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I Constant solution and uniqueness of separable differential eq

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  1. Nov 5, 2016 #1
    Hi,
    I am learning ODE and I have some problems that confuse me.
    In the textbook I am reading, it explains that if we have a separable ODE: ##x'=h(t)g(x(t))##
    then ##x=k## is the only constant solution iff ##x## is a root of ##g##.
    Moreover, it says "all other non-constant solutions are separated by the straight line x=k".

    First, why do we do this separation between finding constant and non-constant solutions?

    Second, I don't understand the quoted sentence. why is that?

    Third, there is an example of finding a solution to the initial value problem ##x'=2tx^3## and ##x(0)=1##. They say that the only constant solution is ##x \equiv 0##, and
    "Therefore if ##x(t)## is a solution such that ##x(0)=1##, then, by uniqueness, ##x(t)## cannot assume the value 0 anywhere. Since ##x(0) =1 >0##, we infer that the solution is always positive."
    But how can ##x \equiv 0## be a constant solution if the solution should satisfy ##x(0)=1##, and how they got that the solution should be positive?

    I am really confused and need some help with this.

    Thank you.
     
  2. jcsd
  3. Nov 10, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Nov 30, 2016 #3

    haruspex

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    Are you sure it says only? Clearly any root of g is a constant solution.
    Suppose x > k at some t. hg might be negative here, so as t increases x decreases. But as x approaches k, g(x) approaches zero, so the slope will level off. It can never reach k unless h goes to infinity.
    x identically zero is a solution to the differential equation without the initial value. From the foregoing, any other solution cannot cross or meet the line x=0. In particular, a solution such that x(0)=1 cannot anywhere go to x=0.
     
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