### 4.8 note p7 added 2/15/17, boundary layer problem solved in details

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This note solves\begin{align*} \varepsilon y^{\prime \prime }\left ( x\right ) +\left ( 1+x\right ) y^{\prime }\left ( x\right ) +y\left ( x\right ) & =0\\ y\left ( 0\right ) & =1\\ y\left ( 1\right ) & =1 \end{align*}

Where $$\varepsilon$$ is small parameter, using boundary layer theory.

##### 4.8.0.1 Solution

Since $$\left ( 1+x\right ) >0$$ in the domain, we expect boundary layer to be on the left. Let $$y_{out}\left ( x\right )$$ be the solution in the outer region. Starting with $$y\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}$$ and substituting back into the ODE gives$\varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\left ( 1+x\right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0$ $$O\left ( 1\right )$$ terms

Collecting all terms with zero powers of $$\varepsilon$$$\left ( 1+x\right ) y_{0}^{\prime }+y_{0}=0$ The above is solved using the right side conditions, since this is where the outer region is located. Solving the above using $$y_{0}\left ( 1\right ) =1$$ gives$y_{0}^{out}\left ( x\right ) =\frac{2}{1+x}$ Now we need to ﬁnd $$y_{in}\left ( x\right )$$. To do this, we convert the ODE using transformation $$\xi =\frac{x}{\varepsilon }$$. Hence $$\frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx}=\frac{dy}{d\xi }\frac{1}{\varepsilon }$$. Hence the operator $$\frac{d}{dx}\equiv \frac{1}{\varepsilon }\frac{d}{d\xi }$$. This means the operator $$\frac{d^{2}}{dx^{2}}\equiv \left ( \frac{1}{\varepsilon }\frac{d}{d\xi }\right ) \left ( \frac{1}{\varepsilon }\frac{d}{d\xi }\right ) =\frac{1}{\varepsilon ^{2}}\frac{d^{2}}{d\xi ^{2}}$$. The ODE becomes\begin{align*} \varepsilon \frac{1}{\varepsilon ^{2}}\frac{d^{2}y\left ( \xi \right ) }{d\xi ^{2}}+\left ( 1+\xi \varepsilon \right ) \frac{1}{\varepsilon }\frac{dy\left ( \xi \right ) }{d\xi }+y\left ( \xi \right ) & =0\\ \frac{1}{\varepsilon }y^{\prime \prime }+\left ( \frac{1}{\varepsilon }+\xi \right ) y^{\prime }+y & =0 \end{align*}

Plugging $$y\left ( \xi \right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y\left ( \xi \right ) _{n}$$ into the above gives\begin{align*} \frac{1}{\varepsilon }\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\left ( \frac{1}{\varepsilon }+\xi \right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) & =0\\ \frac{1}{\varepsilon }\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\frac{1}{\varepsilon }\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\xi \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) & =0 \end{align*}

Collecting all terms with smallest power of $$\varepsilon$$ , which is $$\varepsilon ^{-1}$$ in this case, gives\begin{align*} \frac{1}{\varepsilon }y_{0}^{\prime \prime }+\frac{1}{\varepsilon }y_{0}^{\prime } & =0\\ y_{0}^{\prime \prime }+y_{0}^{\prime } & =0 \end{align*}

Let $$z=y_{0}^{\prime }$$, the above becomes\begin{align*} z^{\prime }+z & =0\\ d\left ( e^{\xi }z\right ) & =0\\ e^{\xi }z & =c\\ z & =ce^{-\xi } \end{align*}

Hence $$y_{0}^{\prime }\left ( \xi \right ) =ce^{-\xi }$$. Integrating $$y_{0}^{in}\left ( \xi \right ) =-ce^{-\xi }+c_{1}\tag{1A}$$ Since $$c$$ is arbitrary constant, the negative sign can be removed, giving$$y_{0}^{in}\left ( \xi \right ) =ce^{-\xi }+c_{1}\tag{1A}$$ This is the lowest order solution for the inner $$y^{in}\left ( \xi \right )$$. We have two boundary conditions, but we can only use the left side one, where $$y_{in}\left ( \xi \right )$$ lives. Hence using $$y_{0}\left ( \xi =0\right ) =1$$, the above becomes\begin{align*} 1 & =c+c_{1}\\ c_{1} & =1-c \end{align*}

The solution (1A) becomes\begin{align*} y_{0}\left ( \xi \right ) & =ce^{-\xi }+\left ( 1-c\right ) \\ & =1+c\left ( e^{-\xi }-1\right ) \end{align*}

