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Solve Riccati (DARE) with MATLAB

  1. Sep 14, 2016 #1

    perplexabot

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    Gold Member

    Hey all, I was wondering if you can help me define the arguments for the dare() function of matlab. Here is the DARE form I have:
    [tex]
    \begin{equation}
    \begin{split}
    \Phi(p_\infty) &= \Phi_\infty = \Phi_{33}-\Phi_{32}(\Phi_\infty+\Phi_{R})^{-1}\Phi_{23}\\
    &=\Phi_{33}+\Phi_{32}\Phi_{R}^{-1}\Phi_\infty\Phi_{R}^{-1}\Phi_{23}-\Phi_{32}\Phi_{R}^{-1}\Phi_\infty(\Phi_\infty+\Phi_{R})^{-1}\Phi_\infty\Phi_{R}^{-1}\Phi_{23}-\Phi_{32}\Phi_{R}^{-1}\Phi_{23}
    \end{split}
    \end{equation}
    [/tex]

    Here is how MATLAB's dare() function works.

    Here is how I defined my variables: [itex]E = I[/itex], [itex]B = I[/itex], [itex]R = \Phi_R[/itex], [itex]Q = \Phi_{33}[/itex], [itex]S=\Phi_{32}[/itex], , [itex]A = 2\Phi_R^{-1}S^T[/itex].

    Is this correct?
    Thank you : )

    PS: Please let me know if this thread belongs elsewhere, as I would like to maximize feedback from the PF community.
     
  2. jcsd
  3. Sep 16, 2016 #2

    jedishrfu

    Staff: Mentor

  4. Sep 16, 2016 #3

    perplexabot

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    Gold Member

  5. Sep 16, 2016 #4

    jedishrfu

    Staff: Mentor

    Sorry I didn't see as it was hidden in the "Here" text.
     
  6. Sep 16, 2016 #5

    perplexabot

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    Gold Member

    No problem! I still appreciate the fact that you are trying to help. Thank you.
     
  7. Sep 16, 2016 #6

    jedishrfu

    Staff: Mentor

    Have you tried to run your code? If so, what was wrong with the answer?

    You could try taking the answer and testing to see if it is the solution to gain confidence in what Matlab returned.
     
  8. Sep 19, 2016 #7

    perplexabot

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    Gold Member

    I have run my code. I got an answer but I am confused with the result as it is not as I expected. I am trying to figure out why the answer is what it is. I was wondering whether my DARE formulation was correct (hence, why I have this post).
     
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