The problem: for $\phi = \phi(x,\varepsilon)$ and $\varepsilon > 0$, $x \geq 0$ we have \begin{equation*} \varepsilon^{2}\phi'' - F(x)\phi = 0,\qquad \phi(0) = a,\qquad \phi'(0) = b \end{equation*} a) Rewrite the problem for $y(x)$ given by \begin{equation*} \phi(x) = ae^{\frac{1}{\varepsilon}\int_{0}^{x}y(z)dz} \end{equation*} Answer: By substitution we find that $\frac{d\phi}{dx} = \frac{1}{\varepsilon}y\phi$ and $\frac{d^{2}\phi}{dx^{2}} = \frac{1}{\varepsilon}(y'\phi + y(\frac{1}{\varepsilon}y\phi))$. After substituting and rewriting everything we find: \begin{equation*} \varepsilon y' + y^{2} -F(x) = 0,\qquad y(0) = \varepsilon \frac{b}{a} \end{equation*} b) Assume that F is sufficiently smooth (analytic), and $F(x) \geq c > 0$ along the interval of interest. Formulate a formal asymptotic solution of $y = y(x,\varepsilon)$ for small $\varepsilon$ up to and including $\mathcal{O}(\varepsilon)$.
- For the outer expansion claim that $y(x,\varepsilon) = y_{0}(x) + \varepsilon y_{1}(x) + ...$ Substituting this in the ode (and including terms up to $\mathcal{O}(\varepsilon)$) results in $\varepsilon y_{0}' + y_{0}^{2} + 2\varepsilon y_{0}y_{1} -F(x) = 0$ and $y_{0}(0) = 0, y_{1}(0) = \frac{b}{a}$. Solving this for the $\mathcal{O}(1)$ terms yields: $y_{0}^{2} = F(x)$, this is a contradiction as $y_{0}(0) = \pm \sqrt{F(0)}$ but we have $F(x) \geq c > 0$. From here I don't know how to proceed.
- For the inner expansion use $x = \varepsilon t$ which results in \begin{equation*} y' + y^{2} -F(\varepsilon t) = 0, \qquad y(0) = \varepsilon \frac{b}{a} \end{equation*} but I don't know how to proceed from this point.