Let $$c=A_{0}$$ to match the book notation.$y_{0}\left ( \xi \right ) =1+A_{0}\left ( e^{-\xi }-1\right )$ To ﬁnd $$A_{0}$$, we match $$y_{0}^{in}\left ( \xi \right )$$ with $$y_{0}^{out}\left ( x\right )$$\begin{align*} \lim _{\xi \rightarrow \infty }1+A_{0}\left ( e^{-\xi }-1\right ) & =\lim _{x\rightarrow 0^{+}}\frac{2}{1+x}\\ 1-A_{0} & =2\\ A_{0} & =-1 \end{align*}

Hence $y_{0}^{in}\left ( \xi \right ) =2-e^{-\xi }$ $$O\left ( \varepsilon \right )$$ terms

We now repeat the process to ﬁnd $$y_{1}^{in}\left ( \xi \right )$$ and $$y_{1}^{out}\left ( x\right ) .$$ Starting with $$y^{out}\left ( x\right )$$$\varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\left ( 1+x\right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0$ Collecting all terms with $$\varepsilon ^{1}$$ now\begin{align*} \varepsilon y_{0}^{\prime \prime }+\left ( 1+x\right ) \varepsilon y_{1}^{\prime }+\varepsilon y_{1} & =0\\ y_{0}^{\prime \prime }+\left ( 1+x\right ) y_{1}^{\prime }+y_{1} & =0 \end{align*}

But we know $$y_{0}=\frac{2}{1+x}$$, from above. Hence $$y_{0}^{\prime \prime }=\frac{4}{\left ( 1+x\right ) ^{3}}$$ and the above becomes\begin{align*} \left ( 1+x\right ) y_{1}^{\prime }+y_{1} & =-\frac{4}{\left ( 1+x\right ) ^{3}}\\ y_{1}^{\prime }+\frac{y_{1}}{1+x} & =-\frac{4}{\left ( 1+x\right ) ^{4}} \end{align*}

Integrating factor $$\mu =e^{\int \frac{1}{1+x}dx}=e^{\ln \left ( 1+x\right ) }=1+x$$ and the above becomes\begin{align*} \frac{d}{dx}\left ( \mu y_{1}\right ) & =-\mu \frac{4}{\left ( 1+x\right ) ^{4}}\\ \frac{d}{dx}\left ( \left ( 1+x\right ) y_{1}\right ) & =-\frac{4}{\left ( 1+x\right ) ^{3}} \end{align*}

Integrating\begin{align*} \left ( 1+x\right ) y_{1} & =-\int \frac{4}{\left ( 1+x\right ) ^{3}}dx+c\\ & =\frac{2}{\left ( 1+x\right ) ^{2}}+c \end{align*}

Hence $y_{1}\left ( x\right ) =\frac{2}{\left ( 1+x\right ) ^{3}}+\frac{c}{1+x}$ Applying $$y\left ( 1\right ) =0$$ (notice the boundary condition now becomes $$y\left ( 1\right ) =0$$ and not $$y\left ( 1\right ) =1$$, since we have already used $$y\left ( 1\right ) =1$$ to ﬁnd leading order). From now on, all boundary conditions will be $$y\left ( 1\right ) =0$$.\begin{align*} 0 & =\frac{2}{\left ( 1+1\right ) ^{3}}+\frac{c}{1+1}\\ c & =-\frac{1}{2} \end{align*}

Hence$y_{1}\left ( x\right ) =\frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2}\frac{1}{\left ( 1+x\right ) }$ Now we need to ﬁnd $$y_{1}^{in}\left ( \xi \right )$$. To do this, starting from\begin{align*} \frac{1}{\varepsilon }\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\left ( \frac{1}{\varepsilon }+\xi \right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) & =0\\ \frac{1}{\varepsilon }\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\frac{1}{\varepsilon }\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\xi \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) & =0 \end{align*}

But now collecting all terms with $$O\left ( 1\right )$$ order, (last time, we collected terms with $$O\left ( \varepsilon ^{-1}\right )$$ ).\begin{align} y_{1}^{\prime \prime }+y_{1}^{\prime }+\xi y_{0}^{\prime }+y_{0} & =0\nonumber \\ y_{1}^{\prime \prime }+y_{1}^{\prime } & =-\xi y_{0}^{\prime }-y_{0}\tag{1} \end{align}

But we found $$y_{0}^{in}$$ earlier which was $y_{0}^{in}\left ( \xi \right ) =1+A_{0}\left ( e^{-\xi }-1\right )$ Hence $$y_{0}^{\prime }=-A_{0}e^{-\xi }$$ and the ODE (1) becomes$y_{1}^{\prime \prime }+y_{1}^{\prime }=\xi A_{0}e^{-\xi }-\left ( 1+A_{0}\left ( e^{-\xi }-1\right ) \right )$ We need to solve this with boundary conditions $$y_{1}\left ( 0\right ) =0$$. (again, notice change in B.C. as was mentioned above). The solution is\begin{align*} y_{1}\left ( \xi \right ) & =-\xi +A_{0}\left ( \xi -\frac{1}{2}\xi ^{2}e^{-\xi }\right ) +A_{1}\left ( 1-e^{-\xi }\right ) \\ & =-\xi +A_{0}\left ( \xi -\frac{1}{2}\xi ^{2}e^{-\xi }\right ) -A_{1}\left ( e^{-\xi }-1\right ) \end{align*}

