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.
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 \begin{equation} y_{0}^{in}\left ( \xi \right ) =-ce^{-\xi }+c_{1}\tag{1A} \end{equation} Since \(c\) is arbitrary constant, the negative sign can be removed, giving\begin{equation} y_{0}^{in}\left ( \xi \right ) =ce^{-\xi }+c_{1}\tag{1A} \end{equation} 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.