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Solve coupled nonlinear differential equations

  1. Nov 16, 2015 #1
    Good evening I have these coupled equations and was wondering if there is any chance solving them analytically. If not, how would you approach it numerically? (shown in attachment)
    Thank you very much
     

    Attached Files:

  2. jcsd
  3. Nov 16, 2015 #2

    BvU

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    Hello eahaidar,

    Don't you think that is a bit terse ? Some explanation of the variables might be in order. Complex variables ? Probably: in particular: g -- if g is real nothing happens.

    Below is the homework template; you case may not be a homework exercise, but the systematic approach might be useful anyway !

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution​
     
  4. Nov 16, 2015 #3
    Thank you for your time. The variables are [/0], [/STOKES], and [/SBS]. As for g, it is a real constant number despite having many variables.
    My attempt well: i only could find a relation between the 3 equations. Clearly equation 1 = 2 + 3 however other than the initial conditions of [/STOKES] and [/SBS] are available while [/0] is found at the end of the length of a waveguide L. I do not know what to do next.
    Hopefully you can help me out
     
  5. Nov 16, 2015 #4

    BvU

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    If ##\ g\ ## is a real number (but a function of many non-involved variables, but not ##z,\ I_0, I_{SBS} {\rm \ or \ } I_{stokes}\ ## -- correct ? ), then what does ##Im(g)## mean ?

    I also have difficulty understanding your "Clearly equation 1 = 2 + 3 " since that implies ##\ \left | A_0 \right |^2 = 1 \ ##, which I haven't seen in the problem description ?

    Whatever, my first approach would be a Runge-Kutta integration and use a shooting method to end up with ##Cst_1## and ##Cst_2## -- which are given constants at ##z = L## -- but then it seems you intend to go the other way with initial conditions for ##\ I_{SBS} {\rm \ and \ } I_{stokes}\ ## at ##\,z = L\,## (?) and have ##\ I_0(0) \ ## at the end of the pipe ?

    --
     
  6. Nov 18, 2015 #5
    Thank you for your time. g is complex and I apologize for that. However, Im(g) is a constant number since it varies with another variable other than z. So lets say Im(g)= 0.5
    What I meant from 1=2 +3 is that notice that d(I0) /dz = d( ISBS)/dz + d(I STOKES)/dz.
    As for your approach, I0 represents the intensity of light introduced into the medium while SBS and STOKES represent 2 light waves counter propagating the forward wave of index 0 which puts me in trouble since initally at z=0 I only have the forward wave.
    With that being said, is your suggestion still valid? If not what other choice do I have?
     
  7. Nov 19, 2015 #6

    BvU

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    I still see an ##|A_0|^2## sitting there. is it constant and equal to 1 ? But even d(I0) /dz = (d( ISBS)/dz)/A02 + d(I STOKES)/dz is enough to reduce this to a two differential equations problem -- doesn't change the suggestion.

    But it is a bit strange that sbs and stokes should both end up at Cst1 for z=L
     
  8. Nov 19, 2015 #7
    Now i see your point. Drop the |A0|^2 since it is inside I0.
    Now what i claimed that first equation is the sum of second and third. SBS and Stokes should be propagate on the other side of the medium and therefore there are initially at z=L that means they are introduced at the end of the medium. Does that make any sense or change anything?
     
  9. Nov 19, 2015 #8

    BvU

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    So is this a set of differential equations in z only, or is there some time dependence that plays a role too ?

