How to Break Coupled ODEs down to first order for Runge-Kutta

Question

My question might seem a bit simple. I am trying to solve a system of ODEs using Runge-Kutta method. I am having difficulty breaking down the equations into a system of first order ones required before applying RK45 to it because there is a third order differential of the same parameter "F" in both equations which I am not used to.

Equation 1 $\left( {\frac {{\rm d}^{2}}{{\rm d}{\eta}^{2}}}F \left( \eta \right) \right) ^{n-1}{\frac {{\rm d}^{3}}{{\rm d}{\eta}^{3}}}F \left( \eta \right) +{\frac {2\,n+2+ \left( 2\,n-1 \right) kF \left( \eta \right) {\frac {{\rm d}^{2}}{{\rm d}{\eta}^{2}}}F \left( \eta \right) }{n+4}}-{\frac { \left( n+2+ \left( n+1 \right) k \right) \left( {\frac {\rm d}{{\rm d}\eta}}F \left( \eta \right) \right) ^{2 }}{n+4}}+G \left( \eta \right) =0$

Equation 2

${\frac { \left( {\frac {{\rm d}^{2}}{{\rm d}{\eta}^{2}}}F \left( \eta \right) \right) ^{n-1} \left( n-1 \right) \left( {\frac {{\rm d}^{3 }}{{\rm d}{\eta}^{3}}}F \left( \eta \right) \right) {\frac {\rm d}{ {\rm d}\eta}}G \left( \eta \right) }{\Pr\,{\frac {{\rm d}^{2}}{{\rm d} {\eta}^{2}}}F \left( \eta \right) }}+ \left( {\frac {{\rm d}^{2}}{ {\rm d}{\eta}^{2}}}F \left( \eta \right) \right) ^{n-1}{\frac { {\rm d}^{2}}{{\rm d}{\eta}^{2}}}G \left( \eta \right) +{\frac {2\,n+2+ \left( 2\,n-1 \right) kF \left( \eta \right) {\frac {\rm d}{{\rm d} \eta}}G \left( \eta \right) }{n+4}}-{\frac {n+2\, \left( n+1 \right) k \left( {\frac {\rm d}{{\rm d}\eta}}F \left( \eta \right) \right) G \left( \eta \right) }{n+4}}=0 $

Boundary Conditions

$ G \left( 0 \right) =1, F \left( 0 \right)=F' \left( 0 \right) =0,G \left( \infty \right) =0,F' \left( \infty \right) =0 $

Any help would be greatly appreciated.

Answer 1

The usual trick is to add more variables that represent the successive derivatives, as in the equations of motion pf physics written as a set of first order ODE of "2" variables: $$\begin{align}\dot{x}&=v\\ \dot{v}&=a(x,t)\end{align}$$ instead of a second order ODE of 1 variable $$\ddot{x}=a(x,t) \ .$$ The dot represents the time derivative.

So, applying to your case, you could have Equation 1 as: $$ \begin{align} x & = \frac{d F}{d \eta}\\ y & = \frac{d x}{d \eta} \\ 0 & = y^{n-1}\frac {d y}{d \eta} + \frac{2n+2 + (2n-1) k F y}{n+4} - \frac { (n+2 + (n+1) k) x^2}{n+4} + G(\eta) \end{align} $$ Note that $y = \frac{d x}{d \eta} = \frac{d^2 F(\eta)}{d \eta^2}$ and that the last equation can easily be written as $\frac{d y}{d \eta} = \ldots$

Equation 2 can have similar substitutions together with the previous ones: $$ \begin{align} u & = \frac{d G}{d \eta}\\ v & = \frac{d u}{d \eta} \\ 0 & = \frac{y^{n-1} (n-1) \frac{d y}{d \eta} u }{\Pr\,y} + y^{n-1} v + \frac {2n+2 + (2n-1) k F u}{n+4} - \frac {n+2 (n+1) k x G}{n+4} \end{align} $$ And the last equation can also be written as $\frac{d y}{d \eta} = \ldots$

This means that you will have 4 virtual new variables: $x$, $y$, $u$ and $v$.

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Edit:

Looking at the number of boundary conditions, since you have 5, you might want to have 5 variables. Depending on your problem, you may write Equation 2 as: $$ \begin{align} u & = \frac{d G}{d \eta}\\ 0 & = \frac{y^{n-1} (n-1) \frac{d y}{d \eta} u }{\Pr\,y} + y^{n-1} \frac{d u}{d \eta} + \frac {2n+2 + (2n-1) k F u}{n+4} - \frac {n+2 (n+1) k x G}{n+4} \end{align} $$ The difference is in the second parcel, where we have $\frac{d u}{d \eta}$ and we no longer need to have $v$.

And then, plug in the expression you got for $\frac{d y}{d \eta}$ in the last equation and solve it for $\frac{d u}{d \eta} = \ldots$

Hence, you'll get a system with 5 (first order) ODE and 5 boundary conditions.