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Thread: Integral with a max function inside

  1. #1
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    Integral with a max function inside

    Hi all !

    I would like to integrate this equation, to check if the values I get by numerical simulation are correct. My initial condition is $\displaystyle e(0)=0$ and I would like to check other values.
    However, I don't know how to deal with the max function so I can't compute the exact solution.
    $\displaystyle \frac{\partial e}{\partial t} = \frac{1}{\eta }max(0,\sigma-\sigma^*) $ with $\displaystyle \sigma=1-e^{-4t}$
    In my specific case, I set the parameters to $\displaystyle \eta = 1/4 $, $\displaystyle \sigma^* = 1/2 $

    Thank you very much for your help !
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Integral with a max function inside

    Quote Originally Posted by Jo37 View Post
    Hi all !

    I would like to integrate this equation, to check if the values I get by numerical simulation are correct. My initial condition is $\displaystyle e(0)=0$ and I would like to check other values.
    However, I don't know how to deal with the max function so I can't compute the exact solution.
    $\displaystyle \frac{\partial e}{\partial t} = \frac{1}{\eta }max(0,\sigma-\sigma^*) $ with $\displaystyle \sigma=1-e^{-4t}$
    In my specific case, I set the parameters to $\displaystyle \eta = 1/4 $, $\displaystyle \sigma^* = 1/2 $

    Thank you very much for your help !
    Split the integral into pieces. One interval will be for $\displaystyle 0 \geq \sigma - \sigma *$ and the other $\displaystyle 0 < \sigma - \sigma *$.

    -Dan
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  3. #3
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    Re: Integral with a max function inside

    Thanks for the reply ! I think my result is still wrong. I don't get the same behavior as the original plot I was given.
    Here is my reasoning :

    $\displaystyle
    e(t)= \int_{-oo}^{0 >=\sigma-\sigma^*} 0dt \; +\int_{0 <\sigma-\sigma^*}^{+oo}4(1-e^{-4t}-0.5)dt \\
    \left\{\begin{matrix}
    e(t)= 2t + e^{-4t} + C& if & 0<\sigma-\sigma^*\\
    & else & e(t)= 0
    \end{matrix}\right.$

    In my case e(0)=0 so C=-1.
    Last edited by Jo37; Jan 10th 2019 at 07:58 AM.
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  4. #4
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    Re: Integral with a max function inside

    $\displaystyle \int \limits_0^t~4 \max\left(0,1-e^{-4\tau}-\dfrac 1 2\right)~d\tau = $

    $\displaystyle \int \limits_{\frac{\ln(2)}{4}}^t~2 - 4e^{-4\tau}~d\tau = $

    $\large \left. 2\tau + e^{-4\tau} \right|_{\frac{\ln(2)}{4}}^t = $

    $2t + e^{-4t} - \left(\dfrac{\ln(2)}{2} + \dfrac 1 2\right) = $

    $2t + e^{-4t} - \dfrac 1 2 - \dfrac{\ln(2)}{2}$
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  5. #5
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    Re: Integral with a max function inside

    Thanks for your help ! Your solution makes sense.
    However, the graph I get when plotting this equation is still weird. If one starts at $\displaystyle e(0)=0$ and the derivative stays at 0 until $\displaystyle ln(2)/4$, there should be a constant plateau before the function increases. This is not what we get when plotting the solution.
    This is the plot I would like to obtain (blue curve epsilon). It represents the equation governing this material : https://en.wikipedia.org/wiki/Bingham_plastic
    Integral with a max function inside-screenshot_2019-01-10-elastic-growth-models-elastic_growth_models-pdf.png
    Last edited by Jo37; Jan 10th 2019 at 10:21 AM.
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