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Differential Equations dy/dx=sqrt(x^2+y^2-9)

  1. Apr 10, 2004 #1
    just a couple of questions:
    given dy/dx=sqrt(x^2+y^2-9)
    and y(a)=b
    i)for what values of a and b would a local solution exist and why
    ii)for what values of a and b would a unique solution exist and why

    for i) i got all values such that a^2+b^2<=9
    for ii) i got all values such that a^2+b^2>9
    i'm not 100% sure this is correct and would be happy if someone could confirm/show otherwise these results.


    secondly
    how would one show that the only solution of
    y''+ty'+(1+t^2)y^2=0
    that touches the t axis at some point (t0,0) is the identically zero solution

    thanks for your help
     
  2. jcsd
  3. Apr 11, 2004 #2

    matt grime

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    If i) were correct then you're allowing the derivative at (a,b) to have (possibly) imaginary values. Is that permitted?

    Not sue what you mean by local, and why a local solution apparently can't also be unique, and is this really a question about Lipschitz and Peano Conditions?
     
  4. Apr 11, 2004 #3

    HallsofIvy

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    Basically, this is a problem about "existance and uniqueness" and involves the "Lipschitz and Peano conditions" as Matt Grime said.

    The d.e. dy/dx= f(x,y) will have a solution (satisfying y(a)= b) as long as f is continuous in some neighborhood of (a,b). It will have a unique solution as long as f is also "Lipschitz" in y in some neighborhood of (a,b).

    In this case f(x,y)= sqrt(x2+ y2- 9) and that will be continuous as long as x2+ y[/sup]2[/sup]>= 9. That is, (a,b) must be outside or on the circle x2+ y[/sup]2[/sup]= 9.
    (In the case that (a,b) is on that circle, the solution would only exist for (x,y) outside it.)

    A function f(x,y) is "Lipschitz" in y on a set as long as |f(x,y1)- f(x,y2)|<= C |y1- y2| for some constant C and (x,y1), (x,y2) in the set. In particular, if the function is differentiable with respect to y on a set, then it is Lipshchitz on that set. Here, f'= y/(x2+y2-9)1/2 which exists as long as (a,b) is strictly outside the circle x2+ y2= 9.

    "how would one show that the only solution of
    y''+ty'+(1+t^2)y^2=0
    that touches the t axis at some point (t0,0) is the identically zero solution?"

    One wouldn't. It's not true. The fact that this was paired with the first question should give you a clue. If we write the equation as y"= -ty'+ (1+t2)y2= f(x,y,y') then f is a "nice" function (infinitely differentiable in all three variables) and so this equation has a unique solution for all "y(a)= b, y'(a)= c" initial conditions. In particular, there must exist a unique solution such that y(t0)= 0 and y'(t0)= 1 so y is NOT identically equal to 0.
     
  5. Apr 12, 2004 #4
    firstly thank you for your responses
    however regarding HallsofIvy's reply i didnt really understand the following

    "If we write the equation as y"= -ty'+ (1+t2)y2= f(x,y,y') then f is a "nice" function (infinitely differentiable in all three variables) and so this equation has a unique solution for all "y(a)= b, y'(a)= c" initial conditions. In particular, there must exist a unique solution such that y(t0)= 0 and y'(t0)= 1 so y is NOT identically equal to 0."

    firstly y''=-ty'-(1+t^2)y^2=f(t,y,y') (*) but i dont think that would change things too much anyway.
    (*) can be written as the following system of equations if we let
    x1=y
    x2=y'
    so x1'=x2=f1
    and x2'=-tx2-(1+t^2)x1^2=f2
    now partial derivative of fi/xj for i=1,2 and j=1,2 exists and are infintely differentiable so we have the lip****z condition and f is continuous everywhere so we must have only 1 solution to (*) namely y=0 and we have proven it.

    sounds ok to me but it goes against what HallsofIvy said, what do you guys think
     
  6. Apr 13, 2004 #5

    HallsofIvy

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    There exist only one solution to the differential equation with given initial values. If you write the equation as two first order equations (or as a single first order vector equation) then you are solving for x1 and x2. The condition "y(t0)= 0" is saying that x1(t0)= 0 but you are still free to assign whatever value you want to x2(t0). In particular, there exist a unique solution to the differential equation with x1(t0)= 0 and x2(t0)= 1. Since x2= y', y'(t0)= 1 means that y is NOT identically 0.
     
  7. Apr 13, 2004 #6
    when you say "there exists only one solution to the differential equation with given initial values" are you saying that for this particular question you have to have y'(t0)=0 as a given initial value?
    i am not sure where or how you plucked the value y'(t0)=1 from either

    thnx
     
  8. Apr 14, 2004 #7

    matt grime

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    Any value for the first derivative that isn't zero would have done. There is some solution satisfying those boundary conditions. As the first derivative is not zero it can't be a constant function (not identically zero).
     
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