Apps

## Apps.BrysonDenhamProblem History

Changed line 9 from:

The parameter u is the acceleration can be adjusted over the time horizon from a starting time of zero to a final time of one. The variable x is the position and v is the velocity.

to:

The parameter u (acceleration) is adjusted over the time horizon from a starting time of zero to a final time of one. The variable x is the position and v is the velocity.

#### Solution

(:html:) <!-- <iframe width="560" height="315" src="https://www.youtube.com/embed/pgJ0jbfFBUE" frameborder="0" allowfullscreen></iframe>

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Changed line 9 from:

The parameter u is the acceleration can be adjusted over the time horizon from a starting time of zero to a time of one. The variable x is the position and v is the velocity.

to:

The parameter u is the acceleration can be adjusted over the time horizon from a starting time of zero to a final time of one. The variable x is the position and v is the velocity.

Changed lines 15-23 from:

$$\quad \frac{dx(t)}{dt} = v$$

$$\quad \frac{dv(t)}{dt} = u$$

$$\quad x(0) \; = \; x(1) \; = \; 0$$

$$\quad v(0) \; = \; -v(1) \; = \; 1$$

$$\quad x(t) \le \ell, \ell=\frac{1}{9}$$

to:

$$\frac{dx(t)}{dt} = v(t)$$

$$\frac{dv(t)}{dt} = u(t)$$

$$x(0) \; = \; x(1) \; = \; 0$$

$$v(0) \; = \; -v(1) \; = \; 1$$

$$x(t) \le \ell, \; \ell=\frac{1}{9}$$

(:title Bryson-Denham Problem:) (:keywords Benchmark, Python, nonlinear control, dynamic programming, optimal control:) (:description Minimize the integral of the control input while meeting certain path and final time constraints.:)

The Bryson-Denham optimal control problem is a benchmark test problem for optimal control algorithms.

#### Problem Statement

The parameter u is the acceleration can be adjusted over the time horizon from a starting time of zero to a time of one. The variable x is the position and v is the velocity.

$$\min J = \frac{1}{2} \; \int_0^1 u^2(t) dt$$

$$\mathrm{subject\;to}$$

$$\quad \frac{dx(t)}{dt} = v$$

$$\quad \frac{dv(t)}{dt} = u$$

$$\quad x(0) \; = \; x(1) \; = \; 0$$

$$\quad v(0) \; = \; -v(1) \; = \; 1$$

$$\quad x(t) \le \ell, \ell=\frac{1}{9}$$

(:html:)

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