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Math Help - long-time behaviour of nonhom. lin. ODE

  1. #1
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    long-time behaviour of nonhom. lin. ODE

    Hi

    given the nonhomogeneous linear differential-equation
    x'(t) + a(t) x(t) = b(t)
    with
    a: [0, \infty[ \rightarrow [0,\infty[,
    b: [0, \infty[ \rightarrow [0,\infty[
    and no explicit solution x(t) of the problem, i like to show that
    x(t) \rightarrow 0.

    in addition for t \rightarrow \infty i'm able to show that:
     a(t) \rightarrow 0 and  b(t) \rightarrow 0

    do you know any theory i'm allowed to use in this case? i just find some results in stabilitytheory for homogeneous case with a constant coefficient a(t).

    thanks for any hints or suggestions in advance.
    kind regards, debelix
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  2. #2
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    Ah, but you do know that, due to the integrating factor method, that

    \displaystyle x(t)=e^{-\int a(s)\,ds}\int e^{\int a(s)\,ds}\,b(s)\,ds+Ce^{-\int a(s)\,ds}.

    Does that get you started?
    Last edited by Ackbeet; March 11th 2011 at 02:04 AM. Reason: Changed dummy variables to s from x to avoid confusion.
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  3. #3
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    thanks for your reply.

    the problem is that for t \rightarrow \infty
    e^{- \int a(s) \ ds} \rightarrow 0

    but

    \int e^{\int a(s) \ ds} b(s) \ ds \rightarrow \infty

    is there any other way instead of analysing these general solution?
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  4. #4
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    You have a confusing statement in the OP that could use some clarification:

    In addition for t\to\infty I'm able to show that:
    a(t)\to 0 and b(t)\to 0.
    Does that mean you have already shown this? Because, given the conditions of the original problem, you definitely cannot conclude that. You need assumptions on the nature of those two functions.
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  5. #5
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    ok, to avoid a confusion (my english isn't the best):
    i have
    x'(t) =  \bigg( 1 / f(t)\bigg)'  ( 1+ x(t)) - 2f(t) e^{\frac{-1}{f(t)}} x(t)

    my assumptitions on f are
    1. f:[0,\infty[ \rightarrow [0,\infty[
    2. f(t) \downarrow 0 and
    3. f(t) \geq \frac{c}{\ln{(2+t)}} \ \mbox{for every} \ t \in [0,\infty[ \ \mbox{and $c>0$ sufficiently large}


    the author of the article i'm working with says: "it's obvious that x(t) \rightarrow0 for t \rightarrow \infty"
    for me, it isn't that obvious
    do you have any ideas?

    EDIT (forgot the c in assumption 3)
    Last edited by Debelix; March 11th 2011 at 03:53 AM.
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  6. #6
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    Quote Originally Posted by Ackbeet View Post



    Does that mean you have already shown this?
    yes, the coefficients go to zero. in addition to that i was able to show that x goes to zero for
    f:=\frac{c}{ln(2+t)} \ \mbox{with $c > 0$ sufficiently large}.

    perhaps by using this information it is easier to proof, that for a continuous differentiable function f with the 3 assumptions above, x(t) goes to zero as well?
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