problem (a) show that if a\left ( x\right ) <0 for 0\leq x\leq 1 then the solution to 9.1.7 has boundary layer at x=1. (b) Find a uniform approximation with error O\left ( \varepsilon \right ) to the solution 9.1.7 when a\left ( x\right ) <0 for 0\leq x\leq 1 (c) Show that if a\left ( x\right ) >0 it is impossible to match to a boundary layer at x=1
solution
Equation 9.1.7 at page 422 is\begin{align} \varepsilon y^{\prime \prime }+a\left ( x\right ) y^{\prime }+b\left ( x\right ) y\left ( x\right ) & =0\tag{9.1.7}\\ y\left ( 0\right ) & =A\nonumber \\ y\left ( 1\right ) & =B\nonumber \end{align}
For 0\leq x\leq 1. Now we solve for y_{in}\left ( x\right ) , but first we introduce inner variable \xi . We assume boundary layer is at x=0, then show that this leads to inconsistency. Let \xi =\frac{x}{\varepsilon ^{p}} be the inner variable. We express the original ODE using this new variable. We also need to determine p. Since \frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx} then \frac{dy}{dx}=\frac{dy}{d\xi }\varepsilon ^{-p}. Hence \frac{d}{dx}\equiv \varepsilon ^{-p}\frac{d}{d\xi }\begin{align*} \frac{d^{2}}{dx^{2}} & =\frac{d}{dx}\frac{d}{dx}\\ & =\left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \\ & =\varepsilon ^{-2p}\frac{d^{2}}{d\xi ^{2}} \end{align*}
Therefore \frac{d^{2}y}{dx^{2}}=\varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}} and (9.1.7) becomes\begin{align*} \varepsilon \varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}}+a\left ( x\right ) \varepsilon ^{-p}\frac{dy}{d\xi }+y & =0\\ \varepsilon ^{1-2p}\frac{d^{2}y}{d\xi ^{2}}+a\left ( x\right ) \varepsilon ^{-p}\frac{dy}{d\xi }+y & =0 \end{align*}
The largest terms are \left \{ \varepsilon ^{1-2p},\varepsilon ^{-p}\right \} , therefore balance gives 1-2p=-p or p=1. The ODE now becomes\begin{equation} \varepsilon ^{-1}\frac{d^{2}y}{d\xi ^{2}}+a\left ( x\right ) \varepsilon ^{-1}\frac{dy}{d\xi }+y=0 \tag{1} \end{equation} Assuming that y_{in}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots And substituting the above into (1) gives\begin{equation} \varepsilon ^{-1}\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +a\left ( x\right ) \varepsilon ^{-1}\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0 \tag{1A} \end{equation} Collecting powers of O\left ( \varepsilon ^{-1}\right ) terms, gives the ODE to solve for y_{0}^{in} as y_{0}^{\prime \prime }\thicksim -a\left ( x\right ) y_{0}^{\prime } In the rapidly changing region, because the boundary layer is very thin, we can approximate a\left ( x\right ) by a\left ( 0\right ) . The above becomes y_{0}^{\prime \prime }\thicksim -a\left ( 0\right ) y_{0}^{\prime } But we are told that a\left ( x\right ) <0, so a\left ( 0\right ) <0, hence -a\left ( 0\right ) is positive. Let -a\left ( 0\right ) =n^{2}, to make it more clear this is positive, then the ODE to solve is y_{0}^{\prime \prime }\thicksim n^{2}y_{0}^{\prime } The solution to this ODE is y_{0}\left ( \xi \right ) \thicksim \frac{C_{1}}{n^{2}}e^{n^{2}\xi }+C_{2} Using y\left ( 0\right ) =A, then the above gives A=\frac{C_{1}}{n^{2}}+C_{2} or C_{2}=A-\frac{C_{1}}{n^{2}} and the ODE becomes\begin{align*} y_{0}\left ( \xi \right ) & \thicksim \frac{C_{1}}{n^{2}}e^{n^{2}\xi }+\left ( A-\frac{C_{1}}{n^{2}}\right ) \\ & \thicksim \frac{C_{1}}{n^{2}}\left ( e^{n^{2}\xi }-1\right ) +A \end{align*}
We see from the above solution for the inner layer, that as \xi increases (meaning we are moving away from x=0), then the solution y_{0}\left ( \xi \right ) and its derivative is increasing and not decreasing since y_{0}^{\prime }\left ( \xi \right ) =C_{1}e^{n^{2}\xi } and y_{0}^{\prime \prime }\left ( \xi \right ) =C_{1}n^{2}e^{n^{2}\xi }.
But this contradicts what we assumed that the boundary layer is at x=0\, since we expect the solution to change less rapidly as we move away from x=0. Hence we conclude that if a\left ( x\right ) <0, then the boundary layer can not be at x=0.
