ S and ξ 1 (9) Fϕλ (ξ) = n Fϕ( ). λ λ LECTURE 5: THE METHOD OF STATIONARY PHASE 3 (2) For any a ∈ Rn , if we let Ta ϕ(x) = ϕ(x + a), then Ta ϕ ∈ S and (FTa ϕ)(ξ) = eia·ξ Fϕ(ξ). (10) Now we can prove Theorem 1.3. F : S → S is an isomorphism with inverse Z 1 −1 (11) F ψ(x) = eix·ξ ψ(ξ)dξ. n (2π) Rn Proof. We have Z Z Z x 1 Fϕ(ξ)ψ(λξ) = n ϕ(x)Fψ( )(x)dx = ϕ(λx)Fψ(x)dx. λ λ Letting λ → 0, we get Z Z Fϕ(ξ)dξ = ϕ(0) ψ(0) In particular, if we take ψ(ξ) = e− (12) Fψ(x)dx. |ξ|2 2 , then we get Z 1 ϕ(0) = Fϕ(ξ)dξ. (2π)n Finally we replace ϕ by Ta ϕ in the above formula, then Z 1 ϕ(a) = Ta ϕ(0) = eia·ξ Fϕ(ξ)dξ. (2π)n As a consequence, we get kϕk2L2 = 1 kϕ̂k2L2 . (2π)n 1 T Qx ¶ The Fourier transform of the Gaussian e− 2 x . Theorem 1.4. Let Q be any real, symmetric, positive definite n × n matrix, then (13) 1 T Qx F(e− 2 x )(ξ) = (2π)n/2 − 1 ξT Q−1 ξ e 2 . (det Q)1/2 Proof. Let’s first assume n = 1, so that Q = q is a positive number. By definition, Z 1 2 − 12 xT qx F(e )(ξ) = e− 2 qx −ixξ dx. R 4 LECTURE 5: THE METHOD OF STATIONARY PHASE Taking ξ derivative, we get Z Z d i d − 1 qx2 −ixξ − 12 qx2 −ixξ − 12 xT qx F(e )(ξ) = −ix e dx = (e 2 )e dx dξ q R dx R Z 1 T 1 ξ −i 2 d (e−ixξ )dx = − F(e− 2 x qx )(ξ). = e− 2 qx q R dx q It follows ξ2 1 T F(e− 2 x qx )(ξ) = Ce− 2q , where r Z 1 T 1 2π 2 C = F(e− 2 x qx )(0) = e− 2 qx dx = . q R Next suppose Q is a diagonal matrix. Then the left hand side of (13) is a product of n one-dimensional integrals, and the result follows from the one-dimensional case. For the general case, one only need to choose an orthogonal matrix O so that OT QO = D = diag(λ1 , · · · , λn ), where λi ’s are the eigenvalues of Q, then if we set y = O−1 x, Z Z 1 T T − 21 xT Qx − 12 xT Qx−ix·ξ F(e )(ξ) = e dx = e− 2 y Dy−iy·O ξ dy Rn Rn n/2 (2π) (2π)n/2 − 1 ξT Q−1 ξ − 12 ξ T OD−1 OT ξ = e e 2 = . (det D)1/2 (det Q)1/2 Now suppose Q is an n × n complex matrix such that Q = QT and Re(Q) is positive definite. We observe that the integral on the left hand side of (13) is an analytic function of the entries Qij = Qji of Q in the region ReQ > 0. The same is true for the right hand side of (13), as long as we choose det1/2 Q to be the branch such that det1/2 Q > 0 for real positive definite Q. So we end up with Theorem 1.5. Let Q be any n × n complex matrix such that Q = QT and Re(Q) is positive definite, then 1 T Qx F(e− 2 x (14) )(ξ) = (2π)n/2 − 1 ξT Q−1 ξ e 2 . (det Q)1/2 Another way to describe this choice of det1/2 : Since Q = QT , for any complex vector w ∈ Cn we have Re(w̄T Qw) = w̄T (ReQ)w. It follows that all eigenvalues λ1 , · · · , λn of Q has positive real part. Then 1/2 (det Q)1/2 = λ1 · · · λ1/2 n , 1/2 where λj is the square root of λj that has positive real part. LECTURE 5: THE METHOD OF STATIONARY PHASE 5 ¶ The Fourier transform on S 0 . Definition 1.6. Any linear continuous map u : S → C is called a tempered distribution. The space of tempered distributions is denoted by S 0 . So S 0 is the dual of S . The topology of the space S 0 is defined to be th weak-* topology, so that uj → u in S 0 if ∀ϕ ∈ S . uj (ϕ) → u(ϕ), For example, • For any a ∈ Rn , the Dirac distribution δa : S → C defined by δa (ϕ) = ϕ(a) is a tempered distribution. • More generally, for any a ∈ Cn and any multi-index α, the map u : S → C defined by u(ϕ) = Dα ϕ(a) is a tempered distribution. • For any bounded continuous function ψ on Rn defines a tempered