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.

  • 1
    $\begingroup$ 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. $\endgroup$ Oct 26, 2022 at 14:53
  • $\begingroup$ @TrevorNorton so generally speaking, in order to have a global solution we need $f$ to be globally Lipschitz, locally in itself is not enough? $\endgroup$ Oct 26, 2022 at 14:56
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 26, 2022 at 14:59
  • $\begingroup$ @TrevorNorton would that also guarantee the uniqueness of the global solution or just its existence? $\endgroup$ Oct 26, 2022 at 15:00
  • $\begingroup$ Yes, I believe so. Lipschitz would imply local uniqueness for any initial condition, and thus the global solutions should also be unique. $\endgroup$ Oct 26, 2022 at 15:10

1 Answer 1


You are missing out on non-autonomous systems.The main difference is that all extended conditions only apply to the state vector, in time only continuity is ever required.

  1. Lipschitz and local Lipschitz conditions are weaker, more general than continuous differentiability. This is mostly encountered in making proofs more streamlined and capturing some cases of piecewise smooth right side functions, such as produced by including absolute values in its evaluation graph.

  2. You get a unique solution that leaves every compact set in the time-space domain. This includes cases of divergence to some state at infinity in finite time.

  3. If the right side is bounded or linearly bounded, the solution is exponentially bounded and divergence can not happen in finite time. A global Lipschitz condition implies a linear bound, but can by itself also be used for a direct proof of global existence, that is, existence of a solution with the whole time line as domain.

  4. Falls under "linearly bounded", but linear systems form a special case that is the basis for many quantitative and geometric explorations.

  • $\begingroup$ Thank you for your answer, unfortunately, we still haven't covered the part about non-autonomous systems in the class, I was only trying to summarize everything that comes before it, thus I find it a bit difficult to follow your reasoning. $\endgroup$ Oct 26, 2022 at 15:20
  • $\begingroup$ Thanks a lot Lutz for the answer (+1) $\endgroup$ Oct 26, 2022 at 15:24

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