# Existences and uniqueness theorem, finding a unique solution of a first order ODE

Consider the initial value problem

$$y′=3x(y−1)^{1/3},y(x_{0})=y_{0}$$

Using the existness and uniqueness theorem, for what points ($x_{0}$,$y_{0}$) imply that the above IVP has a unique solution on some open interval that contains $x_{0}$?

What i tried

Using the seperable equation method, I got a solution of $$y=1+(x^{2}+c)^{1.5}$$

Im unsure of how to find the value of $c$ given the inital conditions and whether is it necessary to find the value of $c$ for this problem. While using the existence and uniqueness theorem, $y_{0}$ must not be equal to $1$ in order for a unique solution to exist. But from here im unsure of how to find the points ($x_{0}$,$y_{0}$), I was thinking that since i already know that $y_{0}$ does not equals to $1$, i must combine this result with the above solution that i got in order to get $x_{0}$ and hence the point ($x_{0}$,$y_{0}$) but im unsure how to do so, especially when there is a $c$ involved. Could anyone explain. Thanks

• @abel: The same equation, a different question. It's a bit more interesting than it seems. Of course, it could have all been asked in one posting. – Orest Bucicovschi Jan 21 '15 at 10:08
• @ys wong: What kind of answers do you look for? – Mhenni Benghorbal Jan 21 '15 at 10:44
• @ys wong: I gave you the idea and how you to solve your problem here and the same with this question I showed you how to proceed! – Mhenni Benghorbal Jan 21 '15 at 11:11

## 2 Answers

The theory tells you about the existence of solution if the function $F(x,y)$ is continuous and existence and uniqueness if the function is $C^1$ ( in fact locally Lipschitz in the second variable is OK). Your function $3x(y-1)^{1/3}$ has not second partial derivative at points with $y=1$. So this will be the problem points.

By separation of variable we get

$$\frac{dy}{dx} = 3 x (y-1)^{\frac{1}{3}}$$ or

$$\frac{2/3}{(y-1)^{1/3}}dy = 2 x dx$$

or $d((y-1)^{2/3} = d x^2$ or $(y-1)^{2/3} = x^2 + c$ . For $(x_0, y_0)$ get $$(y_0-1)^{2/3} = x_0^2 + c$$ so $c = (y_0-1)^{2/3} - x_0^2$. So we get the solutions

$$y(x) = 1 \pm \sqrt{(x^2 - x_0^2 + (y_0-1)^{2/3})^3}$$

If $y_0\ne 1$ then we have to choose the square root that makes $\pm \sqrt{(y_0-1)^2}= y_0-1$. Therefore, for $y_0 <1$ the solution is

$$y(x) = 1 - \sqrt{(x^2 - x_0^2 + (y_0-1)^{2/3})^3}$$

while for $y_0 >1$ the solution is

$$y(x) = 1 + \sqrt{(x^2 - x_0^2 + (y_0-1)^{2/3})^3}$$

However, at points $(x_0, 1)$ , both choices of signs work, however, we have to move on the side on which $x^2 \ge x_0^2$, unless $x_0$ is also $0$, when we can move both ways. We also have the solution $y(x) \equiv 1$. Therefore, at points $(x_0, 1)$ we do not have uniqueness. Quite interesting...

• From the existence and uniqueness theorem, we could work out that $y_{0}$ must not be equal to $1$ in order for the solution to be unique, without even having to work out the solutions? – ys wong Jan 21 '15 at 10:07
• Oh, the solution can hit reach $1$ and has to stop there, or continue with a constant value $1$... a funny thing. Unless it reaches $1$ at time $0$, when surely can pass through it. I am not sure if you can do the analysis without writing the solution, perhaps it is possible. This problem is more intricate than it seems initially. – Orest Bucicovschi Jan 21 '15 at 10:23
• His main concern was about $c$ and I already gave him an answer? So what's new in this answer? – Mhenni Benghorbal Jan 21 '15 at 10:43

Since you have been given an initial condition then you need to use it to find $c$ as

$$y_0 = 1+(x_0^2+c)^{3/2}.$$

Solve the above equation for $c$ and then substitute back the value of $c$ you get in the solution of the differential equation

$$y = 1+(x^2+c)^{3/2}.$$

• +1. The form is correct. I would mention that the root $\sqrt{(x^2 +c)^3}$ can take $2$ values. Something interesting happens at starting conditions $(x_0,1)$, and especially at $(0,1)$. – Orest Bucicovschi Jan 24 '15 at 0:49