LECTURE 5: THE METHOD OF STATIONARY PHASE

LECTURE 5: THE METHOD OF STATIONARY PHASE Some notions.. A crash course on Fourier transform For j =,, n, j = x j. D j = i j. For any multi-index α = (α,, α n ) N n. α = α + + α n. α! = α! α n!. x α = x α x αn n. x n. n. α = α x αn D α = D α D αn For any ϕ C ( ), Gradient vector ϕ = ( ϕ,, n ϕ). Hessian matrix d 2 ϕ = [ i j ϕ] n n. Schwartz functions. Definition.. A function ϕ C ( ) is called a Schwartz function (or a rapidly decreasing function) if () sup x α β ϕ < for all multi-indices α and β. We will denote the set of all Schwartz functions by S ( ), or by S for simplicity. Obviously S is an infinitely dimensional vector space. It is easy to see C 0 ( ) S ( ) L p ( ) for any p. For any λ C with Re(λ) > 0, ϕ(x) = e λ x 2 S ( ). If ϕ is Schwartz, so are x α D β ϕ, D α x β 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, () (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 x α β ϕ.

2 LECTURE 5: THE METHOD OF STATIONARY PHASE We say that ϕ j ϕ in S if ϕ j 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 (3) d(ϕ, ψ) = k=0 2 k ϕ ψ k + ϕ ψ k, where for each k 0, ϕ k = max sup x α D β ϕ α + β k x is a norm on S. Then we have a metric topology on S which coincides with the one described above. Exercise. Under the topology described above, C 0 ( ) is dense in S. Fourier transform on S. Let ϕ S be a Schwartz function. By definition its Fourier transform Fϕ is (4) Fϕ(ξ) = ˆϕ(ξ) = e ix ξ ϕ(x)dx. For example, one can prove (5) F(e x 2 2 ) = (2π) n 2 e ξ 2 2. By defintion we have the obvious inequality Fϕ L ϕ L. It is easy to check that for any ϕ S and any multi-index α, (6) F(x α ϕ) = ( ) α D α ξ (F(ϕ)) and (7) F(D α x ϕ) = ξ α F(ϕ). As a consequence, we see ϕ S = Fϕ S. From definition it is also easy to check (8) Fϕ(ξ)ψ(ξ)dξ = ϕ(x)fψ(x)dx. Lemma.2. Suppose ϕ S. () For any λ R +, if we let ϕ λ = ϕ(λx), then ϕ λ S and (9) Fϕ λ (ξ) = λ n Fϕ( ξ λ ).

LECTURE 5: THE METHOD OF STATIONARY PHASE 3 (2) For any a, if we let T a ϕ(x) = ϕ(x + a), then T a ϕ S and (0) (FT a ϕ)(ξ) = e ia ξ Fϕ(ξ). Now we can prove Theorem.3. F : S S is an isomorphism with inverse () F ψ(x) = e ix ξ ψ(ξ)dξ. (2π) n Proof. We have Fϕ(ξ)ψ(λξ) = λ n ϕ(x)fψ( x λ )(x)dx = ϕ(λx)fψ(x)dx. Letting λ 0, we get ψ(0) Fϕ(ξ)dξ = ϕ(0) Fψ(x)dx. In particular, if we take ψ(ξ) = e ξ 2 2, then we get (2) ϕ(0) = Fϕ(ξ)dξ. (2π) n Finally we replace ϕ by T a ϕ in the above formula, then ϕ(a) = T a ϕ(0) = e ia ξ Fϕ(ξ)dξ. (2π) n As a consequence, we get ϕ 2 L 2 = (2π) n ˆϕ 2 L 2. The Fourier transform of the Gaussian e 2 xt Qx. Theorem.4. Let Q be any real, symmetric, positive definite n n matrix, then (3) F(e 2 xt Qx )(ξ) = (2π)n/2 (det Q) e /2 2 ξt Q ξ. Proof. Let s first assume n =, so that Q = q is a positive number. By definition, F(e 2 xt qx )(ξ) = e 2 qx2 ixξ dx. R

