Find a two term composite asymptotic expansion of second order ODE with boundary conditions

Question

$$4\varepsilon y''+6 \sqrt x y'-3y=-3,\ 0\lt x\lt 1$$ where $y(0)=0$ and $y(1)=3$.

I worked out the first term in this expansion, which composes of an outer solution and inner solution. They are $$y_0^{outer}=1+2e^{\sqrt x -1}$$ $$y_0^{inner}=\frac {1+2e^{-1}}{K} \int_0^{\bar x}e^{-t^{\frac32}}dt$$ where $K=\frac23 \Gamma \left(\frac23\right)$ and $\bar x = \frac x{\varepsilon^{\frac23}}$.

I worked the second outer solution as below, $$y_1^{outer}=\frac13\left( \frac2{\sqrt x}-\frac1x -1 \right) e^{\sqrt x -1}$$ by letting $y_1^{outer}\left(1\right)=0$. However, i realized this solution is a bit problematic since it goes to $-\infty$ when $x\to 0^+$. what can i do with this solution?

I could not work out the second inner solution from the following ODE $$4\varepsilon^ {-\frac13} y''+6\varepsilon^ {-\frac13} \bar x ^\frac12 y'- 3y =-3$$ Can anyone enlighten me on this part?

Answer 1

Let $r=\sqrt{x}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=\dfrac{1}{2\sqrt{x}}\dfrac{dy}{dr}=\dfrac{1}{2r}\dfrac{dy}{dr}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\biggl(\dfrac{1}{2r}\dfrac{dy}{dr}\biggr)=\dfrac{d}{dr}\biggl(\dfrac{1}{2r}\dfrac{dy}{dr}\biggr)\dfrac{dr}{dx}=\biggl(\dfrac{1}{2r}\dfrac{d^2y}{dr^2}-\dfrac{1}{2r^2}\dfrac{dy}{dr}\biggr)\dfrac{1}{2\sqrt{x}}=\biggl(\dfrac{1}{2r}\dfrac{d^2y}{dr^2}-\dfrac{1}{2r^2}\dfrac{dy}{dr}\biggr)\dfrac{1}{2r}=\dfrac{1}{4r^2}\dfrac{d^2y}{dr^2}-\dfrac{1}{4r^3}\dfrac{dy}{dr}$

$\therefore4\varepsilon\biggl(\dfrac{1}{4r^2}\dfrac{d^2y}{dr^2}-\dfrac{1}{4r^3}\dfrac{dy}{dr}\biggr)+6r\dfrac{1}{2r}\dfrac{dy}{dr}-3y=-3$

where $y(0)=0$ and $y(1)=3$

$\dfrac{\varepsilon}{r^2}\dfrac{d^2y}{dr^2}-\dfrac{\varepsilon}{r^3}\dfrac{dy}{dr}+3\dfrac{dy}{dr}-3y+3=0$

where $y(0)=0$ and $y(1)=3$

$\varepsilon r\dfrac{d^2y}{dr^2}+(3r^3-\varepsilon)\dfrac{dy}{dr}-3r^3(y-1)=0$ where $y(0)=0$ and $y(1)=3$

Let $u=y-1$ ,

Then $\varepsilon r\dfrac{d^2u}{dr^2}+(3r^3-\varepsilon)\dfrac{du}{dr}-3r^3u=0$ where $u(0)=-1$ and $u(1)=2$

Answer 2

Here is a way to proceed in the style of WKB theory.

First, notice that $y=1$ is a particular solution to the ODE. Thus WLOG let us instead consider $4\varepsilon z'' + 6 \sqrt{x} z' - 3z = 0,z(0)=-1,z(1)=2$.

Now let $z=fg$ and plug this into the ODE. The goal is to solve a first order ODE for $f$ so that we obtain an ODE for $g$ not involving $g'$.

We have

$$4 \left ( \varepsilon f'' g + 2 \varepsilon f' g' + \varepsilon f g'' \right ) + 6 \sqrt{x} \left ( f' g + f g' \right ) -3 fg = 0.$$

So now group together all terms involving $g'$ and try to set them equal to zero. Thus you want $8\varepsilon f' g' + 6\sqrt{x} fg' = 0$. Assuming $g'$ is never zero, this is a first order ODE that you can solve by an integrating factor: $f' + \frac{3}{4\varepsilon} \sqrt{x} f = 0$ so $f=Ce^{-\int \frac{3}{4\varepsilon} \sqrt{x} dx}=Ce^{-\frac{1}{2\varepsilon} x^{3/2}}$. We can arbitrarily set $C=1$. (Notice that this would not have worked if we hadn't been able to subtract off the particular solution.)

So if we set $f=e^{-\frac{1}{2\varepsilon} x^{3/2}}$ then the ODE for $g$ becomes

$$4\varepsilon f g'' + (4\varepsilon f''+6\sqrt{x} f'-3f) g = 0.$$

I believe you can now proceed using the WKB ansatz $g(x)=e^{\theta(x)/\sqrt{\varepsilon}} h(x)$. Let me know how it goes, if you run into trouble I can expand this answer. Where I think you might encounter a problem is that $f''$ has a singularity at $x=0$ so that you cannot neglect $\varepsilon f''$ relative to $\sqrt{x} f'$ or $f$ at $x=0$.