Since $$A_{1}$$ is arbitrary constant, and to match the book, we can call $$A_{2}=-A_{1}$$ and then rename $$A_{2}$$ back to $$A_{1}$$ and obtain$y_{1}\left ( \xi \right ) =-\xi +A_{0}\left ( \xi -\frac{1}{2}\xi ^{2}e^{-\xi }\right ) +A_{1}\left ( e^{-\xi }-1\right )$ This is to be able to follow the book. Therefore, this is what we have so far\begin{align*} y_{out} & =y_{0}^{out}+\varepsilon y_{1}^{out}\\ & =\frac{2}{1+x}+\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }\right ) \end{align*}

And \begin{align*} y^{in}\left ( \xi \right ) & =y_{0}^{in}+\varepsilon y_{1}^{in}\\ & =\left ( 1+A_{0}\left ( e^{-\xi }-1\right ) \right ) +\varepsilon \left ( -\xi +A_{0}\left ( \xi -\frac{1}{2}\xi ^{2}e^{-\xi }\right ) +A_{1}\left ( e^{-\xi }-1\right ) \right ) \\ & =A_{0}e^{-\xi }-\xi \varepsilon -\varepsilon A_{1}-A_{0}+\varepsilon A_{1}e^{-\xi }+\xi \varepsilon A_{0}-\frac{1}{2}\xi ^{2}\varepsilon A_{0}e^{-\xi }+1 \end{align*}

To ﬁnd $$A_{0},A_{1}$$, we match $$y_{in}$$ with $$y_{out}$$, therefore $\lim _{\xi \rightarrow \infty }y_{in}=\lim _{x\rightarrow 0}y_{out}$ Or\begin{multline*} \lim _{\xi \rightarrow \infty }\left ( A_{0}e^{-\xi }-\xi \varepsilon -\varepsilon A_{1}-A_{0}+\varepsilon A_{1}e^{-\xi }+\xi \varepsilon A_{0}-\allowbreak \frac{1}{2}\xi ^{2}\varepsilon A_{0}e^{-\xi }+1\right ) =\\ \lim _{x\rightarrow 0}\frac{2}{1+x}+\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }\right ) \end{multline*} Which simpliﬁes to$-\xi \varepsilon -\varepsilon A_{1}-A_{0}+\xi \varepsilon A_{0}+1=\lim _{x\rightarrow 0}\frac{2}{1+x}+\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }\right )$ It is easier now to convert the LHS to use $$x$$ instead of $$\xi$$ so we can compare. Since $$\xi =\frac{x}{\varepsilon }$$, then the above becomes$-x-\varepsilon A_{1}-A_{0}+xA_{0}+1=\lim _{x\rightarrow 0}\frac{2}{1+x}+\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }\right )$ Using Taylor series on the RHS\begin{multline*} 1-A_{0}-x+A_{0}x+A_{1}\varepsilon =\lim _{x\rightarrow 0}2\left ( 1-x+x^{2}+\cdots \right ) \\ +2\varepsilon \left ( 1-x+x^{2}+\cdots \right ) \left ( 1-x+x^{2}+\cdots \right ) \left ( 1-x+x^{2}+\cdots \right ) -\frac{\varepsilon }{2}\left ( 1-x+x^{2}+\cdots \right ) \end{multline*} Since we have terms on the the LHS of only $$O\left ( 1\right ) ,O\left ( x\right ) ,O\left ( \varepsilon \right )$$, then we need to keep at least terms with $$O\left ( 1\right ) ,O\left ( x\right ) ,O\left ( \varepsilon \right )$$ on the RHS and drop terms with $$O\left ( x^{2}\right ) ,O\left ( \varepsilon x\right ) ,O\left ( \varepsilon ^{2}\right )$$ to be able to do the matching. So in the above, RHS simpliﬁes to\begin{align*} -x-\varepsilon A_{1}-A_{0}+xA_{0}+1 & =2\left ( 1-x\right ) +2\varepsilon -\frac{\varepsilon }{2}\\ -x-\varepsilon A_{1}-A_{0}+xA_{0}+1 & =2-2x+2\varepsilon -\frac{\varepsilon }{2}\\ -\varepsilon A_{1}-A_{0}+x\left ( A_{0}-1\right ) +1 & =2-2x+\frac{3}{2}\varepsilon \end{align*}

