Differential Equations Question

[roam]

Homework Statement

http://img694.imageshack.us/img694/6672/37517439.jpg

The Attempt at a Solution

For part (a), according to the uniqueness theorem, if f(t,y) and ∂f/∂y are continious in a given interval in which (t₀, y₀) exists, and if y₁(t) and y₂(t) are two functions that solve the initial value problem then

y₁(t) = y₂(t)

So... in my problem the function √y+1 is continious for values greater than or equal to -1. And the partial derivative

\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y+1}}

is continious for y>-1.

So, can I use y(0)=3 (i.e. t₀=0, and y₀=3)? What else do I need to do in order to show that the solution is unique to the IVP?

For part (b), are there any values of t₀ and y₀ such that the initial value problem has more than one solution? The uniqueness theorem says if both f(t,y) and ∂f/∂y are continious, then the uniquesness of the solution is guaranteed. So how is it possible to find an example where the uniqueness fails?

Any help is greatly appreciated.

 

[HallsofIvy]

Homework Statement

http://img694.imageshack.us/img694/6672/37517439.jpg

The Attempt at a Solution

For part (a), according to the uniqueness theorem, if f(t,y) and ∂f/∂y are continious in a given interval in which (t₀, y₀) exists, and if y₁(t) and y₂(t) are two functions that solve the initial value problem then

y₁(t) = y₂(t)

So... in my problem the function √y+1 is continious for values greater than or equal to -1. And the partial derivative

\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y+1}}

is continious for y>-1.

So, can I use y(0)=3 (i.e. t₀=0, and y₀=3)? What else do I need to do in order to show that the solution is unique to the IVP?

Yes, that perfectly good. And showing that f and f_y are continuous is sufficient.

For part (b), are there any values of t₀ and y₀ such that the initial value problem has more than one solution? The uniqueness theorem says if both f(t,y) and ∂f/∂y are continious, then the uniquesness of the solution is guaranteed. So how is it possible to find an example where the uniqueness fails?

Well, how about a place where ∂f/∂y is NOT continuous? That is, what about the initial condition y(0)= -1?

Any help is greatly appreciated.

 

[roam]

Well, how about a place where ∂f/∂y is NOT continuous? That is, what about the initial condition y(0)= -1?

Thank you. But if one of the hypotheses of the Uniqueness Theorem fails (in this case ∂f/∂y fails to exist if y = -1), that simply means that the theorem does not tell us anything about the number of solutions to an initial-value problem of the form y(t₀)=-1. It doesn't prove that there are more than one solutions!

So, how else can we think of a point (t₀, y₀) where the IVP has two or more solutions (without actually calculating any solutions to the DE as the question asks)?

 

[roam]

If the uniqueness theorem fails, then it is inconclusive. It doesn't guarantee that there are more than one solutions, isn't that right?