Let us now see what happens by taking the boundary layer to be at x=1. We repeat the same process as above, but now the inner variable as defined as
\xi =\frac{1-x}{\varepsilon ^{p}} We express the original ODE using this new variable and determine p. Since \frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx} then \frac{dy}{dx}=\frac{dy}{d\xi }\left ( -\varepsilon ^{-p}\right ) . Hence \frac{d}{dx}\equiv \left ( -\varepsilon ^{-p}\right ) \frac{d}{d\xi }\begin{align*} \frac{d^{2}}{dx^{2}} & =\frac{d}{dx}\frac{d}{dx}\\ & =\left ( \left ( -\varepsilon ^{-p}\right ) \frac{d}{d\xi }\right ) \left ( \left ( -\varepsilon ^{-p}\right ) \frac{d}{d\xi }\right ) \\ & =\varepsilon ^{-2p}\frac{d^{2}}{d\xi ^{2}} \end{align*}
Therefore \frac{d^{2}y}{dx^{2}}=\varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}} and equation (9.1.7) becomes\begin{align*} \varepsilon \varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}}-a\left ( x\right ) \varepsilon ^{-p}\frac{dy}{d\xi }+y & =0\\ \varepsilon ^{1-2p}\frac{d^{2}y}{d\xi ^{2}}-a\left ( x\right ) \varepsilon ^{-p}\frac{dy}{d\xi }+y & =0 \end{align*}
The largest terms are \left \{ \varepsilon ^{1-2p},\varepsilon ^{-p}\right \} , therefore matching them gives 1-2p=-p or \fbox{$p=1$} The ODE now becomes\begin{equation} \varepsilon ^{-1}\frac{d^{2}y}{d\xi ^{2}}-a\left ( x\right ) \varepsilon ^{-1}\frac{dy}{d\xi }+y=0 \tag{2} \end{equation} Assuming that y_{in}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots And substituting the above into (2) gives\begin{equation} \varepsilon ^{-1}\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) -a\left ( x\right ) \varepsilon ^{-1}\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0 \tag{2A} \end{equation} Collecting O\left ( \varepsilon ^{-1}\right ) terms, gives the ODE to solve for y_{0}^{in} as y_{0}^{\prime \prime }\thicksim a\left ( x\right ) y_{0}^{\prime } In the rapidly changing region, \alpha =a\left ( 1\right ) , because the boundary layer is very thin, we approximated a\left ( x\right ) by a\left ( 1\right ) . The above becomes y_{0}^{\prime \prime }\thicksim \alpha y_{0}^{\prime } But we are told that a\left ( x\right ) <0, so \alpha <0, and the above becomes y_{0}^{\prime \prime }\thicksim \alpha y_{0}^{\prime } The solution to this ODE is\begin{equation} y_{0}\left ( \xi \right ) \thicksim \frac{C_{1}}{\alpha }e^{\alpha \xi }+C_{2} \tag{3} \end{equation} Using \begin{align*} y\left ( x=1\right ) & =y\left ( \xi =0\right ) \\ & =B \end{align*}
Then (3) gives B=\frac{C_{1}}{\alpha }+C_{2} or C_{2}=B-\frac{C_{1}}{\alpha } and (3) becomes\begin{align} y_{0}\left ( \xi \right ) & \thicksim \frac{C_{1}}{\alpha }e^{\alpha \xi }+\left ( B-\frac{C_{1}}{\alpha }\right ) \nonumber \\ & \thicksim \frac{C_{1}}{\alpha }\left ( e^{\alpha \xi }-1\right ) +B \tag{4} \end{align}
From the above, y_{0}^{\prime }\left ( \xi \right ) =-C_{1}e^{\alpha \xi } and y_{0}^{\prime \prime }\left ( \xi \right ) =C_{1}\alpha e^{\alpha \xi }. We now see that as that as \xi increases (meaning we are moving away from x=1 towards the left), then the solution y_{0}\left ( \xi \right ) is actually changing less rapidly. This is because \alpha <0. The solution is changing less rapidly as we move away from the boundary layer as what we expect. Therefore, we conclude that if a\left ( x\right ) <0 then the boundary layer can not be at x=0 and has to be at x=1.
To find uniform approximation, we need now to find y^{out}\left ( x\right ) and then do the matching. Since from part(a) we concluded that y_{in} is near x=1, then we assume now that y^{out}\left ( x\right ) is near x=0. Let y_{out}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots Substituting this into (9.1.7) gives\begin{equation} \varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\varepsilon ^{2}y_{2}^{\prime \prime }+\cdots \right ) +a\left ( x\right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) +b\left ( x\right ) \left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =0\nonumber \end{equation} Collecting terms of O\left ( 1\right ) gives the ODE a\left ( x\right ) y_{0}^{\prime }+b\left ( x\right ) y_{0}=0 The solution to this ODE is y_{0}\left ( x\right ) =C_{2}e^{-\int _{0}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} Applying y\left ( 0\right ) =A gives\begin{align*} A & =C_{2}e^{-\int _{0}^{1}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}\\ & =C_{2}E \end{align*}
