An argument similar to the above one, but interchanging the indices 1 and 2 shows that qj;z contains no price equilibrium.
In the third case, (Pi, pn belongs to the domain In qj;3, S(P1' pz) is given by (Pz - P1 + 2ad + e)/(4ad + 2e) so that
Some simple calculations then show that pi = pi = 2ad + e while P1(Pi, pn = Pz(Pi, pn = ad + el2. Given pi, we now study the best reply P1 of firm 1 in
respectively. Consider first .s41. The maximum of P1S(P1> pn over [O,oo[ is reached at P1 = 2ad + ae + e/2. Then we have S(fj1, pn > ! - a, from which it follows that there is no best reply against pi in .s41• Let us now come to .s4z. It is easy to check that the maximum of P1S(P1> pn over [O,oo[ is reached at p~ = 2ad + el2 - ae. For e/d ~ 4al(1-2a), s(p~, pn ~ 1 so that P1 = p~. On the other hand, for eld > 4a/(1-2a), s(p~, pi) > 1 which implies that p~ is not the best reply of firm 1 against pi. Actually, in this case, P1 corresponds to the price for which 1 - s(p1> pn = 0, i.e., P1 = e(l - 2a).