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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?

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2 Answers 2

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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 $$ \dddot y=a\ddot y-\dot y+(\ddot y-a\dot y+y)(\dot y-ay-b)-c. $$

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  • $\begingroup$ Thanks for your great answer :) Just to make sure that I got the main point: Under certain conditions a system of first order ODEs can always be transformed into a higher order ODE. So my initial naive guess that first order ODE systems must attain a certain "nice and simple" form if we want to be able to transform them was wrong. Right? $\endgroup$
    – Philipp
    Aug 16, 2021 at 20:12
  • $\begingroup$ @Philipp Yes, that's right. It can be added that sometimes the conditions can be easily checked and the answer is obvious, and sometimes not. $\endgroup$
    – AVK
    Aug 16, 2021 at 20:22
  • $\begingroup$ @AVK How common is it that we can do this? i.e. How likely is it that the various lie derivatives of a $\phi(x)$ will be functionally independent? $\endgroup$
    – Alex Gower
    Nov 3, 2022 at 13:34
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    $\begingroup$ @Alex Gower Likely - in what sense? On the one hand, if we consider linear systems with constant coefficients, we have that the system is equivalent to an equation if the Jordan form of its matrix consists of one block. It's not very common. On the other hand, if the system under consideration is a mathematical model, then reducibility to the equation is a property of the modeled object $\endgroup$
    – AVK
    Nov 4, 2022 at 4:20
  • $\begingroup$ Dear @AVK, do you know of a book that treats this? The (elementary) ODE books I know don't treat the notion of equivalent ODEs. For one I don't know what functionally independent is, and the other things you wrote down don't really ring a bell, but I think this is the way to go, so I'm curious. $\endgroup$
    – Sha Vuklia
    Sep 11, 2023 at 19:19
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It is also possible to transform slight modification of your example into a higher order ODE. For instance: \begin{align} y_1' &= 2y_2, \\ y_2' &= 6y_2 y_1. \end{align} Just define new state variables $z_1 = y_1$ and $z_2 = 2 y_2$. Then in this new variables you have \begin{align} z_1' &= z_2, \\ z_2' &= 3z_2 z_1, \end{align} which gives $$ z_1'' = 3z_1'z_1 \quad \text{and hence}\quad y_1'' = 3y_1' y_1. $$ Clearly you can do less obvious state variable transformations. I hope this solves your confusion.

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  • $\begingroup$ hmm, as far as I have understood the equations of the ODE system of first order must be "interlinked" such that one equation can be derived from another by taking the derivative -except the last one. The last ODE can be any ODE of first order. You have avoided this issue by only considering two ODEs of first order. Maybe my question was too imprecise. But what if we only consider systems of more than $2$ equations? $\endgroup$
    – Philipp
    Aug 16, 2021 at 17:22
  • $\begingroup$ @Philipp I am not sure, if i understand you correctly. You were asking why someone is even asking, if the reverse transformation (system of first oder => single ode of higher order) is possible, because you were thinking that only a very special class of ode systems are worth to consider. However, I wanted to show you that your class is certainly too small, as there are also other systems that can be transformed to a single ode. I don't say that your professor is wrong. I just wanted to point out that you might be too hasty in your reasoning. $\endgroup$ Aug 16, 2021 at 19:03

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