Where E is constant, which is the value of the definite integral E=e^{-\int _{0}^{1}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}. Hence the solution y^{out}\left ( x\right ) can now be written as y_{0}^{out}\left ( x\right ) =\frac{A}{E}e^{-\int _{0}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} We are now ready to do the matching.\begin{align*} \lim _{\xi \rightarrow \infty }y^{in}\left ( \xi \right ) & \thicksim \lim _{x\rightarrow 0}y^{out}\left ( x\right ) \\ \lim _{\xi \rightarrow \infty }\frac{C_{1}}{\alpha }\left ( e^{\alpha \xi }-1\right ) +B & \thicksim \lim _{x\rightarrow 0}\frac{A}{E}e^{-\int _{0}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} \end{align*}
But since \alpha =a\left ( 1\right ) <0 then the above simplifies to\begin{align*} -\frac{C_{1}}{a\left ( 1\right ) }+B & =\frac{A}{E}\\ C_{1} & =-a\left ( 1\right ) \left ( \frac{A}{E}-B\right ) \end{align*}
Hence inner solution becomes \begin{align*} y_{0}^{in}\left ( \xi \right ) & \thicksim \frac{-a\left ( 1\right ) \left ( \frac{A}{E}-B\right ) }{a\left ( 1\right ) }\left ( e^{a\left ( 1\right ) \xi }-1\right ) +B\\ & \thicksim \left ( B-\frac{A}{E}\right ) \left ( e^{a\left ( 1\right ) \xi }-1\right ) +B\\ & \thicksim B\left ( e^{a\left ( 1\right ) \xi }-1\right ) -\frac{A}{E}\left ( e^{a\left ( 1\right ) \xi }-1\right ) +B\\ & \thicksim \left ( B-\frac{A}{E}\right ) \left ( e^{a\left ( 1\right ) \xi }-1\right ) +B \end{align*}
The uniform solution is\begin{align*} y_{\text{uniform}}\left ( x\right ) & \thicksim y^{in}\left ( \xi \right ) +y^{out}\left ( x\right ) -y^{match}\\ & \thicksim \overset{y^{in}}{\overbrace{\left ( B-\frac{A}{E}\right ) \left ( e^{a\left ( 1\right ) \xi }-1\right ) +B}}+\overset{y^{out}}{\overbrace{\frac{A}{E}e^{-\int _{0}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}}}-\frac{A}{E} \end{align*}
Or in terms of x only y_{\text{uniform}}\left ( x\right ) \thicksim \left ( B-\frac{A}{E}\right ) \left ( e^{a\left ( 1\right ) \frac{1-x}{\varepsilon }}-1\right ) +B+\frac{A}{E}e^{-\int _{0}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}-\frac{A}{E}
We now assume the boundary layer is at x=1 but a\left ( x\right ) >0. From part (a), we found that the solution for y_{0}^{in}\left ( \xi \right ) where boundary layer at x=1 is\begin{equation} y_{0}\left ( \xi \right ) \thicksim \frac{C_{1}}{\alpha }\left ( e^{\alpha \xi }-1\right ) +B\nonumber \end{equation} But now \alpha =a\left ( 1\right ) >0 and not negative as before. We also found that y_{0}^{out}\left ( x\right ) solution was y_{0}^{out}\left ( x\right ) =\frac{A}{E}e^{-\int _{0}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} Lets now try to do the matching and see what happens\begin{align*} \lim _{\xi \rightarrow \infty }y^{in}\left ( \xi \right ) & \thicksim \lim _{x\rightarrow 0}y^{out}\left ( x\right ) \\ \lim _{\xi \rightarrow \infty }\frac{C_{1}}{\alpha }\left ( e^{\alpha \xi }-1\right ) +B+O\left ( \varepsilon \right ) & \thicksim \lim _{x\rightarrow 0}\frac{A}{E}e^{-\int _{0}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}+O\left ( \varepsilon \right ) \\ \lim _{\xi \rightarrow \infty }C_{1}\left ( \frac{e^{\alpha \xi }}{\alpha }-\frac{1}{\alpha }\right ) & \thicksim \frac{A}{E}-B \end{align*}
Since now \alpha >0, then the term on the left blows up, while the term on the right is finite. Not possible to match, unless C_{1}=0. But this means the boundary layer solution is just a constant B and that \frac{A}{E}=B. So the matching does not work in general for arbitrary conditions. This means if a\left ( x\right ) >0, it is not possible to match boundary layer at x=1.
Problem Find the leading order uniform asymptotic approximation to the solution of\begin{align} \varepsilon y^{\prime \prime }+\left ( 1+x^{2}\right ) y^{\prime }-x^{3}y\left ( x\right ) & =0\tag{1}\\ y\left ( 0\right ) & =1\nonumber \\ y\left ( 1\right ) & =1\nonumber \end{align}
For 0\leq x\leq 1 in the limit as \varepsilon \rightarrow 0.
solution
