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        riddles >> putnam exam (pure math) >> differential control system.
        
(Message started by: towr on Mar 28^th, 2005, 7:32am)

Title: differential control system.
Post by towr on Mar 28^th, 2005, 7:32am

This shouldn't be hard at all, but is nevertheless proving impossible for me..

given
dx/dt = 2x(t)+u(t), x(0)=x_0
Find the optimal function u(t) which minimizes u(t)^2, such that x(1)=0

I keep getting u(t)=0, which is obviously wrong seeing as x gets pushed away from the origin when there's no input.

Title: Re: differential control system.
Post by william wu on Mar 28^th, 2005, 10:25am

on 03/28/05 at 07:32:43, towr wrote:

Find the optimal function u(t) which minimizes u(t)^2, such that x(1)=0

By minimizing u²(t), do you mean minimizing the quadratic integral \int_0^\infty u²(t) dt ? Or do you mean L-infinity norm?

Title: Re: differential control system.
Post by towr on Mar 28^th, 2005, 10:34am

on 03/28/05 at 10:25:38, william wu wrote:

By minimizing u²(t), do you mean minimizing the quadratic integral \int_0^\infty u²(t) dt ? Or do you mean L-infinity norm?

I mean \int_0^1 u²(t) dt

The point is to get from x(0)=x_0 in t=0 to x(1)=0 in t=1, with a little energy as possible.

Title: Re: differential control system.
Post by william wu on Mar 28^th, 2005, 12:26pm

Well, I can solve the problem if x(0) = 0 and x(1) = x_des (steer zero intial condition to nonzero condition). Then the solution is:

u(tau) = (4 x_des) ( e^2(1-tau)) / (e⁴ - 1 ) for 0 <= tau <= 1

Does that help? At first I thought if I can solve it this way, I should be able to use this to solve the reverse problem. Maybe someone else knows how to use it.

I tried just doing a naive translation of the system, y = x - x_e, but this results in a nonlinear differential equation: ydot = 2x + u = 2y + 2x_e + u.

Title: Re: differential control system.
Post by towr on Mar 28^th, 2005, 12:49pm

Nope, the reverse course doesn't help. Because x is pushed from the origin anyway.

It has soemthing to do with the hamiltonian
H(x,p,u) = p(2x + u) +1/2 (u^2)
which needs to be minimized
So
2px + pu +1/2 (u^2)
=
2px + 1/2 (u + p)^2 - 1/2p^2
Which is minimized wrt u if u=-p

-- the rest is speculation, and may be way off --
So we're left to minimize 2px-1/2p^2 wrt p and x
Now according to the book
I get
dp/dt = -2p, p(1)=x(1)=0 (?)
dx/dt = 2x - p, x(0)=1

but that would make p(t) = C*e^-2t, but C must be 0 if it's ever to reach p(1)=0. But this also makes p(t)=0 and thus u(t)=0. Which is proposterous..

Title: Re: differential control system.
Post by william wu on Mar 28^th, 2005, 1:07pm

just ignored your most recent post for the time being ::)

Is there a way to determine what the minimum energy of the optimal input should be, without determining the input yet?

Stubbornly sticking with my reverse strategy, here's an input that works, in the sense that it meets the target:

u(tau) = - 2x_e + 4(-x_e)*(e^{2(1 - tau)} )/(e⁴ - 1)

(After translating the system to the origin, I designed an input that cancels the nonlinearity to get a nonlinear system, then applied the solution for the reverse problem. Whether it is minimum energy though, I don't know. Probably not.)

Attached an image in case the curve shape tells you anything:

Title: Re: differential control system.
Post by towr on Mar 28^th, 2005, 1:58pm

I think
p = c exp(-2t)
x = 1/4 c exp(-2t) + a exp(2t)
with
a = 1-1/4 c
c = 1/ (1-exp(-4))
might give the solution

But I'm too tired to check it now.. (also still, u=-p)

Title: Re: differential control system.
Post by towr on Mar 29^th, 2005, 4:39am

Hopefully the full solution.

problem description
dx/dt = 2x(t)+u(t), x(0)=x_0, x(1)=0
minimizing \int_0^1 u(t)^2 dt

The Hamiltonian
(note that p functions as langrange multiplier)

H(x,p,u) = p(2x + u) +1/2 (u^2)
= 2px + 1/2 (u + p)^2 - 1/2p^2
=> u = -p

From H by construction
dx/dt = dH/dp = 2x + u = 2x - p
dp/dt = -dH/dx = -2p

solving the new differential equations
dp/dt = -dH/dx = -2p
=>
p = c*exp(-2t)

dx/dt = 2x - p
=>
x = k(t) exp(2t)

using variation of constants
dx/dt = k'(t) exp(2t) + 2 k(t) exp(2t)
dx/dt = 2x - p = 2 k(t) exp(2t) - c*exp(-2t) {from before}
k'(t) exp(2t) = - c*exp(-2t)
k'(t) = - c*exp(-4t)
k(t) = 1/4 c*exp(-4t) + k

x(t) = k(t) exp(2t) = 1/4 c*exp(-2t) + k exp(2t)

Using the boundary conditions
x(0) = x_0 = 1/4 c + k
=> k = x_0 - 1/4 c

x(1) = 0 = 1/4 c*exp(-2) + k exp(2)
=> k = - 1/4 c*exp(-4)
x_0 - 1/4 c = - 1/4 c*exp(-4)
=> c = 4 x_0 / (1- exp(-4))

k= - exp(-4) x_0 / (1- exp(-4)) = x_0 / (1 - exp(4))

(barring mistakes) The solution
x_opt(t) = x_0 * [exp(-2t) / (1 - exp(-4)) + exp(2t) / (1 - exp(4)) ]
p_opt(t) = x_0 * exp(-2t) / (1 - exp(-4))
u_opt(t) = - x_0 * exp(-2t) / (1 - exp(-4))

Follow up question:
Given matrices A and B, assuming the system (A,B) is controllable, solve

dx/dt = Ax(t)+Bu(t), x(0)=x_0, x(1)=0
minimizing \int_0^1 u^T(t)u(t) dt = \int_0^1 ||u(t)||^2 dt

Where x \in R^n and u \in R^m (and the matrices have the appropriate dimensions as well)

[Which is the real problem I was struggling with last night..]