We recently introduced the following theorem in my current lecture:

Theorem (Stability): Let $f(t,u)$ and $g(t,u)$ be two continuous functions on a cylinder $D = I \times \Omega$ where the interval $I$ contains $t_0$ and $\Omega$ is a convex set in $\mathbb{R}^d$. Furthermore, let $f$ admit a Lipschitz condition with constant $L$ on $D$. Let $u$ and $v$ be solutions to the IVP

$$u' = f(t,u)\,\text{ for all }t \in I,\hspace{30pt}v(t_0) = u_0$$ $$v' = g(t,v)\,\text{ for all }t \in I,\hspace{30pt}v(t_0) = v_0$$

Then, there holds:

$$|u(t) - v(t)| \leq e^{L|t-t_0|}\cdot(|u_0 - v_0| + \int_{t_0}^t \max_{x \in \Omega} |f(s,x) - g(s,x)|\,ds)$$

I do understand the theorem per se, but have no clue where to apply this theorems, since $u$ and $v$ are solutions to completely different IVPs. Where is this theorem useful? I can't imagine where this is going..

An intuition why this theorem is useful would be amazing.

  • 2
    $\begingroup$ For instance, if you perturb initial conditions and/or equation a little then the difference between the solution of the original IVP and of the perturbed IVP is bounded, and the bound can be estimated by the given formula. Incidentally, you made a misprint: $\lvert u(t) - v(t)\rvert$, not $\lvert u(t) v(t)\rvert$. $\endgroup$
    – user539887
    Jul 1, 2019 at 11:34
  • 1
    $\begingroup$ Another comment: The property described in your theorem is called not stability, but rather Lipschitz dependence on initial conditions/parameters. By stability one usually means the property that the solutions of the original and of the perturbed IVP are close to each other for all $t\ge 0$. $\endgroup$
    – user539887
    Jul 1, 2019 at 11:38
  • $\begingroup$ @user539887 Thank you for the insight! Just to be clear: In the usual application case: $f(t) \approx g(t)$ and $u_0 \approx v_0$? I didn't chose the name stability. I just copied the name from my lecture. $\endgroup$ Jul 1, 2019 at 15:05
  • $\begingroup$ Yes, that's true. $\endgroup$
    – user539887
    Jul 3, 2019 at 18:20


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy