Thread: existence and uniqueness question of DFQ

1. existence and uniqueness question of DFQ

Hello everybody, thank you very much for reading!

I need to solve the following problem (I'll try my best in English, it's not my native tounge):
The problem:
y' = y^*(1/3) + 1, y(0)=0, and let f(x,y) = y^(1/3) + 1

f(x,y) doesn't satisfy Lipschitz's condition on y=0. (that's a given).
The question is:
Does the problem has one solution or not? Then they add: What is the conclusion on Lipschitz's condition in the existence and uniqueness theorem?

My questions:
1) I know how to prove an equation such as this has one and only solution, but I don't know how to prove an equation has more than one... I mean, there are several conditions for a function to satisfy so that this type of problem will have one solution, but if it doesn't satisfy them - does that mean there isn't one solution?
How do I approach this problem?

2) let D be the area in which f(x,y) is continuous. I know that if f(x,y) satisfies Lipschitz's condition in every partial area to D, then the existence and uniqueness law is being satisfied. I know nothing more than that. What are they implying to?

Thank you very much! I hope I was clear enough...

Tomer.

2. I'm not sure about this, but since nobody else has replied, I'll throw in my two cents' worth.

It seems to me that this equation has two solutions. The reason for suspecting this is the naive one that you can actually derive the solution as an implicit function. In fact, $\displaystyle \int\frac{dy}{y^{1/3}+1} = x + c$. Substitute $\displaystyle u=y^{1/3}$ and integrate:

$\displaystyle \int\frac{3u^2}{u+1}du = \int\Bigl(3u-3+\frac3{u+1}\Bigr)du = {\textstyle\frac32}u^2-3u+3\ln(u+1).$

The initial condition tells you that the constant of integration is 0, so we get the solution as $\displaystyle {\textstyle\frac32}y^{2/3}-3y^{1/3}+3\ln(y^{1/3}+1) = x.$

Here's where I get a bit vague. I suspect that if you have a good enough graphing program you'll find that this curve has two branches through the origin. But I don't know of any formal way to prove that.

3. Thank you very much for the reply.
I have also tried solving this equation of course, but I'm almost certain that's not the way to solve this excerice. It should be a theoretical one.

Other guesses?

Thanks again Oplag!