# Solving coupled system of ODE's

I'm reading a paper on system of ordinary differential equations, and I'm a bit confused with the following definition of a such:

$\dot{\phi} = -A^T\phi-Qx , \quad \dot{x} = \left\{ \begin{array}{ll} Ax-bb^T\phi/r & \mathrm{if}\, \alpha \leq -b^T\phi/r\leq\beta\\ Ax + \alpha b & \mathrm{if}\, -b^T\phi/r < \alpha\\ Ax+\beta b & \mathrm{if}\, -b^T\phi/r > \beta \end{array} \right.$

Let A,Q be some matrices, and b a vector, $\alpha,\beta,r$ some constants, it's not so important. What's important is that this system of ODE's is claimed to be solvable using euler's method. However I'm not sure how this can be done, since the system is coupled this way. I mean the definition of $\dot{x}$ is dependent on $\phi$. How can this be solved, it seems like it's biting its own tail.

Any help is much appreciated!

• You could also write the second equation as $$\dot x = Ax+\max(α,\min(β,-b^Tϕ/r))\,b$$ emphasizing the continuous nature of the right side. – LutzL Jun 29 '17 at 18:47

Euler's method for systems of ODE's is essentially the same as for a single ODE. If your system is $$\dot{X} = F(X)$$ then given a value for $X(t)$, the approximation for $X(t+h)$ is $X(t) + h F(X(t))$. In your case $X$ consists of both $x$ and $\phi$.
• Right, but don't we have a problem, when the definition of $\dot{x}$ depends on part of the solution, $\phi$? – 1233023 Jun 29 '17 at 15:21
• $X$ is the vector $X=\pmatrix{ϕ\\x}$. You apply the Euler method or any other simultaneously to all components. – LutzL Jun 29 '17 at 18:45