4 LECTURE 5: THE METHOD OF STATIONARY PHASE Taking ξ derivative, we get d dξ F(e 2 xt qx )(ξ) = ix e 2 qx2 ixξ dx = i d R q R dx (e 2 qx2 )e ixξ dx = i e d 2 qx2 q dx (e ixξ )dx = ξ q F(e 2 xt qx )(ξ). It follows where R F(e 2 xt qx )(ξ) = Ce ξ2 2q, C = F(e 2 xt qx )(0) = R e 2 qx2 dx = 2π q. Next suppose Q is a diagonal matrix. Then the left hand side of (3) 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 O T QO = D = diag(λ,, λ n ), where λ i s are the eigenvalues of Q, then if we set y = O x, F(e 2 xt Qx )(ξ) = 2 xt Qx ix ξ dx = 2 yt Dy iy O T ξ dy e = (2π)n/2 (det D) e /2 e 2 ξt OD O T ξ = (2π)n/2 (det Q) e 2 ξt Q ξ. /2 Now suppose Q is an n n complex matrix such that Q = Q T and Re(Q) is positive definite. We observe that the integral on the left hand side of (3) is an analytic function of the entries Q ij = Q ji of Q in the region ReQ > 0. The same is true for the right hand side of (3), as long as we choose det /2 Q to be the branch such that det /2 Q > 0 for real positive definite Q. So we end up with Theorem.5. Let Q be any n n complex matrix such that Q = Q T and Re(Q) is positive definite, then (4) F(e 2 xt Qx )(ξ) = (2π)n/2 (det Q) e /2 2 ξt Q ξ. Another way to describe this choice of det /2 : Since Q = Q T, for any complex vector w C n we have Re( w T Qw) = w T (ReQ)w. It follows that all eigenvalues λ,, λ n of Q has positive real part. Then where λ /2 j (det Q) /2 = λ /2 λ /2 n, is the square root of λ j that has positive real part.

LECTURE 5: THE METHOD OF STATIONARY PHASE 5 The Fourier transform on S. Definition.6. Any linear continuous map u : S C is called a tempered distribution. The space of tempered distributions is denoted by S. So S is the dual of S. The topology of the space S is defined to be th weak-* topology, so that u j u in S if For example, u j (ϕ) u(ϕ), ϕ S. For any a, the Dirac distribution δ a : S C defined by δ a (ϕ) = ϕ(a) is a tempered distribution. More generally, for any a C n and any multi-index α, the map u : S C defined by u(ϕ) = D α ϕ(a) is a tempered distribution. For any bounded continuous function ψ on defines a tempered distribution by u(ϕ) = ϕ(x)ψ(x)dx. Note that if ψ S, then the distribution defined by the above formula is non-zero unless ψ = 0. It follows that S S. Definition.7. Let u S. We define () x α u S via (x α u)(ϕ) = u(x α ϕ). (2) D α u S via (D α u)(ϕ) = ( ) α u(d α ϕ). (3) Fu S via (Fu)(ϕ) = u(fϕ). Now let Q be any real, symmetric, non-singular n n matrix. Then the function e i 2 xt 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 (5) sgn(q) = N + (Q) N (Q), where N ± (Q) = the number of positive/negative eigenvalues of Q. Theorem.8. For any real, symmetric, non-singular n n matrix Q, (6) F(e i 2 xt Qx ) = (2π)n/2 e i π 4 sgn(q) e i det Q 2 ξt Q ξ. 2

6 LECTURE 5: THE METHOD OF STATIONARY PHASE Proof. Let s first assume n =, so that Q = q 0 is a non-zero real number. For any ε > 0 we let q ε = q + iε. Applying theorem.5 we get F(e i 2 qεx2 ) = F(e 2 (ε iq)x2 ) = (2π)/2 (ε iq) e 2 (ε iq)ξ2, /2 the square root is described after theorem.5. Note that when ε 0+, we have { qe (ε iq) /2 iπ 4 q > 0 iπ qe 4 q < 0 = q e iπ 4 sgn(q), 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 (7) I = e i ϕ(x) a(x)dx, where ϕ C ( ) is called the amplitude, and a Cc ( ) 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 = e ic/ A, where A = a(x)dx is a constant independent of. Suppose n = and ϕ(x) = x. Then by definition I = F(a)( ). Since a is compactly supported, F(a) is Schwartz. It follows I = O( 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. 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.. Let f = f(). N.