Since a\left ( x\right ) =\left ( 1+x^{2}\right ) is positive, we expect the boundary layer to be near x=0. First we find y^{out}\left ( x\right ) , which is near x=1. Assuming y_{out}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots And substituting this into (1) gives\begin{equation} \varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\varepsilon ^{2}y_{2}^{\prime \prime }+\cdots \right ) +\left ( 1+x^{2}\right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) -x^{3}\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =0\nonumber \end{equation} Collecting terms in O\left ( 1\right ) gives the ODE \left ( 1+x^{2}\right ) y_{0}^{\prime }\thicksim x^{3}y_{0} The ODE becomes y_{0}^{\prime }\thicksim \frac{x^{3}}{\left ( 1+x^{2}\right ) }y_{0} with integrating factor \mu =e^{\int \frac{-x^{3}}{\left ( 1+x^{2}\right ) }dx} . To evaluate \int \frac{-x^{3}}{\left ( 1+x^{2}\right ) }dx, let u=x^{2}, hence \frac{du}{dx}=2x and the integral becomes \int \frac{-x^{3}}{\left ( 1+x^{2}\right ) }dx=-\int \frac{ux}{\left ( 1+u\right ) }\frac{du}{2x}=-\frac{1}{2}\int \frac{u}{\left ( 1+u\right ) }du But \begin{align*} \int \frac{u}{\left ( 1+u\right ) }du & =\int 1-\frac{1}{\left ( 1+u\right ) }du\\ & =u-\ln \left ( 1+u\right ) \end{align*}
But u=x^{2}, hence \int \frac{-x^{3}}{\left ( 1+x^{2}\right ) }dx=\frac{-1}{2}\left ( x^{2}-\ln \left ( 1+x^{2}\right ) \right ) Therefore the integrating factor is \mu =\exp \left ( \frac{-1}{2}x^{2}+\frac{1}{2}\ln \left ( 1+x^{2}\right ) \right ) . The ODE becomes\begin{align*} \frac{d}{dx}\left ( \mu y_{0}\right ) & =0\\ \mu y_{0} & \thicksim c\\ y_{out}\left ( x\right ) & \thicksim c\exp \left ( \frac{1}{2}x^{2}-\frac{1}{2}\ln \left ( 1+x^{2}\right ) \right ) \\ & \thicksim ce^{\frac{1}{2}x^{2}}e^{\ln \left ( 1+x^{2}\right ) ^{\frac{-1}{2}}}\\ & \thicksim c\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}} \end{align*}
To find c, using boundary conditions y\left ( 1\right ) =1 gives\begin{align*} 1 & =c\frac{e^{\frac{1}{2}}}{\sqrt{2}}\\ c & =\sqrt{2}e^{-\frac{1}{2}} \end{align*}
Hence y_{0}^{out}\left ( x\right ) \thicksim \sqrt{2}\frac{e^{\frac{x^{2}-1}{2}}}{\sqrt{1+x^{2}}} Now we find y^{in}\left ( x\right ) near x=0. Let \xi =\frac{x}{\varepsilon ^{p}} be the inner variable. We express the original ODE using this new variable and determine p. Since \frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx} then \frac{dy}{dx}=\frac{dy}{d\xi }\varepsilon ^{-p}. Hence \frac{d}{dx}\equiv \varepsilon ^{-p}\frac{d}{d\xi }\begin{align*} \frac{d^{2}}{dx^{2}} & =\frac{d}{dx}\frac{d}{dx}\\ & =\left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \\ & =\varepsilon ^{-2p}\frac{d^{2}}{d\xi ^{2}} \end{align*}
Therefore \frac{d^{2}y}{dx^{2}}=\varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}} and \varepsilon y^{\prime \prime }+\left ( 1+x^{2}\right ) y^{\prime }-x^{3}y\left ( x\right ) =0 becomes\begin{align*} \varepsilon \varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}}+\left ( 1+\left ( \xi \varepsilon ^{p}\right ) ^{2}\right ) \varepsilon ^{-p}\frac{dy}{d\xi }-\left ( \xi \varepsilon ^{p}\right ) ^{3}y & =0\\ \varepsilon ^{1-2p}\frac{d^{2}y}{d\xi ^{2}}+\left ( 1+\xi ^{2}\varepsilon ^{2p}\right ) \varepsilon ^{-p}\frac{dy}{d\xi }-\xi ^{3}\varepsilon ^{3p}y & =0 \end{align*}
The largest terms are \left \{ \varepsilon ^{1-2p},\varepsilon ^{-p}\right \} , therefore matching them gives 1-2p=-p or p=1. The ODE now becomes\begin{equation} \varepsilon ^{-1}\frac{d^{2}y}{d\xi ^{2}}+\left ( 1+\xi ^{2}\varepsilon ^{2}\right ) \varepsilon ^{-1}\frac{dy}{d\xi }-\xi ^{3}\varepsilon ^{3}y=0 \tag{2} \end{equation} Assuming that y_{in}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots And substituting the above into (2) gives\begin{equation} \varepsilon ^{-1}\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\left ( 1+\xi ^{2}\varepsilon ^{2}\right ) \varepsilon ^{-1}\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) -\xi ^{3}\varepsilon ^{3}\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0 \tag{2A} \end{equation} Collecting terms in O\left ( \varepsilon ^{-1}\right ) gives the ODE y_{0}^{\prime \prime }\left ( \xi \right ) \thicksim -y_{0}^{\prime }\left ( \xi \right ) The solution to this ODE is\begin{equation} y_{0}^{in}\left ( \xi \right ) \thicksim c_{1}+c_{2}e^{-\xi } \tag{3} \end{equation} Applying y_{0}^{in}\left ( 0\right ) =1 gives\begin{align*} 1 & =c_{1}+c_{2}\\ c_{1} & =1-c_{2} \end{align*}
Hence (3) becomes\begin{align} y_{0}^{in}\left ( \xi \right ) & \thicksim \left ( 1-c_{2}\right ) +c_{2}e^{-\xi }\nonumber \\ & \thicksim 1+c_{2}\left ( e^{-\xi }-1\right ) \tag{4} \end{align}
Now that we found y_{out} and y_{in}, we apply matching to find c_{2} in the y_{in} solution.\begin{align*} \lim _{\xi \rightarrow \infty }y_{0}^{in}\left ( \xi \right ) & \thicksim \lim _{x\rightarrow 0^{+}}y_{0}^{out}\left ( x\right ) \\ \lim _{\xi \rightarrow \infty }1+c_{2}\left ( e^{-\xi }-1\right ) & \thicksim \lim _{x\rightarrow 0^{+}}\sqrt{2}\frac{e^{\frac{x^{2}-1}{2}}}{\sqrt{1+x^{2}}}\\ 1-c_{2} & \thicksim \sqrt{\frac{2}{e}}\lim _{x\rightarrow 0^{+}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\\ & \thicksim \sqrt{\frac{2}{e}}\lim _{x\rightarrow 0^{+}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\\ & =\sqrt{\frac{2}{e}} \end{align*}
