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Math Help - Two-term expansion

  1. #1
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    Two-term expansion

    Find a two-term expansion of the solution of

    \frac{dv}{dt}+\epsilon v^2+v=0 for 0<t, v(0)=1.

    I tried using v as a power series in \epsilon, ie v(t)\sim v_0(t)+\epsilon v_1(t)+\epsilon^2 v_2(t)+..., plugging it in, and then separating out terms by powers of \epsilon and got

    O(1): v_0=Ae^{-t}
    O(\epsilon): v_1=Be^{-t}+A^2e^{-2t}

    If I just continue through, I get v(t) \sim Ae^{-t}+\epsilon (Be^{-t}+A^2e^{-2t}), and then applying the BC I get v(0) \sim A+\epsilon(B+A^2)=1. Is it enough to say that B doesn't matter so long as \epsilon is sufficiently small and thus A=1? Was my approach totally wrong?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Two-term expansion

    Quote Originally Posted by davesface View Post
    Find a two-term expansion of the solution of

    \frac{dv}{dt}+\epsilon v^2+v=0 for 0<t, v(0)=1.

    I tried using v as a power series in \epsilon, ie v(t)\sim v_0(t)+\epsilon v_1(t)+\epsilon^2 v_2(t)+..., plugging it in, and then separating out terms by powers of \epsilon and got

    O(1): v_0=Ae^{-t}
    O(\epsilon): v_1=Be^{-t}+A^2e^{-2t}

    If I just continue through, I get v(t) \sim Ae^{-t}+\epsilon (Be^{-t}+A^2e^{-2t}), and then applying the BC I get v(0) \sim A+\epsilon(B+A^2)=1. Is it enough to say that B doesn't matter so long as \epsilon is sufficiently small and thus A=1? Was my approach totally wrong?
    I think that the power series expansion to be used is...

    v(t)= 1+a_{1}\ t + a_{2}\ t^{2} +... (1)

    ... so that is...

    v^{'} (t)= a_{1}+ 2 a_{2}\ t +... (2)

    v^{2}(t)= 1 + (2 a_{2}+a_{1}^{2})\ t +... (3)

    ... where we had taken into account that is v(0)=1. Now we introduce (1), (2) and (3) in the equation...

    v^{'} + \varepsilon v^{2} + v =0

    ... obtaining...

    a_{1} + 1 + \varepsilon = 0 \implies a_{1}=- (1+\varepsilon)

    a_{1} + 2 a_{2} + \varepsilon\ (2 a_{2}+a_{1}^{2})=0 \implies a_{2}= \frac{1-\varepsilon - \varepsilon^{2}}{2}

    ... so that is...

    v(t) = 1 -(1+\varepsilon)\ t + \frac{1 - \varepsilon - \varepsilon^{2}}{2}\ t^{2} + ... (4)

    Of course more coefficients can be computed if necessary...

    Kind regards

    \chi \sigma
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