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

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SUMMARY

The differential equation y'=\sqrt{y} has multiple solutions when the initial condition is y(0) = 0. The solutions include y(t) = \frac{t^2}{4} and y(t) = 0 for all t. Picard's uniqueness theorem fails in this case because the function \sqrt{y} is not Lipschitz continuous at y=0, allowing for the existence of multiple solutions. However, if the initial condition y(0) > 0, the solution becomes unique as long as y remains positive.

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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.
 
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What's wrong with y(t)=0 for all t?
 
And is because \sqrt{y} isn't Lipschitz continuous that there is more than one solution?
 
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?
 
mathman44 said:
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.
 

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