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Differential equation

  1. Feb 14, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm pretty rusty at these. But given the following:

    [tex]\frac{dN_a}{dt}=-\frac{N_a}{\tau_a}[/tex]
    [tex]\frac{dN_b}{dt}=\frac{N_a}{\tau_a}-\frac{N_b}{\tau_b}[/tex]

    3. The attempt at a solution

    The first one, naturally, is easy [tex]N_a(t)=N_a(0)e^\frac{-t}{\tau_a}[/tex]

    The second one is giving me a little trouble.

    I tried it as such:

    [tex]\frac{dN_b}{dt}=\frac{N_a\tau_b-N_b\tau_a}{\tau_a \tau_b}[/tex]

    [tex]\frac{dN_b}{N_a\tau_b-N_b\tau_a}=\frac{dt}{\tau_a \tau_b}[/tex]

    Let [tex]x=N_a\tau_b-N_b\tau_a[/tex]

    [tex]dx=-dN_b\tau_a[/tex]

    [tex]\frac{dx}{x\tau_a}=\frac{dt}{\tau_a \tau_b}[/tex]

    [tex]\frac{dx}{x}=\frac{dt}{\tau_b}[/tex]

    [tex]x(t)=x_0e^\frac{-t}{\tau_b}[/tex]

    [tex]N_a\tau_b-N_b\tau_a=x_0e^\frac{-t}{\tau_b}[/tex]

    [tex]N_b\tau_a=N_a\tau_b-x_0e^\frac{-t}{\tau_b}[/tex]

    [tex]N_b(t)=\frac{N_a\tau_b-x_0e^\frac{-t}{\tau_b}}{\tau_a}[/tex]

    Now, at [tex]t=0[/tex], [tex]N_b(t=0)=N_{b0}[/tex]

    So [tex]x_0=N_a\tau_b-N_{b0}\tau_a[/tex]

    [tex]N_b(t)=\frac{N_a\tau_b-(N_a\tau_b-N_{b0}\tau_a)e^\frac{-t}{\tau_b}}{\tau_a}[/tex]

    But I don't think that's right because it blows up. It's supposed to be a decay problem.

    Did I solve it correctly?
     
  2. jcsd
  3. Feb 14, 2009 #2
    I believe that is right.

    Now I'm trying to solve it using the Euler method...

    [tex]N_b(t+\Delta{t})=N_b(t)-\Delta{t}*\frac{dN_b}{dt}[/tex]

    I get: [tex]N_b(t+\Delta{t})=N_b(t)-\Delta{t}*(\frac{N_a}{\tau_a}-\frac{N_b}{\tau_b})[/tex]

    And THAT blows up.
     
  4. Feb 14, 2009 #3

    HallsofIvy

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    Instead use the solution to the first equation to write the equation as
    [tex]\frac{dN_b}{dt}=\frac{N_a(0)}{\tau_a}e^\frac{-t}{\tau_a}-\frac{N_b}{\tau_b}[/tex]
    [tex]\frac{dN_b}{dt}+ \frac{N_b}{\tau_b}= \frac{N_a(0)}{\tau_a}e^\frac{-t}{\tau_a}[/tex]

    That's a linear non-homogeneous differential equation with constant coefficients. Its characteristic equation is [tex]r+ \frac{r}{\tau_b}= 0[/tex] and you can look for a specific solution of the entire equation of the form [tex]Ae^\frac{-t}{\tau_a}[/tex].

    You will find that has solutions with decreasing exponentials.


     
    Last edited: Feb 14, 2009
  5. Feb 14, 2009 #4
    Finally got it right. I was right with my analytical solution, except for some sign errors. And it checks out with the Euler method solution. Here's a perl script that I wrote to test and illustrate:

    Code (Text):

    #!/usr/bin/perl

    $dt=0.01;
    $t=0;
    $ta=1;
    $tb=2;
    $Na0=100;
    $Nb0=0;
    $Nae=$Na0;
    $Nbe=$Nb0;
    $i=0;
    $imax=20;

    for($i=0;$i<=$imax/$dt;$i++)
    {
        $t=$i*$dt;
        $Na=$Na0*exp(-$t/$ta);
        $Nb=-($Na*$tb-($Na0*$tb+$Nb0*$ta)*exp(-$t/$tb))/$ta;
        if($i > 0)
        {
            $Nae=$Nae-$dt*$Nae/$ta;
            $Nbe=$Nbe+($Nae/$ta-$Nbe/$tb)*$dt;
        }
        printf("%f %f %f %f %f\n",$t,$Na,$Nae,$Nb,$Nbe);
    }
     
    Thanks for your input.
     
  6. Feb 18, 2009 #5
    I guess it wasn't right. If I use different values for the decay constants my analytic solutions differs from the Euler solution.

    I don't know how to solve that.
     
  7. Feb 18, 2009 #6

    djeitnstine

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    Gold Member

    The solution to the characteristic equation posted is the argument of an exponential [tex]e^{\frac{1}{\tau_{b}}}[/tex]. This added with the particular solution of the form [tex]Ae^{\frac{-t}{\tau_{a}}}[/tex] will give you your general solution.

    Solving for the particular. Simply take the solution of the form [tex]Ae^{\frac{-t}{\tau_{a}}}[/tex], differentiate it, add them together and solve for the coefficients.

