Does inequality hold for IVP solutions?

Question

Let $u(t)$, $v(t)$ be solutions of the IVP $$ u'=F(u), u(t_0)=u_0 \\ v'=F(v), v(t_0)=v_0 $$ F is a Lipschitz function with constant $L$. Does this inequality hold? $$ |u(t) − v(t)| \leq |u_0 − v_0|e^{L(t−t_0)} $$ This problem belongs to my homework assignment.

Answer 1

This is easy using Grönwall's inequality. Without loss of generality let $u_0 > v_0$ (so $u(t) > v(t)$ for all $t\geq t_0$). Let $z = u-v$, so $$z' = F(u)-F(v)\leq L(u-v) = Lz$$ and $z(t_0) = u_0-v_0$. Then, by Grönwall, $$u(t)-v(t) = z(t)\leq z(t_0)\exp\left[\int_{t_0}^t L\,\mathrm{d}t\right] = (u_0-v_0)e^{L(t-t_0)}$$

Answer 2

Define $f=u-v$.

Its derivative $f'=u'-v'=F(u)-F(v)$ and we can bound $f$ by: $$f'\leq |F(u)-F(v)|$$ Since $F(\cdot)$ is Lipschitz, we can put: $$f'\leq |F(u)-F(v)|\leq L|u-v|=L|f|$$ Assuming that $f>0\,\forall t$ we can solve: $$\frac{f'}{f}\leq L\quad s.t.\quad f(0) = |u(0)-v(0)|$$ Giving: $$f=|u-v|\leq f(0)\exp{(L(t-t_0)})=|u(0)-v(0)|\exp{(L(t-t_0)})$$