The wave equation is an important tool to study the relation between spectral theory and geometry on manifolds. Let U R n be an open set and let

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1 1. The wave equation The wave equation is an important tool to stuy the relation between spectral theory an geometry on manifols. Let U R n be an open set an let = n j=1 be the Eucliean Laplace operator. Then the wave equation on U is the following ifferential equation. ( ) 2 t u(x,t) = f(x,t), 2 u(x,0) =u 0 (x), Here f,u 0 an u 1 are given functions The wave equation on R n. 2 x 2 j u(x,t) = 0, x U, t > 0. t (x,0) = u 1(x). To unerstan the behavior of the solution of the wave equation we consier first the wave equation on the real line. On R we consier the following equation ( ) 2 t 2 u(x,t) = 0, 2 x 2 where g,h C 2 (R). u(x,0) = g(x), (x,0) = h(x), t The first equation can be factore as follows ( )( ) t + x t u = 0. x Put Then we get ( ) v(x,t): = t u(x,t). x t v(x,t)+ x v(x,t) = 0, x R, t > 0. 1

2 2 This is a transport equation with constant coefficients. To solve this equation, we apply the Fourier transform with respect to x. Let ˆv(ξ,t): = e iξx v(x,t)x. Then we have The solution is given by Hence we have Let a(x) = v(x,0). Then we get R = 0. tˆv(ξ,t)+iξˆv(ξ,t) ˆv(ξ,t) = e itξˆv(ξ,0). v(x,t) = v(x t,0). t u(x,t) u(x,t) = a(x t), x R, t > 0. x this is a non-homogeneous transport equation. It can be solve by a similar metho. The result is u(x,t) = If we use the inatrial conitions t 0 = 1 2 u(x,0) = g(x) an u t (x,0) = h(x), we get a(x+(t s) s) s+u(x+t,0) x+t x t a(y)y+u(x+t,0). a(x) = v(x,0) = u t (x,0) u x (x,0) = h(x) g (x) Inserting this formula for a in the above equation, we obtain the following final form for the solution (6.1) u(x,t) = 1 2 [g(x+t)+g(x t)]+ 1 2 x+t x t h(y)y, x R, t > 0. From this expression for the solution one can erive the following theorem. Theorem 6.1. Assume that g C 2 (R), h C 1 (R), an efine u(x,t) by (6.1). Then the following hols 1) u( C 2 (R [0, ]). ) 2) 2 2 u(x,t) = 0 in R R + t 2 x 2

3 3 3) lim (x,t) (x 0,0) t>0 lim (x,t) (x 0,0) t>0 u(x,t) = g(x 0 ) u t (x,t) = h(x 0 ) Assume that suppg,supph (r,r). Then it follows from (6.1) that suppu ( r t,r + t). This means that the wave equation on R has finite propagation spee. Similar formulars hol for all n 1. Namely consier on R n the wave equation. ( ) 2 t u(x,t) = 0, u(x,0) = g(x), u 2 t (x,0) = h(x). Let n 3 be o. Let m = (n + 1)/2, g C m+1 (R n ) an h C m (R n ). Let γ n = (n 2). Then the unique solution of the wave equation is given by [ ( u(x,t) = 1 1 γ n t t t + )n 3 2 ( t n 2 ( 1 t B(x,t) g S ) )n 3 ( 2 t n 2 hs t B(x,t) This formula also shows that the wave equation satisfies finite propagation spee. )] Energy methos. Energy methos are an important tool to establish finite propagation spee for the wave equation. We illustrate this for the Laplace operator. For x 0 R n, t 0 > 0, let { } C = (x,t): 0 t t 0, x x 0 t 0 t. Theorem 6.2. (Finite propagation spee). Assume that u(x,0) = u t (x,0) 0 on B(x 0,t 0 ). Then u 0 in C. Proof. Define the energy of the solution by (6.2) e(t) = 1 ) (u t (x,t) 2 + u(x,t) 2 x,0 t t 0. 2 B(x 0,t 0 t)

