DBI Analysis of Open String Bound States on Non-compact D-branes

CHAPTER 7. A FIRST EXAMPLE 947.2.4 Perturbed equations <strong>of</strong> motionWe again proceed to computing the equations <strong>of</strong> motion for the perturbation δψ,∂L0 = ∂ µ∂ (∂ µ δψ) = ∂ ∂Lt∂ (∂ t δψ) + ∂ ∂Lr∂ (∂ r δψ)⎡ √ ⎤ [= ∂ t⎣ −kr2 1 − C2( ) 3/2]r 2r 2 (∂ t δψ) ⎦ + ∂ r r 2 1 − C2− 1r 2 (∂ r δψ)= − kr (∂2r 2 − 1 t δψ ) )+(2 + C2r 2 (∂ r δψ) + r(1 − C2r 2 ) (∂2r δψ ) . (7.52)Filling in δψ = X (r)e iEt , this finally give us the following differential equation:)0 = r(1 − C2 (∂2r 2 r X ) )+(2 + C2r 2 (∂ r X) + krE2r 2 X. (7.53)− 17.2.5 Solutions <strong>of</strong> the PDETo transform Eq. 7.53 into a hypergeometric equation, the needed change <strong>of</strong> variablesare almost the same as for the cigar, namelyThe first substition amounts toThe second substitution results inr 2 = w, (7.54)w = ( C 2 − 1 ) z + 1. (7.55)0 = kE2(w − 1) X + 6∂ wX + 4 ( w − C 2) ∂ 2 wX. (7.56)(z − 1)0 = 4(C 2 − 1) ∂2 zX +6(C 2 − 1) ∂ zX +Multiplying through with −z ( C 2 − 1 ) /4 we getThe coefficients are still given bya = 1 (1 + √ )1 + 4kE42resulting in the same general solution,kE 2((C 2 X. (7.57)− 1)z)0 = z(1 − z)∂zX 2 − 3 2 z∂ zX − kE2 X. (7.58)4; b = 1 4(1 − √ 1 + 4kE 2 ); c = 0,u = z m 2F 1 (a + m, b + m; 1 + m; z)( 1= z 2 F 1(5 + √ )1 + 4kE42 , 1 (5 − √ ) )1 + 4kE42 ; 2;z . (7.59)