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Question Concerning Existence/Uniqueness Theorem

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\frac{dy}{dt}[/itex]=y(y-1)(y+3), y(0)=4

    What does the exist/unique theorem say about the solution?

    2. Relevant equations
    Exist Theorem:

    Suppose some continuous function f(t,y) goes through a rectangle. If (to,yo)is a point in the rectanglein the ty-plane, then their exist a [itex]\epsilon[/itex]>0 and func. y(t)
    defined for to-[itex]\epsilon[/itex]<t<to+[itex]\epsilon[/itex]
    that solves the init. val. prob.

    [itex]\frac{dy}{dt}[/itex]=f(t,y), y(to)=yo

    Unique Theorem:

    Suppose f(t,y) and ∂f/∂y are continuous func. in a rectangle in the ty-plane. If (to,yo) is a point in the rectangle and if y1(t) and y2(t) are two functions that solve the initial value prob.

    [itex]\frac{dy}{dt}[/itex]=f(t,y), y(to)=yo

    for all t in to-[itex]\epsilon[/itex]<t<to+[itex]\epsilon[/itex] then
    y1(t)=y2(t) for that interval.

    3. The attempt at a solution

    Their are 3 equilibrium solutions namely y=0,1,-3

    y(t)>3 because why? I know given my initial condition that if Im starting at y=4 when t=0 I can only approach y=3. Also in the book the solution includes that
    y(t) goes to 3 as t decreases and y(t) goes to infinity as t increases. Why cant it be that y(t) goes to infinity as t increases and y(t) goes to 3 as t decreases. How do I know from which way my functions going if I just know the equilib. soln. and the initial cond.?

    Also If I have an initial cond. of y(0)=2 how do I know which equilibrium soln it is approaching as t increases? Thanks
    Last edited: Feb 27, 2013
  2. jcsd
  3. Feb 28, 2013 #2


    User Avatar
    Science Advisor

    Yes, y= 0, y= 1, and y= -3 are equilibrium solutions and, by the existance and uniqueness theorem, no other solution can cross those lines. Now, for any y> 1, each of y, y- 1 and y+ 3 are positive so their product is positive. That means that dy/dt> 0 and so y is increasing in that region. If y(0)= 0, then, as t increases, y increases, going away from y= 1. But as t decreases, y also decreases- but can never cross y= 1. What does that mean?

    What??? Those are the same thing. Grammatically "A and B" is the same as "B and A"! However, your book should say "1" rather than "3". There are equilibrium solutions at x= 1 and y= -3, not at y= 3.

    If you meant "Why cant it be that y(t) goes to infinity as t decreases and y(t) goes to 1 as t decreases" it is because, as I said, for y> 1 dy/dt= y(y- 1)(y+ 3) is positive and so y is an increasing function not decreasing.
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