Existence of one-parameter family of diffeomorphisms

Let $$v(t,x): \mathbb{R}^n \mapsto \mathbb{R}^n$$ be Lipschitz with bounded divergence $$\nabla \cdot v$$. How do I prove that the the solution $$X_t$$ of the initial-value problem

$$\dot{X_t} = v(t, X_t) , \quad X_0 = id ,$$

exists and constitutes a one-parameter family of diffeomorphisms?

• Take $x'=y^2$, $y'=x^2$. The divergence of the vector field is zero. The maximally defined solution satisfying the initial conditions $x(0)=y(0)=1$ equals $(\tfrac{1}{1-t},\tfrac{1}{1-t})$, where $t\in(-\infty,1)$, and it blows up in finite time. Hence, the solution operator does not constitute a one-parameter family of diffeomorphisms. – user539887 Jun 13 at 7:31
• @user539887 That function isn't Lipschitz. – Ben Jun 13 at 8:26
• So, you mean globally Lipschitz. In that case, see, e.g., Corollary 2.6 on p. 41 of Gerard Teschl's Ordinary Differential Equations for the global existence of solutions, and Theorem 2.10 on p. 46 of that same book for the $C^k$ dependence on initial conditions. – user539887 Jun 13 at 12:30
• @user539887 I'm interested in the proof that each $X_t$ is actually a diffeomorphism. Usually, when $v$ is time independent, the proof proceeds by demonstrating that $X_{-t}$ is the inverse of $X_t$. When $v$ is time dependent, this isn't an option. – Ben Jun 13 at 23:12
• Take the vector field $\tilde{v}$ defined on $\mathbb{R}\times\mathbb{R}^n$ by the formula $\tilde{v}(t,x)=(1,v(t,x))$, $t\in\mathbb{R}$, $x\in\mathbb{R}^n$. Incidentally, notice that the original $v$ does not generate a one-parameter family of diffeomorphisms (only $\tilde{v}$ does). – user539887 Jun 14 at 8:33