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Given $f:\mathbb R^n \to \mathbb R^n$ is locally Lipschitz and $\Phi\in C^1 (\mathbb R^n, \mathbb R)$ with $\nabla\Phi(x)=f(x)$ for $x\in \mathbb R^n$, I would like to show that the initial value problem $$\begin{cases}\begin{split} y'(t)=-f(y(t))\\ y(0)=y_0 \end{split}\end{cases}$$

has locally unique solutions. I tried applying the P-L theorem, but it did not help. How do is this done?

I have already considered the $\mathbb R^2$ case, ie if $f:\mathbb R _{\geq 0} \to \mathbb R_{\geq 0}$, $y'=f(y)$, $y(0)=y_0$, then it becomes easy to show that if $f$ is lipschitz, any solution is unique. However, it does not appear to work in this case.

Bernard
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YamahaJacoby
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  • Could you expound a little more on the proof for the 2D case and where you can not apply the same arguments to the higher-dimensional case? The P-L theorem should not make any distinctions, for local existence and uniqueness you should also not need to know that $f$ is a gradient field. – Lutz Lehmann May 18 '21 at 13:04
  • @LutzLehmann You are right, this is from an exercise in a book I am reading, and I copied the first part without thinking. That f is a gradient field is not relevant here. In the 2 dimensional case, I first extended f, and then showed that since $f$ attains a maximum on any interval for some $t_\alpha$ there exists a solution if f is lip. I just did some googling, and it is remarkably similar to an answer you posted on a similar thread, here. This does not seem to work in this case. – YamahaJacoby May 18 '21 at 13:18
  • What do you mean you tried to apply but it did not help? This is fairly vanilla stuff, what went wrong? – copper.hat May 18 '21 at 13:30
  • Could you cite your version of the Picard-Lindelöf theorem and then what conditions you find not satisfied in the assumptions of this task? – Lutz Lehmann May 18 '21 at 13:50
  • Let $f : I\times G \to \mathbb R^n | f\in C^0(I\times G)$ and $f$ be locally Lipschitz with regards to x. Then there exists a $\delta > 0$ and a unique function $u\in C^1 (J_\delta, \mathbb R^n)$ with $J_\delta := [t_0 - \delta, t_0+\delta]$ such that $(t, u(t)) \in (I \times G)$ for all $t \in J_\delta$ such that u solves the IVP on $J_\delta$. Nothing here does not contradict it not working, I just simply cannot find such a $\delta$. – YamahaJacoby May 18 '21 at 14:13
  • There is nothing to find, the theorem states that such a $δ$ exists. For some concrete $f$ one could ask for concrete values of $δ$ and how large they can be chosen, but in this task that is not required. – Lutz Lehmann May 18 '21 at 15:06
  • @LutzLehmann So does this just immediately follow from the P-L theorem? If so, it would seem a bit too easy. Maybe I am overthinking a few things here. – YamahaJacoby May 18 '21 at 16:26
  • Yes, for the task as written it is that easy. Usually the follow-up is to show that any solution can be indefinitely prolonged and will converge to some stationary point if $\Phi$ is bounded below. – Lutz Lehmann May 18 '21 at 16:49
  • @LutzLehmann Thank you. There is one more part of this that perplexes me; it also asks to show that $\Phi(y(t))$ falls monotonically for each solution of the IVP. Intuitively, this would seem somewhat obvious; looking at the derivative of $\Phi(y(t))$ looks like y solves the initial value problem. How does this show that Phi is decreasing? How do I express this formally? – YamahaJacoby May 19 '21 at 09:59
  • This follows quite directly if you apply the chain rule properly and apply what the gradient is. – Lutz Lehmann May 21 '21 at 13:52

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