    I'll get you started:

    [tex]N = Ae^{\frac{-t}{\tau_a}}[/tex]
    [tex]N'= \frac{-1}{\tau_{a}}Ae^{\frac{-t}{\tau_{a}}}[/tex]

    Sum them [tex]N'+\frac{N}{\tau_{b}} = \frac{-1}{\tau_{a}}Ae^{\frac{-t}{\tau_{a}}}+\frac{1}{\tau_{b}}Ae^{\frac{-t}{\tau_{a}}}[/tex]

    Solve for coefficients [tex]A(\frac{-1}{\tau_{a}}+\frac{1}{\tau_{b}})= \frac{N_{a}(0)}{\tau_{b}}[/tex] and
     
    Last edited: Feb 18, 2009
  8. Feb 18, 2009 #7

    HallsofIvy

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    What course is this for then? It is now a single non-homogeneous linear equation with constant coefficients. Much simpler than the original problem and much simpler than what you were trying to do.
     
  9. Feb 18, 2009 #8
    Computational Physics.

    The problem is to approximate the original system of differential equations using the Euler method, which I can do.

    I am now trying to solve it analytically so I can compare it with the Euler solution.
     
  10. Feb 18, 2009 #9

    djeitnstine

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    Take a look at how I solved it...
     
  11. Feb 23, 2009 #10
    Shouldn't it be : [tex]N = Ae^{\frac{-t}{\tau_b}} [/tex]?
     
  12. Feb 23, 2009 #11

    djeitnstine

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    Gold Member

    No. Halls of Ivy showed you this result,

    [tex]
    \frac{dN_b}{dt}=\frac{N_a(0)}{\tau_a}e^\frac{-t}{\tau_a}-\frac{N_b}{\tau_b}
    [/tex]

    Which in turn becomes

    [tex]
    \frac{dN_b}{dt}+ \frac{N_b}{\tau_b}= \frac{N_a(0)}{\tau_a}e^\frac{-t}{\tau_a}
    [/tex]

    using this result, we get a non-homogeneous [tex]2^{nd}[/tex] order differentual equation with a particular solution [tex] Ae^{\frac{-t}{\tau_{a}}}[/tex]

    So we get a particular solution [tex]N_{p} = Ae^{\frac{-t}{\tau_{a}}}[/tex]

    then, [tex]N'=\frac{-1}{\tau_{a}}Ae^{\frac{-t}{\tau_{a}}} [/tex]

    We sum the 2 with your coefficient: [tex]N'+\frac{N}{\tau_{b}} = \frac{-1}{\tau_{a}}Ae^{\frac{-t}{\tau_{a}}}+\frac{1}{\tau_{b}}Ae^{\frac{-t}{\tau_{a}}} [/tex]

    Of course we solve for the coefficient A: [tex] A(\frac{-1}{\tau_{a}}+\frac{1}{\tau_{b}})= \frac{N_{a}(0)}{\tau_{b}} [/tex]

    Lets get a decent looking fraction first [tex] A(\frac{\tau_{a} - \tau_{b}}{\tau_{a} \tau_{b}})= \frac{N_{a}(0)}{\tau_{b}} [/tex]

    [tex] A = \frac{N_{a}(0)\tau_{a} \tau_{b}}{\tau_{b} (\tau_{a} - \tau_{b})} [/tex]

    [tex]N_{p}=\frac{N_{a}(0) \tau_{a} \tau_{b} e^{\frac{-t}{\tau_{a}}}}{\tau_{b} (\tau_{a} - \tau_{b})}[/tex]
    We have only solved the particular equation. Now we have to add it to the homogeneous equation.

    The characteristic is as Ivy showed is [tex] r+ \frac{1}{\tau_b}= 0 [/tex] Thus your [tex]N_{h} [/tex] homogeneous is [tex] N_{h}= C_{1}e^{\frac{-t}{\tau_{b}}[/tex]

    Now summing them to get a general solution...[tex]N_{g}=N_{h}+N_{p}[/tex] gives

    [tex]N_{g}= C_{1} e^{\frac{-t}{\tau_{b}}} + \frac{N_{a}(0) \tau_{a} \tau_{b} }{\tau_{b} (\tau_{a} - \tau_{b})}e^{\frac{-t}{\tau_{a}}}[/tex]

    Now use your initial conditions to find [tex]C_1[/tex]
     
    Last edited: Feb 23, 2009
  13. Feb 25, 2009 #12
    Well, that's quite a bit different than what I just got:

    [tex]N_b(t)=[N_b(0)-\tau_aN_a(0)]e^{-\frac{t}{\tau_b}}-\tau_aN_a(0)e^{-\frac{t}{\tau_a}}[/tex]

    Hmmm
     
  14. Feb 25, 2009 #13
    There's a problem with this. If [tex]\tau_a = \tau_b[/tex], it blows up.
     
  15. Feb 25, 2009 #14

    djeitnstine

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    Yea cus division by 0 doesn't usually go well in math =\
     
  16. Feb 25, 2009 #15

    djeitnstine

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    Oh man I got mixed up with all these tau sub a's and tau sub b's the final answer is
    [tex]N_{g}= C_{1} e^{\frac{-t}{\tau_{b}}} + \frac{N_{a}(0) \tau_{a} \tau_{b} }{\tau_{a} (\tau_{a} - \tau_{b})}e^{\frac{-t}{\tau_{a}}}[/tex]

    (its the little [tex]\tau_{b}[/tex] before the bracket in the denominator that's wrong)
     
  17. Feb 26, 2009 #16
    Final solution that matches the Euler solution:

    [tex]N_b(t)=(N_b(0)-\frac{\tau_b}{\tau_a-\tau_b}N_a(0))e^{-\frac{t}{\tau_b}}+\frac{\tau_b}{\tau_a-\tau_b}N_a(0))e^{-\frac{t}{\tau_A}}[/tex]

    But what can I do about that condition when [tex]\tau_a=\tau_b[/tex]?

    The function should not blow up.
     
  18. Feb 26, 2009 #17

    djeitnstine

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    Check that condition on your original DE, then compare that result with the actual answers. Basically check to see if any other solutions were lost.
     
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