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Math Help - Initial value problem

  1. #1
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    Initial value problem

    Suppose that m is a fixed positive integer. Show that the initial value problem:
    u' = u^\frac{2m}{2m+1}, u(0)=0
    has infinitely many continuously differentiable solutions.

    I have solved the differential equation via the usual method of seperating variables, and have come up with the 2 solutions:
    u=(\frac{t}{2m+1})^{2m+1} and
    u=0.
    However clearly this is not an infinite number of solutions, so could someone point me in the direction of where to find the rest?
    Thanks.
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Initial value problem

    Attached Thumbnails Attached Thumbnails Initial value problem-ode-initial-value.png  
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  3. #3
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    Re: Initial value problem

    "Nice" differential equations have unique solutions once the initial values are provided. This example, because it's non-linear, isn't "nice". Thus it's possible that it fails to have a unique solution. And that is indeed the case. Showing that - the failure of uniqueness - is the purpose of doing this problem.

    Let a \ge 0. Define u_a(t) = \left(\frac{t-a}{2m+1}\right)^{2m+1} \text{ when } t \ge a \text{ and } u_a(t) = 0 \text{ when } t < a.

    Then \frac{du_a}{dt} = \left(\frac{t-a}{2m+1}\right)^{2m} \text{ when } t > a \text{ and } \frac{du_a}{dt} = 0 \text{ when } t < a.

    Examining it at t = a shows the derivative exists at t=a. It should be worked out from the definition, but it's obvious that it's going to work, since it will behave like the derivative of each of those functions depending on the side of approach, and both of those are well behaved (take epsilon to be the minimum of the epsilons you get for each side - the constant 0 function side being especially trival). Therefore it exists everywhere, and the function is then clearly continuous everywhere.

    Obviously both \frac{du_a}{dt} = \left(\frac{t-a}{2m+1}\right)^{2m}, \ t \ge a, \text{ and } \frac{du_a}{dt} = 0 \ t < a,

    satisfy \frac{du_a}{dt} = u_a(t)^{\frac{2m}{2m+1} on their domains (\infty, a) and [a, \infty).

    Thus \frac{du_a}{dt} = u_a(t)^{\frac{2m}{2m+1} holds on all of \mathbb{R}.

    Also have u_a(0) = 0 \ \forall a \ge 0.

    Therefore these represent an (uncountably) infinite number of distinct solutions to the initial value problem: \{u_a(t)\}_{a \ge 0}

    $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

    Note that your solution is not on this list. The solution u(t)=0 for all t is also not on this list.
    Using the same trick I used, only with the "other half" of the natural solution (your solution), and pushing it negative, are also another infinite family of different solutions. Maybe there are even more solutions than all those - I don't know. I suspect one could add your solution, except when it hits zero, it stays at 0 a finite time, and then continues on according to your solution - and all horizontal translations of that. It looks to me like all these that I've now listed, together, constitute the entire solution set for this intial value ODE. I'm not 100% on that, but that's how it seems.

    $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

    How did I find that? Think about u' = f(u). It says the slope is a function only of the height. If you look at the vector field that describes on any vertical line, t = 0, t = 1, t = 46, whatever, they all look the same. Thus any solution - any integral curve of that vector field in the plane - can be translated horizontally and still remain a solution.

    In this particular case, f is an even function (that 2m in the numerator of the power). So think about that vector field along a vertical line - moving upwards, negative towards 0, you have a positive slope that's being less steep as you approach 0. At 0 it's flat (slope = 0). As the height grows positively, it represents ever steeper slopes, starting from flatness. That's the same for every vertical line.

    One consequence of looking at the vector field is that you can see that there's an integral curve just following along the t axis.

    When uniqueness fails, the vector field and its integral curves often permit you to leave one solution (like the horizontal axis integral curve u(t)=0 for all t) and branch out onto another "nearby" solution. That's what happened here and it does in fact work.

    So I found these solutions by seeing how the solution corresponding to u(t) = 0 could "leave" and become the solution you found. Since I know the set of all solutions are horizontally translationally invariant, if it can differentiably leave a place where it's horizontal, on the horizontal line, at t=0 (as in it does in the solution you found), then it can likely do it anywhere, like at t=a. Let a>0 and you solve the ODE initial value problem by following the horizontal integral curve to some time "a" past 0, and then leave on a nearby integral curve that looks like the solution you found.
    That's how I came to my choice for those functions.
    Last edited by johnsomeone; October 12th 2012 at 12:19 PM.
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  4. #4
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    Re: Initial value problem

    Wow, far better answer than I expected. Thank you so much!
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