The Klein-Gordon equation in anti-de Sitter spacetime - Seminario ...

288 K. Yagdjian and A. GalstianOn the other hand, since the function E(z,t;0,b) solves the Klein–Gordon equation,the last integral is equal to∫ e t −10∫ e t −1=0[dz ϕ 0 (x+z)+ϕ 0 (x−z)[dz ϕ 0 (x+z)+ϕ 0 (x−z)Application of (25) and (26) gives]∫ ln(e t −z)0( ∂) 2E(z,t;0,b)dbdb∂b∂b E(z,t;0,ln(et − z))− ∂ ]∂b E(z,t;0,0) .][ ∂[ ]14e − 2 t −116(1+4M 2 )e −2t e 2 t 1 z √e t − z + ∂∂b E(z,t;0,ln(et − z))− ∂ ∂b E(z,t;0,0)]=[14 e− 2 t − 1 16 (1+4M2 )e −2t e 2 t 1z √e t − z + 1 + z(1+4M 2 ))16 e−3t/2(−4et √e t − z− ( 4e t) iM((1+e t ) 2 − z 2) −iM 12[(e t − 1) 2 − z 2 ] √ (1+e t ) 2 − z 2{× (2iM−1) ( e 2t − 1−z 2) F(− 1 2 + iM, 1 2 + ) 2 − z 2 )iM,1,(−1+et(1+e t ) 2 − z 2−2 ( 1−e t + iM ( e 2t − 1−z 2)) (F 12+ iM, 1 2 + ) 2 − z 2 ) }iM,1,(−1+et(1+e t ) 2 .− z 2The terms on the line after the last equality all cancel out, leaving the last three linesthat add up to K 0 (z,t). This completes the proof of Theorem 4.5. The n-dimensional case, n≥2Proof of Theorem 5. Let us consider the case of x∈R n , where first n=2m+1, m∈N.First, for the given function u=u(x,t), we define the spherical means of u about thepoint x:I u (x,r,t) =∫1ω n−1S n−1 u(x+ry,t)dS y,where ω n−1 denotes the area of the unit sphere S n−1 ⊂R n . Then we define an operatorΩ r by( 1 ∂) m−1rΩ r (u)(x,t) :=2m−1 I u (x,r,t).r ∂rOne can show that there are constants c (n)j , j = 0,...,m−1, where n=2m+1, withc (n)0= 1·3·5···(n−2), such that( 1r∂∂r) m−1r 2m−1 ϕ(r)=rm−1∑j=0c (n)j r j ∂ j∂r j ϕ(r).