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:
- I assumed $k=1$; I don't know how this could be generalised to arbitrary dimensions.
- 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.
Best Answer
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.