4 4 Then we have t e(t) = 1 2 = + = B(x 0,t 0 t) B(x 0,t 0 t) B(x 0,t 0 t) B(x 0,t 0 t) B(x 0,t 0 t) (u t u tt + u, u t )x (u 2 t + u 2 )s u t (u tt u)x ν u t s 1 ) 1 (u 2t u 2 S. 2 B(x 0,t 0 t) 2 ( ) ν u t 1 2 u2 t 1 u 2 S. 2 Now note that ν u t u t u 1 2 u t u 2. This implies that e(t) 0. t Thus e(t) e(0) = 0 for 0 t t 0. By (6.2) it follows that u t 0 an u 0 in C. This implies that u c an therefore, u = Graient an ivergence. As preparation for the stuy of the wave equation on manifols we recall some facts about the icergence an the graient on a Riemannian manifol. Let X be a Riemannian manifol. Let f C (X). Then the graient graf C (TX) of f is efine by graf(p),y p = Y(f)(p) = f(y)(p) for all Y C (TY). Let : C (TX) C (T x X TX) be the Levi-Civita connection associate to the Riemannian metric of X. Let Y C (TX). The ivergence ivy of the vector fiel Y is efine by ivy(p) = Tr(ξ T p X ξ Y T p X). In local coorinates graf an Y can escribe as follows. Let x 1,...,x n be local coorinates. Let n g = g ij x i x j i,j=1 be the Riemannian metric in these coorinates. Furthermore, let (g kl ) = (g ij ) 1, ḡ = et(g ij ).

5 5 an Then we have an graf = Y = n k=1 ivy = 1 ḡ n j=1 l=1 f j x j. ( ) n kl f g x l n j=1 x j x k ( ḡfj ) Lemma 6.3. For all f C (X) an Y C (TX) we have (graf,y) = (f,ivy). Proof. Using a partition of unity, the proof can be reuce to the case where suppf is containe in a coorinate chart U. Then ( n n n kl f ḡx n f ḡx (graf,y) = g )g kj f j = f j U x j=1 l k=1 l=1 U x j=1 j ( ) = f ḡ 1 n ḡ ḡx f j = f ivyµ(x) x j U = (f,ivy). j=1 U The Riemannian metric efines an isomorphism. φ: TX = T X. It inuces an isomorphism φ: C (TX) = Λ 1 (X). Lemma 6.4. (1) For all f C (X) we have (2) For all Y C (TX) we have φ(graf) = f. iv(y) = (φ(y)).

6 Symmetric hyperbolic systems. Let X be a Riemannian manifol an E X a Hermitian vector bunle over X. Denote by (, ) the inner prouct in Cc (E) inuce by the Riemannian metric an the fibre metric in E. Let D: C (E) C (E) be an elliptic ifferential operator of orer 1. Assume that D is formally self-ajoint. Example: The basic example is the Dirac operator D: C (S) C (S) on a spin manifol. Let π: T X X the cotangent bunle. Let σ D : π E π E be the principal symbol of D. We recall its efinition. Let x X, ξ T x X, an e E x. We choose f C (X) with f(x) = 0, f(x) = ξ, an ϕ C (E) with ϕ(x) = e. Then σ D (x,ξ)(e) = D(Fϕ)(x). Lemma 6.5. For any f C (X) an ϕ C (E) we have (6.3) D(fϕ) = σ D (f)(ϕ)+fϕ. Note that Definition 6.6. For Ω X let { c(ω): = sup σ D (x,ξ) t = σ (x,ξ). c(ω) is calle the propagation spee of D on Ω. σ D (x,ξ) : ξ T x, ξ, x Ω Now we consier the wave equation (6.4) t = idu, u(x,0) = u 0(x), where u 0 C (E). Proposition 6.7. Let x 0 X an suppose that B r (x 0 ) is a geoesic coorinate system. Let c: = c(b r (x 0 )). Let u C ([ T,T],C (E)) be a solution of t = id u. Then we have u(t) Br ct (x 0 ) u(0) Br(x 0 ) for 0 t < r/c. }.

