# For which initial values does the differential equation have a unique solution

1. May 7, 2012

### sunrah

1. The problem statement, all variables and given/known data
$\frac{dy}{dx} = y^{\frac{2}{3}}$ with y(0) = y0

2. Relevant equations

3. The attempt at a solution
after separating variables:

$\frac{dy}{y^{\frac{2}{3}}} = dx$

then integrating:

$3y^{\frac{1}{3}} = x + c$

and rearranging:

$y(x) = \left( \frac{x + c}{3} \right)^{3}$

after solving for initial value, y(0) = y0:

c = 3y01/3

therefore

$y_{0} = \left( \frac{c}{3}\right)^{3}$

is that what they're asking for? I believe that would give a unique solution, no? also the question is explicit with values is this plurality given by the constant c?

Last edited: May 7, 2012
2. May 7, 2012

### tiny-tim

hi sunrah!
no, from c = 3y01/3,

go back and put that value of c into the previous equation

3. May 7, 2012

### sunrah

hi tiny-tim!
thanks I should have known. I have another question:
when does a solution become no longer unique.

so if we are given new initial conditions of y(0) = y03 != 0

the solution becomes $y(x) = \left(\frac{x - y_{0}}{3}\right)^{3}$

this is no longer unique for y(x) = 0?

4. May 7, 2012

### tiny-tim

(actually $y(x) = \left(\frac{x}{3} - y_{0}\right)^{3}$)

looks unique to me

and where does y(x) = 0 come from ??

5. May 7, 2012

### sunrah

yes sorry c = 3y0.

the questions asks when the solution with initial condition y(0) = y03 != 0 becomes non-unique.

I don't have any ideas I thought maybe when x = -3y0 but I#m just guessing.

6. May 7, 2012

### LCKurtz

If you take $y_0=0$ so $y(0)=0$ and apply that to your solution$$y(x) =\left ( \frac{x+c}{3}\right)^3$$you get $c=0$, giving a perfectly good solution $y=\frac {x^3}{27}$, which you can check solves the DE and $y(0)=0$.

But the question is whether that solution is unique. What does your existence and uniqueness theorem for $y'=f(x,y),\ y(x_0)=y_0$ require in terms of hypotheses to guarantee a unique solution. Are those hypotheses satisfied in this case? If they are you know the solution is unique. If they aren't, the solution isn't guaranteed to be unique, although it might be. If the solution might not be unique, maybe you can find another. This question is about that theorem. What does the theorem say about $y_0$ for this $f(x,y)=y^{\frac 2 3}$ to guarantee uniqueness?

Last edited: May 7, 2012
7. May 7, 2012

### sunrah

Hi LCKurtz,

well y0 != 0 zero because $\frac{df(0,y_{0})}{dy}$ is undefined. so it doesn't guaranty a unique solution.

8. May 7, 2012

### LCKurtz

OK, that much is correct. Are there any other values $y_0$ can't be? Does the theorem work OK if $y_0$ is, for example, a very small nonzero number? You need to know exactly which $y_0$ are OK; it isn't enough to say it can't be zero without checking all the other possibilities.

And since you do know at least that $y_0\ne 0$ that means the solution might not be unique. Can you find another? Hint: It's very easy...

9. May 7, 2012

### sunrah

ok so for y(0) = y03 != 0

f(x,y) = y2/3

f(0,y03) = y02

and

$\frac{d f(0,y_{0}^{3}}{dy} = \frac{2}{3y_{0}}$

but this means that the solution is unique because y0 != 0

so I don't get the second part at all!

10. May 7, 2012

### sunrah

i thought that the unique solution is given when y0 = (c/3)^3

11. May 7, 2012

### LCKurtz

I'm talking about my example solution $y = \frac {x^3}{27}$ satisfying $y(0)=0$. There might be another solution satisfying $y(0)=0$ because the uniqueness theorem doesn't hold in this case. Can you find one by inspection?

I have to leave for a few hours now; will check back in later.