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Is there an analytic solution to this system of equations?

  1. Apr 19, 2015 #1

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    I have the following system of equations with variables ##a,m##, and I'm wondering—can this system be solved symbolically/analytically?

    \begin{align}
    m &= 100 + \frac{ \left( 200 \frac{\ln{\frac{1}{2}}}{26.8} \right) }{\left(\dfrac{\ln{\frac{1}{2}}}{26.8} + a \right)}
    \\ \\
    50 &= me^{-a\left( 19.9 \right)}- \frac{ \left( 200 \frac{\ln{\frac{1}{2}}}{26.8} \right) \exp{ \left(\dfrac{\ln{\frac{1}{2}}}{26.8} \cdot 19.9 \right) }}{\left(\dfrac{\ln{\frac{1}{2}}}{26.8} + a \right)}
    \end{align}

    (What are some ways to recognize that a problem does or does not have an analytic solution?)

    For this specific example, I understand the answers can be numerically approximated to

    \begin{align}
    a &\approx 0.092409
    \\
    m &\approx 22.2674
    \end{align}

    Via such numerical methods as a graph:
    L9sjZ.png

    or other computational device such as WolframAlpha [(link to this problem). For this specific problem, WolframAlpha only provided a numerical approximation.

    Note: The variables in the WolframAlpha link are ## a, n ## respectively
     
  2. jcsd
  3. Apr 19, 2015 #2

    jedishrfu

    Staff: Mentor

    If you replace the factor ln(0.5)/26.8 with K and use x and y for a and m you get for the first equation:

    y = 100 + 200K/(x+K)

    and for the second you get:

    50/e^19.9 = y - 200Ke^K/(x+K)

    which rearranges to:

    y = 50/e^19.9 + 200e^K/(x+K)

    From here you can set the equations equal to each other and solve for x noting that x=/= -K
     
    Last edited: Apr 19, 2015
  4. Apr 19, 2015 #3
    I don't see an analytical solution. For a numerical solution, consider the following:

    With
    \begin{align*}
    m &= 100 + 200 \cdot \frac{\beta}{\beta+a} \\
    50 &= m \cdot e^{-19.9a} - 200 \cdot \frac{\beta}{\beta+a} \cdot e^{19.9 \cdot \beta}
    \end{align*}
    Where [itex]\beta = \ln(1/2) / 26.8 \approx -0.02586[/itex].

    Multiply the second equation by [itex]\exp{(19.9 \cdot a)}[/itex]
    \begin{equation*}
    50 \cdot e^{19.9 \cdot a} = m - 200 \cdot \frac{\beta}{\beta + a} \cdot e^{19.9 \cdot \beta} \cdot e^{19.9 \cdot a}
    \end{equation*}

    Let [itex]w= \beta + a[/itex] and substitute first equation into second to yield
    \begin{equation*}
    c \cdot e^{19.9 \cdot w} = \left[100 + 200 \cdot \frac{\beta}{w}\right] - 200 \cdot \frac{\beta}{w} \cdot e^{19.9 \cdot w}
    \end{equation*}
    where [itex]c = 50 \cdot e^{-19.9 \cdot \beta} \approx 83.6558[/itex].

    This is a non-linear scalar equation in one parameter [itex]w[/itex]. A numerical solution will be necessary. Before we go down that path, re-scale and introduce a non-parametric variable for [itex]w[/itex]. Let [itex]\alpha = a / \beta[/itex], [itex]\zeta = w / \beta = 1 + \alpha[/itex] and re-formulate into [itex]F(\zeta) = 0[/itex].

    \begin{equation*}
    F(\zeta) = 0 = 100 \cdot \left[ \left(\gamma + \frac{2}{\zeta}\right) \cdot e^{\lambda \cdot \zeta} - \left(1 + \frac{2}{\zeta}\right) \right]
    \end{equation*}
    where [itex]\gamma = c / 100[/itex] and [itex]\lambda = 19.9 \cdot \beta \approx -.51469[/itex].

    This could be solved numerically, but we don't know the magnitude or an applicable domain bracket that contains [itex]a[/itex] (ie, [itex]\zeta[/itex]). If we look at the above equation for [itex]F[/itex], we see it is equivalent to
    \begin{equation*}
    F(\zeta) = 100 \cdot \left[
    \left(1 + \frac{2}{\zeta}\right) \cdot \left(e^{\lambda \cdot \zeta} -1\right) - \mu \cdot e^{\lambda \cdot \zeta}\right]
    \end{equation*}
    where [itex]\mu = 1 - \gamma \approx .167[/itex].

    If we ignored the last term, we would underestimate [itex]F[/itex] (since [itex]\mu > 0[/itex] and the exponential is positive). But if we did, we would see that two conditions could result in [itex]F= 0[/itex]:

    First: [itex]\exp{(\lambda \cdot \zeta)} - 1 = 0[/itex]. This would imply [itex]\zeta = 0[/itex] or [itex]a = -\beta \approx 0.02586[/itex]. The scaled error term in [itex]F[/itex], which is the last term in the expression above, is [itex]E = -\mu \cdot e^{\lambda \cdot \zeta} = -\mu = -0.167[/itex]. However, this is probably not a good idea since we have the [itex]1/\zeta[/itex] term which contradicts this solution.

    Second: [itex]1 + 2 / \zeta = 0[/itex]. This would imply [itex]\zeta = -2[/itex] or [itex]a = -3 \cdot \beta \approx 0.07759[/itex]. The scaled error term in [itex]F[/itex] is now [itex]E = -\mu \cdot e^{-2 \cdot \lambda} \approx -0.456[/itex]. That's probably a safer bet. However, since we underestimate [itex]F[/itex], we should probably bump up this value somewhat (ie, decrease [itex]\zeta[/itex] somewhat away from [itex]-2[/itex]).

    Selecting [itex]\zeta = -3[/itex] yields a scaled value [itex](1/100) \cdot F(-3) \approx 0.4301[/itex]. We have now bracketed the root. A suggested initial value for an iteration routine would be [itex]\zeta^{(0)} \approx -2.5[/itex] (ie, [itex]a^{(0)} = 0.09052[/itex]).
     
  5. Apr 19, 2015 #4

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    Thank you for the in-depth explanation to a numerical approach, TheoMcCloskey! The technique you outlined is just beyond my current mathematical level of understanding, but it's educating to see a technique like that in action, and it's definitely brought up some points to research.

    Does your approach have a specific name? I can't wait to study these concepts in the future.
     
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