# reduce differential equations system of first order and using Euler

Given differential equations $$\ddot x=Gm_1\frac{y-x}{|y-x|^3}\hspace{2cm}\ddot y=Gm_2\frac{x-y}{|y-x|^3}$$ with constant $G,m_1,m_2$ I want to solve them with the Euler method. I know I have to reduce the equations to a system of first order. So I did for $\ddot x$

$$\frac{d}{dt}\begin{pmatrix}v_0\\v_1\end{pmatrix}=\begin{pmatrix}v_1\\Gm_1\frac{y-v_0}{|v_0-y|^3}\end{pmatrix}$$and the same for $\ddot y$. But now my problem is that the right side depends on the other equation and the other way round. How can you get a system of first order so you can apply Euler to it?

So, before doing it, first write them down. \begin{align} \dot x & = u \\ \dot u & = G m_1 \frac {y-x}{|y-x|^3} \\ \dot y & = v \\ \dot v & = G m_1 \frac {x-y}{|x-y|^3} \end{align} So, now it's well formed (from Euler method standpoint) system of ODEs of general type $$\dot{\mathbf y} = \mathbf{f}(\mathbf y, t)$$ where $$\mathbf y = [x,u,y,v]^T$$ and $$\mathbf f = \left [ \begin{array}{c} x \\ G m_1 \frac {y-x}{|y-x|^3} \\ y \\ G m_2 \frac {x-y}{|x-y|^3} \end{array} \right ]$$ Now, you can proceed to the Euler method from this point. $$\mathbf y_{n+1} = \mathbf y_n + h \mathbf f(\mathbf y_n)$$ or, explicitly $$\left [ \begin{array}{c} x_{n+1} \\ u_{n+1} \\ y_{n+1} \\ v_{n+1} \end{array} \right ] = \left [ \begin{array}{c} x_n \\ u_n \\ y_n \\ v_n \end{array} \right ]+ h \cdot \left [ \begin{array}{c} u_n \\ G m_1 \frac {y_n-x_n}{|y_n-x_n|^3} \\ v_n \\ G m_2 \frac {x_n-y_n}{|x_n-y_n|^3} \end{array} \right ]$$ So, if you do it iteratively with sufficiently small $h$, you'll get $1^{st}$ order approximation of solution of the ODEs given.
You need some initial values $[x_0, u_0, y_0, v_0]^T$ as a starting point, and those are usually know as physical initial conditions. They are of the form \begin{align} x(0) &= something \\ u(0) &= \dot x(0) = something \\ y(0) &= something \\ v(0) &= \dot y(0) = something \end{align}
• thanks. But I am still a little bit confused. I need $y=[x,z,y,v]^T$, right, but what's explicitly in the vector now? – sheldoor May 12 '13 at 14:14