I have the IVP:

$x'=x\ln(1+\lVert x \rVert^2)$ with $x(0)=x_0\in\mathbb{R}^n$

and have to prove that it has an unique solution that can be extended on $[t_0,\infty)$. ($t_0$ is not defined anywhere in the problem so I think it refers to the initial condition $x(t_0)=x_0$)

To prove the uniqueness I have defined $f(x)=x\ln(1+\lVert x \rVert^2)$. Since the $\ln$ is continuous for positive numbers and $\lVert x \rVert^2$ is always positive, $f(x)$ is contionuos. And because the derivative $f'(x)=\ln(x^2+1)+\frac{2x^2}{x^2+1}$ is also continuous, because $x^2$ is always positive, $f(x)$ is locally Lipschitz at the variable x. Therefore the uniqueness and existence theorem applies and the IVP has a unique solution.

For the extension of the solution I am not sure how to proceed.

In the script from where I have this problem there is a theorem that states that if $f(x)$ is continuous and locally Lipschitz at the variable x, then the IVP has a unique solution on the maximum interval of existence $(T_-,T_+)$.

Should I try to prove that $[t_0,\infty)$ is the maximum interval of existence?

  • $\begingroup$ $x$ is a vector, your $f'$ has $xx^T$ in the second term. $\endgroup$ Jan 29 at 14:40
  • $\begingroup$ Could you explain that a bit further, please? And is the reasoning for existence and uniqueness otherwise okey? $\endgroup$
    – user1142625
    Jan 29 at 16:17
  • $\begingroup$ Yes, apart from that everything is correct. You have to compute a Jacobi matrix, it should be clear that with $f_i(x)=x_ih(\|x\|^2)$ you get mixed partial derivatives $f_{i;j}=2x_ix_jh'(\|x\|^2)$. $\endgroup$ Jan 29 at 16:29

1 Answer 1


Use, as usual $v(t)=V(x(t))=\|x(t)\|^2$. Then $$ v'(t)=2x^Tx'(t)=2v(t)\ln(1+v(t))\le 2(1+v)\ln(1+v). $$ Using separation of variables, one gets $$ \ln(\ln(1+v(t)))-\ln(\ln(1+v_0))\le 2t $$ This means that the solution remains bounded (by a doubly exponential upper bound, but still, without singularities) over all finite intervals $[0,T]$ and can thus be extended to $[0,\infty)$.


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