Introduction to Pseudodifferential Operators

Transcription

1 Introduction to Pseudodifferential Operators Mengxuan Yang Directed by Prof. Dean Baskin May, 206. Introductions and Motivations Classical Mechanics & Quantum Mechanics In classical mechanics, the status of a particle can be represented by position and momenta (x, ξ). We call the curve t (x(t), ξ(t)) phase space trajectory or classical trajectory of the particle, which lies in R 3 R 3. The energy is defined by E 2m ξ2 (t) + V (x(t)), and by Newton s Law, it is independent of t, and only denpends on (x, ξ). Note that the energy of the particle is called the energy observable at (x(t), ξ(t)). More general, one calls a classical observable any real smooth function a a(x, ξ) defined on phase space R 3 R 3 : Its value along the curve gives information about the particle at time t. In quantum mechanics, a particle is defined by a function R R 3 (t, x) ψ(t, x), which is called wave function of the particle and satisfied ψ t (x) : x ψ(t, x) L 2 (R 3 ) and ψ t L 2 (R 3 ). And ψ is typically satisfies a Schrödinger equation. We can define the average position of the particle as and also the so called average impulse x ψt : xψ t, ψ t ξ ψt : h i ψ t, ψ t. Now we can define the h-fourier transform in R 3 F h ψ t (ξ) : ˆψ t (ξ) : e ixξ/ h ψ (2π h) 3/2 t (x)dx, This work is based on Chapter, 2 of An Introduction to Semiclassical and Microlocal Analysis by Andre Martinez.

2 we have F h ( h i ψ t)(ξ) ξf h ψ t (ξ). Thus, the average impulse can be written as ξ ψt : ξ ˆψ t, ˆψ t, which is similar to the average position of the particle. Here we observe that the wave-particle duality of particle is given by the correspondence between ψ t (x) and ˆψ t (x) via h-fourier transform. The quantum parts respect to the classical ones can be obtained by the following operations: x : ψ xψ and hd x : ψ h i ψ, and this is the correspondence between a classical observable and a quantum observable. A Classical Example Consider the differential equation ( h 2 )u f, which is a simplified form of Schrödinger equation (i h t + H)ψ f, with H h 2. We can obtain the former equation by taking ψ e iwt/ h u(x) with w in Schrödingerequation. By taking the h-fourier transform of both side, we can get a simpler formation: ( + ξ 2 )û ˆf, which is the corresponding classical observable and is easy to be solved. Hence, by taking the inverse h-fourier transform, we can get the solution u F h (û) F h ( ˆf + ξ 2 ). Thus we can get the inverse operator of h 2 by the following. u(x) F h ( f(y)eiyξ/ h (2π h) n/2 + ξ 2 dy) (2π h) n e i(x y)ξ/ h + ξ 2 f(y)dydξ Thus operator A a e i(x y)ξ/ h + ξ 2 dξ, can be considered as the inverse operator of h 2. We can give the general concept of pseudodifferential operator similarly, which is the main purpose of this work. 2

3 General theory about Pseudodifferential Operator Definition.(Pseudodifferential Operator). For a S 3n ( ξ m ), and u C0 (R n ), we set Op h (a)u(x; h) e i(x y)ξ/h a(x, y, ξ)u(y)dydξ. Then Op h (a)u C (R n ), and for any ν R, the operator h ν Op h (a) is called the semiclassical pseudodifferential operator for symbol h ν a. In this case, h ν Op h (a) is said to be of degree m and of order ν. Remarks. In this definition, a S 3n ( ξ m ) represents a symbol which will be introduced in the following section. 2. Based on basic knowledge of mathematical analysis, this integral may not be convergent. We will define the equavalent form later to make the integral converge. A natural question arises: given a classical observable a(x, ξ), is there a natural way to obtain the corresponding quantum observable, which could reasonably be denoted by a(x, hd x ). This is exactly one of the main purpose of pseudodifferential calculus. Examples Kinetic energy: 2m ξ2 h2 2m Total energy: 2m ξ2 + V (x) H h2 2m + V (x) In order to make more sufficient work on psuedodifferential operator, we introduce some notions in the following sections. 2. Spaces of Symbols What we called symbols is just what we have called classical observable. Definition 2.. Let g C (R d, R + ) satisfying α x g O(g), for any α N d and uniformly on R d, such function is called an order function on R d. Remark f O(g) means that f g is bounded. 3

