# Are these ODEs equivalent?

I have the following set of ordinary differential equations: \begin{equation} \left\{ \begin{array}{l} \dot{a} = f_1(a, b, c, d) \\ \dot{b} = f_2(a, b, c, d) \\ \dot{c} = f_1(c, d, a, b) \\ \dot{d} = f_2(c, d, a, b) \end{array} \right. \end{equation}

where $f_1$ and $f_2$ are two functions from $\mathbb{R}^4$ to $\mathbb{R}$. Also, I have the following initial conditions: \begin{equation} \left\{ \begin{array}{l} a(0) = \alpha \\ b(0) = \beta \\ c(0) = \alpha \\ d(0) = \beta \end{array} \right. \end{equation} and hence, under some suitable conditions, I have the unique solutions $a(t)$, $b(t)$, $c(t)$ and $d(t)$.

Then, I pose that: $$x(t) = \frac{a(t) + c(t)}{2} ~ \text{and} ~ y(t) = \frac{b(t) + d(t)}{2}$$

Can I prove that there exists two function, $g_1$ and $g_2$, from $\mathbb{R}^2$ to $\mathbb{R}$, such that the solution of the Cauchy problem

\begin{equation} \left\{ \begin{array}{l} \dot{x} = g_1(x, y) \\ x(0) = \alpha \\ \dot{y} = g_2(x, y)\\ y(0) = \beta \end{array} \right. \end{equation}

has the properties that $x(t) = a(t) = c(t)$ and $y(t) = b(t) = d(t)$?

If you can express $f_1(a,b,c,d) + f_1(c, d, a, b)$ and $f_2(a, b, c, d) + f_2(c, d, a, b)$ as a function of $(a+c)$ and $(b+d)$, then yes. If you can't, then no.
Of course if $f_1(a, b, c, d)$ and $f_2(a, b, c, d)$ are both function of $(a+c)$ and $(b+d)$ then this condition will be satisfied.
• What if $f_1(a, b,c, d) + f_1(c, d, a, b)$ can be expressed as function of $(a+c)$ and $(b+d)$? – the_candyman May 16 '13 at 10:47