Problem for Theorem of Uniqueness

[peace-Econ]

Homework Statement

Check if the given initial value problem has a unique solution

Homework Equations

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

The Attempt at a Solution

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.

I want to make sure if this way is correct. Some help please.

Thanks!

 

[hunt_mat]

The usual way you tackle this kind of thing is proof by contradiction and examine the integral:
$$\int \left| \frac{dw}{dx}\right|^{2}dx$$
where $w=u-v$ and u and v solve the original equation.

I think in this case though, you can solve the general equation (without using the iniitial/boundary condition) and show that it's solution depends upon a single parameter. That parameter is determined with the condition you were given.

 

[peace-Econ]

Did you mean that my way and answer is wrong? This problem should be verified by using the theorem of uniqueness.

I got y(t)=(4x^2)-64. Then can i say that this is a unique solution?

 

[hunt_mat]

As I don't know the theorem you're using I can't say either way.

 

[peace-Econ]

sorry, you're right. The theorem I was trying to use is that if f and its partial derivative is continuous on the rectangle containing the given initial value problem, we can say that it has a unique theorem.

 

[HallsofIvy]

Homework Statement

Check if the given initial value problem has a unique solution

Homework Equations

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

The Attempt at a Solution

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.

It would help if you actually stated the theorem you are using. I suspect you are using
"The initial value problem y'= f(x,y), $y(x_0)= y_0$, has a unique solution in some neighborhood of the point $(x_0, y_0)$ if both f(x,y) and $f_y(x,y)$ are continuous in some neighborhood of that point."
No, you cannot "take any rectangle R containing the initial value point (4, 0)" because that will necessarily include points where y<= 0, where the functions are NOT continuous.

I want to make sure if this way is correct. Some help please.

Thanks!