- #1
Number2Pencil
- 208
- 1
Homework Statement
Find the optimal trajectory x*(t) that minimizes:
[tex]
J = \int_{0}^{1} \left( \frac{\dot{x}(t)^2}{2} + 3x(t) \dot{x}(t) + 2x^2(t) + 4x(t) \right) dt
[/tex]
with x(0) = 1 and x(1) = 4
Homework Equations
Euler's equation:
[tex]
\frac{\partial g}{\partial x} - \frac{d}{dt} \left[ \frac{\partial g}{\partial \dot{x}} \right] = 0
[/tex]
where g is the portion inside the integral.
The Attempt at a Solution
Implementing Euler's equation on g:
[tex]
-\ddot{x}(t) + 4x(t) + 4 = 0
[/tex]
In which the question becomes, how do I solve this differential equation? In other classes, I would take the laplace transform, solve for X(s), and then do an inverse laplace transform. The problem is that I need to know the initial condition of x'(0) to do this, which I wasn't provided.
I looked at my notes from class, and in a similar example problem, my professor wound up with the following diff-eq:
[tex]
\ddddot{x}(t) = 16x(t)
[/tex]
He said he took the laplace transform and wound up with:
[tex]
s^4 - 16 = 0
[/tex]
But where did the X(s) go, and why didn't he use the initial conditions? His next step was to solve for s:
[tex]
s = \pm 2, \pm j2
[/tex]
Then he did some kind of generic form of the answer:
[tex]
x^*(t) = A_1 e^{2t} + A_2 e^{-2t} + A_3 e^{j2t} + A_4 e^{-j2t}
[/tex]
at this point, he created 4 simultaneous equations using the original problem's provided points to solve for A1,A2,A3, and A4. So when I try to do the same steps on my problem:
[tex]
-\ddot{x}(t) + 4x(t) + 4 = 0
[/tex]
I take the laplace, ignoring initial conditions:
[tex]
-s^2X(s) + 4X(s) + \frac{4}{s} = 0
[/tex]
But right here, can I really just "drop" the X(s) term and solve for s? Is that mathematically legal?
If I just solve for X(s) as I normally would and then do an inverse laplace, the answer does not satisfy the bounded points. ...and trying to do the laplace transform without the initial x'(0) point is confusing as well.
How do I solve this?