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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
up vote 2 down vote accepted

The general idea of proving that two solutions is to take their difference and prove that it's zero. For nonlinear equations this is less straightforward because the difference of two solutions will not solve the same equation. But it will solve some other equation - one in which zero is also a solution. Ultimately, we want to be in the position to apply the Gronwall lemma.

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.

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Nice answer +1. – Shuhao Cao Jul 16 '13 at 4:32

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