4 Examples x m : ( + x 2 ) m/2, m R, x (x,..., x k ) with k d Since α x ( x m ) C α,m ( + x ) m 2 α x α, we have thus x m is an order function α x ( x m ) x m C α,mx α ( + x 2 ) α C α,m, e α x with α R, or generally, e f(x), f bounded together with all its derivatives. Definition 2.2. A function a a(x; h) defined on R d (0, h 0 ] for some h 0 is said to be in S d (g) with g an order function if. a depends smoothly on x, and 2. for α N d, α x a(x; h) O(g(x)) uniformly with respect to (x, h) R d (0, h 0 ]. In particilar, S d () is the set of symbols a s.t. all its derivatives are uniformly bounded. Examples χ C 0 (R n ) S d () x m S d ( x m ) Proposition 2.3. a S d (g) a g S d() Proof of Proposition 2.3. By Leibniz s Rule, it is obvious that g is order function g is order function. Thus we have, a S d (g) α x a O(g), α N d α x ( a g ) O(), α Nd a g S d() the first and third are by definition, while for the second, using the fact that x ( a g ) a xg ag x g 2 O(g)O(g) g 2 O(g) C g and by induction, we can get, and also with the fact that if a S d (g ), b S d (g 2 ), then ab S d (g g 2 ), we get the other direction. 4

5 Definition 2.4. A symbol a S d (g) is said to be elliptic if there exists a positive constant C 0 such that a C 0 g uniformly on R (0, h 0 ] (for some h 0 > 0). Then we have the following proposition. Proposition 2.5. If a S d (g) is elliptic, then /a S d (/g). Proof. Set b /a, then by differentiating iteratively ab and using the Leibniz formula we can get the result. 3. Oscillatory Integrals First, let s recall the Fourier Transform. Let S(R n ) denote the space of Schwartz functions in R n, define the (semiclassical) Fourier transform of u as the function F h u(ξ) û(ξ) e ixξ/h u(x)dx, /2 where ξ R n and xξ is the product of x and ξ. Same as (usual) Fourier transform, F h is an isomorphism on S(R n ), and its inverse is given by F h v(x) e ixξ/h v(ξ)dξ. /2 Moreover, F h can be extended by duality to an isomorphism of S (R n ), which denotes the tempered distribution on R n, and is an isometric map from L 2 (R n ) to itself. By definition, we can get the following properties of Fourier transform:. F h (hd x u) ξf h u 2. F h (xu) hd ξ F h u By generalizing the example in Section, let m R and a a(x, y, ξ) S 3n ( ξ m ) in the sense of Definition 2.2. Consider integral I(a) e i(x y)ξ/h a(x, y, ξ)dξ. If m < n, this integral is absolutely convergent and hence well-defined. To define it when m n, we are going to interpret it as the distribution kernel of an operator based on the following result. 5

6 Theorem 3.(Schwartz Kernel Theorem). If A : C 0 (R n ) D (R n ) is linear and continuous, then there exists a unique K D (R n R n ) such that for any u, v C 0 (R n ) we has Au, v D,D K, v u D,D. K is called the distribution kernel of operator A. We will not discuss the complete proof of this theorem but instead give the main idea of the proof. Since S(R n ) S(R n ) dense in S(R n R n ), to prove K is continuous, only need to prove that φ ψ L(φ ψ) Aφ, ψ D,D is continuous in topology of S(R n R n ), which is generated by φ k,m x m φ C k (R n ) sup αx( x mφ) Thus we need to find some k, m, s.t. x R n α k L(φ ψ) C φ ψ k,m. Since this equation holds if and only if L extends to S(R n R n ) and L(Φ) C Φ k,m for Φ S(R n R n ). Then K : S(R n R n ) C is continuous, so K S (R n R n ) with L(Φ) K, Φ D,D Since A is continuous, by definition we have for some k, l, m, r Au, v D,D C ψ l,r φ k,m C φ ψ k+l,r+m. And this is exactly what we need. Remarks. D (R n ) is the space of distribution on R n. 2. v u C 0 (R 2n ) denotes the tensor product between v and u. 3., D,D represents the duality bracket between D and D C 0. For u C0 (R n ) and a S 3n ( ξ m ) with m < n, we set A a u(x, h) e i(x y)ξ/h a(x, y, ξ)u(y)dydξ, which is well defined. 6

