# An ODE is linear if and only if the maximal solutions are a linear space?

Let $$I$$ be a non trivial interval of $$\mathbb R$$, let $$f : I \times \mathbb R^n \to \mathbb R^n$$ and consider the following ordinary differential equation (ODE): $$$$\tag{\mathscr E}\label{ode} y'(t) = f\big(t,y(t)\big)$$$$

Suppose that:

1. all the maximal solutions of \eqref{ode} are global (defined on $$I$$) ;
2. and the set $$S$$ of such global solutions is a linear space.

Is it true that \eqref{ode} is linear ? i.e. that it exists $$A : I \to \mathrm M_n(\mathbb R)$$ (the square matrices of size $$n \times n$$) such that every differentiable function $$y: I \to \mathbb R^n$$ is a solution of \eqref{ode} if and only if it is a solution of the linear ODE $$$$\tag{\mathscr L}\label{ode2} y'(t) = A(t)\,y(t)$$$$

Remark. I did not make any assumptions on $$f$$ and on the dimension of $$S$$ but if needed, we can assume for $$f$$ that the Picard–Lindelöf theorem holds ($$f$$ is continuous in $$t$$ and Lipschitz continuous in $$y$$, or even one can assume that $$f$$ is $$\mathscr C^1$$) and we can assume that $$\dim S = n$$, but maybe 1 and/or 2 implies these assumptions.

This is a partial solution, assuming that $$f$$ is analytic on $$I\times R$$, and so solutions are analytic as well, and that the space $$E$$ of solutions is of the same dimension $$n$$ as vectors $$y$$ and $$f$$.
We represent solutions $$y$$ and the function $$f$$ in the right hand side as column vectors, as usual. Let $$y_1,\ldots,y_n$$ be a basis in $$E$$. Then every vector $$y\in E$$ can be written as a linear combination $$y=c_1y_1+\ldots+c_ny_n.\quad\quad\quad\quad\quad\quad\quad\quad (1)$$ Solving this by Cramer's rule we obtain $$c_k$$ as a ratio of two determinants made of coordinates of $$y$$ and $$y_j$$. Since the coordinates of $$y$$ make one column in the numerator, it is clear that this ratio is a linear function of coordinates of $$y$$, so we can write $$c_j=a_j(t)y,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (2)$$ for some row vector $$a_j$$ which depends on $$t$$. Now, since all elements of $$E$$ satisfy your non-linear equation we must have for every $$y\in E$$ of the form (1): $$y'=\sum_{j=1}^n c_jy_j^\prime=\sum_{j=1}^nc_jf(t,y_j),$$ Substituting our formula for $$c_j$$ we obtain that $$y'=A(t)y,\quad\mbox{where}\quad A(t)=\sum_jf(t,y_j(t))a_j(t)$$ (column vectors $$f(t,y_j(t))$$ times row vectors $$a_j$$ give you $$n\times n$$ matrices.) There is a little problem with the denominator in the Cramer Rule: one has to show that it cannot be zero at some point. If all functions $$f,y_j$$ are analytic, this can only happen at isolated points. Away from these isolated points $$t_k$$ both equations $$y'=f(t,y)$$ and $$y'=A(t)y$$ have the same set of solutions, therefore $$f(t,y)=A(t)y$$ on some open set of $$(t,y)$$, and since all functions are analytic they must coincide.
• Thanks a lot ! The case $f$ analytic already covers a lot of cases, including almost all in applications. May 15, 2022 at 1:32