# 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. 1

2. 2

3. 3

4. 4

5. 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?

• Three answers, all different! So far we have $2, 4,$ and $\infty$. My money is on $\infty$. – TonyK Nov 13 '14 at 12:27

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.

• Well...What happens for the problem $\dfrac{dy}{dx}=60y^{\dfrac{2}{5}}; y(0)=0$? Is this problem similar to the given problem? If yes then I have no question. If not then what is the different between two problems & what is solution(/s) of this problem? – Empty Nov 13 '14 at 18:20
• @Panja.S. Yes, this problem is similar, and solutions can be constructed by the same way. There will be infinitely many solutions. – Voliar Nov 13 '14 at 19:04
• If we impose the extra condition $x\gt 0$ on the problem given in comment box, then what changes in the solution? – Empty Nov 13 '14 at 19:08
• @Panja.S. Nothing will be changed. In the construction of the solutions above, $y_s(x)$ is always equal to $0$ for $x < 0$. And in your situation it will be the same. Hence, you can just forget about the part $x < 0$, and work only with $x > 0$. – Voliar Nov 14 '14 at 7:11
• But, for the problem in comment box, the answer is only two solutions. I am confused. It is a question of CSIR-NET-2012. – Empty Nov 17 '14 at 6:08

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 $y<0$).

If we allow $x$ and $y$ to be complex there are arguably 3 nonzero solutions, with $x$ equated to the 3 complex roots of $y^{\tfrac{1}{3}}$, bringing the total number of solutions to 4.

• Please explain in details.... – Empty Nov 11 '14 at 9:37