7 7 Proof. We efine a smooth vector fiel Y t on X by Y t (f)(x) = i u(x,t),σ D (f x,x)(u(x,t)) x Let f Cc (X). Then by Lemma 6.4 we have ivy t (x) f(x)x = (ivy t,f) = (Y t,graf) = Y t (f). X By efinition of Y t (f) an (6.3) we get ivy t (x) f(x)x = i ( u(t),d(fu(t)) fdu(t) ) X = i ( Du(t),fu(t) ) i ( u(t),fdu(t) ) ( = i u(x,t),u(x,t) x u(x,t),du(x,t) x ) f(x)x. Since this equality hols for every f C c (X), it follows that X (6.5) ivy t (x) = i Du(x,t),u(x,t) x i u(x,t),du(x,t) x. Now u(x,t) 2 x t B r ct (x 0 ) ( = B r ct (x 0 ) t u(x,t),u(x,t) + u(x,t), ) x t u(x,t) x x c u(x,t) 2 x S(x) = i c B r ct (x 0 ) B r ct (x 0 ) B r ct (x 0 ) ( Du(x,t),u(x,t) u(x,t) 2 x S(x). x u(x,t),du(x,t) x For the last equality we use that u(x,t) satisfies the wave equation (6.6). Using (6.5) an the ivergence theorem it follows that u(x,t) 2 x t x = ivy t (x) u(x,t)x B r ct (x 0 ) B r ct (x 0 ) c u(x,t) 2 x S(x) = B r ct (x 0 ) B r ct (x 0 ) ) x Y t (x),ν(x) x S(x) c u(x,t) 2 x S(x), B r ct (x 0 )

8 8 where ν(x) enotes the exterior unit normal vector fiel. Now observe that by the efinition of c = c(b r (x 0 )) This implies that Thus we obtain Y t (x),ν(x) x = u(x,t),σ D (ν(x),x)(u(x,t)) x c u(x,t) 2. which conclues the proof. Let c = c(b r (x 0 )) an let u(x,t) 2 xx 0. t B r ct (x 0 ) u(t) Br ct (x 0 ) u(0) Br (x 0 ), C = {(t,x): t 0, (x,x 0 ) r ct} Corollary 6.8. Let u C ([ T,T],C (E)) be a solution of the equation t = idu on C. Suppose that u(0) = 0 on B r (x 0 ). Then u = 0 on C. Let U = B r (x 0 ) X be a normal coorinate chart an let φ: E U = U C N be a trivialization of E U. Let D U be restriction of D to C (U,E U ). Then t u = id U(u), u(0,x) = u 0 (x) is a hyperbolic system oforer 1 inr n. The usual theory for such systems implies existence of solutions with smooth initial conitions. In this way we get Proposition 6.9. For every x 0 X there exists r > 0 such that for u 0 C (B r (x 0,E)) there exists a unique solution of t = id(u), u(0,x) = u 0(x) on C 0 = {(x,t): t 0, (x,s 0 ) r ct}, where c = c(b r (x 0 )). The next proposition extens the above result to a larger region. Proposition Let S = B R (x 0 ) be a compact ball in X. Let c: c(s) an C 0 = {(x,t): t 0, (x,x 0 ) r ct} Let u 0 C (S,E). Then there is in C 0 a unique smooth solution of the equation t = id(u), u(0) = u 0.

9 Proof. Since S is compact, there exists r > 0 such for all y S the injectivity raius i(y) r. Thusforally S, B r (y)isanormalcoorinatechart. ItfollowsfromProposition (6.9) that for all y B R r (x 0 ) an u 0 C (B r (y),e), the wave equation t = id(u), u(0) = u 0, has a unique C -solution on the truncate cone C y = {(x,t): (x,y) r ct,0 t r/2c}. Byuniqueness, solutions agreeonc y C z. Therefore, we obtainasolutiononthetruncate cone {(x,t): (x,x 0 ) R ct, 0 t r/2c}. The solution u at time t = r/2c serves as initial conition on B(x 0,R r/2). If we repeat the above argument, we get a solution on the truncate cone an therefore, a solution on {(x,t): (x,x 0 ) R ct, r/2c t r/c} {(x,t): (x,x 0 ) R ct, 0 t r/c}. After a finite number of steps, we obtain a smooth solution on the cone with base S. Let x 0 X. Put c(r): = c(b r (x 0 )), r > 0. Theorem Let X be a complete Riemannian manifol. Suppose that r c(r) =. (6.6) 1) Uniqueness: Suppose that u C (X,E) is a solution of t = idu on [0,T] X with u(0) = 0. Then u 0. 2) Existence: Let u 0 C (X,E). Then the wave equation 0 t = idu, u(0) = u 0, has a unique solution on R X. Moreover, for fixe t, u(,t) has compact support. Proof. We first establish uniqueness. Let R > 0. Put S R = B R (x 0 ). We shall show that u(t) vanishes on S R. Of course, we also have that u(t ) vanishes for T < T. Let t R = c(r+1) 1,R > 0. By Proposition (6.10), the wave equation (6.6) on {(x,t): (x,x 0 ) R c(r+1)t,0 t t R } 9