7 Notice that We can define We have ( hξd y )(e i(x y)ξ/h ) ( + ξ 2 )e i(x y)ξ/h. L + ξ 2 ( hξd y) L(ξ, hd y ). L(e i(x y)ξ/h ) e i(x y)ξ/h. Now we can use this property to construct the case m n. Since k N, one has L k (e i(x y)ξ/h ) e i(x y)ξ/h. We obtain by making intergration by parts k times in the equation of A a u(x, h), A a u(x, h) e i(x y)ξ/h ( t L(ξ, hd y )) k (au)dydξ : I k u(x), where ( t L) k (au) ( + hξd y + ξ 2 ) k (au) O ξ m k uniformly as ξ +. Thus, I k u(x) is absolutely convergent as m k < n, i.e. m < k n, Since A a u(x, h) e i(x y)ξ/h a(x, y, ξ)u(y)dydξ L(e i(x y)ξ/h )a(x, y, ξ)u(y)dydξ + ξ 2 ( hξd y)e i(x y)ξ/h a(x, y, ξ)u(y)dydξ + ξ 2 ei(x y)ξ/h ( + hξd y )(a(x, y, ξ)u(y))dydξ e i(x y)ξ/h tl(a(x, y, ξ)u(y))dydξ. By induction, we have I k u I k+l u for all l 0, thus it is reasonable to make following definition: Definition 3.2. For all m R, a S 3n ( ξ m ), and u C0 (R n ), we define A a u(x) e i(x y)ξ/h ( t L(ξ, hd y )) k (au)dydξ, where k is any nonnegative integer greater than m + n. Then we have the following theorem. 7

8 Theorem 3.3. A a defines a continuous linear operator from C 0 (R n ) to C (R n ). Thus, by Theorem 3., we can define the distribution kernel of A a. Definition 3.4. We defined by I(a) e i(x y)ξ/h a(x, y, ξ)dξ D (R n R n ) the distribution kernel of A a. 4. Pseudodifferential Operators First recall the definition of pseudodifferential operator. Definition 4.(Pseudodifferential Operator). For a S 3n ( ξ m ), and u C0 (R n ), we set Op h (a)u(x; h) e i(x y)ξ/h a(x, y, ξ)u(y)dydξ. Then Op h (a)u C (R n ), and for any ν R, the operator h ν Op h (a) is called the semiclassical pseudodifferential operator for symbol h ν a. In this case, h ν Op h (a) is said to be of degree m and of order ν. Example: Semiclassical differential operator If a is in particular of the form a(x, y, ξ) α m b α (x)ξ α with b α S n (), we get Op h ( α m b α (x)ξ α ) α m b α (x)(hd x ) α. If instead one obtains Op h ( α m a(x, y, ξ) α m b α (y)ξ α, b α (y)ξ α ) α m(hd x ) α b α (x). More generally, if a(x, y, ξ) α m b α (x, y)ξ α with b α (x, y) S 2n (), then for any u C 0 (R n ) Op h ( α m b α (x, y)ξ α )u(x) α m(hd x ) α (b α (x, x)u(x)) x x. Since C 0 (R n ) is dense in S(R n ), it is reasonable to have following theorem. 8

