1. ## 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?

2. ## Re: Two-term expansion

Originally Posted by davesface
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

$\chi$ $\sigma$