Ultra-Hyperbolic Equations

Transcription

1 Ultra-Hyperbolic Equations Minh Truong, Ph.D. Fontbonne University December 16, 2016 MSI Tech Support (Institute) Slides - Beamer 09/ / 28

2 Abstract We obtained a classical solution in Fourier space to the modified Cauchy problem in p + q = n = 6, where p = q = 3, with signature (+, +, +,,, ). The two extra time-like dimensions along with an additional bounded condition u M, where M R is an arbitrary constant. We also obtained upper-bound value to the solution. f (p, ξ) = M. (1) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

3 Abstract We obtained a classical solution in Fourier space to the modified Cauchy problem in p + q = n = 6, where p = q = 3, with signature (+, +, +,,, ). The two extra time-like dimensions along with an additional bounded condition u M, where M R is an arbitrary constant. We also obtained upper-bound value to the solution. f (p, ξ) = M. (1) We seek the distributional solutions or generalized functions to the following equation, L k u = f (x). (2) The elementary solutions obtained E = lim e 3 2 πi ε +0 e + lim 2 3 πi ε 0 (P + iε x 2 4π x 6 2 4π 3 (P iε x x 2 6 ) 2 ) 2 MSI Tech Support (Institute) Slides - Beamer 09/ / 28

4 Issues with Theories with Extra Time-Like Dimensions Closed time-like loops violations of causality[1][thorne]. Motion of extra time-like dimensions must be non-regular or chaotically quantized low probability for creating a closed loop. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

5 Issues with Theories with Extra Time-Like Dimensions Closed time-like loops violations of causality[1][thorne]. Motion of extra time-like dimensions must be non-regular or chaotically quantized low probability for creating a closed loop. Violations of Unitarity compactified time dimensions effectively leads to the imaginary correction to the potential violations of conservation of probability [3][Yndurain] MSI Tech Support (Institute) Slides - Beamer 09/ / 28

6 Issues with Theories with Extra Time-Like Dimensions Closed time-like loops violations of causality[1][thorne]. Motion of extra time-like dimensions must be non-regular or chaotically quantized low probability for creating a closed loop. Violations of Unitarity compactified time dimensions effectively leads to the imaginary correction to the potential violations of conservation of probability [3][Yndurain] Tachyonic modes= instabilities of a given system[2] MSI Tech Support (Institute) Slides - Beamer 09/ / 28

7 Ultra-Hyperbolic Equations (Modified Cauchy Problem) [6][7][5] 2 u x u x u x 2 3 with a bounded condition on the solution, = 2 u t u t u t2 2 for t 0,and an arbitrary constant M R. (3) u(x, t, 0) = f (x, t), (4) u t (x, t, 0) = g(x, t), (5) u M, (6) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

8 Ultra-Hyperbolic Equations (Modified Cauchy Problem) [6][7][5] 2 u x u x u x 2 3 with a bounded condition on the solution, = 2 u t u t u t2 2 for t 0,and an arbitrary constant M R. (3) u(x, t, 0) = f (x, t), (4) u t (x, t, 0) = g(x, t), (5) u M, (6) Where u = u(x, t, t), t = (t 1, t 2 ), t =physical time, and x = (x 1, x 2, x 3 ). Using bold variables to represent vector quantities... MSI Tech Support (Institute) Slides - Beamer 09/ / 28

9 Ultra-Hyperbolic Equations (Modified Cauchy Problem) Fourier transform over the x and t variables, û (p, ξ, t) = u(x, t, t) exp i (p x + ξ t) dxdt. (7) R 5 Inverse Fourier transform over p and ξ, MSI Tech Support (Institute) Slides - Beamer 09/ / 28

10 Ultra-Hyperbolic Equations (Modified Cauchy Problem) Fourier transform over the x and t variables, û (p, ξ, t) = u(x, t, t) exp i (p x + ξ t) dxdt. (7) R 5 Inverse Fourier transform over p and ξ, u(x, t, t) = 1 (2π) 5 û (p, ξ, t) exp i (p x + ξ t) dpdξ. (8) R 5 MSI Tech Support (Institute) Slides - Beamer 09/ / 28

11 Ultra-Hyperbolic Equations (Modified Cauchy Problem) Fourier transform over the x and t variables, û (p, ξ, t) = u(x, t, t) exp i (p x + ξ t) dxdt. (7) R 5 Inverse Fourier transform over p and ξ, u(x, t, t) = 1 (2π) 5 û (p, ξ, t) exp i (p x + ξ t) dpdξ. (8) R 5 Where p = (p 1, p 2, p 3 ), ξ = (ξ 1, ξ 2 ), dx = dx 1 dx 2 dx 3 dt = dt 1 dt 2,dξ = dξ 1 dξ 2 and dp = dp 1 dp 2 dp 3. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

