Consider the differential equation $\dot{x} = f(t, x)$ with $f$ being a continuously differentiable function from $G \to \mathbb{R}^n$, with $G \subset \mathbb{R} \times \mathbb{R}^n$ an open subset. Let $f$ be (globally) Lipschitz on $G$ with respect to $x$, with a Lipschitz constant $L > 0$.
I want to prove: if $x_1$ and $x_2$ are two solutions of the differential equation $\dot{x} = f(t, x)$ that are defined on an interval $[a, b] \subset \mathbb{R}$, then we have that:
$$||x_1(t) - x_2(t)|| ≤ ||x_1(a) - x_2(a)||e^{L(t-a)}$$
for all $t \in [a, b]$.
I was looking at using the Grönwall's inequality to solve this.
I see that, in order to use Grönwall's inequality, I need to show that
$$ ||x_1(t) - x_2(t)|| ≤ ||x_1(a) - x_2(a)|| + L\left\lvert\int_a^t ||x_1(t) - x_2(t)||ds\right\rvert$$
for all $t \in [a, b]$. However, that's where I got stuck. This inequality is obviously true for $t = a$, but how can I show this for $t > 0$? I know that I need to somehow use the fact that $f(t, x)$ is Lipschitz continuous with the constant $L > 0$; but why does that tell me that the integral on the RHS always "grows fast enough"?
First, the absolute value around the integral in unnecessary, since norms are nonnegative, the integral will be nonnegative.
Let's start by creatively adding $0$ and using the Triangle Inequality: $$\|x_1(t) - x_2(t)\| = \|x_1(t) - x_1(a) + x_1(a) -x_2(a) - (x_2(t) - x_2(a))\|$$ $$ \leq \|x_1(a) - x_2(a)\| + \|x_1(t) - x_1(a) - (x_2(t) - x_2(a))\| $$ By the Fundamental Theorem of Calculus: $$ = \|x_1(a) - x_2(a)\| + \left\| \int_a^t x_1'(s) - x_2'(s)\,ds \right\| $$ Since $x_1$ and $x_2$ are solutions to the ODE: $$= \|x_1(a) - x_2(a)\| + \left\| \int_a^t f(s,x_1(s)) - f(s,x_2(s)) \,ds \right\|$$ Now we use the fact given here and we get $$\leq \|x_1(a) - x_2(a)\| + \int_a^t \|f(s,x_1(s)) - f(s,x_2(s))\| \,ds$$ By the Lipschitz condition and monotonicity of the integral: $$ \leq \|x_1(a) - x_2(a)\| + \int_a^t L \|x_1(s) - x_2(s)\|\,ds $$ and the result follows.