Izabella Ponce

2022-06-20

We know that every differential equation is equivalent to a first-order system. I am trying to prove or disprove the converse. For example in ${\mathbb{R}}^{2}$, if we have a system $\dot{x}=f(x,y)$, $\dot{y}=g(x,y)$. Can we always convert it to one differential equation (for example, only in terms of $x$)? Under what condition, this is possible?

Abigail Palmer

Beginner2022-06-21Added 30 answers

Consider your example. In order to make this a second-order equation in $x$, you want to solve $\dot{x}=f(x,y)$ for $y$ as a function of $x$ and $\dot{x}$, say $y=a(x,\dot{x})$. This may or may not be possible, (and in most cases even if it is possible in principle it can't be done in closed form). Then we get

$\ddot{x}={f}_{1}(x,y)\dot{x}+{f}_{2}(x,y)\dot{y}={f}_{1}(x,a(x,\dot{x}))\dot{x}+{f}_{2}(x,a(x,\dot{x}))g(x,a(x,\dot{x}))$

where ${f}_{1}$ and ${f}_{2}$ are the partial derivatives of $f$.

$\ddot{x}={f}_{1}(x,y)\dot{x}+{f}_{2}(x,y)\dot{y}={f}_{1}(x,a(x,\dot{x}))\dot{x}+{f}_{2}(x,a(x,\dot{x}))g(x,a(x,\dot{x}))$

where ${f}_{1}$ and ${f}_{2}$ are the partial derivatives of $f$.