12 Cauchy Problem in Fourier Space û tt (p, ξ, t) + (p 2 ξ 2 )û (p, ξ, t) = 0, (9) û(p, ξ, 0) = f (p, ξ), (10) û t (p, ξ, 0) = ĝ (p, ξ), (11) where f (p, ξ) and ĝ (p, ξ) are functions on R 5, and (p, ξ) R 5. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

13 Cauchy Problem in Fourier Space û tt (p, ξ, t) + (p 2 ξ 2 )û (p, ξ, t) = 0, (9) û(p, ξ, 0) = f (p, ξ), (10) û t (p, ξ, 0) = ĝ (p, ξ), (11) where f (p, ξ) and ĝ (p, ξ) are functions on R 5, and (p, ξ) R 5. Let E R 5, E = {(p, ξ) R 5 ξ > p } Elliptic region H R 5, H = {(p, ξ) R 5 ξ < p } Hyperbolic region N R 5, N = {(p, ξ) R 5 ξ = p } Linear region MSI Tech Support (Institute) Slides - Beamer 09/ / 28

14 Solution to the Reduced Cauchy Problem If (p, ξ) N, the null region. Then the solution to û tt (p, ξ, t) = 0, û N (p, ξ, t) = ĝ(p, ξ) t + f (p, ξ). (12) The bounded condition on û N (p, ξ, t) = f (p, ξ) M, û N (p, ξ, t) = f (p, ξ) M. (13) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

15 Solution to the Reduced Cauchy Problem If (p, ξ) N, the null region. Then the solution to û tt (p, ξ, t) = 0, û N (p, ξ, t) = ĝ(p, ξ) t + f (p, ξ). (12) The bounded condition on û N (p, ξ, t) = f (p, ξ) M, û N (p, ξ, t) = f (p, ξ) M. (13) If (p, ξ) H, hyperbolic region. Then the solution to û tt (p, ξ, t) + (p 2 ξ 2 )û (p, ξ, t) = 0. û H (p, ξ, t) = f (p, ξ) cos(λ H t) + ĝ(p, ξ) sin(λ H t) λ H, (14) where λ H = ± p 2 ξ 2. The solutions are oscillatory... MSI Tech Support (Institute) Slides - Beamer 09/ / 28

16 Solutions to the Reduced Cauchy Problem If (p, ξ) E, Elliptic region and the bounded condition, u M, then the solution to û tt (p, ξ, t) + (p 2 ξ 2 )û (p, ξ, t) = 0 û E (p, ξ, t) = f (p, ξ) exp( λ E t) (15) where λ E = ± ξ 2 p 2, and λ E = i λ H. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

17 Solutions to the Reduced Cauchy Problem If (p, ξ) E, Elliptic region and the bounded condition, u M, then the solution to û tt (p, ξ, t) + (p 2 ξ 2 )û (p, ξ, t) = 0 û E (p, ξ, t) = f (p, ξ) exp( λ E t) (15) where λ E = ± ξ 2 p 2, and λ E = i λ H. The bounded condition on the solution u(p, ξ, t) M t 0 and a bit of analysis, one yields the value of the upper-bound to be M = f (x, t). (16) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

18 Solution to the Reduced Cauchy Problem The solution to the modified Cauchy problem in Fourier space, for all (p, ξ, t) R 6 t 0. û(p, ξ, t) = û H (p, ξ, t) + û E (p, ξ, t) + û L (p, ξ, t). (17) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

19 Preliminaries on Distributions (Basics of Differential Geometry)[8][9] Tangent bundle of manifold M, dim M = n TM := T x M. (18) x M MSI Tech Support (Institute) Slides - Beamer 09/ / 28

20 Preliminaries on Distributions (Basics of Differential Geometry)[8][9] Tangent bundle of manifold M, dim M = n TM := x M T x M. (18) Vector bundle: Let K = R or C. Let E be manifold of dimension 2n, let π : E M be a surjective map. Let the fiber E x := π 1 (x) preserve a structure K-vector space V x for each x M. (E, π, M, {V x } x M ) is called a K-vector bundle if for x M, U of x and a diffeomorphism Φ or a local trivialization of the vector bundle E. Φ : π 1 (U) U K n. (19) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

