# Asymptotic matching of a first order ODE

Consider the following ODE: $$$$\frac{\mathrm{d}y(t)}{\mathrm{d}t} = \varepsilon y(1-y)+y^2-\varepsilon \frac{y^2}{1-y} \quad \quad \text{for} \quad \quad t>0, \tag{1}$$$$ where $$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.$$

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 $$$$\frac{\mathrm{d}y_0(t)}{\mathrm{d}t} = y_0^2, \quad \quad \text{where} \quad \quad y_0(0)=c. \tag{2}$$$$ Solving yields $$$$y_0(t) = \frac{c}{1-ct}. \tag{3}$$$$ 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 $$$$\tau = \frac{t-t_0}{\varepsilon^{\alpha}} \quad \quad \text{which provides} \quad \quad \frac{\mathrm{d}}{\mathrm{d}t}= \varepsilon^{-\alpha}\frac{\mathrm{d}}{\mathrm{d}\tau},\tag{4}$$$$ for $$\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).$$ Substituting this into (1) yields $$$$-\varepsilon^{1-\alpha}\frac{\mathrm{d}Y_1}{\mathrm{d}\tau} = -2\varepsilon y_{\infty} Y_1+\mathcal{O}(\varepsilon^2). \tag{5}$$$$

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.

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...

• 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? Mar 15, 2021 at 12:09
• Is your $y$ the same as my $Y_1$ ? Mar 15, 2021 at 12:10
• 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 $y(t)=\frac{1-ε}{1+ε^αU(t/ε)}$. Mar 15, 2021 at 12:13
• 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 $y(t)=\frac{y_{\infty}-ε}{1+ε^αU(t/ε)}$ Mar 15, 2021 at 12:37
• 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 $α=1$. Mar 15, 2021 at 12:51