7

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?

Calvin Khor
  • 34,903
Empty
  • 13,012
  • 1
    Three answers, all different! So far we have $2, 4,$ and $\infty$. My money is on $\infty$. – TonyK Nov 13 '14 at 12:27

3 Answers3

9

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 $s \geq 0$.

enter image description here

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.

Voliar
  • 1,651
  • 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 07:11
  • 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 06:08
  • @Panja.S. This means that the answer which you saw problably here http://epaper.jagranjosh.com/187586/UGC/Joint-CSIR-UGC-NET-Answer-Key-December-2012-Mathematical-Sciences-#dual/1/2 is incorrect. If you construct the solution by the method as above, there will be infinitely many of solutions, and you can easily check it. Person how gave the answer "2" probably didn't know about this trick, and found two obvious solutions 0 and $C x^{5/3}$. – Voliar Nov 17 '14 at 07:04
1

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 $y<0$).

Christian Blatter
  • 226,825
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.

MartinG
  • 978