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

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%width=350px%Attach:bryson_denham_solution.png

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.

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

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

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

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

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

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

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

----

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