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!

For which initial values does the differential equation have a unique solution

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\frac{dy}{dx} = y^{\frac{2}{3}}[/itex] with y(0) = y0

    2. Relevant equations

    3. The attempt at a solution
    after separating variables:

    [itex] \frac{dy}{y^{\frac{2}{3}}} = dx [/itex]

    then integrating:

    [itex] 3y^{\frac{1}{3}} = x + c [/itex]

    and rearranging:

    [itex]y(x) = \left( \frac{x + c}{3} \right)^{3}[/itex]

    after solving for initial value, y(0) = y0:

    c = 3y01/3

    therefore

    [itex]y_{0} = \left( \frac{c}{3}\right)^{3}[/itex]

    is that what they're asking for? I believe that would give a unique solution, no? also the question is explicit with values is this plurality given by the constant c?
     
    Last edited: May 7, 2012
  2. jcsd
  3. May 7, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi sunrah! :smile:
    no, from c = 3y01/3,

    go back and put that value of c into the previous equation :wink:
     
  4. May 7, 2012 #3
    hi tiny-tim!
    thanks I should have known. I have another question:
    when does a solution become no longer unique.

    so if we are given new initial conditions of y(0) = y03 != 0

    the solution becomes [itex]y(x) = \left(\frac{x - y_{0}}{3}\right)^{3}[/itex]

    this is no longer unique for y(x) = 0?
     
  5. May 7, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (actually [itex]y(x) = \left(\frac{x}{3} - y_{0}\right)^{3}[/itex])

    looks unique to me :confused:

    and where does y(x) = 0 come from ??
     
  6. May 7, 2012 #5
    yes sorry c = 3y0.

    the questions asks when the solution with initial condition y(0) = y03 != 0 becomes non-unique.

    I don't have any ideas I thought maybe when x = -3y0 but I#m just guessing.
     
  7. May 7, 2012 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you take ##y_0=0## so ##y(0)=0## and apply that to your solution$$
    y(x) =\left ( \frac{x+c}{3}\right)^3$$you get ##c=0##, giving a perfectly good solution ##y=\frac {x^3}{27}##, which you can check solves the DE and ##y(0)=0##.

    But the question is whether that solution is unique. What does your existence and uniqueness theorem for ##y'=f(x,y),\ y(x_0)=y_0## require in terms of hypotheses to guarantee a unique solution. Are those hypotheses satisfied in this case? If they are you know the solution is unique. If they aren't, the solution isn't guaranteed to be unique, although it might be. If the solution might not be unique, maybe you can find another. This question is about that theorem. What does the theorem say about ##y_0## for this ##f(x,y)=y^{\frac 2 3}## to guarantee uniqueness?
     
    Last edited: May 7, 2012
  8. May 7, 2012 #7
    Hi LCKurtz,

    well y0 != 0 zero because [itex]\frac{df(0,y_{0})}{dy}[/itex] is undefined. so it doesn't guaranty a unique solution.
     
  9. May 7, 2012 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, that much is correct. Are there any other values ##y_0## can't be? Does the theorem work OK if ##y_0## is, for example, a very small nonzero number? You need to know exactly which ##y_0## are OK; it isn't enough to say it can't be zero without checking all the other possibilities.

    And since you do know at least that ##y_0\ne 0## that means the solution might not be unique. Can you find another? Hint: It's very easy...
     
  10. May 7, 2012 #9
    ok so for y(0) = y03 != 0

    f(x,y) = y2/3

    f(0,y03) = y02

    and

    [itex]\frac{d f(0,y_{0}^{3}}{dy} = \frac{2}{3y_{0}}[/itex]

    but this means that the solution is unique because y0 != 0

    so I don't get the second part at all!
     
  11. May 7, 2012 #10
    i thought that the unique solution is given when y0 = (c/3)^3
     
  12. May 7, 2012 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm talking about my example solution ##y = \frac {x^3}{27}## satisfying ##y(0)=0##. There might be another solution satisfying ##y(0)=0## because the uniqueness theorem doesn't hold in this case. Can you find one by inspection?

    I have to leave for a few hours now; will check back in later.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: For which initial values does the differential equation have a unique solution
Loading...