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Math Help - Integrating factor in optimal control theory

  1. #1
    Newbie AmberLamps's Avatar
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    Integrating factor in optimal control theory

    Hey guys,

    This is an optimal control problem for an econ class but it is the maths behind it I am struggling with - so have left the context out. I have derived two differential equations in my state and costate variables:

    costate:  \frac {dy} {dt} = Ae^{(r+\beta)t}

    state: \frac {dk} {dt} = \beta k - (Ae^{(r+\beta)t})^2

    where A is some arbitrary constant.

    Now i need to solve \frac {dk} {dt} for k

    Using the integrating factor I get:

    K = Ce^{\beta t} - \frac {A^2e^{(4r+4\beta)t}} {4r+3\beta}

    Where C is another arbitrary constant. Is this right?

    I then have to apply the transversality condition:

     e^{-rt} k(t) y(t) = 0

    Which using my solutions for k and y is:

    ACe^{2\beta t} - \frac {A^3e^{(4r+5\beta)t}} {4r+3\beta} = 0

    I won't go any further with the problem here because I am convinced I have gone wrong somewhere since the equation isn't very pretty can anyone check my value for k please?
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  2. #2
    Super Member
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    This is what I get:

    \frac{dk}{dt}=\beta k-A^{2}e^{(2r+2\beta)t} \Rightarrow \frac{dk}{dt}-\beta k= -A^{2}e^{(2r+2\beta)t}

    \mu = e^{-\int \beta dt} \Rightarrow \mu = e^{-\beta t}

    \left(K \cdot e^{-\beta t} \right)' = -A^{2}e^{(2r+2\beta)t}\cdot e^{-\beta t} \Rightarrow \left(K \cdot e^{-\beta t} \right)' = -A^{2}e^{(2r+\beta)t}

     \int \left(K \cdot e^{-\beta t} \right)' = -A^{2}\int e^{(2r+\beta)t}

    K \cdot e^{-\beta t} = -A^{2}\frac{e^{(2r+\beta)t}}{2r+\beta}+C

    K = -A^{2}\frac{e^{(2r+2\beta)t}}{2r+\beta}+C_1e^{\beta t}

    When you integrate y(t), I think you are leaving off a constant:

    Y = \frac{Ae^{(r+\beta)t}}{r+\beta}+C_2
    Last edited by ANDS!; June 4th 2010 at 05:36 PM.
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  3. #3
    Newbie AmberLamps's Avatar
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    thanks - have gone through it myself and got the same - my dy/dy should have just read y(t) but thanks anyway!
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