21 Preliminaries on Distributions (Basics of Differential Geometry) If U = M, then Φ : π 1 (M) M K n. (20) Then a vector bundle E is called trivial if it admits a global trivialization (i.e. U = M). MSI Tech Support (Institute) Slides - Beamer 09/ / 28

22 Preliminaries on Distributions (Basics of Differential Geometry) If U = M, then Φ : π 1 (M) M K n. (20) Then a vector bundle E is called trivial if it admits a global trivialization (i.e. U = M). Let π 2 : U K n K n, then we have a vector space isomorphism that maps element from the fiber E x := π 1 (U) to the K n -vector space π 2 Φ : π 1 (U) K n (21) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

23 Sections and Densities on Vector Bundle A section in E is a map s : M E such that π s = id M The sections in E = TM are called vector fields on M and spanned by { } x i i=1,..,n The sections in E = T M are called the 1-forms on M and spanned by {dx i } i=1,..n MSI Tech Support (Institute) Slides - Beamer 09/ / 28

24 Sections and Densities on Vector Bundle A section in E is a map s : M E such that π s = id M The sections in E = TM are called vector fields on M and spanned by { } x i i=1,..,n The sections in E = T M are called the 1-forms on M and spanned by {dx i } i=1,..n A section in E = Λ k T M is called k-forms dx i 1 dx i k. When k = n the bundle Λ n T M has rank 1 yields = smooth section dx 1 dx n = M being orientable. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

25 Sections and Densities on Vector Bundle A section in E is a map s : M E such that π s = id M The sections in E = TM are called vector fields on M and spanned by { } x i i=1,..,n The sections in E = T M are called the 1-forms on M and spanned by {dx i } i=1,..n A section in E = Λ k T M is called k-forms dx i 1 dx i k. When k = n the bundle Λ n T M has rank 1 yields = smooth section dx 1 dx n = M being orientable. For each x M, let ΛM x be a set of all functions ρ : Λ n Tx M R with ρ(λξ) = λ ρ(ξ) ξ Λ n Tx M ΛM x is 1D vector space = vector bundle ΛM of rank 1 over M. Sections in ΛM are called densities are characterized by dx (dx 1 dx n ) = 1 (22) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

26 Sections and Densities on Vector Bundle A section in E is a map s : M E such that π s = id M The sections in E = TM are called vector fields on M and spanned by { } x i i=1,..,n The sections in E = T M are called the 1-forms on M and spanned by {dx i } i=1,..n A section in E = Λ k T M is called k-forms dx i 1 dx i k. When k = n the bundle Λ n T M has rank 1 yields = smooth section dx 1 dx n = M being orientable. For each x M, let ΛM x be a set of all functions ρ : Λ n Tx M R with ρ(λξ) = λ ρ(ξ) ξ Λ n Tx M ΛM x is 1D vector space = vector bundle ΛM of rank 1 over M. Sections in ΛM are called densities are characterized by M : D(M, ΛM ) R dx (dx 1 dx n ) = 1 (22) M f dx = ϕ(u) (f ϕ 1 )(x 1,..., x n )dx 1...dx n MSI Tech Support (Institute) Slides - Beamer 09/ / 28

27 Linear Connection on K n -Vector-Bundle E The K-bundle E M. Let D(M, E ) space of compactly support sections in E. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

28 Linear Connection on K n -Vector-Bundle E The K-bundle E M. Let D(M, E ) space of compactly support sections in E. Equip E with connection, induces a connection on the dual vector bundle E by ( X θ)(s) := X (θ(s)) θ( X s) for all X C (M, TM), θ C (M, E ),s C (M, E ), and θ(s) C (M) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

29 Linear Connection on K n -Vector-Bundle E The K-bundle E M. Let D(M, E ) space of compactly support sections in E. Equip E with connection, induces a connection on the dual vector bundle E by ( X θ)(s) := X (θ(s)) θ( X s) for all X C (M, TM), θ C (M, E ),s C (M, E ), and θ(s) C (M) For a continuously differentiable section ϕ C 1 (M, E ), the covariant derivative is a continuous section in T M E, ϕ C 0 (M, T M E ) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

30 Linear Connection on K n -Vector-Bundle E The K-bundle E M. Let D(M, E ) space of compactly support sections in E. Equip E with connection, induces a connection on the dual vector bundle E by ( X θ)(s) := X (θ(s)) θ( X s) for all X C (M, TM), θ C (M, E ),s C (M, E ), and θ(s) C (M) For a continuously differentiable section ϕ C 1 (M, E ), the covariant derivative is a continuous section in T M E, ϕ C 0 (M, T M E ) In general, if ϕ C k (M, E ) k ϕ C 0 (M, T M... T M E ) k times MSI Tech Support (Institute) Slides - Beamer 09/ / 28

