# Lipschitz condition on a second order nonlinear ODE?

Preliminaries: Let the matrix norm be $$\sqrt{\sum_{j=1}^n\sum_{i=1}^n a_{ij}^2}=||\mathbf A||.$$

I am trying to prove uniqueness and existence of a second order nonlinear ODE (Ordinary Differential Equation).

So I need to show f a function is continuous and satisfies a Lipschitz condition.

Take for example the second order nonlinear ODE $$x''=-\cos(x).$$ Now simplify to a first order ODE by letting $$x'=y$$ and $$y'=x''=-\cos(x).$$ So how do I set up the matrix $\mathbf A$ is it $$\begin{pmatrix} 0 & 1 \\ -\cos(x) & 0 \end{pmatrix}$$ How do I show the lipschitz condition for this. Do I just take the $${||\mathbf Ax_1 - \mathbf Ax_2 || \over |x_1-x_2|} \le L$$ where L is you Lipschitz constant.

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There is no matrix $A$ here, since the equation is nonlinear. Matrices help to represent linear equations. The system can be written as $\dot{\vec{x}}=F(\vec x)$ where $F(x_1,x_2)=(x_2,-\cos x_1)$. (By the way, you should use \cos rather than cos, for better typography).
The Lipschitz condition for $F$ can be checked one component at a time: a vector- valued function is Lipschitz if and only if every component is a Lipschitz function. The only question here is whether $\cos x$ is a Lipschitz function of $x$. And the answer is yes, because it is differentiable with bounded derivative. Any such function is Lipschitz, thanks to the Mean Value Theorem.