Numer of solutions for IVP
Consider the initial value problem
$\dfrac{dy}{dx}=3y^{{2}/{3}}$ with initial condition $y(0)=0$.
How many solutions are there for this IVP?
1
2
3
4
infinitely many.
Clearly, $f(x,y)=3y^{\dfrac{2}{3}}$ does not satisfy Lipschitz's condition & so the solution of the IVP is not unique.
Solving the equation with initial condition we get $y=x^{3}$. Again $y=0$ is the trivial solution. So I get two solutions.Are there any other solution(/s)? I want to know all the solutions & how we find their?
Solution 1:
There are infinitely many solutions, of course. Indeed, all the functions of the form $$ y_s(x) = \left\{ \begin{aligned} &0 &&\text{ if } x\leq s\\ &(x-s)^3 &&\text{ if } x > s \end{aligned} \right. $$ are solutions of your Cauchy problem for any $s \geq 0$.
It is easy to check. Zero is a solution everywhere (in particular, for $x \leq s$); and $(x-s)^3$ is a solution of the Cauchy problem $$ \frac{dy}{dx} = 3 y^{2/3} \quad \text{with} \quad y(s) = 0 $$ for $x \geq s$.
Now, if we "glue" zero and $(x-s)^3$, then it will be solution of your problem, except, probably, the point $x = s$. However, at $x = s$ the function $y_s(x)$ is continuously differentiable, and hence, $y_s(x)$ is the correct solution of $$ \frac{dy}{dx} = 3 y^{2/3} \quad \text{with} \quad y(0) = 0. $$
Thus, there are infinitely many of solutions, since $s \geq 0$ is arbitrary.
Solution 2:
There is a slight subtlety about the definition of $y^{2/3}$ when $y<0$. If it is left undefined then I opt for $2$ solutions, if it is defined as $\bigl({\root 3\of{-| y|}}\bigr)^2=\bigl(-{\root 3\of {|y|}}\bigr)^2=|y|^{2/3}$ I opt for $4$ of them.
My reason for this is that I would count two solutions $x\mapsto\phi_1(x)$ and $x\mapsto\phi_2(x)$ which agree in an open neighborhood of $x=0$ as representants of the same solution germ. Since we can choose between $x\mapsto0$ and $x\mapsto x^3$ independently for $x\leq0$ and $x\geq0$ there are $4$ different germs in all (resp., $2$ of them if we leave $y^{2/3}$ undefined for $y<0$).