31 Distributions as Sections of the Vector Bundles [8] K-linear map T : D(M, E ) W is called a distribution in E with values in W. Where D (M, E, W ) space of all W -valued distributions in E. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

32 Distributions as Sections of the Vector Bundles [8] K-linear map T : D(M, E ) W is called a distribution in E with values in W. Where D (M, E, W ) space of all W -valued distributions in E. Let E and F be two K-vector bundles over M, The linear differential operator L L : C (M, E ) C (M, F ), ϕ D(M, E ) L : C (M, F ) C (M, E ), ψ D(M, F ) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

33 Distributions as Sections of the Vector Bundles [8] K-linear map T : D(M, E ) W is called a distribution in E with values in W. Where D (M, E, W ) space of all W -valued distributions in E. Let E and F be two K-vector bundles over M, The linear differential operator L L : C (M, E ) C (M, F ), ϕ D(M, E ) L : C (M, F ) C (M, E ), ψ D(M, F ) Defined by smooth maps A α : U (M, Hom K (E, F )) such that on U x Ls = α k A α s α x where α = (α α 1,..., α d ) N d with α := d r=1 α r k and α x = α α α d x α x α d d M ψ(lϕ)dv = M (L ψ)ϕdv MSI Tech Support (Institute) Slides - Beamer 09/ / 28

34 Distributions as Sections of the Vector Bundles [8] K-linear map T : D(M, E ) W is called a distribution in E with values in W. Where D (M, E, W ) space of all W -valued distributions in E. Let E and F be two K-vector bundles over M, The linear differential operator L L : C (M, E ) C (M, F ), ϕ D(M, E ) L : C (M, F ) C (M, E ), ψ D(M, F ) Defined by smooth maps A α : U (M, Hom K (E, F )) such that on U x Ls = α k A α s α x where α = (α α 1,..., α d ) N d with α := d r=1 α r k and α x = α α α d x α x α d d M ψ(lϕ)dv = M (L ψ)ϕdv If L is of order zero, then the operators are just sections of the K -vector bundle homomorphism L C (M, Hom K (E, F )). MSI Tech Support (Institute) Slides - Beamer 09/ / 28

35 The Principle Symbol of Linear Differential Operators The principle symbol of L is the map σ L : T M Hom K (E, F ). For every ξ = d r=1 ξ r dx r Tx M we get σ L (ξ) = α =k ξ α A α (x), x M and ξ α = ξ α ξα d d MSI Tech Support (Institute) Slides - Beamer 09/ / 28

36 The Principle Symbol of Linear Differential Operators The principle symbol of L is the map σ L : T M Hom K (E, F ). For every ξ = d r=1 ξ r dx r Tx M we get σ L (ξ) = α =k ξ α A α (x), x M and ξ α = ξ α ξα d d grad : C (M, R) C (M, TM) with principal symbol σ grad (ξ)f = f ξ div : C (M, TM) C (M, R) with principal symbol σ div (ξ)x = ξ(x ) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

37 The Principle Symbol of Linear Differential Operators The principle symbol of L is the map σ L : T M Hom K (E, F ). For every ξ = d r=1 ξ r dx r Tx M we get σ L (ξ) = α =k ξ α A α (x), x M and ξ α = ξ α ξα d d grad : C (M, R) C (M, TM) with principal symbol σ grad (ξ)f = f ξ div : C (M, TM) C (M, R) with principal symbol σ div (ξ)x = ξ(x ) d : C (M, Λ k T M) C (M, Λ k+1 T ) with principle symbol σ d (ξ)ω = ξ ω : C (M, E ) C (M, T M E ) σ (ξ)e = ξ e MSI Tech Support (Institute) Slides - Beamer 09/ / 28

38 Complex Vector Bundle and Complex manifolds [11][12] If E is a complex vector bundle, a positive definite quadratic form is given on each fiber E x. Let θ : E Ω Ω C r be a trivialization and let (e 1,.., e r ) be the corresponding frame field for the fiber E Ω. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

39 Complex Vector Bundle and Complex manifolds [11][12] If E is a complex vector bundle, a positive definite quadratic form is given on each fiber E x. Let θ : E Ω Ω C r be a trivialization and let (e 1,.., e r ) be the corresponding frame field for the fiber E Ω. The associated inner product is called Hermitian matrix, h λµ (x) = e λ (x), e µ (x), (23) x Ω. A sesquilinear map C p (M, E ) C q (M, E ) C p+q(m, E ) (s, t) {s, t}. If s = σ λ e λ and t = τ µ e µ, and we have {s, t} = 1 λ,µ r σ λ τ µ eλ, e µ. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

