Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.