Hence c_{2}=1-\sqrt{\frac{2}{e}} Therefore the y_{0}^{in}\left ( \xi \right ) becomes\begin{align*} y_{0}^{in}\left ( \xi \right ) & \thicksim 1+\left ( 1-\sqrt{\frac{2}{e}}\right ) \left ( e^{-\xi }-1\right ) \\ & \thicksim 1+\left ( e^{-\xi }-1\right ) -\sqrt{\frac{2}{e}}\left ( e^{-\xi }-1\right ) \\ & \thicksim e^{-\xi }-\sqrt{\frac{2}{e}}\left ( e^{-\xi }-1\right ) \\ & \thicksim e^{-\xi }-\sqrt{\frac{2}{e}}e^{-\xi }+\sqrt{\frac{2}{e}}\\ & \thicksim e^{-\xi }\left ( 1-\sqrt{\frac{2}{e}}\right ) +\sqrt{\frac{2}{e}}\\ & \thicksim 0.858+0.142e^{-\xi } \end{align*}
Therefore, the uniform solution is\begin{equation} y_{uniform}\thicksim y_{in}\left ( x\right ) +y_{out}\left ( x\right ) -y_{match}+O\left ( \varepsilon \right ) \tag{4} \end{equation} Where y_{match} is y_{in}\left ( x\right ) at the boundary layer matching location. (or y_{out} at same matching location). Hence \begin{align*} y_{match} & \thicksim 1-c_{2}\\ & \thicksim 1-\left ( 1-\sqrt{\frac{2}{e}}\right ) \\ & \thicksim \sqrt{\frac{2}{e}} \end{align*}
Hence (4) becomes\begin{align*} y_{uniform} & \thicksim \overset{y_{in}}{\overbrace{e^{-\xi }\left ( 1-\sqrt{\frac{2}{e}}\right ) +\sqrt{\frac{2}{e}}}}+\overset{y_{out}}{\overbrace{\sqrt{2}\frac{e^{\frac{x^{2}-1}{2}}}{\sqrt{1+x^{2}}}}}-\overset{y_{match}}{\overbrace{\sqrt{\frac{2}{e}}}}\\ & \thicksim e^{-\frac{x}{\varepsilon }}\left ( 1-\sqrt{\frac{2}{e}}\right ) +\sqrt{2}\frac{e^{\frac{x^{2}-1}{2}}}{\sqrt{1+x^{2}}}+O\left ( \varepsilon \right ) \end{align*}
This is the leading order uniform asymptotic approximation solution. To verify the result, the numerical solution was plotted against the above solution for \varepsilon =\left \{ 0.1,0.05,0.01\right \} . We see from these plots that as \varepsilon becomes smaller, the asymptotic solution becomes more accurate when compared to the numerical solution. This is because the error, which is O\left ( \varepsilon \right ) , becomes smaller. The code used to generate these plots is
The following are the three plots for each value of \varepsilon
To see the effect on changing \varepsilon on only the asymptotic approximation, the following plot gives the approximation solution only as \varepsilon changes. We see how the approximation converges to the numerical solution as \varepsilon becomes smaller.
Problem Consider initial value problem y^{\prime }=\left ( 1+\frac{x^{-2}}{100}\right ) y^{2}-2y+1 With y\left ( 1\right ) =1 on the interval 0\leq x\leq 1. (a) Formulate this problem as perturbation problem by introducing a small parameter \varepsilon . (b) Find outer approximation correct to order \varepsilon with errors of order \varepsilon ^{2}. Where does this approximation break down? (c) Introduce inner variable and find the inner solution valid to order 1 (with errors of order \varepsilon ). By matching to the outer solution find a uniform valid solution to y\left ( x\right ) on interval 0\leq x\leq 1. Estimate the accuracy of this approximation. (d) Find inner solution correct to order \varepsilon (with errors of order \varepsilon ^{2}) and show that it matches to the outer solution correct to order \varepsilon .
solution
Since \frac{1}{100} is relatively small compared to all other coefficients, we replace it with \varepsilon and the ODE becomes\begin{equation} y^{\prime }-\left ( 1+\frac{\varepsilon }{x^{2}}\right ) y^{2}+2y=1 \tag{1} \end{equation}
Assuming boundary layer is on the left side at x=0. We now solve for y_{out}\left ( x\right ) , which is the solution near x=1. y_{out}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots Substituting this into (1) gives\begin{equation} \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) -\left ( 1+\frac{\varepsilon }{x^{2}}\right ) \left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) ^{2}+2\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =1\nonumber \end{equation} Expanding the above to see more clearly the terms gives\begin{equation} \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) -\left ( 1+\frac{\varepsilon }{x^{2}}\right ) \left ( y_{0}^{2}+\varepsilon \left ( 2y_{0}y_{1}\right ) +\varepsilon ^{2}\left ( 2y_{0}y_{2}+y_{1}^{2}\right ) +\cdots \right ) +2\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =1 \tag{2} \end{equation} The leading order are those terms of coefficient O\left ( 1\right ) . This gives
