The First one
$$\frac{du}{dt}=u^{1/3},\quad u(0)=0.$$
I showed that the above IVP does not satisfy the Lipschitz condition. Then the question asked if the initial condition is changed to
$$u(0)= u_0,\quad \text{where}\; u_0\neq 0$$
then the Lipschitz condition is satisfied ? I am not sure about this part.
The Second one
$$\frac{du}{dt}=u^{4/3}, \quad u(t_0)=u_0.$$
Here the function $u^{4/3}$ being continuous assures us the existence of the solution but in order to get the uniqueness it needs to satisfy the Lipschitz condition which I tried but was unable to get to a decisive point.
How to show this satisfies the Lipschitz condition ?
Best Answer
For the first, note that the slope becomes unbounded at $0$, so as long as you can find an interval avoiding $0$, you may make sure that the vector field is Lipschitz. Indeed, by the mean value theorem, $$ |u_1^{1/3}-u_2^{1/3}|=|u_1-u_2|\frac{1}{3c^{2/3}} $$ for $c$ between $u_1$ and $u_2$. Provided we have $u(0)=u_0\ne 0$, we can make sure that $\frac{1}{3c^{2/3}}$ is bounded by a constant independent of the $u_1$ and $u_2$ we pick.
Indeed, suppose without a loss of generality that $u(0)=u_0>0$, then for any $u_1,u_2$ inside $(u_0/2,+\infty)$ we have $$ |u_1^{1/3}-u_2^{1/3}|=|u_1-u_2|\frac{1}{3c^{2/3}}\leq |u_1-u_2|\frac{1}{3}\frac{2^{2/3}}{u_0^{2/3}} $$ since $\frac{1}{c^{2/3}}$ is no larger than when $c$ is smallest. Thus, for nonzero $u(0)$, we can find an interval on which $f(t,u)=u^{1/3}$ is Lipschitz.
Can you apply similar analysis for the second problem?