I am trying to solve a boundary layer problem using matched expansion $$\epsilon y'' + xy' + y = 0$$ where the boundary condition is $$y(0) = 1, y(1) = 1$$ and $x\in (0,1)$.

So far, I have the outer solution by solving $xy'+y=0$, it gives $$y_0=\frac{1}{x}$$ and introducing $x=\epsilon X$ to get the inner solution $$Y'' + \epsilon XY' + Y = 0$$ Apparently, for $x>0$, there is a boundary layer at $x=0$. The inner solution I get is $$Y_0 = C(X-1)$$ Obviously, classical matching doesn't work since both solution will go to infinity. I also tried intermediate matching, but everything looks weird. Just wondering is there another right boundary layer?

So could anyone give me some hints? Thanks.

  • $\begingroup$ Please explain in more detail why you chose $x=ϵX$ and why the resulting equation is not the unbalanced $Y''+ϵXY'+ϵY=0$ $\endgroup$ – LutzL Mar 24 at 11:51

I suppose that the OP expect a method of solving of the Physicists kind. That is not what I intend to do. So, what follows might be considered not as an answer but as a comment (Too long to be edited in the comments section).

By the way, that is the opportunity for me to express my admiration for the Physicist's methods to easily find very good approximate solutions, which would require arduous purely analytical calculus. The below purely mathematical approach is a play but is certainly not recommended for concrete application.

$$\epsilon y'' + xy' + y = 0$$ The general solution is : $$y(x)=e^{-x^2/(2\epsilon)}\big(c_1+c_2\:\text{erfi}(x/\sqrt{2\epsilon}) \big)$$ Function erfi : http://mathworld.wolfram.com/Erfi.html

With conditions $y(0)=y(1)=1$ : $$\boxed{y(x)=e^{-x^2/(2\epsilon)}\left(1+(e^{1/(2\epsilon)}-1)\frac{\text{erfi}(x/\sqrt{2\epsilon})}{\text{erfi}(1/\sqrt{2\epsilon})} \right)}$$

This is the exact solution, valid on the whole range $0\leq x \leq 1$.

$$ $$

APPOXIMATE on $x<\epsilon\ll 1$ :

For $\epsilon$ small, that is $X=1/\sqrt{2\epsilon}$ large :

erfi$(X)\simeq \frac{e^{X^2}}{\sqrt{\pi}\:X}\left(1+\frac{1}{2X^2}+O(\frac{1}{X^4}) \right)$

erfi$(1/\sqrt{2\epsilon})\simeq \sqrt{\frac{2}{\pi}}e^{1/(2\epsilon)}\sqrt{\epsilon} \left(1+\epsilon+O(\epsilon^2) \right)$

$$y(x)\simeq e^{-x^2/(2\epsilon)}\big(1+\sqrt{\frac{\pi}{2\epsilon}} \left(1-\epsilon+O(\epsilon^2) \right)\text{erfi}(x/\sqrt{2\epsilon})\big)$$

For $\frac{x^2}{2\epsilon}<1$ :

$e^{-x^2/(2\epsilon)}\simeq 1-\frac{x^2}{2\epsilon}+O\left(x^4/\epsilon^2 \right)$

erfi$(x/\sqrt{2\epsilon})\simeq \frac{2}{\sqrt{\pi}}\frac{x}{\sqrt{2\epsilon}}\left(1+\frac13\frac{x^2}{2\epsilon}+O(x^4/\epsilon^2) \right)$

$$y(x)\simeq \left(1-\frac{x^2}{2\epsilon}+O\left(x^4/\epsilon^2 \right)\right)\left(1+\sqrt{\frac{\pi}{2\epsilon}} \left(1-\epsilon+O(\epsilon^2) \right) \frac{2}{\sqrt{\pi}}\frac{x}{\sqrt{2\epsilon}}\left(1+\frac13\frac{x^2}{2\epsilon}+O(x^4/\epsilon^2) \right)\right)$$

After simplification :

$$y(x)\simeq 1+\frac{x}{\epsilon}(1-\epsilon+...)\left(1+\frac{x^2}{6\epsilon}+...\right) \quad\text{in}\quad 0\leq x<\epsilon\ll 1.$$ This is a better approximate than $y(x)\simeq 1+\frac{x}{\epsilon}$.

$$ $$

APPROXIMATE on $\quad \epsilon\ll x\leq 1$ :

$X=x/\sqrt{2\epsilon})$ is large.

erfi$(X)\simeq \frac{e^{X^2}}{\sqrt{\pi}\:X}\left(1+\frac{1}{2X^2}+O(\frac{1}{X^4}) \right)$

erfi$(x/\sqrt{2\epsilon})\simeq \frac{e^{x^2/(2\epsilon)}}{\sqrt{\pi}\:x}\sqrt{2\epsilon}\left(1+\frac{\epsilon}{x^2}+O(\frac{\epsilon^2}{x^4}) \right)$

$y(x)\simeq e^{-x^2/(2\epsilon)}\left(1+\sqrt{\frac{\pi}{2\epsilon}} \left(1-\epsilon+O(\epsilon^2) \right)\frac{e^{x^2/(2\epsilon)}}{\sqrt{\pi}\:x}\sqrt{2\epsilon}\left(1+\frac{\epsilon}{x^2}+O(\frac{\epsilon^2}{x^4}) \right)\right)$

After simplification : $$y(x)\simeq \frac{1}{x}\left(1-\epsilon+...\right)\left(1+\frac{\epsilon}{x^2}+... \right)$$ We find again the result $y(x)\simeq \frac{1}{x}$ .

A more accurate approximate in the range close to $x=1$ would require the change of variable $\xi=1-x$ and the series expansion of the above exact solution for $\xi\to 0$.

  • $\begingroup$ Like you said, I am looking for a "Physicists" way to solve this. Your solution makes me feel like a "reverse-engineering". $\endgroup$ – TurbJet Mar 14 at 13:51
  • $\begingroup$ @TurbJet : You still get the full equation $Y''(X)+XY'(X)+Y(X)=0\implies Y'(X)+XY(X)=C$ for the inner solution at the boundary layer at $x=0$, thus you can not avoid the $\operatorname{erfi}$ terms. $\endgroup$ – LutzL Mar 24 at 10:59

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