distribution by Z ϕ(x)ψ(x)dx. u(ϕ) = Rn Note that if ψ ∈ S , then the distribution defined by the above formula is non-zero unless ψ = 0. It follows that S ⊂ S 0 . Definition 1.7. Let u ∈ S 0 . We define (1) xα u ∈ S 0 via (xα u)(ϕ) = u(xα ϕ). (2) Dα u ∈ S 0 via (Dα u)(ϕ) = (−1)|α| u(Dα ϕ). (3) Fu ∈ S 0 via (Fu)(ϕ) = u(Fϕ). Now let Q be any real, symmetric, non-singular n × n matrix. Then the function i T e 2 x Qx is not a Schwartz function, but we can think of it as a distribution. We would like to calculate its Fourier transform. Recall that the signature of Q is sgn(Q) = N + (Q) − N − (Q), (15) where N ± (Q) = the number of positive/negative eigenvalues of Q. Theorem 1.8. For any real, symmetric, non-singular n × n matrix Q, π (16) i F(e 2 x T Qx )= (2π)n/2 ei 4 sgn(Q) | det Q| 1 2 i T −1 Q ξ e− 2 ξ . 6 LECTURE 5: THE METHOD OF STATIONARY PHASE Proof. Let’s first assume n = 1, so that Q = q 6= 0 is a non-zero real number. For any ε > 0 we let qε = q + iε. Applying theorem 1.5 we get i 2 1 2 F(e 2 qε x ) = F(e− 2 (ε−iq)x ) = (2π)1/2 − 1 (ε−iq)ξ2 e 2 , (ε − iq)1/2 the square root is described after theorem 1.5. Note that when ε → 0+, we have √ − iπ p iπ qe 4 q > 0 1/2 = |q|e− 4 sgn(q) , (ε − iq) −→ √ iπ −qe 4 q < 0 so the conclusion follows. For general n, one just proceed via the diagonalization method as before. 2. The method of stationary phase ¶ Oscillatory integrals. The method of stationary phase is a method for estimating the small ~ behavior of the oscillatory integrals Z ϕ(x) (17) I~ = ei ~ a(x)dx, Rn where ϕ ∈ C ∞ (Rn ) is called the amplitude, and a ∈ Cc∞ (Rn ) is called the phase. For simplicity we will assume ϕ is real-valued. To illustrate, let’s start with two extremal cases: • Suppose ϕ(x) = c is a constant. Then I~ = eic/~ A, R 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 ), ∀N. 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. We will use the following notions: Definition 2.1. Let f = f (~). LECTURE 5: THE METHOD OF STATIONARY PHASE (1) We say f ∼ P∞ k=0 ak ~k if for each nonnegative integer N , f− (18) 7 N X ak ~k = O(~N +1 ), ~ → 0. k=0 (2) We say f = O(~∞ ) if f ∼ 0, i.e. f = O(~N ), (19) ~→0 for all N . So in the second extremal case we discussed above, we can write I~ = O(~∞ ). P k Remark. We don’t require the series ∞ k=0 ak ~ to be convergent! ¶ Non-stationary phase. Proposition 2.2. Suppose the phase function ϕ has no critical point in a neighborhood of the support of a. Then I~ = O(~∞ ). (20) Proof. Let χ be a smooth cut-off function which is identically one on the support of a so that ϕ has no critical point on the support of χ. Then a(x) = χ(x)a(x), and χ(x) is a smooth function. Define an operator L by |∇ϕ(x)|2 Lf (x) = χ(x) X ∂j ϕ(x)∂j f (x). |∇ϕ(x)|2 j Then it is easy to check L(ei ϕ(x) ~ )(x) = χ X i∂j ϕ i ϕ(x) i i ϕ(x) ~ ∂ ϕ e = χe ~ . j |∇ϕ|2 j ~ ~ It follows ~ I~ = i Z L(e Rn i ϕ(x) ~ ~ )a(x)dx = i Z ei ϕ(x) ~ (L∗ a)(x)dx. Rn Repeating this N times, we get N Z ϕ(x) ~ I~ = ei ~ (L∗ )N a(x)dx = O(~N ). i Rn 8 LECTURE 5: THE METHOD OF STATIONARY PHASE ¶ Morse lemma. Recall that a point p is called a critical point of a smooth function ϕ if ∇ϕ(p) = 0. (21) A critical point p of ϕ is called non-degenerate if the matrix d2 ϕ(p) is non-degenerate: 2 ∂ ϕ 2 (22) det d ϕ(p) = det (p) 6= 0. ∂xi ∂xj A smooth function all of whose critical points are non-degenerate is called a Morse function. Theorem 2.3 (Morse lemma, form 1). Let ϕ ∈ C ∞ (Rn ). 