LECTURE 5: THE METHOD OF STATIONARY PHASE 7 () We say f k=0 a k k if for each nonnegative integer N, (8) f N a k k = O( N+ ), 0. k=0 (2) We say f = O( ) if f 0, i.e. (9) f = O( N ), 0 for all N. So in the second extremal case we discussed above, we can write I = O( ). Remark. We don t require the series k=0 a k k 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 (20) I = O( ). 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 is a smooth function. Define an operator L by χ(x) ϕ(x) 2 Then it is easy to check It follows Lf(x) = L(e i ϕ(x) )(x) = χ I = i L(e i χ(x) ϕ(x) 2 j ϕ(x) j f(x). ϕ 2 j j j ϕ i jϕ ϕ(x) ei = i ϕ(x) χei. ϕ(x) )a(x)dx = e i ϕ(x) (L a)(x)dx. i Repeating this N times, we get ( ) N I = e i ϕ(x) (L ) N a(x)dx = O( N ). i

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 (2) ϕ(p) = 0. A critical point p of ϕ is called non-degenerate if the matrix d 2 ϕ(p) is non-degenerate: [ ] (22) det d 2 2 ϕ ϕ(p) = det (p) 0. x i x j A smooth function all of whose critical points are non-degenerate is called a Morse function. Theorem 2.3 (Morse lemma, form ). Let ϕ C ( ). 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 (23) f ϕ(x) = ϕ(p) + 2 (x2 + + x 2 r x 2 r+ x 2 n), where r is the number of positive eigenvalues of the Hessian matrix d 2 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 ϕ 0 and ϕ be smooth functions on such that ϕ i (0) = 0, ϕ 0 (0) = ϕ (0) = 0. d 2 ϕ 0 (0) = d 2 ϕ (0) is non-degenerate. Then there exist neighborhoods U 0 and U of 0 and a diffeomorphism f : U 0 U such that f(0) = 0 and f ϕ = ϕ 0. Proof. Let ϕ t = ( t)ϕ 0 + tϕ. Applying Moser s trick, it is enough to find a time-dependent vector field X t such that X t (0) = 0 and (24) L Xt ϕ t = ϕ 0 ϕ. Let X t = A j (x, t) x j, where A j (0, t) = 0. According to the second and the third conditions, [ 2 ϕ t x i x j (0)] is non-degenerate. It follows that the system of functions { ϕ t x i i =, 2,, n} form a system of coordinates near 0 with ϕt x i (0) = 0. According to the next lemma, one can find functions A ij (x, t) with A ij (0, t) = 0 so that A j (x, t) = A ij (x, t) ϕ t x i.

LECTURE 5: THE METHOD OF STATIONARY PHASE 9 Inserting this into the equation (24), we get (25) Aij (x, t) ϕ t ϕ t = ϕ 0 ϕ. x i x j On the other hand, the three conditions implies that ϕ 0 ϕ vanishes to 2nd order at 0, so according the next lemma, one can find functions B ijk (x, t) so that ϕ 0 ϕ = B ijk (x, t) ϕ t x i ϕ t x j ϕ t x k. In other words, the equation (25) is solvable with A ij (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 ϕ α so that (26) ϕ(x) = x α ϕ α (x). α =k Proof. For k =, one just take ϕ ϕ i (x) = (tx)dt. 0 x i For larger k, apply the above formula and induction. By exactly the same method, one can prove the following Morse lemma with parameters: Theorem 2.6 (Morse lemma with parameter). Let ϕ s i C ( ) be two family of smooth functions, depending smoothly on the parameter s R k. Suppose ϕ s i (0) = 0, ϕ s 0(0) = ϕ s (0) = 0, d 2 ϕ s 0(0) = d 2 ϕ s (0) are non-degenerate. Then there exist an ε > 0, a neighborhood U of 0 and for all s < ε an open embedding f s : U, depending smoothly on s, so that f s (0) = 0 and f s ϕ s = ϕ s 0. The stationary phase formula for quadratic phase. Associated to any polynomial p(ξ) = p α ξ α we have a constant coefficient differential operator p(d) = p α D α. According to the Fourier transform formula that for any a S, we immediately get F(D α a) = ξ α Fa

