1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Differential Equations Question

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data

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

    3. 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.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 14, 2012 #2


    User Avatar
    Science Advisor

    Yes, that perfectly good. And showing that [itex]f[/itex] and [itex]f_y[/itex] are continuous is sufficient.

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

    Last edited by a moderator: May 5, 2017
  4. Mar 17, 2012 #3
    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:
  5. Mar 17, 2012 #4
    If the uniqueness theorem fails, then it is inconclusive. It doesn't guarantee that there are more than one solutions, isn't that right?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook