Consider the following ODE: \begin{equation} \frac{\mathrm{d}y(t)}{\mathrm{d}t} = \varepsilon y(1-y)+y^2-\varepsilon \frac{y^2}{1-y} \quad \quad \text{for} \quad \quad t>0, \tag{1} \end{equation} where $y(0)=c$ and $\varepsilon\ll1.$ I would like to find an asymptotic solution that is valid for all $t>0.$ I have attached a figure of the numerical solution for $\varepsilon = 0.01$ and $c=0.5.$

enter image description here

For the outer solution (the solution that satisfies the initial conditions), I take $y(t) \sim y_0(t) +\varepsilon y_1(t)$ and substitute this into equation (1), which at leading order, gives \begin{equation} \frac{\mathrm{d}y_0(t)}{\mathrm{d}t} = y_0^2, \quad \quad \text{where} \quad \quad y_0(0)=c. \tag{2} \end{equation} Solving yields \begin{equation} y_0(t) = \frac{c}{1-ct}. \tag{3} \end{equation} Im relatively comfortable that this is correct outer solution.

Now, I need to find the inner solution which matches with this outer solution. Taking $y_{\infty}$ to be the steady state solution to (1), I make the variable transform \begin{equation} \tau = \frac{t-t_0}{\varepsilon^{\alpha}} \quad \quad \text{which provides} \quad \quad \frac{\mathrm{d}}{\mathrm{d}t}= \varepsilon^{-\alpha}\frac{\mathrm{d}}{\mathrm{d}\tau},\tag{4} \end{equation} for $\alpha>0.$ Here $t_0$ is the overlap between the outer and inner solutions; this can be estimated by seeing when $y_0(t)$ first meets $y_{\infty}$, meaning we just solve $y_0(t_0)=y_{\infty}.$ Doing so yields $t_0 = (y_{\infty}-c)/y_{\infty}c.$ For the inner region, the numerical simulations suggest that we take $Y(\tau) \sim y_{\infty}-\varepsilon Y_1(\tau). $ Substituting this into (1) yields \begin{equation} -\varepsilon^{1-\alpha}\frac{\mathrm{d}Y_1}{\mathrm{d}\tau} = -2\varepsilon y_{\infty} Y_1+\mathcal{O}(\varepsilon^2). \tag{5} \end{equation}

This is where I'm stuck. There is no possible balance except from taking $\alpha = 0$, which is not allowed. Furthermore, say if $\alpha=0$, equation (5) solves to give $Y_1 = k \exp (2y_{\infty}\tau),$ which is clearly secular.

Am I using the wrong variable transform? I'm clearly doing something wrong, since $Y_1\rightarrow \infty$ as $\tau \rightarrow \infty$.

Thanks for any help in advanved.


1 Answer 1


As you approach $y=1$, or better $y=1-ε$, the next term to become "active" is the third one. So try $y=\frac1{1+εu}$ with $u>1+O(ε)$ to get $$ -εu'=ε^2u+1-\frac{1+εu}{u}=\frac{ε^2u^2+(1-ε)u-1}{u}\tag1 $$ Thus approximately $$ -εu'=1-\frac1u\iff u'+\frac{u'}{u-1}=-\frac1ε \\ u(t)+\ln(u(t)-1)=-\frac{t}ε+d \\ (u(t)-1)e^{u(t)-1}=De^{-\frac{t}ε}\iff u(t)=1+W(De^{-\frac{t}ε}) ,~~~ y(t)=\frac1{1+ε+εW(De^{-\frac{t}ε})}\tag2 $$ using the Lambert-W function

This is a rather short segment of width $ε$ around the $t_0$ given in the question.

Then even closer to the asymptote, set $u=1+δv$ and insert in (1) $$ -εδv'=ε^2(1+δv)+1-\frac{1}{1+δv}-ε=ε^2(1+δv)+\frac{δv}{1+δv}-ε. $$ This balances on the right for $δ=ε$ to give in the leading order terms $$ -εv'=v-1\implies v(t)=1+Fe^{-tε},~~~ y(t)=\frac1{1+ε+ε^2+ε^2Fe^{-tε}} $$

Now select the constants so that all fits together...

  • $\begingroup$ Thanks for the reply. When you say ‘try $y= \frac{1}{1+\varepsilon u}$’, is this your suggestion for the inner solution? Is my outer solution okay? $\endgroup$ Mar 15, 2021 at 12:09
  • $\begingroup$ Is your $y$ the same as my $Y_1$ ? $\endgroup$ Mar 15, 2021 at 12:10
  • $\begingroup$ Yes, there is no fault with the outer solution. No, $y$ is the full perturbation series. $Y_1(\tau)$ would be equal to $u(ε\tau)$. One could even more generally try $y(t)=\frac{1-ε}{1+ε^αU(t/ε)}$. $\endgroup$ Mar 15, 2021 at 12:13
  • $\begingroup$ but the asymptote is $y=y_{\infty}$, not $y=1?$ does this not need to be reflected in the $y(t)$ you have proposed? say $y(t)=\frac{y_{\infty}-ε}{1+ε^αU(t/ε)}$ $\endgroup$ Mar 15, 2021 at 12:37
  • 1
    $\begingroup$ The asymptotic value is the root of the right side close to $1$, which refines to close to $1-ε$. Any further refinements are due to the asymptotic value of $U(\infty)$. Insertion gives $$ε^{α-1}U'=ε^2(1+ε^{α-1}U)+(1-ε)^2\frac{ε^{α-1}U}{1+ε^{α-1}U},$$ which seems to balance at $α=3$ resulting in the leading terms $U'=1+U$. This gives the asymptotic value more in line with the second section of the answer. The first section is compatible with $α=1$. $\endgroup$ Mar 15, 2021 at 12:51

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