9 Theorem 4.2. For all a S 3n ( ξ m ), Op h (a) can be extended in a unique way to linear continuous opreator S(R n ) S(R n ). Proof. For any α, β N, writing x β x α I k u(x) I +I 2 ( ) + x β α x y 2 x x y 2 x x [e i(x y)ξ/h ( t L) k (au)]dydξ with L hξdy +ξ 2, since for any γ > 0 x β ξ m+ α k y γ O( ξ m+ α k y β γ ) is uniformly on { x y 2 x }, we have for k > m + n + α and γ > β + n I is integrable. Let L + x y 2 ( + h(x y)d ξ). Thus by integrating by parts with respect to ξ that for any N N, I 2 x β α x y 2 x x [e i(x y)ξ/h ( t L) k (au)]dydξ x β α x y 2 x x [(L ) N (e i(x y)ξ/h )( t L) k (au)]dydξ C α,α x β e i(x y)ξ/h ( t L ) N [ξ α x y 2 x α +α α x α ( t L) k (au)]dydξ and therefore is O() if we take N > β. Theorem 4.3. For all a S 3n ( ξ m ), Op h (a) can be extended in a unique way to a linear continuous operator S (R n ) S (R n ). Idea of the proof. Set A Op h (a), we can define t Av(y) e i(y x)ξ/h a(x, y, ξ)v(x)dxdξ which by Theorem 4.2 is in S(R n ). And we have for u, v S(R n ) u, t Av S,S Au, v S,S. We can extend this to u S (R n ). By density of C 0 (R n ) in S (R n ), we get the conclusion. 9

10 5. Composition By Theorem 4.3, We can see that a pseudodifferential operator can be extended to linear operator on S (R). Thus it is natural to consider the composition of these operators. Here arises the questions. Is the composition still a pseudodifferential differential operator? Theorem 5. (Theorem of Composition). For all a S( ξ m ) and b S ξ m ), there exist c S( ξ m+m ) such that Op h (a) Op h (b) Op h (c) Moreover, a possible choice for c is given by the oscillatory integral c(x, y, ξ) e i(x z)(η ξ)/h a(x, z, η)b(z, y, ξ)dzdη : a #b(x, y, ξ) which satisfies a #b(x, y, ξ) α 0 h α i α α! α z η α (a(x, z, η)b(z, y, ξ)) in S 3n ( ξ m+m ). zx,ηξ To prove this theorem, the main tool we need is called Stationary Phase Theorem. Theorem 5.2 (Stationary Phase Theorem). For d N, let Q be a d d nondegenerate real matrix. Then for all u C 0 (R d ) and for all N, one has with e ix Qx/2h u(x)dx N k0 S N (u, h) ChN+d/2 2 N N! detq (2π) d/2 hk + d/2e i π 4 sgnq ((D x Q D (2i) k detq /2 x ) k u)(0)+s N (u, h) k! α d+ α x (D x Q D x ) N u L (R d ) where C > 0 depends only on d, and sgnq denotes the signature of Q. Idea of the proof. Consider F : d-dimensional Fourier transform with h, we have F (e ix Qx/2h ) h d/2 e i π 4 sgnq detq /2 e ih(ξ Q ξ)/2 And this formula transform the rapid oscillations of e ix Qx/2h into the slow ones of e ih(ξ Q ξ)/2. Apply the definition of Fourier transform we get e ix Qx/2h u(x)dx F (eix Qx/2h ), F u S,S Use the former equation and apply Taylor Formula to the term e ih(ξ Q ξ)/2, we can get the result. 0