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 f : U → V so that f (0) = p and 1 (23) f ∗ ϕ(x) = ϕ(p) + (x21 + · · · + x2r − x2r+1 − · · · − x2n ), 2 where r is the number of positive eigenvalues of the Hessian matrix d2 ϕ(p). We shall prove the following form of the Morse lemma which is obviously equivalent to the above statement: Theorem 2.4 (Morse lemma, form 2). Let ϕ0 and ϕ1 be smooth functions on Rn such that • ϕi (0) = 0, • ∇ϕ0 (0) = ∇ϕ1 (0) = 0. • d2 ϕ0 (0) = d2 ϕ1 (0) is non-degenerate. Then there exist neighborhoods U0 and U1 of 0 and a diffeomorphism f : U0 → U1 such that f (0) = 0 and f ∗ ϕ1 = ϕ0 . Proof. Let ϕt = (1 − t)ϕ0 + tϕ1 . Applying Moser’s trick, it is enough to find a time-dependent vector field Xt such that Xt (0) = 0 and LXt ϕt = ϕ0 − ϕ1 . (24) Let Xt = P Aj (x, t) ∂x∂ j , where Aj (0, t) = 0. According to the second and the third 2 ϕt conditions, [ ∂x∂ i ∂x (0)] is non-degenerate. It follows that the system of functions j { ∂ϕt | i = 1, 2, · · · , n} ∂xi t form a system of coordinates near 0 with ∂ϕ (0) = 0. According to the next lemma, ∂xi one can find functions Aij (x, t) with Aij (0, t) = 0 so that X ∂ϕt Aj (x, t) = Aij (x, t) . ∂xi LECTURE 5: THE METHOD OF STATIONARY PHASE 9 Inserting this into the equation (24), we get X ∂ϕt ∂ϕt (25) Aij (x, t) = ϕ0 − ϕ1 . ∂xi ∂xj On the other hand, the three conditions implies that ϕ0 − ϕ1 vanishes to 2nd order at 0, so according the next lemma, one can find functions Bijk (x, t) so that X ∂ϕt ∂ϕt ∂ϕt . ϕ0 − ϕ1 = Bijk (x, t) ∂xi ∂xj ∂xk In other words, the equation (25) is solvable with Aij (0, t) = 0. This proves the existence of the diffeomorphism f . Lemma 2.5. Let ϕ be a smooth function with ϕ(0) = 0 and ∂ α ϕ(0) = 0 for all |α| < k. Then there exists smooth functions ϕα so that X (26) ϕ(x) = xα ϕα (x). |α|=k Proof. For k = 1, one just take Z 1 ∂ϕ (tx)dt. 0 ∂xi For larger k, apply the above formula and induction. ϕi (x) = By exactly the same method, one can prove the following Morse lemma with parameters: Theorem 2.6 (Morse lemma with parameter). Let ϕsi ∈ C ∞ (Rn ) be two family of smooth functions, depending smoothly on the parameter s ∈ Rk . Suppose • ϕsi (0) = 0, • ∇ϕs0 (0) = ∇ϕs1 (0) = 0, • d2 ϕs0 (0) = d2 ϕs1 (0) are non-degenerate. Then there exist an ε > 0, a neighborhood U of 0 and for all |s| < ε an open embedding fs : U → Rn , depending smoothly on s, so that fs (0) = 0 and fs∗ ϕs1 = ϕs0 . ¶ The stationary phase formula for quadratic phase. Associated to any polynomial X p(ξ) = pα ξ α we have a constant coefficient differential operator X p(D) = pα D α . According to the Fourier transform formula that for any a ∈ S , F(Dα a) = ξ α Fa we immediately get 10 LECTURE 5: THE METHOD OF STATIONARY PHASE Lemma 2.7. For any a ∈ S , F(p(D)a)(ξ) = p(ξ)Fa(ξ). (27) In particular, associated to any n × n symmetric matrix A = (aij ) one has a quadratic function X pA (ξ) = ξ T Aξ = aij ξi ξj and thus a second order constant coefficient differential operator X (28) pA (D) := akl Dk Dl . It follows that F(pA (D)a)(ξ) = (ξ T Aξ)Fa(ξ). More generally, for any nonnegative integer k, F(pA (D)k a)(ξ) = (ξ T Aξ)k Fa(ξ). (29) Suppose the phase function is a quadratic function, i.e. ϕ(x) = 12 xT Qx for some symmetric real n × n matrix Q. Then the only critical point of ϕ is the origin x = 0. More over, the non-degeneracy condition