Prove uniqueness theoremn via Gronwall inequality

Question

A question says:

Prove Theorem 1.7 (Uniqueness). Hint: suppose that $x$ and $x^*$ are distinct solutions to the same IVP (from the same initial point). Consider the function $\nu(t)=||x(t)-x^*(t)||_1$ to he integral representation of the solutions $x$ and $x^*$ and apply the Gronwall inequality.

The gronwall inequality is given in my notes:

$u(t) <= a + b \int_0^t u(s) ds$

then we have

$u(t) <= ae^{bt}$

My attempt:

Consider the ode: $\dot{x} = f(t, x)$

Consider: $\nu(t)=||x(t)-x^*(t)||$

$x(t) - x^*(t) <= a + b\int_0^t[x(s)-x^*(s)]ds$

$\implies x(t) - x^*(t) <= ae^{bt}$

but $x(t) = x_0 + \int_{t_0}^t f(s,x(s)) ds$

and $x^*(t) = x_0 + \int_{t_0}^t f(s,x^*(s)) ds$

$x(t) - x^*(t) = \int_{t_0}^t [f(s,x(s))-f(s,x^*(s))] ds$

$||x(t) - x^*(t)|| = ||\int_{t_0}^t [f(s,x(s))-f(s,x^*(s))] ds||$

But: $||\int_{t_0}^t [f(s,x(s))-f(s,x^*(s))] ds||<= \int_{t_0}^t ||f(s,x(s))-f(s,x^*(s))|| ds$

I don't know how I can change from $f$ to $x(t)$ now... Maybe I've gone off on a wrong tangent?

Answer 1

First of all, an important correction. It should be:

If $u(t) \le a + b \int_{t_0}^t u(s) \,ds$ for all $t\in[t_0,T]$, then $u(t) \le ae^{bt}$ for all $t\in[t_0,T]$.

What you wrote is not true in general.

On any compact set you can write $$\int_{t_0}^t \|f(s,x(s))-f(s,x^*(s))\| \,ds\le L\int_{t_0}^t \|x(s)-x^*(s)\| \,ds$$ for some constant $L$. More precisely, in some neighborhood of $t_0$ your solutions $x$ and $x^*$ are controlled by Gronwall's lemma and so they will have a uniform bound; this is what allows you to consider a compact set.

After this technical issue you can apply once more Gronwall's lemma after noting that $$ \nu(t)\le L\int_{t_0}^t \nu(s) \,ds $$ in that neighborhood of $t_0$.