2
$\begingroup$

Let $A(t)\in \mathcal M_{n\times n}(\mathbb R)$ periodic, i.e. $$\exists \quad T>0:\forall t\in \mathbb R, A(t+T)=A(t).$$ Consider the system $$ \dot x(t)=A(t)x(t),\quad x(t)\in \mathbb R^n.$$ It's written in my course that if $t\mapsto A(t)$ is continuous, the existence and uniqueness theorem for ODE implies the existence of $n$ linearly independents solutions.


I don't understand in what this "existence and uniqueness theorem for ODE" implies existence of $n$ linearly independents solution. Could someone explain why ?

This theorem says (more or less) :

If $F:\mathbb R\times \mathbb R^n\to \mathbb R^n$ is locally Lipschitz w.r.t. the second variable, then the Cauchy problem $$\begin{cases}\dot x(t)=F(t,x(t))\\ x(0)=x_0\end{cases}$$ has a unique (local) solution.

So indeed, we take $F(t,x)=A(t)x$ which is obviously globally Lipschitz. But in one case we have a Cauchy problem, and in the other case we have an ODE without initial condition.

$\endgroup$

1 Answer 1

3
$\begingroup$

A basis for the solution set is given by the $n$ functions $g_k(t)$ (where $1 \le k \le n$), where $x(t)=g_k(t)$ is the unique solution of the initital value problem $$ \dot x = A x ,\qquad x(0) = e_k , $$ where $e_k$ is the $k$th standard basis vector for $\mathbf{R}^n$.

Proof: If $x=z(t)$ is any solution of $\dot x = Ax$, let $c=z(0)$. Then $x=\sum c_k g_k(t)$ is also a solution, with the same initial value as $z$ at $t=0$, so by uniqueness $z=\sum c_k g_k$. This shows that any solution is a linear combination of the functions $g_k$. And if the linear combination $\sum c_k g_k(t)$ is the zero function, then in particular it's zero at $t=0$, $\sum c_k e_k = 0$, so all $c_k=0$. This shows that the functions $g_k$ are linearly independent. QED.

(By the way, the periodicity of $A(t)$ plays no role here, only the continuity.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .