The IVP, $\dot x(t)=x^{2/3};x(0)=0$ in an interval arount $t=0$ has a
(a)No solution
(b)Unique solution
(c)Finitely many solutions
(d)Infinitely many solutions.
Solution:- Applying Picard's -Lindelof Uniqueness and existence theorem. $f(x,t)=3x^{3/2}$, Will it be continuous at $(0,0)$? When $x<0$, $f(x,t)$ no more real valued. So, Discontinuous at $(0,0)$. So, I can not judge from the theorem. I solved using variable separable form.
I got the solution, $3x(t)^{1/3}=t+c$. When I apply initial condition, I get $3x(t)^{1/3}=t$. A unique solution. But in the answer key it is written that equation has infinitely many solutions. How it is possible? Please explain.
Best Answer
I'm reading the given ODE as $x'=|x|^{2/3}$, so that it is defined in the full $(t,x)$-plane. The standard separation of variables technique gives the solutions $$x_c(t)={1\over27}(t-c)^3\qquad(-\infty<t<\infty),\qquad c\in{\mathbb R}\ .\tag{1}$$ Furthermore it is obvious that $x_*(t)\equiv0$ is a solution as well; but it is at first sight unclear what it has to do with $(1)$. Now the RHS $f(t,x):=|x|^{2/3}$ of the given ODE does along the line $x=0$ not fulfill the assumption that $f$ should be locally Lipschitz with respect to the variable $x$. As a consequence IVPs starting at points $(t_0,0)$ may have several solutions, and this is indeed the case here.
The IVP $x'=|x|^{2/3}$, $\>x(0)=0$, so far has the solutions $x_0(t)={1\over27}t^3$ and $x_*(t)\equiv0$. But we may as well consider $x_+(t)=x_0(t)$ $(t>0)$ and $=0$ otherwise, and similarly $x_-(t)=x_0(t)$ $(t<0)$ and $=0$ otherwise, consider as solutions. In my view there are just these four solution germs; but in a more liberal counting you can also count the functions $x_{c, +}(t):=x_c(t)$ $(t>c)$ and $=0$ otherwise ($c>0)$, and similarly for $c<0$, as new solutions of the given IVP. In this way you would obtain an infinity of solutions. All solutions described here are $C^1$ (continuously differentiable) but the patched ones are not twice differentiable at the patching point.