its x -dependence cancel. We need to solve the large- x equations from which This is a minor modification of the differential equation for the Airy functions.[32] Since we require decaying solutions at our starting point x max for the inwards integrations, we discard the runaway Bi solution for g ( x ). The BC’s at large x = x max are then Despite appearances, g ( x max ) and f ( x max ) still start off with the opposite slopes, as in Eq. (16), since, for t > 0, Ai ( t ) > 0 and Ai $^{′}$( t ) < 0. For calculations, we simplified the displacement between the two potentials by setting x$_{v}$ = 0, giving us one less parameter to worry about. The first wave function we want to compute is the ground state for the massless quark case, m$_{q}$ = 0, starting out with α$_{s}$ = 0 and then adjusting both x$_{d}$ = x$_{s}$ and α$_{s}$ until we get the desired eigenenergy, noted above in Eq. (31). It turns out that to do so is trickier than in the V$_{s}$ -only case, presumably because of the cancellation between V$_{s}$ ( x ) and V$_{v}$ ( x ) in the differential equation for g $^{′}$( x ). Our first iterative search with α$_{s}$ = 0 found a highly excited state with ˜ E = 7 . 975, not the desired 0.7263, for which ψ$_{a}$ ( x ) had six nodes! After some hand-searching on x$_{s}$ , we found we could get a ground state solution, i.e., a ψ$_{a}$ ( x ) with no nodes, looking like the b -quark ψ$_{a}$ ( x ) of Fig. 7. This solution started with x$_{s}$ = 4 . 0 (very large!) and converged, in six iterations, to ˜ E = 5 . 735 (about eight times too large). From this α$_{s}$ = 0 solution we were able to march down, in small steps in x$_{s}$ (still with α$_{s}$ = 0), to x$_{s}$ = 0 . 20, with the energy now smaller, ˜ E = 2 . 779, but still a factor of four too large. (For smaller values of x$_{s}$ the iterative procedure would not converge, with the matrix of Eq. (22) becoming singular.) The wave function ψ$_{a}$ ( x ) for this solution was broader than that for the x$_{s}$ = 4 . 0 case, and the ψ$_{b}$ ( x ) was relatively larger in comparison with ψ$_{a}$ . At this point we began marching the value of α$_{s}$ in small steps up from zero, i.e., we began turning on the Coulomb attraction. As expected this attraction lowers the energy, 23 g ( x max ) = a 1 Ai [2 1 / $^{3}$( ˜ E + x$_{d}$ + ˜ m ) 1 / $^{3}$x max ] , f ( x max ) = a 1 [ 2 ( ˜ E + x$_{d}$ + ˜ m ) 2 ] 1 / 3 Ai ′ [ 2 1 / $^{3}$( ˜ E + x$_{d}$ + ˜ m ) 1 / $^{3}$x max ] . (54) g $^{′}$( x ) − ( ˜ E + x$_{d}$ + ˜ m ) f ( x ) = 0 , f $^{′}$( x ) − 2 x g ( x ) = 0 , (52) g $^{′′}$( x ) − 2( ˜ E + x$_{d}$ + ˜ m ) x g ( x ) = 0 . (53)