10 10 with initial conition u 0 C (S R+1,E) has a unique solution. Hence u(t) on S R is etermine by u(t t R ) on S R+1. By the same argument u(t t R ) is etermine by u(t t R t R+1 ) ons R+2. Since thesets S R arecompact, this maybecontinue inefinitely. Now we have (6.7) t R+n =, n=0 because c(r) is monoton an r 1 c(r) = by assumption. Hence there exists N N such that T : = t R +t R+1 + +t R+N < T an T +t R+N+1 > T. From our consierations above follows that u(t) SR is etermine by u(t T ) SR+N. Let ε = (R+N +1)(T T ). Then it follows as above that u(t T ) SR+N is etermine by u(0) SR+N+ε. Note that T T t R+N+1. But u(0) = 0. Hence u(t T ) SR+N = 0. This implies that u(t) SR = 0. Next we establish existence. The argument is like the uniqueness proof run in reverse. Let u 0 C c (X,E). Let R > 0 such that suppu 0 S R. Since X is complete, S R is compact. By Proposition (6.10) there exists a C -solution of (6.6) in C = {(x,t): 0 t t R+2,(x,x 0 ) R+3 c(r+3)t} Moreover, it follows from the uniqueness part of Proposition (6.10) that the solution vanishes outsie S R+1. So we can exten it by 0 to a global solution on X which exists for time 0 t t R+2. Now we iterate this process. The solution at time t = t R+2 is supporte in S R+1. By the above argument, it extens to a solution for 0 t t R+2 + t R+3 with support in S R+2. Using again (6.7), we can exten the solution to any time t an for fixe t, u(t) has compact support. by For each t efine a map where u is the unique solution of U t : C c (X,E) C (X,E) U t (u 0 ) = u(t), t = idu, u(0) = u 0. Corollary Uner the assumptions of Theorem (6.11), {U t } is a one-parameter group. Moreover, if w C c (X,E), we have t (U t(u 0 ),w) = (idu t (u 0 ),w).

11 Let ψ H an suppose that ((A±i)(ϕ),ψ ) = 0, ϕ D. 11 Finally DU t (u 0 ) = U t (Du 0 ). Proof. Let u(t): = U t (u 0 ). Then u(t) C (X,E) an t = idu. Since w Cc (X,E), we can ifferentiate unter the integral which gives t (U t(u 0 ),w) = ( t,w ) = (idu t (u 0 ),w) Moreover, U s+t = U s U t an DU t = U t D follows from uniqueness. Note that U t is unitary. Inee we have ( u ) ( t u(t) 2 = t (t),u(t) + u(t), u ) t (t) ( ) ( ) = idu(t), u(t) + u(t), idu(t) ( ) ( ) = idu(t), u(t) id(t),u(t) = Essential self-ajointness. In this section we apply the results obtaine in the previous section to establish the essential self-ajointness of geometric operators. We begin with an abstract result. Lemma Let T be a symmetric operator in a Hilbert space H with ense omain D H. Suppose that T(D) D). Furthermore suppose that there is a one-parameter group U t of unitary operators on H such that U t (D) D, U t T = TU t on D an t U t(u) = itu t (u) for u D. Then every power of T is essentially self-aoint. Proof. Let n N an A: = T n. Then A: D H is symmetric. To show that A is essentially self-ajoint, it suffices to verify that (A±iI)(D) = H Then it follows that ψ D(A ) an A ψ = iψ.