0 LECTURE 5: THE METHOD OF STATIONARY PHASE Lemma 2.7. For any a S, (27) F(p(D)a)(ξ) = p(ξ)fa(ξ). In particular, associated to any n n symmetric matrix A = (a ij ) one has a quadratic function p A (ξ) = ξ T Aξ = a ij ξ i ξ j and thus a second order constant coefficient differential operator (28) p A (D) := a kl D k D l. It follows that F(p A (D)a)(ξ) = (ξ T Aξ)Fa(ξ). More generally, for any nonnegative integer k, (29) F(p A (D) k a)(ξ) = (ξ T Aξ) k Fa(ξ). Suppose the phase function is a quadratic function, i.e. ϕ(x) = 2 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 ( ), one has (30) e i 2 xt Qx a(x)dx (2π) n/2 ei π 4 sgn(q) k ( i ) k R det Q n /2 k! 2 p Q (D) a(0). Proof. We start with the formula Fϕ(ξ)ψ(ξ)dξ = Recall from theorem.5 that if ϕ(x) = e i 2 xt Qx, then k ϕ(x)fψ(x)dx. Fϕ(ξ) = (2π)n/2 e i π 4 sgn(q) e i det(q) /2 2 ξt Q ξ. It follows e i 2 xt Qx a(x)dx = n/2 e i π 4 sgn(q) e i R (2π) n n/2 det Q /2 2 ξt Q ξ Fa(ξ)dξ. Using the Taylor s expansion formula for the exponential function, we see that for any non-negative integer N, the difference is bounded by e i 2 ξt Q ξ Fa(ξ)dξ N k=0 k! ( i 2 )k (ξ T Q ξ) k Fa(ξ)dξ N+ 2 N+ (N + )! (ξ T Q ξ) k Fa(ξ) dξ.

LECTURE 5: THE METHOD OF STATIONARY PHASE Note that so by the Fourier inversion formula, The conclusion follows. (ξ T Q ξ) k Fa(ξ) = F(p Q (D) k a)(ξ), p Q (D) k a(0) = (2π) n (ξ T Q 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 I = e i ϕ(x) a(x)dx. In what follows we will assume that ϕ C ( ) 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 {U i i N + } of the support of a so that each U i, i N, contains exactly one critical point p i of ϕ, and U N+ 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 I p i = e i ϕ(x) U i χi (x)a(x)dx, where χ i is a function compactly supported in U i and equals identically near p i. According to the Morse lemma, for each U i (one can always shrink U i in the above integral if necessary) one can find a diffeomorphism f : V i U i so that f(0) = p i and f ϕ(x) = ϕ(p i ) + 2 (x2 + + x 2 r i x 2 r i + x 2 n) =: ψ pi (x), where r i is the number of positive eigenvalues of the matrix d 2 ϕ(p i ). It follows from the change of variable formula that = a(f(x))χ i (f(x)) det df(x) dx. I p i e i ψp i (x) V i Moreover, since ψ pi (x) has a unique non-degenerate critical point at 0, modulo O( ) one has (3) I p i = a(f(x)) det df(x) dx + O( ). e i ψp i (x) 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

2 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 (32) df T (0)d 2 ϕ(p)df(0) = d 2 ψ(0). Proof. We start with the equation ϕ f = ψ. Taking derivatives of both sides one gets ϕ (f(x)) f k = ψ. y k x j x j Taking the second order derivatives at 0 of both sides, and using the conditions f(0) = p and ϕ(p) = 0, one gets in other words It follows that in our case, 2 ϕ y k y l (p) f l x i (0) f k x j (0) = 2 ψ x i x j (0), df T (0)d 2 ϕ(p)df(0) = d 2 ψ(0). (33) det df(0) = det d 2 ϕ(p i ) /2. In conclusion, we have Theorem 2.0. As 0, the oscillatory integral has the asymptotic expansion (34) I I p i, where dϕ(p i )=0 (35) I p i (2π)n/2 e i ϕ(p i) e iπ 4 sgn(d2 ϕ(p i )) a(p i ) j L det d 2 ϕ(p i ) /2 j (a)(p i ), where L j = L j (x, D) is a differential operator of order 2j (which depends on the phase function ϕ) with L 0 =. In particular, the leading term is (36) I = (2π) n/2 e i ϕ(p i ) e iπ 4 sgn(d2 ϕ(p i )) a(p i ) det d 2 ϕ(p i ) + O( n /2 2 + ). dϕ(p i )=0 Remark. There are more complicated version of the stationary phase method where the phase function is allowed to have degenerate critical points. j 0