Comparing, we see that \begin{align*} A_{0}-1 & =-2\\ A_{0} & =-1 \end{align*}

We notice this is the same $$A_{0}$$ we found for the lowest order. This is how it should always come out. If we get diﬀerent value, it means we made mistake. We could also match $$-A_{0}+1=2$$ which gives $$A_{0}=-1$$ as well. Finally \begin{align*} -\varepsilon A_{1} & =\frac{3}{2}\varepsilon \\ A_{1} & =-\frac{3}{2} \end{align*}

So we have used matching to ﬁnd all the constants for $$y_{in}$$. Here is the ﬁnal solution so far

\begin{align*} y_{out}\left ( x\right ) & =\overset{y_{0}}{\overbrace{\frac{2}{1+x}}}+\varepsilon \overset{y_{1}}{\overbrace{\left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }\right ) }}\\ y_{in}\left ( \xi \right ) & =\overset{y_{0}}{\overbrace{1+A_{0}\left ( e^{-\xi }-1\right ) }}+\varepsilon \overset{y_{1}}{\overbrace{\left ( -\xi +A_{0}\left ( \xi -\frac{1}{2}\xi ^{2}e^{-\xi }\right ) +A_{1}\left ( e^{-\xi }-1\right ) \right ) }}\\ & =1-\left ( e^{-\xi }-1\right ) +\varepsilon \left ( -\xi -\left ( \xi -\frac{1}{2}\xi ^{2}e^{-\xi }\right ) -\frac{3}{2}\left ( e^{-\xi }-1\right ) \right ) \\ & =\frac{3}{2}\varepsilon -e^{-\xi }-2\xi \varepsilon -\frac{3}{2}\varepsilon e^{-\xi }+\frac{1}{2}\xi ^{2}\varepsilon e^{-\xi }+\allowbreak 2 \end{align*}

In terms of $$x$$, since Since $$\xi =\frac{x}{\varepsilon }$$ the above becomes\begin{align*} y_{in}\left ( x\right ) & =\frac{3}{2}\varepsilon -e^{-\frac{x}{\varepsilon }}-2x-\frac{3}{2}\varepsilon e^{-\frac{x}{\varepsilon }}+\frac{1}{2}\frac{x^{2}}{\varepsilon }e^{-\frac{x}{\varepsilon }}+\allowbreak 2\\ & =2-2x+\frac{3}{2}\varepsilon +e^{-\frac{x}{\varepsilon }}\left ( \frac{1}{2}\frac{x^{2}}{\varepsilon }-\frac{3}{2}\varepsilon -1\right ) \end{align*}

Hence $y_{uniform}=y_{in}+y_{out}-y_{match}$ Where \begin{align*} y_{match} & =\lim _{\xi \rightarrow \infty }y_{in}\\ & =2-2x+\frac{3}{2}\varepsilon \end{align*}

Hence\begin{align*} y_{uniform} & =2-2x+\frac{3}{2}\varepsilon +e^{-\frac{x}{\varepsilon }}\left ( \frac{1}{2}\frac{x^{2}}{\varepsilon }-\frac{3}{2}\varepsilon -1\right ) +\frac{2}{1+x}+\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }\right ) -\left ( 2-2x+\frac{3}{2}\varepsilon \right ) \\ & =e^{-\frac{x}{\varepsilon }}\left ( \frac{1}{2}\frac{x^{2}}{\varepsilon }-\frac{3}{2}\varepsilon -1\right ) +\frac{2}{1+x}+\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }\right ) \\ & =\left ( \frac{2}{1+x}-e^{-\frac{x}{\varepsilon }}+\frac{1}{2}\frac{x^{2}}{\varepsilon }e^{-\frac{x}{\varepsilon }}\right ) +\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }-\frac{3}{2}e^{-\frac{x}{\varepsilon }}\right ) \end{align*}

Which is the same as\begin{align} y_{uniform} & =\left ( \frac{2}{1+x}-e^{-\xi }+\frac{1}{2}\frac{x^{2}}{\varepsilon }e^{-\xi }\right ) +\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }-\frac{3}{2}e^{-\xi }\right ) \nonumber \\ & =\left ( \frac{2}{1+x}-e^{-\xi }\right ) +\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }-\frac{3}{2}e^{-\xi }+\frac{1}{2}\xi ^{2}e^{-\xi }\right ) \nonumber \\ & =\left ( \frac{2}{1+x}-e^{-\xi }\right ) +\varepsilon \left ( \frac{2}{\left ( 1+x\right ) ^{3}}-\frac{1}{2\left ( 1+x\right ) }+\left ( \frac{1}{2}\xi ^{2}-\frac{3}{2}\right ) e^{-\xi }\right ) \tag{1} \end{align}

Comparing (1) above, with book result in ﬁrst line of 9.3.16, page 433, we see the same result.

##### 4.8.0.2 References

1.
Advanced Mathematica methods, Bender and Orszag. Chapter 9.
2.
Lecture notes. Feb 16, 2017. By Professor Smith. University of Wisconsin. NE 548