Well, it is indeed a differential equation system (velocity is the time derivative of position, acceleration is the time derivative of velocity), but in one dimension we can certainly look at it as an algebraic system … almost. It’s probably much more efficient to look at it as differential equations.
The original statement didn’t include the minimum-time constraint, but that one certainly simplifies the problem, so let’s have at it it one dimension. Relative to the frame of the initial state of the pursuing ship, the pursued ship starts Pi away from the pursuer, moving at velocity Vi and accelerating at Ai. The maximum acceleration of the pursuer is A, and the pursuer always accelerates as much as it can. We see intutitively that the pursuer should always accelerate.
There are five unkown quantities:
Pr (the position at the time of interception)
Vr (the velocity at the time of interception)
T1 (the initial acceleration time of the pursuer)
T2 (the final deceleration time of the pursuer)
T (the total time)
And let’s start writing equations:
The pursued ship arrives at Pr after time T:
Pr = Pi + ViT + Ai(T)^2/2
The pursuer arrives at Pr after time T:
Pr = A*(T1^)^2/2 - A*(T2)^2/2
At interception, the pursued ship and the pursuer have the same velocity:
Vr = Vi + Ai*T (the pursued ship)
Vr = AT1 - AT2 (the pursuer)
The pursuer is always accelerating:
T1 + T2 = T
So we have five equations in five unknowns, and there may be a solution (since three of the variables apppear non-linearly, a solution is not guaranteed, and there may be spurious solutions). Note that we don't know the *sign[/] of A *a priori*, although we can figure it out from the sign of Pi. I don't have MathCAD here, or I'd take a shot at a solution. Unfortunately, there doesn't appear to be a linear subset of the equations.
OK, three dimensions. We have fifteen unknowns (Prx, Pry, Prz, Vrx, Vry, Vrz, T1, T2, and T plus the three initial pursuer accelerations A1x, A1y, and A2z and the three final pursuer accelerations A2x, A2y, and A2z). I don’t think that we can assume that A1x = -A2x and so on. We have fifteen equations:
Prx = Pix + VixT + Aix(T)^2/2
Pry = Piy + ViyT + Aiy(T)^2/2
Prz = Piz + VizT + Aiz(T)^2/2
Prx = A1x*(T1^)^2/2 + A2x*(T2)^2/2
Pry = A1y*(T1^)^2/2 + A2y*(T2)^2/2
Prz = A1z*(T1^)^2/2 + A2z*(T2)^2/2
Vrx = Vix + AixT
Vry = Viy + AiyT
Vrz = Viz + Aiz*T
Vrx = A1xT1 + A2xT2
Vry = A1yT1 + A2yT2
Vrz = A1zT1 + A2zT2
(these are just the 3D vector versions of the first four equations of the one dimensional case)
T1 + T2 = T
sqrt(A1x^2 + A1y^2 + A1z^2) = A (the maximum possible)
sqrt(A2x^2 + A2y^2 + A2z^2) = A
(the last two are just 3D Pythagoras), so it looks conceivably solvable … I’d prefer not to do it! I’d probably try to do it numerically, but I don’t see immediately how.
jrf