Optimal control problem (constant magnitude acceleration)
$begingroup$
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?
functional-analysis control-theory optimal-control
$endgroup$
add a comment |
$begingroup$
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?
functional-analysis control-theory optimal-control
$endgroup$
1
$begingroup$
I believe you can relax the control to be in the convex hull of the unit ball and since you are looking for a $min$ time solution, the solution will end up on the boundary ae. (cf. bang bang control). It is a fairly standard problem, but I am not sure it has a common name other than steering.
$endgroup$
– copper.hat
Jun 20 '18 at 2:22
$begingroup$
Does your problem involve angular motions (yaw) too? The it is not a linear problem.
$endgroup$
– Arash
Jun 21 '18 at 4:49
$begingroup$
It reminds me my own NLP. But it is not exactly the same a your problem. Your problem has a terminal constraint too.
$endgroup$
– Arash
Jun 21 '18 at 4:52
add a comment |
$begingroup$
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?
functional-analysis control-theory optimal-control
$endgroup$
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?
functional-analysis control-theory optimal-control
functional-analysis control-theory optimal-control
asked Jun 20 '18 at 2:03
Alexander YoungAlexander Young
1
1
1
$begingroup$
I believe you can relax the control to be in the convex hull of the unit ball and since you are looking for a $min$ time solution, the solution will end up on the boundary ae. (cf. bang bang control). It is a fairly standard problem, but I am not sure it has a common name other than steering.
$endgroup$
– copper.hat
Jun 20 '18 at 2:22
$begingroup$
Does your problem involve angular motions (yaw) too? The it is not a linear problem.
$endgroup$
– Arash
Jun 21 '18 at 4:49
$begingroup$
It reminds me my own NLP. But it is not exactly the same a your problem. Your problem has a terminal constraint too.
$endgroup$
– Arash
Jun 21 '18 at 4:52
add a comment |
1
$begingroup$
I believe you can relax the control to be in the convex hull of the unit ball and since you are looking for a $min$ time solution, the solution will end up on the boundary ae. (cf. bang bang control). It is a fairly standard problem, but I am not sure it has a common name other than steering.
$endgroup$
– copper.hat
Jun 20 '18 at 2:22
$begingroup$
Does your problem involve angular motions (yaw) too? The it is not a linear problem.
$endgroup$
– Arash
Jun 21 '18 at 4:49
$begingroup$
It reminds me my own NLP. But it is not exactly the same a your problem. Your problem has a terminal constraint too.
$endgroup$
– Arash
Jun 21 '18 at 4:52
1
1
$begingroup$
I believe you can relax the control to be in the convex hull of the unit ball and since you are looking for a $min$ time solution, the solution will end up on the boundary ae. (cf. bang bang control). It is a fairly standard problem, but I am not sure it has a common name other than steering.
$endgroup$
– copper.hat
Jun 20 '18 at 2:22
$begingroup$
I believe you can relax the control to be in the convex hull of the unit ball and since you are looking for a $min$ time solution, the solution will end up on the boundary ae. (cf. bang bang control). It is a fairly standard problem, but I am not sure it has a common name other than steering.
$endgroup$
– copper.hat
Jun 20 '18 at 2:22
$begingroup$
Does your problem involve angular motions (yaw) too? The it is not a linear problem.
$endgroup$
– Arash
Jun 21 '18 at 4:49
$begingroup$
Does your problem involve angular motions (yaw) too? The it is not a linear problem.
$endgroup$
– Arash
Jun 21 '18 at 4:49
$begingroup$
It reminds me my own NLP. But it is not exactly the same a your problem. Your problem has a terminal constraint too.
$endgroup$
– Arash
Jun 21 '18 at 4:52
$begingroup$
It reminds me my own NLP. But it is not exactly the same a your problem. Your problem has a terminal constraint too.
$endgroup$
– Arash
Jun 21 '18 at 4:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $t_f$
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $t_f$
$endgroup$
add a comment |
$begingroup$
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 $t_f$
$endgroup$
add a comment |
$begingroup$
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 $t_f$
$endgroup$
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 $t_f$
edited Dec 20 '18 at 23:17
answered Dec 13 '18 at 11:29
CesareoCesareo
8,6693516
8,6693516
add a comment |
add a comment |
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I believe you can relax the control to be in the convex hull of the unit ball and since you are looking for a $min$ time solution, the solution will end up on the boundary ae. (cf. bang bang control). It is a fairly standard problem, but I am not sure it has a common name other than steering.
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– copper.hat
Jun 20 '18 at 2:22
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Does your problem involve angular motions (yaw) too? The it is not a linear problem.
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– Arash
Jun 21 '18 at 4:49
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It reminds me my own NLP. But it is not exactly the same a your problem. Your problem has a terminal constraint too.
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– Arash
Jun 21 '18 at 4:52