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

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286 K. Yagdjian and A. GalstianProof. We set u(x,t)=w(x,t)+ϕ 0 (x), thenw tt − e 2t w xx + M 2 w=e 2t ϕ (2)0 (x)−M2 ϕ 0 (x), w(x,0)=0, w t (x,0)=0.Next we plug f(x,t)= e 2t ϕ (2)0 (x)−M2 ϕ 0 (x) into the formula given by Theorem 3 andobtainw(x,t) = ˜w(x,t)−∫ t ∫ x+e t −e b(30)db dyM 2 ϕ 0 (y)E(x−y,t;0,b),0 x−(e t −e b )where we have denoted˜w(x,t) :=∫ t0∫ x+e te 2b −e bdb dyϕ (2)x−(e t −e b 0 (y)E(x−y,t;0,b).)Next we integrate by parts and apply (19):∫ t (˜w(x,t) = e 2b db ϕ (1)0 (x+et − e b )E(−e t + e b ,t;0,b)0)−ϕ (1)0 (x−et + e b )E(e t − e b ,t;0,b)∫ t ∫ x+e t− e 2b −e bdb dyϕ (1)0 x−(e t −e b 0 (y) ∂ ) ∂y E(x−y,t;0,b).On the other hand,ϕ (1)0 (x+et − e b )=−e −b ∂∂b ϕ 0(x+e t − e b ), ϕ (1)0 (x−et + e b )=e −b ∂ ∂b ϕ 0(x−e t + e b )implies that˜w(x,t) =∫ t(e b db − ∂0 ∂b ϕ 0(x+e t − e b )E(−e t + e b ,t;0,b)− ∂)∂b ϕ 0(x−e t + e b )E(e t − e b ,t;0,b)∫ t ∫ x+e t− e 2b −e bdb dyϕ (1)0 x−(e t −e b 0 (y) ∂ ) ∂y E(x−y,t;0,b).One more integration by parts leads to()˜w(x,t)+ϕ 0 (x) = 1 2 e− 2t ϕ 0 (x+e t − 1)+ϕ 0 (x−e t + 1)∫ t(+ db ϕ 0 (x+e t − e b ) ∂0∂b(e b E(−e t + e b ,t;0,b))( ) )e b E(e t − e b ,t;0,b)+ϕ 0 (x−e t + e b ) ∂∂b∫ t ∫ x+e t− e 2b −e bdb dyϕ (1)0 x−(e t −e b 0 (y) ∂ ) ∂y E(x−y,t;0,b),
Klein-Gordon equation in anti-de Sitter spacetime 287where E(0,t;0,t)=e −t /2, and E(e t − 1,t;0,0)= E(1−e t ,t;0,0)=e − 2/2 t have beenused. Next we apply (21) of Proposition 1 and an integration by parts to obtain()˜w(x,t)+ϕ 0 (x) = 1 2 e− 2t ϕ 0 (x+e t − 1)+ϕ 0 (x−e t + 1)∫ t ()+ db 1 4 e 2 b e − 2t ϕ 0 (x+e t − e b )+ϕ 0 (x−e t + e b )−+0∫ t0∫ t0e 2b db[ϕ 0 (y) ∂ ] y=x+e t∂y E(x−y,t;0,b) −e b∫ x+e te 2b −e b ( ∂db dyϕ 0 (y)x−(e t −e b ) ∂yFrom (23) and (24) of Proposition 1, we have[]˜w(x,t)+ϕ 0 (x) = 1 2 e− 2t ϕ 0 (x+e t − 1)+ϕ 0 (x−e t + 1)∫ t+ db 1 4 e b 2 e− 2t0∫ t− e 2b db 1 16 (1+4M2 )e −2(b+t) e b/2 e t/2 (e t − e b )0y=x−(e t −e b )) 2E(x−y,t;0,b).[ϕ 0 (x+e t − e b )+ϕ 0 (x−e t + e b )[]ϕ 0 (x+e t − e b )+ϕ 0 (x−(e t − e b ))∫ t ∫ x+e t+ e 2b −e b ( ∂) 2E(x−y,t;0,b),db dy ϕ 0 (y)0 x−(e t −e b ) ∂yThen the last equation together with (30) proves the desired representation. The propositionis proven.Completion of the proof of Theorem 4. We make the change z=e t − e b , dz=−e b db,and b=ln(e t −z) in the second term of the representation given by the previous proposition:[14e 2 b e − 2 t +16 1 (1+4M2 )e −2t e b 2 e 2(e t b − e )]t=∫ t0∫ e t −10[]× ϕ 0 (x+e t − e b )+ϕ 0 (x−e t + e b ) db[ ]14e − 2 t −116(1+4M 2 )e −2t e 2 t 1[ ]z √ ϕ e t 0 (x−z)+ϕ 0 (x+z) dz.− zNext we apply (19) to the last term of that representation, and then we change the orderof integration:∫ t∫ x+e t −e bdb0 x−(e t −e b )∫ e t −1 [=0dz[ ( ∂ 2 ]dyϕ 0 (y) e∂y) 2b E(x−y,t;0,b)−M 2 E(x−y,t;0,b)ϕ 0 (x+z)+ϕ 0 (x−z)]∫ ln(e t −z)0db[ ( ∂ 2 ]× e∂z) 2b E(z,t;0,b)−M 2 E(z,t;0,b) .]
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