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Differential equation uniqueness

  1. Sep 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG

    2. Relevant equations
    Leibniz notation: dy/dx = f(x) g(y)
    integral 1/g(y) dy = integral f(x) dx

    3. The attempt at a solution
    integral 1/y dy = integral sqrt (abs x) dx

    ln (y) = ? because sqrt (abs x) is not integrable at x =0
    Then my thought is that y=0 is not unique
     
  2. jcsd
  3. Sep 12, 2016 #2

    Math_QED

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    Why don't you continue to solve the equation using change of variables?

    ##\frac{dy}{dx}= y \sqrt{|x|}
    \Rightarrow \frac{dy}{y} = \sqrt{|x|}dx
    \Rightarrow \int \frac{dy}{y} = \int\sqrt{|x|}dx
    \Rightarrow log_e y = \int\sqrt{|x|}dx##

    Now, ##\int\sqrt{|x|}dx = \dots##

    If you have trouble integrating this function due to the absolute value, use the definition of absolute value to integrate it for ##x \leq 0## and/or ##x \geq 0##. Then, use some algebra to find the function ##y##.

    Can you say why you think that?
     
    Last edited: Sep 12, 2016
  4. Sep 13, 2016 #3
    So I have to consider two cases, x>0 and x<= 0?
    I really don't know how to integrate sqrt (abs x)
    :(
     
  5. Sep 13, 2016 #4

    Math_QED

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    Yes, split it up in apart cases (which is not nessecary, but is easier)

    ##\sqrt{|x|} = \sqrt{x}## if ##x \geq 0##
    ## \sqrt{|x|} = \sqrt{-x}## if ## x \leq 0##

    Those should be standard to integrate.
     
  6. Sep 13, 2016 #5

    LCKurtz

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    Well, surely if ##x>0## so ##|x| = x## you can integrate ##\sqrt{|x|}##, right? As another hint, if the case ##x<0## confuses you, you might use the fact that if ##f(x)## is an even function (which ##\sqrt{|x|}## is), then$$
    H(x) = \int_0^x f(u)~du$$ is an odd function. Draw a graph using that and see what it looks like.
     
    Last edited: Sep 13, 2016
  7. Sep 13, 2016 #6
    √-x = -(-x)^3/2 ??
    so i have two solutions then I can solve for y in each case
     
  8. Sep 13, 2016 #7

    LCKurtz

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    That is true for ##x < 0##. So you have a two piece formula for ##y##. That is not the same thing as two solutions to your DE. So you have a couple of things left to do. The first is to check whether$$
    y = \left \{ \begin{array}{l}
    \frac 2 3 x^{\frac 3 2},~ x \ge 0\\
    -\frac 2 3 (-x)^{\frac 3 2},~ x < 0
    \end{array}\right .$$is a solution to your DE, including at ##x=0##. If it is, then you have found one solution. But your question was about uniqueness of solutions to the given problem. Is there another solution?
     
  9. Sep 14, 2016 #8

    Math_QED

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    Uhm, that's not a solution to the DE. That's the solution to the integral you need to solve to find a solution. Also, I must say to the OP that I made a small mistake in post #2.

    The last step should be: ##\log_e |y| = \int \sqrt{|x|} dx## with the absolute values around the y.
     
  10. Sep 14, 2016 #9
    So, y(0)=0 is unique, cuz both solutions lead to it
     
  11. Sep 14, 2016 #10

    Math_QED

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    Show your work please.
     
  12. Sep 14, 2016 #11
    for x>0
    ln abs y = 2/3x^3/2
    y = A e^(2/3x^3/2) where A is any number

    for x<=0

    ln abs y = -2/3x^3/2
    y = A e^(-2/3x^3/2) where A is any number

    for y(0) =0
    Any of those equation = 0 where A have to be 0..
    so solution is 0 in both case
     
  13. Sep 14, 2016 #12

    Math_QED

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    You made a mistake:

    Recalculate:

    ##\int\sqrt{-x}dx##
     
  14. Sep 14, 2016 #13
    isn't it -2/3 x^3/2 ??
     
  15. Sep 14, 2016 #14

    Math_QED

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    No:

    ##\int\sqrt{-x}dx##
    ##-x = u \Rightarrow -du = dx##
    ##= - \int\sqrt{u}du##
    ##= - \frac{2}{3}u^{\frac{3}{2}} + c##
    ##= - \frac{2}{3}(-x)^{\frac{3}{2}} + c##

    Anyway, I think you see that you will have a function of the form ##y = Ae^{\dots}##. Your job is to find a constant ##A## such that ##y(0) = 0##
     
    Last edited: Sep 14, 2016
  16. Sep 14, 2016 #15

    LCKurtz

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    Correct. I mis-spoke there.

    It makes no sense to say "y(0)=0 is unique". You are being asked about uniqueness of solutions to a given DE with boundary condition. In other words, is there more than one solution to the given initial value problem. It is trivial to see in this problem that ##y \equiv 0## is a solution. The question is whether there is another different solution. I would think this problem arose in a section where you have an existence and uniqueness theorem that is being discussed, and the question is whether or not it applies to this problem. Do you have such a theorem? Does this problem satisfy its hypotheses? So what can you or can't you say about uniqueness?
     
  17. Sep 14, 2016 #16
    I found A is 0 in both cases
     
  18. Sep 14, 2016 #17

    Math_QED

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    Okay. That is correct. So, you found the solution ##y = 0##, the trivial solution it seems. Now answer to @LCKurtz's questions.
     
  19. Sep 14, 2016 #18
    cuz the equation is continuous and differentiable, so it is unique, based on Picard’s theorem on existence and uniqueness
     
  20. Sep 14, 2016 #19

    LCKurtz

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    That is a very glib and unsatisfactory response. You need to actually address the question. What are the hypotheses of Picard's theorem? State them. Then explain why it does or doesn't apply to this problem.
     
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