implies that Q is non-singular. Theorem 2.8. Let Q be a real, symmetric, non-singular n × n matrix. For any a ∈ Cc∞ (Rn ), one has k Z π i T ei 4 sgn(Q) X k 1 i x Qx n/2 (30) e 2~ a(x)dx ∼ (2π~) ~ − pQ−1 (D) a(0). | det Q|1/2 k k! 2 Rn Proof. We start with the formula Z Z Fϕ(ξ)ψ(ξ)dξ = ϕ(x)Fψ(x)dx. i Recall from theorem 1.5 that if ϕ(x) = e 2 x Fϕ(ξ) = T Qx , then i π4 sgn(Q) (2π)n/2 e | det(Q)|1/2 i T −1 Q ξ e− 2 ξ . It follows Z π ~n/2 ei 4 sgn(Q) −~ 2i ξ T Q−1 ξ e a(x)dx = e Fa(ξ)dξ. (2π)n/2 | det Q|1/2 Rn Rn Using the Taylor’s expansion formula for the exponential function, we see that for any non-negative integer N , the difference Z Z N X 1 i~ k −~ 2i ξ T Q−1 ξ e Fa(ξ)dξ − (− ) (ξ T Q−1 ξ)k Fa(ξ)dξ k! 2 Rn k=0 Z i T x Qx 2~ is bounded by N +1 ~ 1 N +1 2 (N + 1)! Z Rn |(ξ T Q−1 ξ)k Fa(ξ)|dξ. LECTURE 5: THE METHOD OF STATIONARY PHASE 11 Note that (ξ T Q−1 ξ)k Fa(ξ) = F(pQ−1 (D)k a)(ξ), so by the Fourier inversion formula, 1 pQ−1 (D) a(0) = (2π)n k The conclusion follows. Z (ξ T Q−1 ξ)k Fa(ξ)dξ. ¶ 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 Z ϕ(x) I~ = ei ~ a(x)dx. Rn In what follows we will assume that ϕ ∈ C ∞ (Rn ) 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 no critical points of ϕ. A partition of unity arguments converts the asymptotic behavior of I~ to the sum of the asymptotic behavior of Z ϕ(x) pi I~ = ei ~ χi (x)a(x)dx, Ui 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 f : Vi → Ui so that f (0) = pi and 1 f ∗ ϕ(x) = ϕ(pi ) + (x21 + · · · + x2ri − x2ri +1 − · · · − x2n ) =: ψpi (x), 2 where ri is the number of positive eigenvalues of the matrix d2 ϕ(pi ). It follows from the change of variable formula that Z ψpi (x) pi I~ = ei ~ a(f (x))χi (f (x))| det df (x)|dx. Vi Moreover, since ψpi (x) has a unique non-degenerate critical point at 0, modulo O(~∞ ) one has Z ψpi (x) pi (31) I~ = ei ~ a(f (x))| det df (x)|dx + O(~∞ ). Rn The asymptotic of this integral is basically given by theorem (2.8). In particular, to get the leading term in the asymptotic expansion above, one only need to evaluate | det df (0)|, which follows from 12 LECTURE 5: THE METHOD OF STATIONARY PHASE Lemma 2.9. Let ψ, ϕ be smooth functions defined on V and U respectively, and suppose 0 is a non-degenerate critical point of both ψ and p a non-degenerate critical point of ϕ. If f : V → U a diffeomorphism so that f (0) = p and f ∗ ϕ = ψ. Then df T (0)d2 ϕ(p)df (0) = d2 ψ(0). (32) Proof. We start with the equation ϕ ◦ f = ψ. Taking derivatives of both sides one gets ∂ϕ ∂fk ∂ψ (f (x)) = . ∂yk ∂xj ∂xj Taking the second order derivatives at 0 of both sides, and using the conditions f (0) = p and ∇ϕ(p) = 0, one gets ∂fl ∂fk ∂ 2ψ ∂ 2ϕ (p) (0) (0) = (0), ∂yk ∂yl ∂xi ∂xj ∂xi ∂xj in other words df T (0)d2 ϕ(p)df (0) = d2 ψ(0). It follows that in our case, | det df (0)| = | det d2 ϕ(pi )|−1/2 . (33) In conclusion, we have Theorem 2.10. As ~ → 0, the oscillatory integral has the asymptotic expansion X p (34) I~ ∼ I~ i , dϕ(pi )=0 where (35) I~pi ∼ (2π~)n/2 ei ϕ(pi ) ~ iπ 2 ϕ(p )) i e 4 sgn(d X a(pi ) ~j Lj (a)(pi ), 2 1/2 | det d ϕ(pi )| j≥0 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 X ϕ(pi ) iπ n a(pi ) 2 +1 2 + O(~ (36) I~ = (2π~)n/2 ei ~ e 4 sgn(d ϕ(pi )) ). | det d2 ϕ(pi )|1/2 dϕ(pi )=0 Remark. There are more complicated version of the stationary phase method where the phase function is allowed to have degenerate critical points.