    So far "Initial conditions" to me meant conditions at z = 0 and "boundary conditions" meant conditions at z = L

    Are you doing things like Y Zhu describes in his thesis (e.g. (2.27) ) ? (stumbled on it while trying to decipher the term SBS)
     
  10. Nov 19, 2015 #9
    Yes it only relies on z.
    SBS stands for stimulated Brillouin scattering and so far yes the initial conditions are as you mentioned
     
  11. Nov 20, 2015 #10

    BvU

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    So, if I summarize: you want to solve
    $$ \begin{cases}
    {dI_0\over dz} & = \ \ \ \ Im(g) I_0(I_{sbs}-I_{stokes})\\ {dI_{sbs}\over dz} &=\ \ \ \ Im(g) I_0 I_{sbs}\\ {dI_{stokes}\over dz} &= - Im(g) I_0 I_{stokes}
    \end{cases}
    $$ With initial condition ##\ I_0(0) = C_2,\ ## and boundary conditions ##\ I_{sbs} (L) = I_{stokes}(L) = C_1##

    Since 1 = 2 + 3 ## {d\over dz}( I_0 - I_{sbs}-I_{stokes}) = 0 \Rightarrow I_0 - I_{sbs}-I_{stokes} = Const ##, an unknown constant

    I leave out the Im(g) (It's an over-all scale factor ) and give it to an integrator with initial conditions C2, u and v, respectively. Then I fumble with u and v until both end up at ##C_1##. In the picture below ##C_2 = 0.1, \ \ C_1 = 0.0312## u and v come out 0.0135, 0.072

    SBS.jpg

    I did the fumbling (shooting) by hand, but you can let the computer do it.


    Since 1 = 2 + 3, ## \ \ {d\over dz}( I_0 - I_{sbs}-I_{stokes}) = 0 \ \ \Rightarrow \ \ I_0 - I_{sbs}-I_{stokes} = Const ##, an unknown constant, you don't really have to integrate all three (##Const = C_2 - u - v## )

    No joy finding something analytical (lack of talent for that kind of thing ..:smile:).

    --
     
    Last edited: Nov 20, 2015
  12. Nov 20, 2015 #11
    Very interesting my friend. How did you get the plots? What program did you use?
    I know analytical stuff is hard but they come into handy. Another condition i found was that ISBS * I Stokes = CST if you divide equation 2 from 3.
    I am not sure how you got the curves to be honest. Is is just fumbling with the numbers or what exactly?
    I really want to thank you for your help
     
  13. Nov 23, 2015 #12

    BvU

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    (Sorry, I missed this alert during the weekend).
    Program is a dynamic simulation program for the chemical industry, called aspen custom modeler. But all I use is the integrator, so I didn't mention it. It has lousy graphics, so picture is made by moving the data to excel. Also easier to do further analysis: take logarithm, fit that to a polynomial, to see if it's a simple exponential. It is not:
    (##\ \ln(I_{\rm stokes} ) = -2.6356 - 0.0939 z + 0.0011 z^2\ ## )

    Got the curves by doing the forward integration from z = 0 to 10. Chose C2 = I0(0) = 0.1 and played around with ISBS(0) and Istokes(0) until they ended up at the same value. I called that C1. Normally you would let the computer do the guessing (shooting method) with a reasonable efficiency.

    So still no joy finding something analytical.

    What tools do you have at your disposal for this ?

    --
     
  14. Nov 23, 2015 #13
    I usually use matlab for any simulation. I went through the equations. With the help of a colleague, we managed to reduce the coupled equations to a single equation. I will update you on that once i get to the bottom of this equation which includes the 3 variables. Again thanks for your help and would love to learn your integrator. Sounds fun to use.
     
  15. Nov 23, 2015 #14

    BvU

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    Yes, ACM is fun, but if you already have matlab: there's integrators available by the truckload and they are free. And shooting isn't a problem either! Well documented and built-in too, I see from the nice example on his p. 174.

    I understand you are still eager to get an analytical solution (e.g. in the other thread), but I can't really help there.

    [edit] I browsed through some earlier posts by eahaidar and found you were already pointed to ode45 in May 2014 ! Any luck getting that to work since then ? I mean, you can do a lot of work numerically and keep an eye open for analytical solution opportunities at the same time !

    By the way, did you find anything useful in the Zhu thesis from link in post #8 ?
     
    Last edited: Nov 23, 2015
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