# Uniqueness theorem in non authonomous ODE

I am reading (the italian version of) Arnold's book on Ordinary Differential Equation. On page 29 (Chapter 1, paragraph 2: vectorial fields on the real line) problem 2 says: Given the differential equation $\dot x = v(x,t)$, where $v$ is a differentiable function, if there exists a solution $x = \varphi(t)$ which verifies the intial condition $\varphi(t_0)=x_0$, prove its unicity.

Arnold gives an indication: let $y = x-\varphi(t)$, confront $y$ with a convenient equation of the form $\dot x = kx$, $k\ne 0$.

My question is: how does the substitution $y=x-\varphi(t)$ works? Isn't $x$ equal to $\varphi(t)$? Should this substitution transform the non-authonomous equation in an authonomous one? How?

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I just looked through Arnold's (English) version. I think what he is saying is to compare with a suitable equation to the solution of the given x' and validate uniqueness. A way of doing this is to follow his setup using successive approximations (Picard iterates) and verifying that those can only lead to a unique solution (called "Completion of Proof" in my English version). "40 votes" gave a nice answer too, but Arnold did not introduce those concepts 9at least yet). Regards –  Amzoti Jul 16 '13 at 2:54
Arnold writes about comparing $y=x-\varphi(t)$ with a convenient equation in the form of $\dot y=ky$. Can we set up a suitable inequality using Picard iterates? –  zar Jul 16 '13 at 7:56
In my copy of the book, the page right before where this problem is written is the approach. He shows using $\phi_1(t)$, $\phi_2(t)$, ... using Successive Approximations (Picard iterates) and how that converges to the solution given by $y'=ky$. Do you see that in your book? –  Amzoti Jul 16 '13 at 12:41
Yes, but my problem is the uniqueness, not the convergence. I understand how the inequality works, but I can't see it in the non-authonomous case. I think 40 votes answer is in the righ direction, even if Arnold introduces those concepts later. Maybe in the real case things are simpler. –  zar Jul 17 '13 at 8:07

So, letting $x(t)=y(t)+\varphi(t)$ we transform the given equation into $$\dot y=v(y+\varphi(t),t)-\varphi'(t),\quad y(0)=0\tag1$$ One solution of (1) is $y\equiv 0$. To show there is no other, we need an inequality of the form $|\dot y|\le k|y|$; then Gronwall finishes off the problem thanks to $y(0)=0$.
Since $\varphi'(t)=v(\varphi(t),t)$, the mean value theorem helps: $$|\dot y|= |v(y+\varphi(t),t)-v(\varphi(t),t)| \le |y| \sup \left|\frac{\partial v}{\partial x}\right|$$ So, if we can control the derivative of $v$ with respect to the first variable, we are done. The derivative isn't really necessary: the Lipschitz condition suffices.