Solution. The strategy here is different from the other problems because $\varepsilon$ is in two coefficients. For this problem, we need to factor out $\varepsilon x$. Doing this yields $\varepsilon x(x^3+\varepsilon)-1=0$. Since we are looking for $x$ such that $|x|\gg 1$ and since $0<\varepsilon \ll 1$, $x^3+\varepsilon\sim x^3$. It follows that $\varepsilon x^4-1\sim 0$, and so $x=\mathcal O(\varepsilon^{-1/4})$. Thus, we take $z=\varepsilon^{1/4}x$. To actually obtain approximations for the roots, we replace $x$ in the polynomial by $z$ to get $z^2 +\varepsilon^{7/4}z -1=0$. The perturbation expansion for the solution would have the form $x=a\varepsilon^{-1/4}+b\varepsilon^{3/2} +\mathcal O(\varepsilon^{13/4})$. Not particularly nice, but not horrible either.
Solution. We will break this problem into steps.
\[ \frac{\partial^2\! u}{\partial \sigma^2} + u + \bigg(\frac{\partial^2\! v}{\partial \sigma^2} + v + 2\frac{\partial^2\! u}{\partial \sigma \partial \tau} + \frac{\partial u}{\partial \sigma} + u^3 \bigg) \varepsilon + \mathcal O(\varepsilon^2) =0.\]
In addition, the inital conditons are $u+\varepsilon v +\mathcal O(\epsilon^2) = 0$ and $ 1 = \frac{\partial u}{\partial \sigma}+ \bigg(\frac{\partial u}{\partial \tau} + \frac{\partial v}{\partial \sigma}\bigg)\varepsilon + \mathcal O(\varepsilon^2)$. Equating to the coefficients of $\varepsilon^0$ and $\varepsilon^1$ to 0 results in these initial value problems:
\[\frac{\partial^2\! u}{\partial \sigma^2} + u =0,\ u(0,0)=0,\ \frac{\partial u}{\partial \sigma}(0,0) = 1. \] \[ \frac{\partial^2\! v}{\partial \sigma^2} + v + 2\frac{\partial^2\! u}{\partial \sigma \partial \tau} + \frac{\partial u}{\partial \sigma} + u^2 =0,\ v(0,0)=0,\ \frac{\partial v}{\partial \sigma}(0,0) + \frac{\partial u}{\partial \tau}(0,0) = 0. \]
\[ \frac{\partial^2\! v}{\partial \sigma^2} + v + 2A'(\tau)\cos(\sigma) - 2B'(\tau)\sin(\sigma) + A(\tau)\cos(\sigma) - B(\tau)\sin(\sigma) +(A(\tau)\sin(\tau)+B(\tau)\cos(\sigma))^2=0\] Since $(A\cos(\sigma)+B\cos(\sigma))^2= A^2\sin^2(\sigma)+2AB\sin(\sigma)\cos(\sigma) + B^2\cos^2(\sigma)= \frac12\big(A^2+B^2)+ \frac12\big(B^2-A^2)\cos(2\sigma)+ AB\sin(2\sigma)$, we see that, after a little algebra, the equation above becomes
\[ \frac{\partial^2\! v}{\partial \sigma^2} + v +(2A'+A)\cos(\sigma)-(2B'+B)\sin(\sigma)+\frac12\big(A^2+B^2)+ \frac12\big(B^2-A^2) \cos(2\sigma)+ AB\sin(2\sigma) \]
Solution. For $x\approx 1$, drop the $\varepsilon$ term and work with $y' +\frac y{x+1} =2$, $y(1)=3$. This is a first order linear equation with integrating factor (see an ODE book) $\mu = \exp\big(\int (x+1)^{-1}dx\big) = \exp(\log(x+1)) = x+1$. Hence $[(x+1)y]'=2(x+1)$ and so $(x+1)y=x^2+2x +C$. Since $y(1)=3$, $C=3$. The outer solution is thus $y_{out}=x+1+2(x+1)^{-1}$. To find the inner solution, we go back to the orginal equation $\varepsilon y'' + y' +\frac y{x+1} =2$ and change variables to $z=x/\varepsilon$. After simplifying, we obtain the equation $d^2y/dz^2+dy/dz+4\varepsilon y(1+\varepsilon z)^{-1}=2\varepsilon$. Dropping $\mathcal O(\varepsilon)$ terms, we arrive at $d^2y/dz^2+dy/dz=0$. The solution to this equation is $y=A+Be^{-z}$. The boundary condition is $y(x=0)=y(z=0)=0$. Using this in the previous equation we see that $B=-A$ and $y=A(1-e^{-z})$. Putting back $x$ then gives us the boundary layer solution $y_{BL}=A(1-e^{-x/\varepsilon})$. The next step is to find $A$ and $y_{overlap}$. To do this, let $w=x/\sqrt{\varepsilon}$ and then note that, with $w$, $y_{BL} =A(1-e^{-w/\sqrt{\varepsilon}})$ and $y_{out} = \sqrt{\varepsilon}w + 1+2(\sqrt{\varepsilon}w+1)^{-1}$. We remember that $\lim_{\varepsilon \downarrow 0} y_{BL} = \lim_{\varepsilon \downarrow 0}y_{out}= y_{overlap}$. Doing the limits the implies $A=3=y_{overlap}$. The uniform solution is then $y_{unif}=y_{BL}+y_{out}-y_{overlap} = x+1+2(x+1)^{-1}-3e^{-x/\varepsilon}$.
Answer. $y_{unif} = 2+x-e^{-4x/\varepsilon}$.