# Summary of the solutions of ODE IVP

I have been studying IVP of ODEs, and after reading through and learning all the necessary theorems and proofs, I was trying to compile a summary of certain conditions (where would be implied by theorems) that would imply the existence and uniqueness of the solutions.

I would appreciate help with checking if my reasoning and assumptions are correct.

So if we consider IVP:

$$\dot{x}=f(x);$$

$$x(t_0)=x_0$$

with $$f:\mathbb{R^n}\rightarrow\mathbb{R^n}$$, $$\; t_0 \in \mathbb{R}$$, $$\; x_0\in\mathbb{R^n}$$.

1. If $$f$$ is continuous, for all $$t\in\mathbb{R}; \; x_0\in\mathbb{R^n}$$, then there is at least one local solution to the IVP. (Due to Peano existence theorem)
2. If $$f$$ is only continuous, that is insufficient to imply the uniqueness of the local solution. ($$f$$ would need to be continuously differentiable, i.e. locally Lipschitz continuous for that)
3. If $$f$$ is continuously differentiable, for all $$t\in\mathbb{R}; \; x_0\in\mathbb{R^n}$$, then there is a unique local solution to the IVP. (since $$f$$ is continuously differentiable, then $$f$$ is locally Lipschitz continuous, which implies the uniqueness of the solution, which is implied by Peano existence theorem)
4. If $$f$$ is continuously differentiable, for all $$t\in\mathbb{R}; \; x_0\in\mathbb{R^n}$$, then there is a global solution to the IVP.(I am really not sure if this would be true at all, and if it were true, would it be unique?)
5. If $$f$$ is continuous and bounded, for all $$t\in\mathbb{R}; \; x_0\in\mathbb{R^n}$$, then there is a global solution to the IVP. (I would assume that since $$f$$ is bounded and continuous then it would be locally Lipschitz, but I am not sure if this holds? Also would it imply uniqueness?)
6. If $$f$$ is linear, for all $$t\in\mathbb{R}; \; x_0\in\mathbb{R^n}$$, then there is a global solution to the IVP. (I believe this holds since $$f$$ is linear, there is an explicit solution to IVP: $$x = \exp((t-t_0) A)x_0$$, would this solution be unique and why? I assume due to the Rxistence and Uniqueness theorem, but not sure how?)

So basically, I would need help with the 4th, 5th and 6th and to check the 1st, 2nd, and 3rd. I appreciate any comment and help.

• Number 4 isn't correct. Take $\dot x = x^2$ for example. You can solve it using separation of variables to get $x(t) = 1/(c-t)$ if $x_0 \neq 0$, which is not a global solution. You would need control over the derivative of $f$ to get global solutions (e.g. $f$ is globally Lipschitz). I doubt 5 is correct, but that might be harder to show. You could look for $f$ which are not globally Lipschitz but continuous and bounded. Oct 26, 2022 at 14:53
• @TrevorNorton so generally speaking, in order to have a global solution we need $f$ to be globally Lipschitz, locally in itself is not enough? Oct 26, 2022 at 14:56
• Yes. Locally Lipschitz can guarantee a unique local solution, but that's not enough to get a global solution (as the example I gave shows). Typically you can guarantee global solutions by showing $f$ is Lipschitz. Oct 26, 2022 at 14:59
• @TrevorNorton would that also guarantee the uniqueness of the global solution or just its existence? Oct 26, 2022 at 15:00
• Yes, I believe so. Lipschitz would imply local uniqueness for any initial condition, and thus the global solutions should also be unique. Oct 26, 2022 at 15:10