40 Complex Vector Bundle and Complex manifolds [11][12] If E is a complex vector bundle, a positive definite quadratic form is given on each fiber E x. Let θ : E Ω Ω C r be a trivialization and let (e 1,.., e r ) be the corresponding frame field for the fiber E Ω. The associated inner product is called Hermitian matrix, h λµ (x) = e λ (x), e µ (x), (23) x Ω. A sesquilinear map C p (M, E ) C q (M, E ) C p+q(m, E ) (s, t) {s, t}. If s = σ λ e λ and t = τ µ e µ, and we have {s, t} = 1 λ,µ r σ λ τ µ eλ, e µ. The metric tensor on M h C given by G = i,j g ij dz i dz j, G is just a section of E R E. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

41 Complex Vector Bundle and Complex manifolds [11][12] If E is a complex vector bundle, a positive definite quadratic form is given on each fiber E x. Let θ : E Ω Ω C r be a trivialization and let (e 1,.., e r ) be the corresponding frame field for the fiber E Ω. The associated inner product is called Hermitian matrix, h λµ (x) = e λ (x), e µ (x), (23) x Ω. A sesquilinear map C p (M, E ) C q (M, E ) C p+q(m, E ) (s, t) {s, t}. If s = σ λ e λ and t = τ µ e µ, and we have {s, t} = 1 λ,µ r σ λ τ µ eλ, e µ. The metric tensor on M h C given by G = i,j g ij dz i dz j, G is just a section of E R E. I think we are ready to find the generalized solutions... MSI Tech Support (Institute) Slides - Beamer 09/ / 28

42 Invariant Quadratic Form under Linear Transformations[10] Since the quadratic form is invariant under linear transformations P = x x 2 p x 2 p+1... x 2 p+q (24) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

43 Invariant Quadratic Form under Linear Transformations[10] Since the quadratic form is invariant under linear transformations P = x xp 2 xp xp+q 2 (24) we seek the elementary solution E (x) = f (P) L k E = δ(x) (25) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

44 Invariant Quadratic Form under Linear Transformations[10] Since the quadratic form is invariant under linear transformations P = x x 2 p x 2 p+1... x 2 p+q (24) we seek the elementary solution E (x) = f (P) L k E = δ(x) (25) f (P) = (P + i0) λ + (P i0) λ where (P ± i0) λ = lim (P ± iε x x n 2 ) λ are homogeneous ε +0 generalized functions to P. The solutions are in C-vector bundle TM with λ = 1 2 n + k, f (P) is the solution to Lk E = δ(x). MSI Tech Support (Institute) Slides - Beamer 09/ / 28

45 Generalized Functions with Complex Coefficients Let P = i,j g ij x i x j be a quadratic form whose coefficients may be complex wherep 2 is positive definite. P = P 1 + ip 2 (26) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

46 Generalized Functions with Complex Coefficients Let P = i,j g ij x i x j be a quadratic form whose coefficients may be complex wherep 2 is positive definite. P = P 1 + ip 2 (26) The P λ can be written as P λ = exp λ(ln P + i arg P), where 0 < arg P < π,for single valued analytic function of λ. Start with P = i,j g ij x i x j = ip 2 lies on positive imaginary axis g ij = ia ij for i, j = 1,...n. We find singular points of P λ on the upper-half-plane, MSI Tech Support (Institute) Slides - Beamer 09/ / 28

47 Generalized Functions with Complex Coefficients ( ) P λ, φ ( ) λ = e 1 2 πλi g ij x i x j φdx. (27) i,j We can always find a linear transformations x i = n j=1 α ij x j such that i,j g ij x i x j transforms into r 2 = x x n 2, and φ E dual to K-vector bundle and is a test function. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

48 Generalized Functions with Complex Coefficients ( ) P λ, φ ( ) λ = e 1 2 πλi g ij x i x j φdx. (27) i,j We can always find a linear transformations x i = n j=1 α ij x j such that i,j g ij x i x j transforms into r 2 = x x n 2, and φ E dual to K-vector bundle and is a test function. ( ) P λ, φ = e 1 2 πλi 2λ φdx = a e 1 2 πλi ( i) n g r 2λ φdx (28) where g is determinant of coefficients of P. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

49 Generalized Functions with Complex Coefficients res P λ = λ= 1 2 n e 4 1 πqi π n 2 ( i) n δ(x) g Γ(( n (29) 2 )) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