y_{0}^{\prime }-y_{0}^{2}+2y_{0}\thicksim 1 With boundary conditions y\left ( 1\right ) =1. \frac{dy_{0}}{dx}\thicksim y_{0}^{2}-2y_{0}+1 This is separable\begin{align*} \frac{dy_{0}}{y_{0}^{2}-2y_{0}+1} & \thicksim dx\\ \frac{dy_{0}}{\left ( y_{0}-1\right ) ^{2}} & \thicksim dx \end{align*}
For y_{0}\neq 1. Integrating\begin{align} \int \frac{dy_{0}}{\left ( y_{0}-1\right ) ^{2}} & \thicksim \int dx\nonumber \\ \frac{-1}{y_{0}-1} & \thicksim x+C\nonumber \\ \left ( y_{0}-1\right ) \left ( x+C\right ) & \thicksim -1\nonumber \\ y_{0} & \thicksim \frac{-1}{x+C}+1 \tag{3} \end{align}
To find C, from y\left ( 1\right ) =1, we find\begin{align*} 1 & \thicksim \frac{-1}{1+C}+1\\ 1 & \thicksim \frac{C}{1+C} \end{align*}
This is only possible if C=\infty . Therefore from (2), we conclude that \fbox{$y_0\left ( x\right ) \thicksim 1$} The above is leading order for the outer solution. Now we repeat everything to find y_{1}^{out}\left ( x\right ) . From (2) above, we now keep all terms with O\left ( \varepsilon \right ) which gives y_{1}^{\prime }-2y_{0}y_{1}+2y_{1}\thicksim \frac{1}{x^{2}}y_{0}^{2} But we found y_{0}\left ( x\right ) \thicksim 1 from above, so the above ODE becomes\begin{align*} y_{1}^{\prime }-2y_{1}+2y_{1} & \thicksim \frac{1}{x^{2}}\\ y_{1}^{\prime } & \thicksim \frac{1}{x^{2}} \end{align*}
Integrating gives y_{1}\left ( x\right ) \thicksim -\frac{1}{x}+C The boundary condition now becomes y_{1}\left ( 1\right ) =0 (since we used y\left ( 1\right ) =1 earlier with y_{0}). This gives\begin{align*} 0 & =-\frac{1}{1}+C\\ C & =1 \end{align*}
Therefore the solution becomes \fbox{$y_1\left ( x\right ) \thicksim 1-\frac{1}{x}$} Therefore, the outer solution is y_{out}\left ( x\right ) \thicksim y_{0}+\varepsilon y_{1} Or \fbox{$y\left ( x\right ) \thicksim 1+\varepsilon \left ( 1-\frac{1}{x}\right ) +O\left ( \varepsilon ^2\right ) $} Since the ODE is y^{\prime }-\left ( 1+\frac{\varepsilon }{x^{2}}\right ) y^{2}+2y=1, the approximation breaks down when x<\sqrt{\varepsilon } or x<\frac{1}{10}. Because when x<\sqrt{\varepsilon }, the \frac{\varepsilon }{x^{2}} will start to become large. The term \frac{\varepsilon }{x^{2}} should remain small for the approximation to be accurate. The following are plots of the y_{0} and y_{0}+\varepsilon y_{1} solutions (using \varepsilon =\frac{1}{100}) showing that with two terms the approximation has improved for the outer layer, compared to the full solution of the original ODE obtained using CAS. But the outer solution breaks down near x=0.1 and smaller as can be seen in these plots. Here is the solution of the original ODE obtained using CAS
In the following plot, the y_{0} and the y_{0}+\varepsilon y_{1} solutions are superimposed on same figure, to show how the outer solution has improved when adding another term. But we also notice that the outer solution y_{0}+\varepsilon y_{1} only gives good approximation to the exact solution for about x>0.1 and it breaks down quickly as x becomes smaller.
Now we will obtain solution inside the boundary layer y_{in}\left ( \xi \right ) =y_{0}^{in}\left ( \xi \right ) +O\left ( \varepsilon \right ) . The first step is to always introduce new inner variable. Since the boundary layer is on the right side, then \xi =\frac{x}{\varepsilon ^{p}} And then to express the original ODE using this new variable. We also need to determine p in the above expression. Since the original ODE is y^{\prime }-\left ( 1+\varepsilon x^{-2}\right ) y^{2}+2y=1, then \frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx}=\frac{dy}{d\xi }\left ( \varepsilon ^{-p}\right ) , then the ODE now becomes\begin{align*} \frac{dy}{d\xi }\varepsilon ^{-p}-\left ( 1+\frac{\varepsilon }{\xi \varepsilon ^{p2}}\right ) y^{2}+2y & =1\\ \frac{dy}{d\xi }\varepsilon ^{-p}-\left ( 1+\frac{\varepsilon ^{1-2p}}{\xi ^{2}}\right ) y^{2}+2y & =1 \end{align*}
Where in the above y\equiv y\left ( \xi \right ) . We see that we have \left \{ \varepsilon ^{-p},\varepsilon ^{\left ( 1-2p\right ) }\right \} as the two biggest terms to match. This means -p=1-2p or p=1 Hence the above ODE becomes \frac{dy}{d\xi }\varepsilon ^{-1}-\left ( 1+\frac{\varepsilon ^{-1}}{\xi ^{2}}\right ) y^{2}+2y=1 We are now ready to replace y\left ( \xi \right ) with \sum _{n=0}^{\infty }\varepsilon ^{n}y_{n} which gives \begin{align} \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) \varepsilon ^{-1}-\left ( 1+\frac{\varepsilon ^{-1}}{\xi ^{2}}\right ) \left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) ^{2}+2\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) & =1\nonumber \\ \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) \varepsilon ^{-1}-\left ( 1+\frac{\varepsilon ^{-1}}{\xi ^{2}}\right ) \left ( y_{0}^{2}+\varepsilon \left ( 2y_{0}y_{1}\right ) +\cdots \right ) +2\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) & =1\tag{3} \end{align}
