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DE Question

  1. Sep 17, 2011 #1
    The given problem for multiple differential equations (not in a system) is: "Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0,y0) in the region."

    I'm not entirely sure what to do with this. Looking at the function:

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

    I found the unknown function to be:
    [tex]y=(\frac{x}{3})^3[/tex]
    And I have verified that it is a solution.

    Now the books says the answer is: "half-planes defined by either y>0 or y<0"
    I'm not sure how this solution works. Could someone please explain it to me?

    I also don't understand why y=0 isn't valid. I can see y=0 satisfying this equation:

    0=0
     
  2. jcsd
  3. Sep 18, 2011 #2

    Dick

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    Yes, y=0 is a solution that passes through (0,0). y=(x/3)^3 is also a solution that passes through (0,0). So there is no unique solution if the initial point (x0,y0)=(0,0). What about (x0,y0) being a point besides (0,0)? I thinks that's what they are asking.
     
  4. Sep 18, 2011 #3

    lanedance

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    the DE is separable
    [tex] \frac{dy}{dx}=y^{\frac{2}{3}} [/tex]

    integrating both sides gives
    [tex] \int dyy^{-\frac{2}{3}}=\int dx [/tex]
    [tex] \frac{1}{3}y^{\frac{1}{3}}=x+c [/tex]
    [tex] \frac{1}{3}y^{\frac{1}{3}}=(\frac{x+c}{3})^3 [/tex]

    this is a family of solutions given by c

    now consider whether the following is a solution:
    [itex]for \ \ \ \ x \leq 0, \ \ \ \ \ \ y(x) = (\frac{x}{3})^3 [/itex]
    [itex]for \ \ 0< x \leq 1, \ \ y(x) = 0 [/itex]
    [itex]for \ \ \ \ 1< x, \ \ \ \ \ \ y(x) = (\frac{x-1}{3})^3 [/itex]
     
  5. Sep 18, 2011 #4
    So you are only using solutions which do not make the equation equal to zero? Is the term homogenous applicable to that case?
     
  6. Sep 18, 2011 #5

    HallsofIvy

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    On the contrary, Lancelot59's solution explicitely includes y= 0. That's why he can construct an infinite number of solutions such that y(0)= 0.

    The "fundamental existence and uniqueness theorem for first order differential equations" says that the problem dy/dx= f(x,y), y(x0)= y0 has a unique solution if there is some neighborhood about (x0,y0) such that f(x,y) is continuous and f(x,y) is "Lipschitz" in y ("differentiable with respect to y is a simpler, sufficient but not necessary condition) in that neighborhood.

    Here, [itex]f(x,y)= y^{2/3}[/itex] is continuous in any neighborhood of (0,0) but [itex]f_y= (2/3)y^{-1/3}[/itex] does not exist at (0, 0) so the function is not Lipschtz there.

    We can guarentee a unique solution on any neighborhood that does NOT include the x-axis but not in any neighborhood that does.
     
  7. Sep 18, 2011 #6
    I see, so the trick is to differentiate the original differential equation with respect to the "dependent" (Y in this case) variable, and see where solutions for it are valid?
     
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