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Thread: Proving that one solution always lies above the other

  1. #1
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    Proving that one solution always lies above the other

    Hi all,


    I'd be very happy if you could help me solve a problem in my research.


    I need to prove the following:
    $\displaystyle H'(r) = -y(r) - k H(r) $


    k is a constant.


    y is strictly increasing, but not continuous.


    Let $\displaystyle (a,b]\subset R $.


    $\displaystyle (H_x, y_x) $ denotes solution x.


    $\displaystyle H_1(a)<H_0(a)<0 $.


    $\displaystyle H_0(s)<0, H_1(s)<0 $ for all $\displaystyle s\in(a,b] $.


    $\displaystyle y_1(s)>y_0(s) $ for all $\displaystyle s \in (a,b] $.


    Show:


    $\displaystyle H_1(r)<H_0(r) $ for all $\displaystyle r \in (a,b] $.
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  2. #2
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    Re: Proving that one solution always lies above the other

    What does "solution x" mean? What is "x" here?
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  3. #3
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    Re: Proving that one solution always lies above the other

    x is just an indicator. A solution is a pair of functions y and H.

    Ultimately, I need to prove it also for:

    $\displaystyle H'(r)=-y(r)-H(r) k(-H(r))$,

    where k is a strictly increasing positive function.

    Thanks a lot!
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  4. #4
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    Re: Proving that one solution always lies above the other

    The solutions of a differential equation obviously must be "differentiable" so must be continuous. So if $\displaystyle H_1(x_0)< H_2(x_0)$ for some $\displaystyle x_0$ and $\displaystyle H_1(x_1)> H_2(x_1)$ for some $\displaystyle x_1$, there must exist $\displaystyle x_2$ between $\displaystyle x_0$ and $\displaystyle x_1$ such that $\displaystyle H_1(x_2)= H_2(x_2)$.

    But by the "existence and uniqueness" theorem, since there can be only one solution to this first order differential equation having a given value at $\displaystyle x_2$.
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