11 We gives a direct corollary which will be( used in the) proof of Composition 0 I Theorem by taking d 2n and Q Q. I 0 Corollary 5.3. For all u C 0 (R 2n ) and N, one has with e ixy/h u(x, y)dxdy S N ChN N! where C > 0 depends only on n. α+β 2n+ N h k i k k! ( n k0 j xj yj ) k u(0, 0) + S N α x β y ( x y ) N u L (R 2n ) In order to clarify the theorem and the proof, we introduce another notion called semiclassical expansion of symbols. Definition 5.4. Let a S d (g) and let (a j ) j N be a sequence of symbols of S d (g). Then we say that a is asymptotically equivalent to the formal sum h j a j in S d (g), and we write a h j a j j0 if for any N N and for any α N d there exists h N,α > 0 and C N,α > 0 such that α (a h j a j ) C N,α h N g j0 uniformly on R d (0, h n,α ]. In particular case where all a j s are zero, we write a O(h ) in S d (g) if a 0 in S d (g). Now let s prove the Theorem of Composition. Proof of Theorem 5.. Making integration by parts and using the same decomposition as in (2.5.), we see that for u C0 we have Op h (b)u(z) lim ɛ 0 +,δ 0 +,S(R n ) e i(z y)ξ/h ɛ ξ δ z u(y)b(z, y, ξ)dydξ where lim S(R n ) means that the limits is taken on the topology of S(R n ).Thus by continuity of Op h (a), we have (2πh) 2n Op h (a) Op h (b)u(x) ( ) lim e i(x z)η/h a(x, z, η) e i(z y)ξ/h ɛ ξ δ z u(y)b(z, y, ξ)dydξ dzdη ɛ 0 +,δ 0 + lim e i(x z)η/h+i(z y)ξ/h ɛ ξ δ z δ η a(x, z, η)b(z, y, ξ)u(y)dydξdzdη ɛ 0 +,δ 0 + j0

12 Therefore, Op h (a) Op h (b)u(x) lim lim ɛ 0 + δ 0 + ei(x z)η/h ɛ ξ c δ (x, y, ξ)u(y)dydξ where c δ (x, y, ξ) e i(x z)(η ξ)/h δ z δ η a(x, z, η)b(z, y, ξ)dzdη. Thus, by dominated convergence theorem, only need to show that c δ O( ξ m+m ) uniformly with respect to δ, that for all (x, y, ξ) R 3n, c δ (x, y, ξ) has a limit c 0 (x, y, ξ) as δ 0 + (so that the first limit δ 0 + above can be taken), and that c 0 S 3n ( ξ m+m ) (so that the second limit ɛ 0 + above can be taken). To do so, the main idea is to split the integral into three parts (by means of cut off functions), such that x z (respectively η ξ ) remains away from zero in the first (respectively second) part, while both x z and η ξ remain bounded in the last one. We can show that the first two parts are essentially negligible, while the stationary phase theorem can be applied to the third one. We give the detail of the proof below. Let χ C 0 (R) with χ (s) for s, χ (s) 0 for s 2. For x, y R n, define χ(x, y) χ ( x y ). Also define we have L ( + η ξ 2 h 2 + ) ( x z 2 h 2 (η ξ) D z + h L (e i(x z)(η ξ)/h ) e i(x z)(η ξ)/h ) (x z) D η h thus for k m + 2n + one has c δ (x, y, ξ) e i(x z)(η ξ)/h δ z δ η a(x, z, η)b(z, y, ξ)dzdη (L ) k (e i(x z)(η ξ)/h ) e δ z δ η a(x, z, η)b(z, y, ξ)dzdη e i(x z)(η ξ)/h ( t L ) k (e δ z δ η a(x, z, η)b(z, y, ξ))dzdη d δ (x, y, ξ) + e δ (x, y, ξ) + f δ (x, y, ξ), 2

