Is there a unique solution to y'=\sqrt{y} with y(0) = 0?

[mathman44]

Homework Statement

Show that the DE $$y'=\sqrt{y}$$ has more than one solution when y(0) = 0 by finding two of them. Why does Picard's uniqueness theorem not apply?

The Attempt at a Solution

By standard ODE techniques, $$y=\frac{1}{4}[2c+c^2+t^2]$$

but if y(0)=0 then $$y=\frac{t^2}{4}$$ only.

I can't find another solution.

 

[Dick]

What's wrong with y(t)=0 for all t?

 

[mathman44]

And is because $$\sqrt{y}$$ isn't Lipschitz continuous that there is more than one solution?

 

[mathman44]

For example if y(0) > 0 and $$y=\frac{1}{4}[2c_1+c_1^2+t^2]$$, then is the solution unique for all t > 0 since the function y' is Lipschitz continuous for all t > 0?

 

[Dick]

For example if y(0) > 0 and $$y=\frac{1}{4}[2c_1+c_1^2+t^2]$$, then is the solution unique for all t > 0 since the function y' is Lipschitz continuous for all t > 0?

Well, yes, sqrt(y) is not Lipschitz continuous at y=0. So Picard doesn't prove there is a unique solution with y(0)=0. And yes, if y(0)>0 then it does imply there is a unique solution as long as y stays greater than zero.