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

$$4\varepsilon y''+6 \sqrt x y'-3y=-3,\ 0\lt x\lt 1$$ 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?

• @epiliam I have worked out the inner solution for the order $O(\varepsilon ^ {-\frac13})$, the second term will be solving $O(1)$ right. This term is very troublesome since it involves $y_0 ^{inner}$ and it is in integral form. I am stuck there. Commented Sep 4, 2020 at 6:13
• Sorry, I thought the problem was a lot simpler than it was, thus my earlier comments. I did all the work up to where you got to and got the same results. That is a very nasty integral of a Gamma function that you get from the second term inner expansion. I don't think there is anyway around it.
– user765629
Commented Sep 4, 2020 at 7:03

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 $$u(1)=2$$

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 $$x=0$$.