I know that you can always transform a higher order ODE into a system of first order ODEs.
In our lecture the professor made the statement that the reverse, namely transforming a given system of first order ODEs into one ODE of a higher order, is not always possible.
I was a bit confused by this statement because as far as I have understood the reverse is only possible in the very rare case where
\begin{align}
y_0'&=y_1\\
y_1'&=y_2\\
&\;\vdots\\
y_{n-1}'&=y_n\\
y_n'&=f(x,y_0,y_1, \cdots, y_n).
\end{align}
If the given system doesn't have this form you can't transform it into an ODE of $n$-th order.
Is this correct? Did I miss something?
Best Answer
An autonomous system of differential equations $$ \dot x= f(x),\quad x\in\Omega\subseteq\mathbb R^n,\quad f\in C^n(\Omega) $$ can be transformed to the form of a n-th order differential equation iff there is a $C^{n}(\Omega)$ function $\varphi:\; \mathbb R^n\to\mathbb R$ such that the system of functions $$ \Phi(x)=(\varphi(x),\mathbf F \varphi(x),\ldots,\mathbf F^{n-1} \varphi(x)) $$ is functionally independent on $\Omega$. Here $$ \mathbf F \varphi(x)=\sum_{i=1}^{n} f_i(x)\frac{\partial \varphi}{\partial x_i}, \;\mathbf F^2 \varphi(x)= \mathbf F( \mathbf F\varphi(x)),\;\ldots,\; \mathbf F^{n-1} \varphi(x)= \underbrace{\mathbf F(\ldots \mathbf F(\mathbf F}_{n-1}\varphi(x))\ldots). $$ The mapping $\mathbf F \varphi(x)$ (or $L_f \varphi(x)$) is sometimes called the Lie derivative and sometimes the derivative along the trajectories of the system (depending on the field of science involved).
If such a function $\varphi(x)$ is known, then the equivalent differential equation is $$\tag{1} y^{(n)}=\mathbf F^n \varphi(x){\Large|}_{x=\Phi^{-1}(\bar y)},\quad \bar y= (y,\dot y,\ldots,y^{(n-1)}). $$ It can be obtained by using the change of variables $$ y= \varphi(x),\;\dot y= \mathbf F \varphi(x),\;\ddot y= \mathbf F^2 \varphi(x),\;\ldots\; y^{(n-1)}=\mathbf F^{n-1} \varphi(x). $$
For example, consider the Rossler system $$ \left\{ \begin{array}{rcl} \dot x_1&=&-x_2-x_3,\\ \dot x_2&=&x_1+ax_2,\\ \dot x_3&=&x_1x_3-bx_3+c. \end{array} \right. $$ If we guess that we can choose $\varphi(x)=x_2$, then we can get the differential equation. The change of variables is $$ y=x_2, $$ $$ \dot y= \mathbf F y= \mathbf F x_2= (-x_2-x_3)\cdot 0+ (x_1+ax_2)\cdot1+ (x_1x_3-bx_3+c)\cdot0= x_1+ax_2, $$ $$ \ddot y= \mathbf F^2 x_2= \mathbf F (x_1+ax_2)= ax_1+(a^2-1)x_2-x_3; $$ we can also express $x_1,x_2,x_3$ in terms of $y,\dot y,\ddot y$: $$\tag{2} x_1=\dot y-ay,\quad x_2=y,\quad x_3=-\ddot y+a\dot y-y. $$ Finally, we can obtain the right part of the differential equation (1): $$ \mathbf F^3 x_2= \mathbf F(ax_1+(a^2-1)x_2-x_3) $$ $$ =a(-x_2-x_3)+(a^2-1)(x_1+ax_2)-(x_1x_3-bx_3+c). $$ Applying (2) to express $x_1,x_2,x_3$ in terms of $y,\dot y,\ddot y$ , we obtain the equation $$ \dddot y=a\ddot y-\dot y+(\ddot y-a\dot y+y)(\dot y-ay-b)-c. $$