# Diff Equations: Theorem of Uniqueness

1. Aug 14, 2011

### peace-Econ

1. The problem statement, all variables and given/known data

Check if the given initial value is a unique solution.

2. Relevant equations

y'=y^(1/2), y(4)=0

3. The attempt at a solution

I got y(t)=(t/2)^2 and 0 at t=4

So, we have two solutions to i.v.p.; therefore, it's not a unique solution.

Is it correct?

2. Aug 14, 2011

### HallsofIvy

I have no idea what you mean by "y(t)= (t/2)^2 and 0 at t= 4". If you mean "y(t)= (t/2)^2 if t is not 4, y(4)= 0", it is not a solution because it is not differentiable at t= 4. If you mean that y(t)= (t/2)^2 is one solution and y= 0 for all t is another solution, that is not correct because y(t)= (t/2)^2 give y(4)= 4, not 0. Please show exactly what the problem said and what you have tried to do.

3. Aug 14, 2011

### peace-Econ

I guess I needed to mention about theorem of uniqueness. f=y^(1/2) and its partial derivative 1/2(root of y) are continuous except where y<=0. We can take any rectangle R containing the initial value point (4,0). Then the hypothesis of theorem of uniqueness is satisfied. How about this way???