Ytukyg

A particle in $mathbb R^2$ begins at initial position $(x_0, y_0)$ and velocity $(u_0, v_0)$. It must eventually reach a target position $(x_1, y_1)$ and velocity $(u_1, v_1)$.

The acceleration of the particle is a vector of constant magnitude $1$. The only control for this system is the direction of this acceleration, as a (not necessarily continuous) function of time.

What path will bring the particle to the target position and velocity in the least amount of time?

I believe I already have a solution to this problem, but what I'm wondering is: does this problem already have a name? Does the curve? What work has already been done to solve and generalize this problem?

After considering $x = x_1, y = x_3$, given the dynamics
$$
begin{array}{rcl}
dot x_1 & = & x_2\
dot x_2 & = & U_0cos u\
dot x_3 & = & x_4\
dot x_4 & = & U_0sin u
end{array}
$$

with $|u| = U_0$ we have

$$
H(x,u,lambda) = lambda_1 x_2+lambda_2 U_0cos u +lambda_3 x_4+lambda_4 U_0sin u
$$

from which we obtain the adjoint dynamics

$$
left{
begin{array}{rcl}
dotlambda_1 & = & 0\
dotlambda_2 & = & -lambda_1\
dotlambda_3 & = & 0\
dotlambda_4 & = & -lambda_3
end{array}
right.
Rightarrow
left{
begin{array}{rcl}
lambda_1 & = & c_1\
lambda_2 & = & -c_1t+c_2\
lambda_3 & = & c_3\
lambda_4 & = & -c_3t+c_4
end{array}
right.
$$

and

$$
frac{partial H}{partial u} = -lambda_2 sin u +lambda_4 cos u = 0
$$

or

$$
sin u = frac{pmlambda_2}{sqrt{lambda_2^2+lambda_4^2}}, cos u = frac{pmlambda_4}{sqrt{lambda_2^2+lambda_4^2}}
$$

The right sign is found by applying the maximum principle then

$$
sin u^* = frac{lambda_2}{sqrt{lambda_2^2+lambda_4^2}}, cos u^* = frac{lambda_4}{sqrt{lambda_2^2+lambda_4^2}}
$$

now according to the transversality conditions

$$
dt_f-sum_{i=1}^4 lambda_i f_i vert_{t_i}^{t_f}dt_f+sum_{i=1}^4 lambda_i dx_i vert_{t_i}^{t_f}dt_f = 0
$$

but at $t = t_i$

$$
begin{array}{rcl}
x_1 & = & x_0\
x_2 & = & u_0\
x_3 & = & y_0\
x_4 & = & v_0\
end{array}
$$

and at $t = t_f$

$$
begin{array}{rcl}
x_1 & = & x_1\
x_2 & = & u_1\
x_3 & = & y_1\
x_4 & = & v_1\
end{array}
$$

so a free terminal time requires

$$
1-sum_{i=1}^4 lambda_i f_i vert_{t_i}^{t_f}=0
$$

the movement reads now

$$
begin{array}{rcl}
dot x_1 & = & x_2\
dot x_2 & = & frac{(c_4-c_3 t)U_0}{sqrt{(c_2-c_1 t)^2+(c_4-c_3 t)^2}}\
dot x_3 & = & x_4\
dot x_4 & = & frac{(c_2-c_1 t)U_0}{sqrt{(c_2-c_1 t)^2+(c_4-c_3 t)^2}}
end{array}
$$

Now resuming, the movement equations and the adjoint movement equations require $8$ constants plus $t_f$ definition. In theory then can be solved because we have $8$ initial-final conditions plus the transversality condition involving $t_f$