# Solve coupled nonlinear differential equations

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1. Nov 16, 2015

### eahaidar

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

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2. Nov 16, 2015

### BvU

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​

3. Nov 16, 2015

### eahaidar

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

4. Nov 16, 2015

### BvU

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 ?

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5. Nov 18, 2015

### eahaidar

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?

6. Nov 19, 2015

### BvU

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

7. Nov 19, 2015

### eahaidar

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?

8. Nov 19, 2015

### BvU

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)

9. Nov 19, 2015

### eahaidar

Yes it only relies on z.
SBS stands for stimulated Brillouin scattering and so far yes the initial conditions are as you mentioned

10. Nov 20, 2015

### BvU

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

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 ..).

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Last edited: Nov 20, 2015
11. Nov 20, 2015

### eahaidar

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

12. Nov 23, 2015

### BvU

(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 ?

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13. Nov 23, 2015

### eahaidar

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.

14. Nov 23, 2015

### BvU

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.

 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