Solutions of an ODE when the function is continuous but not a lipschitz function.

Question

$$x' = 1+x^{2/3} \\ x(0) = 0$$

Does this differential equation have infinite solutions since it doesn't satisfy lipschitz condition? Is the solution set similar in shape to $x' = x^{2/3}$ with its characteristic paintbrush shape? Is there any analytic solution for the differential equation after separating variables? I'm stuck with $3(x+\arctan(x)) = t$

Answer 1

In this case the solution is unique and globally defined. If you consider the (strictly increasing) function $$ F(x) := \int_0^x \frac{1}{1+s^{2/3}}\, ds = 3(\sqrt[3]{x} - \arctan \sqrt[3]{x}), \qquad x\in\mathbb{R}, $$ then the solution is implicitly defined by $$ F(x) = t, $$ i.e. $$ x(t) = F^{-1}(t), \qquad t\in\mathbb{R}. $$

Answer 2

In general, an initial value problem of the form $$ x'=f(x), \quad x(t_0)=x_0, $$ where $f$ is continuous and positive for all $x\in\mathbb R$ (or positive for all $x\in\mathbb R$), then it enjoys global uniqueness.

Justification. Setting, $$ F(x)=\int_{x_0}^x\frac{dw}{f(w)}, $$ one observes that $$ \frac{d}{dt}F(x)=\frac{x'}{f(x)}=1, $$ and hence $F\big(x(t)\big)=t-t_0$ and finally $$ x(t)=F^{-1}(t-t_0). $$ This provides UNIQUENESS. Note that $F$ has an inverse as $F'>0$.