Collecting terms with O\left ( \varepsilon ^{-1}\right ) gives y_{0}^{\prime }\thicksim \frac{1}{\xi ^{2}}y_{0}^{2} This is separable\begin{align*} \int \frac{dy_{0}}{y_{0}^{2}} & \thicksim \int \frac{1}{\xi ^{2}}d\xi \\ -y_{0}^{-1} & \thicksim -\xi ^{-1}+C\\ \frac{1}{y_{0}} & \thicksim \frac{1}{\xi }-C\\ \frac{1}{y_{0}} & \thicksim \frac{1-C\xi }{\xi }\\ y_{0}^{in} & \thicksim \frac{\xi }{1-\xi C} \end{align*}
Now we use matching with y_{out} to find C. We have found before that y_{0}^{out}\left ( x\right ) \thicksim 1 therefore\begin{align*} \lim _{\xi \rightarrow \infty }y_{0}^{in}\left ( \xi \right ) +O\left ( \varepsilon \right ) & =\lim _{x\rightarrow 0}y_{0}^{out}\left ( x\right ) +O\left ( \varepsilon \right ) \\ \lim _{\xi \rightarrow \infty }\frac{\xi }{1-\xi C} & =1+O\left ( \varepsilon \right ) \\ \lim _{\xi \rightarrow \infty }\left ( -C\right ) +O\left ( \xi ^{-1}\right ) +O\left ( \varepsilon \right ) & =1+O\left ( \varepsilon \right ) \\ -C & =1 \end{align*}
Therefore \begin{equation} y_{0}^{in}\left ( \xi \right ) \thicksim \frac{\xi }{1+\xi }\tag{4} \end{equation} Therefore, \begin{align*} y_{uniform} & =y_{0}^{in}+y_{0}^{out}-y_{match}\\ & =\overset{y_{in}}{\overbrace{\frac{\xi }{1+\xi }}}+\overset{y_{out}}{\overbrace{1}}-1 \end{align*}
Since y_{match}=1 (this is what \lim _{\xi \rightarrow \infty }y_{0}^{in} is). Writing everything in x, using \xi =\frac{x}{\varepsilon } the above becomes\begin{align*} y_{\text{uniform}} & =\frac{\frac{x}{\varepsilon }}{1+\frac{x}{\varepsilon }}\\ & =\frac{x}{\varepsilon +x} \end{align*}
The following is a plot of the above, using \varepsilon =\frac{1}{100} to compare with the exact solution.,
Now we will obtain y_{1}^{in} solution inside the boundary layer. Using (3) we found in part (c), reproduced here \begin{equation} \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) \varepsilon ^{-1}-\left ( 1+\frac{\varepsilon ^{-1}}{\xi ^{2}}\right ) \left ( y_{0}^{2}+\varepsilon \left ( 2y_{0}y_{1}\right ) +\cdots \right ) +2\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =1 \tag{3} \end{equation} But now collecting all terms of order of O\left ( 1\right ) , this results in y_{1}^{\prime }-y_{0}^{2}-\frac{2}{\xi ^{2}}y_{0}y_{1}+2y_{0}\thicksim 1 Using y_{0}^{in} found in part (c) into the above gives\begin{align*} y_{1}^{\prime }-\frac{2}{\xi }\left ( \frac{1}{1+\xi }\right ) y_{1} & \thicksim 1-2\left ( \frac{\xi }{1+\xi }\right ) +\left ( \frac{\xi }{1+\xi }\right ) ^{2}\\ y_{1}^{\prime }-\left ( \frac{2}{\xi \left ( 1+\xi \right ) }\right ) y_{1} & \thicksim \frac{1}{\left ( \xi +1\right ) ^{2}} \end{align*}
This can be solved using integrating factor \mu =e^{\int \frac{-2}{\xi +\xi ^{2}}d\xi } using partial fractions gives \mu =\exp \left ( -2\ln \xi +2\ln \left ( 1+\xi \right ) \right ) or \mu =\frac{1}{\xi ^{2}}\left ( 1+\xi \right ) ^{2}. Hence we obtain\begin{align*} \frac{d}{dx}\left ( \mu y_{1}\right ) & \thicksim \mu \frac{1}{\left ( \xi +1\right ) ^{2}}\\ \frac{d}{dx}\left ( \frac{1}{\xi ^{2}}\left ( 1+\xi \right ) ^{2}y_{1}\right ) & \thicksim \frac{1}{\xi ^{2}} \end{align*}
Integrating\begin{align*} \frac{1}{\xi ^{2}}\left ( 1+\xi \right ) ^{2}y_{1} & \thicksim \int \frac{1}{\xi ^{2}}d\xi \\ \frac{1}{\xi ^{2}}\left ( 1+\xi \right ) ^{2}y_{1} & \thicksim \frac{-1}{\xi }+C_{2}\\ \left ( 1+\xi \right ) ^{2}y_{1} & \thicksim -\xi +\xi ^{2}C_{2}\\ y_{1} & \thicksim \frac{-\xi +\xi ^{2}C_{2}}{\left ( 1+\xi \right ) ^{2}} \end{align*}
Therefore, the inner solution becomes\begin{align*} y^{in}\left ( \xi \right ) & =y_{0}+\varepsilon y_{1}\\ & =\frac{\xi }{1+\xi C_{1}}+\varepsilon \frac{\xi ^{2}C_{2}-\xi }{\left ( 1+\xi \right ) ^{2}} \end{align*}
