Case I (w=0).
Here are considered Figure 4 and the formulae of potential energy v and its derivatives. Substituting w=0 in the formula (33) we have

It can be noticed that minimal value of the force p amounts to p_k0=5 and it occurs with a=0.
- If pk0, then the function v has the only extremum for a=0 and it is its minimum ( dv / da = 0 , for a=0).
- If p=p_k0, then the function v also has its minimum for a=0, but it is more flat in this point (for a=0 all differential coefficients of the function v with respect to a up to the third degree inclusive are equal to zero, while ).
- In case when p>p_k0, then the derivative when a >0 has already two zero points:
one for a=0, the other for a=am defined by the formula

It can be checked that for a=0 the derivative , so in this point there is the maximum of potential energy v. On the other hand, for a=am the derivative , so there is the minimum of potential energy v.

To conclude, for p< p_k0 there exists only the rectilinear form of equilibrium, that is the stable position is only for a=0.
However if p>p_k0, there are two positions of equilibrium. First one, for a=0 is unstable, whereas the other, for am defined by the formula (36) is the position of stable equilibrium.

Case II (w>0)
Like above, here are considered the formulae of potential energy v and its derivatives.

Basing on Figure 4 it can be seen that the minimal value of the force p, which for further consideration will be marked as pk, is greater than it was in the previous situation for w=0 (pk>pk0=5).
It can be also seen that the minimum occurs with a>0. Let this point be designated as ak like in Figure 5. To determine the values pk and ak the minimum of the function p given by the formula (34) must be found. After appropriate transformation, we obtain