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Initial Condition?

  1. Sep 14, 2008 #1
    This question asks to solve the initial value problem and determine the interval in which it is defined.

    [tex]y'=2x/(1+2y) , y(2)=0[/tex] and after solving, we get the equation [tex] y= -0.5 + \sqrt{x^{2} - 15/4}[/tex](discarded the negative squareroot since governed by initial condition).

    Therefore, the interval is found to be [tex] x > or = \sqrt{15/4}[/tex], but my question is, why can't the interval also be defined at [tex] x < \sqrt{15/4}[/tex]? How does it violate the initial condition? The I.C. only points out that when x=2, y=0.

    I think i'm missing some basic knowledge for initial value problems...

    Thanks in advance
     
  2. jcsd
  3. Sep 14, 2008 #2

    Defennder

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    Is this restricted to the reals? If so, there's your explanation.
     
  4. Sep 14, 2008 #3
    Oh sorry, I was typing this question very late last night, i meant to mean [tex]x=-\sqrt{15/4}[/tex]
     
  5. Sep 14, 2008 #4

    Defennder

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    Did you instead mean to say that why [tex]x \neq \pm \sqrt{15/4}[/tex] y' wouldn't be defined at that point, would it?
     
  6. Sep 15, 2008 #5
    Well I wasn't really considering whether y' played a role in defining the interval. I was just lost when the solution to that question was any x-value where [tex]x \geq + \sqrt{15/4} [/tex] and the negative of that wasn't part of the answer. I plugged in [tex]x=- \sqrt{15/4}[/tex] to y(x) and it gave me zero also. So why shouldn't [tex] x < -\sqrt{15/4} [/tex] be a solution as well?
     
  7. Sep 15, 2008 #6

    HallsofIvy

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    ?? Well, again, if [itex]x< -\sqrt{15/4}[/itex], then [itex]x^2> 15/4[/itex] and your formula will involve a complex number.
     
  8. Sep 18, 2008 #7

    Redbelly98

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    Maybe 2RIP means to ask: why isn't the interval

    [tex]|x| \geq \sqrt{15/4}
    [/tex]

    ?
     
  9. Sep 19, 2008 #8

    Defennder

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    Did you mean [tex]|x| \leq \sqrt{15/4}[/tex] instead?
     
  10. Sep 19, 2008 #9

    Redbelly98

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    x2 needs to be greater or equal to 15/4 to avoid having a negative number under the radical.

    [tex]
    y= -0.5 + \sqrt{x^{2} - 15/4}
    [/tex]
     
  11. Sep 19, 2008 #10

    HallsofIvy

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    Okay, [itex]\sqrt{x^2- 15/4}[/itex] is defined only for [itex]x>\sqrt{15}/2[/itex] or for [itex]x< -\sqrt{15}/2[/itex]. You are asking why the solutions is not defined for [itex]x< -\sqrt{15}/2[/itex], right?

    If you were to find the general solution to the differential equation it would involve a single undetermined constant, C say. The value of that constant can be determined by the initial value, y(2)= 0. Since [itex]x= 2> \sqrt{15}/2[/itex], that C applies to all [itex]x>\sqrt{15}/2[/itex]. However, there is no way to "connect" values of the function for [itex]x>\sqrt{15}/2[/itex] and values of the function for [itex]x< -\sqrt{15}/2[/itex]. There is no way of determining the "C" for [itex]x< -\sqrt{15}/2[/itex].
     
  12. Sep 20, 2008 #11
    hi,
    i have problem to determine initial condition for this problem. the system models a coupled oscillator
    m1x''=-k1x+k2(y-x),
    m2y''=-k2(y-x).
    Imagine a spring (with constant spring k1) attached to a hook in the ceiling.Mass m1 is attached to the spring, and a second spring (with constant spring k2) is attached to the bottom of mass m1. by attaching mass m2 to the second spring, i have a coupled oscillator where x and y represent the displacements of masses m1 and m2 from their respective equilibrium positions. I have set x1-x, x2=x',x3=y and x4=y1 and i get
    x1'=x2,
    x2'=(-k1/m1)x1+(k2/m1)(x3-x1),
    x3'=x4,
    x4'=(-k2/m2)(x3-x1).
    Assume k1=k2=1 and m1=m2=1.
    Suppose that the first mass is displaced upward two units, the second mass downward two units, and both masses are released from rest, therefore the initial conditions for this problem are
    x(0)=-2,x'(0)=0;y(0)=2;y',(0)=0;

    Can anyone verify whether my stated initial conditions right or wrong...:frown:
     
  13. Sep 20, 2008 #12
    Does anyone can verify these initial conditions?
     
  14. Sep 20, 2008 #13

    Redbelly98

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    Yes, looks good, as long as you meant to use downward as the positive direction.
     
  15. Sep 20, 2008 #14

    HallsofIvy

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    Yes, those initial conditions are correct.


    However, in the future, please do not "hijack" someone else's thread to ask a new question. It is very easy to start your own thread. Click on the "new topic" button at the top of the list of threads.
     
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