Using Gronwall to prove bi-Lipschitz

Question

I am working through a proof which, for fixed $x \in \mathbb{R}^k$ considers an initial value problem of the form:

$$\frac{d}{d t}u_t(x)=v_t(u_t(x)), \quad u_t(0)=x$$

where $u_t:\mathbb{R}^k \rightarrow \mathbb{R}^k$ and $t \in [0,1]$. The function $v_t$ also satisfies a lipschitz condition of the form:

$|v_t(u_1)-v_t(u_2)|\leq A|u_1-u_2| \quad \text{ for all } u_1,u_2\in \mathbb{R}^k$,

where $A \in \mathbb{R}$ is constant.

The proof I am working through claims that the function $u_t:\mathbb{R} \rightarrow \mathbb{R}^k$ defined by this initial value problem must be bi-Lipschitz with respect to $x$, but does not explain why. E.g. it claims there must exist constants $B_1,B_2 \in \mathbb{R}_+$ such that:

$B_1|x_1-x_2|\leq |u_t(x_1)-u_t(x_2)|\leq B_2|x_1-x_2| \quad \text{ for all } x_1,x_2\in \mathbb{R}^k$.

Can someone prove it must be bi-lipschitz please?

My attempt at this is below:

---

Suppose for ease $k=1$ and assume wlog $u_t(x_1) \geq u_t(x_2)$. As $|v_t(u_t(x_1))-v_t(u_t(x_2))|\leq A|u_t(x_1)-u_t(x_2)|$ we have:

$$- A(u_t(x_1)-u_t(x_2)) \leq v_t(u_t(x_1))-v_t(u_t(x_2))\leq A(u_t(x_1)-u_t(x_2))$$

Or, similarly: $$v_t(u_t(x_1))-v_t(u_t(x_2))\leq A(u_t(x_1)-u_t(x_2))$$ $$-[v_t(u_t(x_1))-v_t(u_t(x_2))]\leq A[-(u_t(x_1)-u_t(x_2))]$$

So applying the Gronwall inequality, we get: $$u_t(x_1)-u_t(x_2)\leq (u_0(x_1)-u_0(x_2))\text{exp}\bigg(\int_0^t A ds\bigg)= (x_1-x_2)e^{At} \leq (x_1-x_2)e^{A}$$ $$-(u_t(x_1)-u_t(x_2))\leq -(u_0(x_1)-u_0(x_2))\text{exp}\bigg(\int_0^t -A ds\bigg)= -(x_1-x_2)e^{-At} \leq -(x_1-x_2)e^{-A}$$

So:

$$e^{-At}|x_1-x_2|\leq |u_t(x_1)-u_t(x_2)|\leq e^{At}|x_1-x_2| $$

I think the proof must look something like this, but there are two problems with my approach that I can't see how to fix:

1. I assumed $k=1$; I don't know how this could be generalised to arbitrary dimensions.
2. I started by saying wlog $u_t(x_1) > u_t(x_2)$, but to apply Gronwall I guess this assumption should hold for all $t\in[0,1]$, and this may not be reasonable to assume.

Please could someone either give me hints on how to adjust my proof or provide an alternative proof.

Answer 1

Let us integrate the IVP to obtain $$ u_t(x) = x + \int_0^t v_s(u_s(x))\ \mathrm{d}s. $$ Thus for $x_1, x_2 \in \mathbb{R}^k$, we have $$ \begin{align*} |u_t(x_1) - u_t(x_2)| &= \left|x_1 - x_2 + \int_0^t v_s(u_s(x_1))-v_s(u_s(x_2))\ \mathrm{d}s\right| \\ &\le |x_1 - x_2| + \int_0^t|v_s(u_s(x_1))-v_s(u_s(x_2))|\ \mathrm{d}s\\ &\le |x_1 - x_2| + \int_0^t A|u_s(x_1)- u_s(x_2)|\ \mathrm{d}s \end{align*}. $$ Consequently, Grönwall's inequality implies $$ |u_t(x_1) - u_t(x_2)| \le |x_1-x_2|e^{At} \le |x_1-x_2|e^{A}, $$ which proves that $u_t(x)$ is Lipschitz in $x$.

Using now the reverse triangle inequality, the above and the Lipshcitz property of $v_t$, we get $$ \begin{align*} |u_t(x_1) - u_t(x_2)| &= \left|x_1 - x_2 + \int_0^t v_s(u_s(x_1))-v_s(u_s(x_2))\ \mathrm{d}s\right| \\ &\ge |x_1 - x_2| - \int_0^t|v_s(u_s(x_1))-v_s(u_s(x_2))|\ \mathrm{d}s\\ &\ge |x_1 - x_2| - \int_0^t A|u_s(x_1)- u_s(x_2)|\ \mathrm{d}s \\ &\ge |x_1 - x_2| - \int_0^t Ae^{As} |x_1 - x_2| \mathrm{d}s = (2-e^{At}) |x_1 - x_2| \end{align*}. $$ Thus if $A$ is so small that $e^A < 2$ (or for small enough $t$, $e^{At} < 2$), we see that $u_t$ is Bi-Lipschitz.