Hello. We have a potential V= (x-3) / (x^2 + 16) And the question is if a particle is released at x=2 with initial velocity u find the values of u such that x(t) -> minus infinity as t-> infinity.
My attempt: I sketched the graph first of all. Then I was thinking that if we have a negative initial velocity then the particle moves to the left and to the right if we have a positive initial velocity. Then I said that if initial velocity is positive then 0<=E<1/16 (where 1/6 is the maximum turning point of V and E is the total energy) so the particle will move to the right reach the left turning point of V=E and then returns back so x(t) -> minus infinity as t->infinity as we wanted. Now if initial velocity is negative then we must have E>=0. Then we solve the inequalities w.r.t the speed v (which is always positive) since at x=2 E=0.5*v^2+V(2) and deduce from there the values of velocity (which can be negative or positive).
Am I thinking too complicated? Is there an easier way? My attempt was based on the sketch of V(x), which I couldn't upload, but the sketch is easy. [maximum at (8, 1/16) minimum at (-2, -1/4), crosses the axes at (3,0) and ( 0, -3/16), and also as x->infinity V-> 0 from above and as x-> minus infinity V->0 from below.]
Any help and guidance will be greatly appreciated. Thank you very much!