13 where d δ (x, y, ξ) e i(x z)(η ξ)/h ( t L ) k( ( χ(ξ, η))e δ z δ η a(x, z, η)b(z, y, ξ) ) dzdη ( ) ( ) k( η ξ O + η ξ + η ξ 2 h + x z 2 h D z + x z h D η) k (e δ z δ η ab) 2 h ( 2 ) η m ξ m O η ξ ( + h η ξ + h x z k ) ( ) η m ξ m O ( + + η ξ 2h ) k n /2 and thus in the case m 0, ) d δ (x, y, ξ) O (h k 2n /2 ( ξ m + η ξ m ) ξ m η ξ k n /2 ( O h k 2n /2 ξ m+m ). We get the third equality by the estimation x + a 2 + x 2 + a2 + x C 2 + a + x for a. In the case m < 0, the integral can be split into two rigions { η ξ /2} and { η ξ /2}, while in the first region we have η m O( ξ m ) and the second one we have ( η ξ ξ /C for some ) C. Thus the second part can also estimated by O h k 2n /2 ξ m (k 2n ) The second part can be taken the cut off function similarly: e δ (x, y, ξ) e i(x z)(η ξ)/h ( t L) k[ χ(ξ, η)( χ(x, z))e δ z δ η a(x, z, η)b(z, y, ξ) ] dzdη O (h k 2n /2 ξ m+m ) for all k > m +2n+, and uniformly with respect to (x, y, ξ) R 3n and δ > 0. And the same argument holds for all the derivatives of d δ and e δ For f δ, we can rewrite it by integration by parts. f δ (x, y, ξ) e i(x z)(η ξ)/h χ(ξ, η)χ(x, z)e δ z δ η a(x, z, η)b(z, y, ξ)dzdη Making change of variables { z z x, η η ξ, 3

14 we have with f δ (x, y, ξ) e iz η /h u δ x,y,ξ(z, η )dz dη u δ x,y,ξ(z, η ) χ(ξ, η + ξ)χ(x, z + x)e δ z +x δ η +ξ a(x, z + x, η + ξ)b(z + x, y, ξ) C 0 (R n z Rn η ) Then we can apply stationary phase theorem to this integral. We have for all N, f δ (x, y, ξ) α N h α i α α! α z α η u δ x,y,ξ(z, η) z0,η0 + S N with S N ChN z α η β ( z η ) N u δ N! x,y,ξ L (R n z R n η ) α+β 2n+ ) O (h N η m ξ m dzdη η ξ 2, x z 2 O (h N ξ m+m ) uniformly. And we can apply this deduction to all derivatives of f δ exactly same result. Moreover, since for k m + 2n + one has and get ( t L ) k (e δ z δ η a(x, z, η)b(z, y, ξ)) L (R n z R n η ) O x,y,ξ () uniformly with respect to δ, by dominated convergence theorem we get: c δ (x, y, ξ) c 0 (x, y, ξ) as δ 0 +, where c 0 (x, y, ξ) e i(x z)(η ξ)/h ( t L ) k (a(x, z, η)b(z, y, ξ))dzdη. Since all the estimates are uniform with respect to δ, we can get c 0 (x, y, ξ) S 3n ( ξ m+m ) And by taking δ 0 + is the expansion, we can get the semiclassical asympotic expansion of c 0 (x, y, ξ). Thus we complete the proof. We can give an application of the Composition Theorem. 4

15 Proposition 5.5. Let m R and let a S 3n ( ξ m ) be an elliptic symbol in the sense of Definition 2.4. Then there exist b S 3n ( ξ m ) such that { Op h (a) Op h (b) + Op h (r) Op h (b) Op h (a) + Op h (r ) with r, r O(h ) in S 3n (). Proof. By Proposition 2.5, we know that a S 3n( ξ m ). Then, setting b 0 a and using the expansion of a #b given in the theorem, we can define b j S 3n ( ξ m ) (j, 2 ) recursively, in such a way that if b h j b j, then a #b + O(h ) in S 3n () Similarly, there exists b S 3n ( ξ m ) such that b #a + O(h ) in S 3n (). By theorem 5., this implies { Op h (a) Op h (b) + Op h (r) Op h (b ) Op h (a) + Op h (r ) with r, r O(h ) in S 3n (). Since we have ( + Op h (r )) Op h (b) Op h (b ) Op h (a) Op h (b) Op h (b ) ( + Op h (r)), using Theorem 5. again,we have Op h (b) Op h (b ) + Op h (b )Op h (r) Op h (r )Op h (b) Op h (b ) + Op h (r ) with r O(h ) in S 3n ( ξ m ). Using again Theorem 5. to compose Op h (b) with Op h (a), we get the conclusion that b solve the problem. 5