A. Negative region In the study of the discrete spectrum it is convenient to use a different variable. Now we choose y − 1 = 1 + e − a ( x + L $^{)}$, with d/dx = ay (1 − y ) d/dy , V ( y ) = − W y and m ( y ) = m$_{0}$ (1 − y ) and using the parametric transformation φ = y $^{σ}$(1 − y ) $^{ϵ}$h ( y ), we obtain from Eq. (7) The most general condition that we can impose on the term multiplying 1 / [ y (1 − y )] in order that the equation be that of the hypergeometric function is that it be equal to a constant ζ . Therefore we get three equations: From Eq. (46a) we can solve for σ , while summing Eq. (46b) and Eq. (46c) we obtain the equation for ϵ So σ = ν . A solution of the second one is ϵ = − i ( E + W ) /a , which is the same of µ in the scattering problem but with the replacement W → − W . In determining the parameters a and b ( c = 1 + 2 ν ) of the hypergeomtric equation we make use of the relation in Eq. (46c), where we define λ = i $^{√}$( W 2 − m 2 $_{0}$) /a 2 and finally obtain: 16 y (1 − y ) d $^{2}$f ( y ) dy 2 + [1 + 2 ν − (1 + 2 ϵ + 2 ν ) y ] d f ( y ) dy 2 − ( ϵ + ν + λ )( ϵ + ν − λ ) f ( y ) = 0 . (48) ( E + W ) 2 a 2 − i W + E a − ϵ + ϵ 2 = 0 (47b) σ = $^{√}$m 2 0 − E 2 a (47a) σ 2 + ϵ 2 − ϵ + W 2 a 2 − m 2 0 a 2 = − ζ (46c) 2 m 2 0 a 2 + 2 EW a 2 − i W + E a − 2 σ 2 = ζ (46b) σ 2 + E 2 a 2 − m 2 0 a 2 = 0 (46a) y (1 − y ) d $^{2}$h ( y ) dy 2 + [1 + 2 σ − y (1 + 2 σ + 2 ϵ )] d h ( y ) dy 2 + + 1 a $^{2}$y (1 − y ) [( E + W y ) 2 − m 2 $_{0}$(1 − y ) 2 − iaW y (1 − y ) − iay ( E + W y )] h ( y ) + 1 y (1 − y ) [ σ ( σ − 1)(1 − y ) 2 + ϵ ( ϵ − 1) y 2 + σ (1 − y ) $^{2}$] h ( y ) + ( − 2 ϵσ − ϵ ) h ( y ) = 0 (45)