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I have system of coupled differential equations:

$x_0 x_0'=\dfrac{32}{(r^4-2r^2)}$

$(x_o x_1)'=a x_0'$

$(x_o x_2)'=-x_1x_1'-a (-x_1'-\dfrac{x_0'}{x_0})$

$(x_o x_3)'=-(x_1x_2)' +a( x_2'+\dfrac{x_1'}{x_0}+\dfrac{2x_1'x_0'}{x_0^2}+\dfrac{x_1 x_0'}{x_0^2})$

Where I need to find values for $x_0, x_1, x_2, x_3$.

In equations $'$ means that is derivative $\dfrac{d}{dz}$, on the other side values of $a$ and $r$ are dependent on $z$ coordinate ($a=a(z), r=r(z)$). I have values of $x$ at the outlet boundary, I think that is boundary value problem than.

It is necessary to solve this system with Runge Kutta 4th order method, I don't know where to start when I have system of equations of this type on left side?

If I know boundary value, not initial values, does it mean that I need to rotate my geometry and equations on that way where my zero coordinate will be on the boundary, not at he inlet? Is it necessary for solution of system of this type?

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  • $\begingroup$ I've not worked with equations of this type before, so I'm not 100% sure, but my impression is -- If you want to integrate $(x_i)_{i=0}^3$ on $[a, b]$, and you only have the initial conditions $x_i(b)$, then it's still an initial value problem. To integrate using Runge-Kutta, you could either take the step-size $dz < 0$ (if your implementation of R-K lets you do that) or make a change of variables $z = b(1-s) + as$ (with $0 \leq s \leq 1$) and pick the step-size $ds > 0$. Does that make sense? $\endgroup$ – Kyle Mar 28 '18 at 3:02

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