# Show that an IVP has a unique solution that can be extended on $[t_0, \infty)$.

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?

• $x$ is a vector, your $f'$ has $xx^T$ in the second term. Jan 29 at 14:40
• Could you explain that a bit further, please? And is the reasoning for existence and uniqueness otherwise okey?
– user1142625
Jan 29 at 16:17
• 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 $f_{i;j}=2x_ix_jh'(\|x\|^2)$. Jan 29 at 16:29

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 $$[0,\infty)$$.