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Solving coupled differential equations

  1. Jan 18, 2012 #1
    Ok guys, I've got an issue with a coupled differential equation and I just can't get to solve it:

    [itex]\frac{\partial r}{\partial t} = Q\frac{\partial c}{\partial t}[/itex]

    Obviously, r depends on c and visa versa, but they both depend on time. Is there a way to uncouple these variables and solve the equation, so that I get r(t) and c(t)?

    To me, this really seemed trivial at first, but I've spent hours solving this using decoupling recipes I found all over the internet, but I just can't get it to work (which is frustrating). Could you please help me out?
     
  2. jcsd
  3. Jan 18, 2012 #2

    fluidistic

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    With such an equation I think the best you can reach is that [itex]r(t)=Qc(t)+K[/itex] where K is a constant. There's not enough information to get r(t) or c(t). They could in principle be any differentiable function with respect to time.
    I'd wait a more experienced opinion on this though, but this is what I think (as a student).
     
  4. Jan 18, 2012 #3

    lanedance

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    i think you can just integrate this w.r.t. t
    [tex]
    \frac{\partial r}{\partial t} = Q\frac{\partial c}{\partial t}
    [/tex]
    [tex]
    \int_0^t \frac{\partial r}{\partial t} dt= \int_0^t Q\frac{\partial c}{\partial t}dt
    [/tex]
    [tex]
    r(t)|_0^t dt= Qc(t)_0^t
    [/tex]
    [tex]
    r(t)-r(0)= Qc(t)-Qc(0)
    [/tex]

    Then we can roll up all the constants into a single constant b=-Qc(0)+r(0)
    [tex]
    r(t)= Qc(t)-Qc(0)+r(0)
    [/tex]

    Though i note you use partial differntiation sign rather than total - is that intended?
     
    Last edited: Jan 18, 2012
  5. Jan 18, 2012 #4
    Well, problem still is that I don't know either r(t) or c(t). But I do know r(0) and c(0), and that dc/dt = k @ t = 0, perhaps that helps?
     
  6. Jan 19, 2012 #5

    lanedance

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    yeah that helps

    For example, the following is a solution to your problem
    c(t)=c(0)e^(kt/c(0))
    r(t)=c(0)Q(e^(kt/c(0))-1)+r(0)

    not unique though, as here is another
    c(t)=kt+c(0)
    r(t)=Qkt+r(0)
     
    Last edited: Jan 19, 2012
  7. Jan 19, 2012 #6

    HallsofIvy

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    This is not, by the way, "coupled differential equations". "Coupled differential equations" would be two ordinary differential equations of two dependent variables with derivatives with respect to one independent variable. This is a "partial differential equation" with a single dependent variable depending on two independent variables.

    In general, where the general solution to a single first order ordinary differential equation involves a single undetermined constant, the solution to a single first order partrial differential equation involves a single undetermined function.
     
  8. Jan 19, 2012 #7

    lanedance

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    actually I wouldn't call this a partial differential equation...

    Not too sure what it is, but wouldn't partial imply differention by at least 2 different & independent variables?

    Not really sure what this is besides related rates
     
  9. Jan 21, 2012 #8
    Oh I'm sorry, I thought coupled would be appropiate terminology, since one is related to the other and visa verse. But thanks for putting that straight :)

    How did you find those solutions? (Btw, c(o) = 0 in this particular case)

    I have the teacher's numerical solution attached as a graph, so I guess the analytical solutions should look like that aswell.
     

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    Last edited: Jan 21, 2012
  10. Jan 21, 2012 #9

    HallsofIvy

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    Oh, I see. your use of "[itex]\partial[/itex]" rather than "d" fooled me. I see now that both derivatives are with respect to t. Here, the difficulty is that you have only one equation. Just as you cannot solve one algebraic equation for two unknown numbers, so you cannot solve one differentrial equation for two unknown functions. The best you can do is arrive at a relation between the two functions- the relation that fluidistic and lanedance got.
     
  11. Jan 21, 2012 #10
    So it is a coupled differential equation then?

    Such a shame :(. I should have known...

    Thanks everyone!
     
  12. Jan 21, 2012 #11
    Ok guys, this problem keeps bugging me and I've revised the whole thing (again, sigh...) and came to the conclusion that I've made some errors along the way (again, sigh...). Now I have these sets of equations:

    [itex]\frac{dr}{dt} = -\frac{M}{ρ}k(1-\frac{c}{c_{eq}})[/itex]

    and

    [itex]\frac{dc}{dt} = \frac{4π}{V}r^{2}k(1-\frac{c}{c_{eq}})[/itex]

    in which r is a function of c and t, and c a function of r and t; M, ρ, k, [itex]c_{eq}[/itex], π and V are constants.

    I appear to have two independent equations now, so could this be solved?

    (btw, I'm sorry to have bugged you with the wrong equation before, I hope these ones are correct)
     
  13. Jan 22, 2012 #12

    lanedance

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    the second equation is separable so you can integrate directly to find c(t) and r will follow
     
  14. Jan 22, 2012 #13
    When I separate the 2nd equation, I get the following (if I'm correct):

    [itex]\frac{dc}{1-\frac{c}{c_{eq}}} = \frac{4πk}{V}r^{2}dt[/itex]

    Which solves to:

    [itex]-c_{eq}ln(1-\frac{c}{c_{eq}}) = \frac{4πk}{V}\int r^{2}dt + constant[/itex]

    But I do not yet know what [itex]\int r^{2}dt[/itex] is, only [itex]\frac{dr}{dt}[/itex]. Did I do something wrong?
     
  15. Jan 25, 2012 #14
    Could anyone still give me a hand on my post above?
     
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