Differential Equations Question

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Homework Statement



http://img694.imageshack.us/img694/6672/37517439.jpg [Broken]

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

[itex]\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y+1}}[/itex]

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.
 
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Answers and Replies

  • #2
HallsofIvy
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Homework Statement



http://img694.imageshack.us/img694/6672/37517439.jpg [Broken]

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

[itex]\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y+1}}[/itex]

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 [itex]f[/itex] and [itex]f_y[/itex] 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.
 
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  • #3
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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)? :confused:
 
  • #4
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If the uniqueness theorem fails, then it is inconclusive. It doesn't guarantee that there are more than one solutions, isn't that right?
 

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