12 12 We consier the case where A ψ = iψ. For u D efine f(t) = (U t (u),ψ), t R. Since U t is unitary, f(t) is boune. Furthermore we have n t nf(t) = (itn U t (u),ψ) = (iau t (u),ψ) = (i n U t (u),a ψ) = i n+1 (U t (u),ψ) = i n+1 f(t). Thus f(t) satisfies the orinary linear ifferential equation. n f (6.8) t (t) = n in+1 f(t). Let α j, j = 1,...,n, be the ifferent roots of the equation z n = i n+1. Then e αjt, j = 1,...,n, is a basis for the space of solutions of (6.8). Therefore f(t) can be written as n f(t) = c j e α jt j=1 for some constants c j C. Now observe that Re(α j ) 0 for j = 1,...,n. Since f is boune this implies that f 0. Hence we get (u,ψ) = f(0) = 0, u D. Since D H is ense, it follows that ψ = 0. The case A ψ = iψ can be treate in the same way. We can now state the main result about essential self-ajointness. Theorem Let X be a complete Riemannian manifol an E X a Hermitian vector bunle over X. Let D: C (X,E) C (X,E) be an elliptic ifferential operator of orer 1 which is formally self-ajoint. Assume that 1 r c(r) =. Let T : C c (X,E) L2 (X,E) be the operator which is efine by D. Then every power of T is essentially self-ajoint. Proof. Let U t : C c (X,E) C c (X,E) be the 1-parameter group, efine by Corollary For each t R we have U t (u) = u, u C c (X,E).

13 Inee, for u,w C c (X,E), we have t (U t(u),u t (w)) = (idu t (u),u t (w))+(u t (u),idu t (w)) ((id id)u t (u),u t (w)) = 0. Hence U t extens by continuity to a one-parameter family U t : L 2 (X,e) L 2 (X,E) of unitary operators. The assumptions of Lemma 6.13 are satisfie. This implies the theorem Applications. Now we are reay to apply the results of the previous section to geometric situations. Let X be a complete Riemannian manifol. Let D = + : Λ (X) Λ (X). Then D is formally self-ajoint. To etermine its principal symbol, fix p X,ξ T p X an v Λ T p X. Let f C (X) an ϕ Λ (X) be such that f(p) = 0,f p = ξ an ϕ(p) = v. Then we have σ (p,ξ)v = D(fϕ)(p) = (+ )(fϕ)(p) = f p ϕ(p) (f ϕ)(p) = ξ v i ξ (v), where i ξ : Λ Tp X Λ Tp X enotes interior multiplication by ξ. This implies that σ D (x,ξ) = ξ. Hence we have c(x) = 1, i.e. D has unite propagation spee. By Theorem 6.14 it follows that for all n N, the operator (+ ) n : Λ c (X) L2 Λ (X) is essentially self-ajoint. Now recall that the laplace operator is given by Thus it follows that for all n N, = (+ ) 2. n : Λ c(x) L 2 Λ (X) is essentially self-ajoint. The Laplace operator preserves Λ p (X) for every p. Therefore is essentially self-ajoint for all p = 0,...,n. n : Λ p c(x) L 2 Λ p (X) Next we consier a complex manifol equippe with a Hermitian metric, so that X, equippe with the associate Riemannian metric is complete. Let E X be a holomorphic Hermitian vector bunle over X. Then we efine the space of (p,q)-forms with values in E as the space of C -sections of Λ p T (1.0) (X) Λ q T (0,1) (X) E. The operator : Λ p,q (X,E) Λ p,q+1 (X,E) 13

14 14 is uniquely efine by emaning that (ω ϕ) = ( ω) ϕ for every ω Λ p,q (X) an every holomorphic section of E. In this way we get the Dolbeault complex Let D = ( + ). Then we have where π: T x Λ p,q (X,E) Λ p,q+1 (X,E) σ D (x,ξ)(ω ϕ) = (π(ξ) ω i π(ξ) (ω)) ϕ X C T (0,1) x X is the canonical projection. It follows that an hence, c(x) = 1/ 2. By Theorem 6.14, σ D (X,ξ) = π(ξ) ( + ) n : Λ p, c (X,E) L 2 Λ p, (X,E) is essentially self-ajoint for all p an all n N. References [Ch] Chernoff, Paul R. Essential self-ajointness of powers of generators of hyperbolic equations. J. Functional Analysis 12 (1973), [Ev] Evans, Lawrence C., Partial ifferential equations. Grauate Stuies in Mathematics, 19. American Mathematical Society, Provience, RI, 1998.