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Two ODEs which are coupled

  1. Mar 26, 2015 #1
    Hello :)

    I have a system which consists of two coupled ODEs for which I want to solve.

    F'' *(1/b²) - α²*F = Ra*(1/b³)*G

    G'' *(1/b²) - α*²G + Ra*(1/b²)*G = F'(1/b)

    In these two equations F(z) and G(z) both depend on z. b is a constant, Ra is the rayleigh number which I need to keep as Ra in my answere.

    I suspect that my answere for the expressions of F and G will be expressions including hyperbolic functions, but have not managed to find it!

    Does anybody have an idea?:)

    Last edited: Mar 26, 2015
  2. jcsd
  3. Mar 26, 2015 #2


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    Please check the equations. Especially a*2G (should it be a2*G) and F'(1/b) (should it be F'*(1/b)).
  4. Mar 26, 2015 #3
    If I multiply by [itex]b^2[/itex], is it then this system below?
  5. Mar 26, 2015 #4
    There are 2 ways to solve this. First you can introduce dummy variables [itex]F'=H [/itex] and [itex]G'=I [/itex]. This will allow you to covert the system of 2 2nd order ODES into a system of 4 first order ODEs. If you know how to solve systems of ODE's, then you can write this as a martix equations of the form [itex]Ax=0[/itex] and the eigenvalues/eigenvectors of [itex] A[/itex] give you the solution.

    If you don't know what I'm talking about thats okay. Treat the first equation as a defination for [itex]G [/itex] .You can differentiate it two times to get a defination for [itex]G'' [/itex]. Substitute these two definitions into the second equation and it will give you a forth order ODE for [itex] F [/itex] (there we be no terms that depend on [itex]G [/itex]). The coefficients are all constants, and its straight forward to solve. Once you know [itex]F [/itex] you can use the first equation to find [itex]G [/itex].
  6. Mar 27, 2015 #5
    Svein: Yes, what you suggested is correct

    Bigfooted: Yes it is the system then

    The_wolfman: I have tried your solutions but cannot solve it that way. I have read some other similar papers where their soulution to both F and G are functions of sinh and cosh, but I cannot seem to find such a solution to my system, eventhough I suspect that it excist!
  7. Mar 27, 2015 #6
    Can you show some work? Which method did you use? I listed 2 different methods. It's impossible to provide feedback if you don't tell us what you did.

    Hyperbolic trig functions are nothing more than the sum and difference of exponentials. If your solution is composed of multiple exponential functions then its possible that you have the correct answer.
  8. Mar 28, 2015 #7
    Bianca: I have used the_wolfman's suggestions and rewritten the second order system to a system of first order equations of the form Ix' = Ax, and this system can be solved in terms of exponential functions. Maybe you can show us your system of first order ODE's, preferably in Latex format?
  9. Apr 6, 2015 #8
    So my system is

    F'' - (α^2 * b^2 * F ) = (Ra / b)*G'
    G'' - (α^2 * b^2 +Ra)*G = F'*b

    If I say that F'=H and G'=I, I have the following system:

    H' - α^2*b^2*F = (Ra/b)*I
    I' - (α^2*b^2 - Ra)*G =b*H

    If I write my system as a matrix looking like: x'=Ax, I have:

    F' 0 0 1 0 F
    G' = 0 0 0 1 G
    H' α^2*b^2 0 0 Ra/b H
    I' 0 α^2*b^2-Ra b 0 I

    (I do not know how to make this look pretty in this reply, but this is suppose to look like x'=Ax, where x= [F,G,H,I] )

    If I try to solve this by finding eigenalues and eigenvectors, I find that λ1/λ2= +/- ( √(α^2*b^2 + α*b*Ra)) and λ3/λ4= +/- ( √(α^2*b^2 - α*b*Ra)). None f these eigenvalues give any reasonable eigenvectors other than v1=v2=v3=v4=0, which is not the solution I am looking for. Does this mean that the only solution to this is the zero solution?

    Thank you all for your replies:)
  10. Apr 7, 2015 #9
  11. Apr 7, 2015 #10
    I see now that in my first thread I miss the derivative sign on the G, it is suppose to be (Ra/b³) *G' !
    So the system I have written in my latest thread(the one over here) is the correct one!
  12. Apr 8, 2015 #11
    So your system is like this (If you want to see how to make this pretty, then quote this message in your reply and you will see the code):


    You introduce new variables and your system becomes (please check):
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\

    F' \\ G' \\ H' \\ I'


    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
    a^2b^2 & 0 & 0 & Ra/b \\
    0 & (a^2b^2+Ra) & b & 0 \\

    F \\ G \\ H \\ I

    computing the eigenvectors and eigenvalues gives me nonzero eigenvectors. How do you compute your eigenvectors?
  13. Apr 9, 2015 #12
    I have also the same as you, only not with the identity matrix on the left side( do I need this?) and with a²b²-Ra on the left bottom in the matrix. I think it is correct with minus instead of plus(like you have).
    Then I get the following eigenvalues and eigenvectors:

    Squareroot= SQ

    λ1=SQ(α²*b² + α*b*SQ(Ra))

    λ2=SQ(α²*b² - α*b*SQ(Ra))

    λ3=-SQ(α²*b² + α*b*SQ(Ra))

    λ4=-SQ(α²*b² - α*b*SQ(Ra))

    This gives me that F and G is as follows:

    F=C1*e^(SQ(α²*b² + α*b*SQ(Ra))*z + C2*e^(SQ(α²*b² - α*b*SQ(Ra))*z + C3*e^-(SQ(α²*b² + α*b*SQ(Ra))*z + C4*e^-(SQ(α²*b² - α*b*SQ(Ra))*z

    G=[C1*SQ(α²b²+αbSQ(Ra))*(αb²SQ(Ra)-bRa)*e^(SQ(α²*b² + α*b*SQ(Ra))*z ]/[(α²b²-R)*R]
    -[C2*SQ(α²b²-αbSQ(Ra))*(αb²SQ(Ra)+bRa)*e^(SQ(α²*b² - α*b*SQ(Ra))*z ]/[(α²b²-R)*R]
    -[C3*SQ(α²b²+αbSQ(Ra))*(αb²SQ(Ra)-bRa)*e^-(SQ(α²*b² + α*b*SQ(Ra))*z ]/[(α²b²-R)*R]
    +[C4*SQ(α²b²-αbSQ(Ra))*(αb²SQ(Ra)+bRa)*e^-(SQ(α²*b² - α*b*SQ(Ra))*z ]/[(α²b²-R)*R]

    I have tried to put these two into my first two coupled ODE's and it works. But I think they (F and G) looks very messy! Does anyone see how one can write this in another way, for example in hyperbolic functions?

    Last edited: Apr 9, 2015
  14. Apr 9, 2015 #13
    I think you can only write this in a more compact form if you substitute the boundary conditions. Or you keep using [itex]\lambda[/itex], as in [itex]F=c_i e^{\lambda_i z}[/itex], where repeated index means summation.
  15. Apr 13, 2015 #14
    Okay, thank you for replies:D
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