50 Generalized Functions with Complex Coefficients res P λ = λ= 1 2 n e 4 1 πqi π n 2 ( i) n δ(x) g Γ(( n (29) 2 )) To calculate the residues of P λ at others singular points, we consider L P = g ij 2 x i x j (30) defined by n i=1 g ri g is = δ r s where δ r s = 1 for r = s and δ r s = 0 otherwise. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

51 Generalized Functions with Complex coefficients We have L P P λ+1 = 4(λ + 1)(λ + n 2 )Pλ L k P Pλ+k = 4 k (λ + 1)...(λ + k)...(λ + n 2 + k 1)Pλ. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

52 Generalized Functions with Complex coefficients We have L P P λ+1 = 4(λ + 1)(λ + n 2 )Pλ L k P Pλ+k = 4 k (λ + 1)...(λ + k)...(λ + n 2 + k 1)Pλ. Yields P λ = 1 4 k (λ+1)...(λ+k)...(λ+ n 2 +k 1)Lk P Pλ+k. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

53 Generalized Functions with Complex coefficients We have L P P λ+1 = 4(λ + 1)(λ + n 2 )Pλ L k P Pλ+k = 4 k (λ + 1)...(λ + k)...(λ + n 2 + k 1)Pλ. Yields P λ 1 = 4 k (λ+1)...(λ+k)...(λ+ n 2 +k 1)Lk P Pλ+k. With this res P λ 1 = λ= 1 2 n k 4 k (λ+1)...(λ+k)...(λ+ n 2 +k 1) λ= 1 2 n k res L k λ= 2 1 n P Pλ We get res P λ = λ= 1 2 n k e 4 1 πqi π n 2 4 k k! ( i) n g Γ( n 2 + k)lk Pδ(x). (31) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

54 Generalized Functions with Complex coefficients To find residue at other singular points e 1 4 πqi π n 2 ( i) n g Γ(( n 2 ))Lk P res P λ = δ(x). This is equation assume P lies λ= 1 2 n k on the positive imaginary axis. We continue it analytically to the entire upper half-plane. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

55 Generalized Functions with Complex coefficients To find residue at other singular points e 1 4 πqi π n 2 ( i) n g Γ(( n 2 ))Lk P res P λ = δ(x). This is equation assume P lies λ= 1 2 n k on the positive imaginary axis. We continue it analytically to the entire upper half-plane. P = P 1 + ip 2 where P 1 and P 2 are quadratic forms with real coefficients and P 2 is positive definite. a nonsingular transformation x i = n j=1 b ij y j such that P 1 = λ 1 y λ ny 2 n, P 2 = y y 2 n thus we have g = b 2 (λ 1 + i)... (λ n + i) ( i) n g = b 2 (1 iλ 1 )... (1 iλ n ) ( i) n g = b (1 iλ 1 ) (1 iλ n ) 1 2 MSI Tech Support (Institute) Slides - Beamer 09/ / 28

56 Generalized Functions with Complex coefficients P λ is regular analytic function of λ the upper-half-plane except for λ = 1 2n k, k is non-negative integer. We need to calculate the residues at the singular points by MSI Tech Support (Institute) Slides - Beamer 09/ / 28

57 Generalized Functions with Complex coefficients P λ is regular analytic function of λ the upper-half-plane except for λ = 1 2n k, k is non-negative integer. We need to calculate the residues at the singular points by res P λ = λ= 2 1 n k ( ) e 1 4 πqi π n k 2 ( i) n g Γ(( n 2 )) g ij 2 δ(x) (32) i,j=1 x i x j MSI Tech Support (Institute) Slides - Beamer 09/ / 28

58 Generalized Functions with Complex coefficients The residue res (P 1 + i0) λ = lim res (P 1 + ip 2 ) λ where λ= 2 1 n k P 2 0λ= 1 2 n k P 2 = ɛ x x n 2 ɛ > 0, P 1 = P MSI Tech Support (Institute) Slides - Beamer 09/ / 28

59 Generalized Functions with Complex coefficients The residue res (P 1 + i0) λ = lim res (P 1 + ip 2 ) λ where λ= 2 1 n k P 2 0λ= 1 2 n k P 2 = ɛ x x n 2 ɛ > 0, P 1 = P lim ( i) n g = b (ɛ iλ 1 ) (ɛ iλ n ) 1 2 now recall that there ɛ 0 are p of eigenvalues of P 1 are positive and q eigenvalues are negative. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

