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

CHAPTER 8. COMPUTATIONS 104see what comes out anyway. Concentrating on the ∼ ∂ 2 ω term, we see we can put thisterm in the right form by applying the substitutionThe second order term hereby becomesω −→ −z ( 1 − C 2) + 1. (8.41)4 ( ω − C 2) (ω − 1) ( ∂ 2 ωf ) −→ −4 (1 − z)z. (8.42)The only thing left to do is multiply through by −4 −1 . Going through the same motionsfor the other terms, the fully transformed equation becomes (courtesy <strong>of</strong> Mathematica):0 ={ [2 − ( −5C 2 + 5 + kE 2) z + ( C 2 − 1 ) ( 6C 2 − 3 − 2kE 2) z 2 − ( C 2 − 1 ) 2kE 2 z 3+ L 2 (z − 1) ( 1 + ( C 2 − 1 ) z ) ] [2× 4z ( 1 + ( C 2 − 1 ) z ) ] } −12f−{ [− 2 + 6C 2 + 2z − 6C 2 z + 4C 4 z − C 2 ( 1 + ( C 2 − 1 ) z ) 3/2+ ( 1 + ( C 2 − 1 ) z ) ] [5/2× 2 ( C 2 − 1 )( 1 + ( C 2 − 1 ) z ) ] } −1∂ z f+ z (1 − z)∂ 2 zf. (8.43)This is decidely not a hypergeometric equation.