60 Generalized Functions with Complex coefficients The residue res (P 1 + i0) λ = lim res (P 1 + ip 2 ) λ where λ= 2 1 n k P 2 0λ= 1 2 n k P 2 = ɛ x x n 2 ɛ > 0, P 1 = P lim ( i) n g = b (ɛ iλ 1 ) (ɛ iλ n ) 1 2 now recall that there ɛ 0 are p of eigenvalues of P 1 are positive and q eigenvalues are negative. ( i) n g = λ 1...λ n ( i) p 2 (i) q 2. lim ɛ 0 MSI Tech Support (Institute) Slides - Beamer 09/ / 28

61 Generalized Functions with Complex coefficients The residue res (P 1 + i0) λ = lim res (P 1 + ip 2 ) λ where λ= 2 1 n k P 2 0λ= 1 2 n k P 2 = ɛ x x n 2 ɛ > 0, P 1 = P lim ( i) n g = b (ɛ iλ 1 ) (ɛ iλ n ) 1 2 now recall that there ɛ 0 are p of eigenvalues of P 1 are positive and q eigenvalues are negative. ( i) n g = λ 1...λ n ( i) p 2 (i) q 2. lim ɛ 0 For λ = 1 2n k, we have res (P ± i0) λ = λ= 1 2 n k e 1 2 πqi π n 2 4 k k! λ Γ( n 2 + k) ( n i,j=1 2 x i x j ) k δ(x). (33) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

62 We seek the distributional solutions or generalized functions where L = 2 x x 2 p L k u = f (x) (34) 2... xp+1 2 2, when k = 1, we have xp+q 2 ultra-hyperbolic PDE equation. When either p = 1 or q = 1, we have the wave equation, and either p = 0 or q = 0, we have the Laplace s equation MSI Tech Support (Institute) Slides - Beamer 09/ / 28

63 We seek the distributional solutions or generalized functions where L = 2 x x 2 p L k u = f (x) (34) 2... xp+1 2 2, when k = 1, we have xp+q 2 ultra-hyperbolic PDE equation. When either p = 1 or q = 1, we have the wave equation, and either p = 0 or q = 0, we have the Laplace s equation Since the quadratic form is invariant under any linear transformations P = x x 2 p x 2 p+1... x 2 p+q (35) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

64 We seek the distributional solutions or generalized functions where L = 2 x x 2 p L k u = f (x) (34) 2... xp+1 2 2, when k = 1, we have xp+q 2 ultra-hyperbolic PDE equation. When either p = 1 or q = 1, we have the wave equation, and either p = 0 or q = 0, we have the Laplace s equation Since the quadratic form is invariant under any linear transformations P = x x 2 p x 2 p+1... x 2 p+q (35) We seek the elementary solution E (x) = f (P) to the L k E = δ(x). MSI Tech Support (Institute) Slides - Beamer 09/ / 28

65 We seek the distributional solutions or generalized functions where L = 2 x x 2 p L k u = f (x) (34) 2... xp+1 2 2, when k = 1, we have xp+q 2 ultra-hyperbolic PDE equation. When either p = 1 or q = 1, we have the wave equation, and either p = 0 or q = 0, we have the Laplace s equation Since the quadratic form is invariant under any linear transformations P = x x 2 p x 2 p+1... x 2 p+q (35) We seek the elementary solution E (x) = f (P) to the L k E = δ(x). f (P) = (P + i0) λ + (P i0) λ where (P ± i0) λ = lim (P ± iε x xn 2 ) λ are homogeneous ε +0 generalized functions to P. The solutions are in C-vector bundle TM with λ = 1 2 n + k, f (P) is the solution to Lk E = δ(x). MSI Tech Support (Institute) Slides - Beamer 09/ / 28

66 Distributionsor Generalized Functions L k (P + i0) λ+k = 4 k (λ + 1)...(λ + k)(λ + n 2 )...(λ + n 2 + k 1) = P + i0) λ. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

67 Distributionsor Generalized Functions L k (P + i0) λ+k = 4 k (λ + 1)...(λ + k)(λ + n 2 )...(λ + n 2 + k 1) = P + i0) λ. L k (P + i0) n 2 +k = 4 k (1 n 2 )...(k n 2 )( n 2 + n 2 )...(k 1)! e 1 2 πqi π n 2 Γ(( n 2 )) δ(x). MSI Tech Support (Institute) Slides - Beamer 09/ / 28

68 Distributionsor Generalized Functions L k (P + i0) λ+k = 4 k (λ + 1)...(λ + k)(λ + n 2 )...(λ + n 2 + k 1) = P + i0) λ. L k (P + i0) n 2 +k = 4 k (1 n 2 )...(k n 2 )( n 2 + n 2 )...(k 1)! e 1 2 πqi π n 2 Γ(( n 2 )) L k (P + i0) n 2 +k e 1 2 πqi Γ( n 2 k) 4 k (k 1)!π n 2 ( 1)k = δ(x). δ(x). MSI Tech Support (Institute) Slides - Beamer 09/ / 28

69 Distributionsor Generalized Functions Hence E 1 = ( 1) k e 2 1 πqi Γ( n 2 k) (P + i0) n 4 k (k 1)!π n 2 +k (36) 2 MSI Tech Support (Institute) Slides - Beamer 09/ / 28

70 Distributionsor Generalized Functions Hence Similarly, E 1 = ( 1) k e 2 1 πqi Γ( n 2 k) (P + i0) n 4 k (k 1)!π n 2 +k (36) 2 E 2 = ( 1 k ) e 2 1 πqi Γ( n 2 k) (P i0) n 4 k (k 1)!π n 2 +k. (37) 2 MSI Tech Support (Institute) Slides - Beamer 09/ / 28

71 Distributionsor Generalized Functions Hence Similarly, E 1 = ( 1) k e 2 1 πqi Γ( n 2 k) (P + i0) n 4 k (k 1)!π n 2 +k (36) 2 E 2 = ( 1 k ) e 2 1 πqi Γ( n 2 k) (P i0) n 4 k (k 1)!π n 2 +k. (37) 2 Thus the solution E = E 1 E 2 res (P ± i0) λ = λ= 1 2 n k e 1 2 πqi π n 2 4 k k! λ Γ( n 2 + k) ( n i,j=1 2 x i x j ) k δ(x) (38) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

72 Elementary Solutions The elementary solution to L k u = f (x) MSI Tech Support (Institute) Slides - Beamer 09/ / 28

73 Elementary Solutions The elementary solution to L k u = f (x) = ( 1) k e 1 2 πqi Γ( n 2 k) 4 k (k 1)!π n 2 (P + i0) n 2 +k ( 1) k e 1 2 πqi Γ( n 2 k) 4 k (k 1)!π n 2 (P i0) n 2 +k MSI Tech Support (Institute) Slides - Beamer 09/ / 28

74 Elementary Solutions The elementary solution to L k u = f (x) = ( 1) k e 1 2 πqi Γ( n 2 k) 4 k (k 1)!π n 2 (P + i0) n 2 +k ( 1) k e where L = 2 x x x 2 3 n = 6, p = q = 3,and k = 1. The solutions reduces to 2 x x πqi Γ( n 2 k) 4 k (k 1)!π n 2 2, for x6 2 E = e 3 2 πi 4π 3 (P + i0) 2 + e 3 2 πi (P i0) n 2 +k 4π 3 (P i0) 2. (39) Therefore, our homogeneous generalized functions or distributions are sections of the K 6 -vector bundle. MSI Tech Support (Institute) Slides - Beamer 09/ / 28

75 References: [1] K. Thorne, Closed Timelike Curves. Submitted to Proceeding of the 13 th Internationnal Conference on General Relativity and Gravitation, ed. C Kozameh, (1993) [2] G. R. Dvali, G. Gabadadze, G. Senjanovic, Many Faces of the Superworld: Yuri Golfand Memorial Volume, World Scientific (1999) [2] F. J. Yndurain,Phys. Lett. B 256, 15 (1991) [3] G. Owens, Amer. Math. J. 69, 184 (1947) [4] F. John, Duke Math. J. 4, 300 (1938) [5] R. Courant. Methods of Mathematical Physics, Vol. II: Partial Differential Equations. Interscience, New York, (1962) [6] F. John. Partial Differential Equations. Springer-Verlag, fourth edition,1991. [7] C. Bar, N. Ginoux and F. Pfaffle. Wave Equations on Lorentzian Manifolds and Quantization. European Mathematical Society, MSI Tech Support (Institute) Slides - Beamer 09/ / 28

76 [8] M. Fecko. Differential Geometry and Lie Groups for Physicists, Cambridge University Press, 2006 [9] I. M. Gel fand and G.E. Shilov, Generalized Functions, Volume I, AMS Chelsea Publishing, (1958) [10] J. P. D ly, Complex Analytic and Differential Geometry, Universit e de Grenoble I Institut Fourier, UMR 5582 du CNRS Saint-Martin d H eres, France [11] R. Greene, Complex Differential Geometry, Springer Publishing (2006) MSI Tech Support (Institute) Slides - Beamer 09/ / 28