To find C_{1},C_{2} we do matching with with y^{out} that we found in part (a) which is y_{out}\left ( x\right ) \thicksim 1+\varepsilon \left ( 1-\frac{1}{x}\right ) \lim _{\xi \rightarrow \infty }\left ( \frac{\xi }{1+\xi C_{1}}+\varepsilon \frac{\xi ^{2}C_{2}-\xi }{\left ( 1+\xi \right ) ^{2}}\right ) \thicksim \lim _{x\rightarrow 0}1+\varepsilon \left ( 1-\frac{1}{x}\right ) Doing long division \frac{\xi }{1+\xi C_{1}}=\frac{1}{C_{1}}-\frac{1}{\xi C_{1}^{2}}+\frac{1}{\xi ^{2}C_{1}^{3}}+\cdots and \frac{\xi ^{2}C_{2}-\xi }{\left ( 1+\xi \right ) ^{2}}=C_{2}-\frac{2C_{2}+1}{\xi }-\cdots , hence the above becomes\begin{align*} \lim _{\xi \rightarrow \infty }\left ( \left ( \frac{1}{C_{1}}-\frac{1}{\xi C_{1}^{2}}+\frac{1}{\xi ^{2}C_{1}^{3}}+\cdots \right ) +\left ( \varepsilon C_{2}-\varepsilon \frac{2C_{2}+1}{\xi }+\cdots \right ) \right ) & \thicksim \lim _{x\rightarrow 0}1+\varepsilon \left ( 1-\frac{1}{x}\right ) \\ \frac{1}{C_{1}}+\varepsilon C_{2} & \thicksim \lim _{x\rightarrow 0}1+\varepsilon \left ( 1-\frac{1}{x}\right ) \end{align*}
Using x=\xi \varepsilon on the RHS, the above simplifies to\begin{align*} \frac{1}{C_{1}}+\varepsilon C_{2} & \thicksim \lim _{\xi \rightarrow \infty }1+\varepsilon \left ( 1-\frac{1}{\xi \varepsilon }\right ) \\ & \thicksim \lim _{\xi \rightarrow \infty }1+\left ( \varepsilon -\frac{1}{\xi }\right ) \\ & \thicksim 1+\varepsilon \end{align*}
Therefore, C_{1}=1 and C_{2}=1. Hence the inner solution is \begin{align*} y^{in}\left ( \xi \right ) & =y_{0}+\varepsilon y_{1}\\ & =\frac{\xi }{1+\xi }+\varepsilon \frac{\xi ^{2}-\xi }{\left ( 1+\xi \right ) ^{2}} \end{align*}
Therefore \begin{align*} y_{\text{uniform}} & =y_{in}+y_{out}-y_{match}\\ & =\overset{y_{in}}{\overbrace{\frac{\xi }{1+\xi }+\varepsilon \frac{\xi ^{2}-\xi }{\left ( 1+\xi \right ) ^{2}}}}+\overset{y_{out}}{\overbrace{1+\varepsilon \left ( 1-\frac{1}{x}\right ) }}-\left ( 1+\varepsilon \right ) \end{align*}
Writing everything in x, using \xi =\frac{x}{\varepsilon } the above becomes\begin{align*} y_{\text{uniform}} & =\frac{\frac{x}{\varepsilon }}{1+\frac{x}{\varepsilon }}+\varepsilon \frac{\frac{x^{2}}{\varepsilon ^{2}}-\frac{x}{\varepsilon }}{\left ( 1+\frac{x}{\varepsilon }\right ) ^{2}}+1+\varepsilon \left ( 1-\frac{1}{x}\right ) -\left ( 1+\varepsilon \right ) \\ & =\frac{x}{\varepsilon +x}+\frac{x^{2}-x\varepsilon }{\varepsilon \left ( 1+\frac{x}{\varepsilon }\right ) ^{2}}+1+\varepsilon -\frac{\varepsilon }{x}-1-\varepsilon \\ & =\frac{x}{\varepsilon +x}+\frac{x^{2}-x\varepsilon }{\varepsilon \left ( 1+\frac{x}{\varepsilon }\right ) ^{2}}-\frac{\varepsilon }{x}+O\left ( \varepsilon ^{2}\right ) \end{align*}
The following is a plot of the above, using \varepsilon =\frac{1}{100} to compare with the exact solution.
Let us check if y_{\text{uniform}}\left ( x\right ) satisfies y\left ( 1\right ) =1 or not. \begin{align*} y_{\text{uniform}}\left ( 1\right ) & =\frac{1}{\varepsilon +1}+\frac{1-\varepsilon }{\varepsilon \left ( 1+\frac{1}{\varepsilon }\right ) ^{2}}-\varepsilon +O\left ( \varepsilon ^{2}\right ) \\ & =\frac{1-\varepsilon ^{3}-3\varepsilon ^{2}+\varepsilon }{\left ( \varepsilon +1\right ) ^{2}} \end{align*}
Taking the limit \varepsilon \rightarrow 0 gives 1. Therefore y_{\text{uniform}}\left ( x\right ) satisfies y\left ( 1\right ) =1.
problem Use boundary layer methods to find an approximate solution to initial value problem
\begin{align} \varepsilon y^{\prime \prime }+a\left ( x\right ) y^{\prime }+b\left ( x\right ) y & =0\tag{1}\\ y\left ( 0\right ) & =1\nonumber \\ y^{\prime }\left ( 0\right ) & =1\nonumber \end{align}
And a>0. Show that leading order uniform approximation satisfies y\left ( 0\right ) =1 but not y^{\prime }\left ( 0\right ) =1 for arbitrary b. Compare leading order uniform approximation with the exact solution to the problem when a\left ( x\right ) ,b\left ( x\right ) are constants.
Solution
Since a\left ( x\right ) >0 then we expect the boundary layer to be at x=0. We start by finding y_{out}\left ( x\right ) . y_{out}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots Substituting this into (1) gives\begin{equation} \varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\varepsilon ^{2}y_{2}^{\prime \prime }+\cdots \right ) +a\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) +b\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =0\nonumber \end{equation} Collecting terms with O\left ( 1\right ) results in\begin{align*} ay_{0}^{\prime } & \thicksim -by_{0}\\ \frac{dy_{0}}{dx} & \thicksim -\frac{b}{a}y_{0} \end{align*}
This is separable\begin{align*} \int \frac{dy_{0}}{y_{0}} & \thicksim -\int \frac{b\left ( x\right ) }{a\left ( x\right ) }dx\\ \ln \left \vert y_{0}\right \vert & \thicksim -\int \frac{b\left ( x\right ) }{a\left ( x\right ) }dx+C\\ y_{0} & \thicksim Ce^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} \end{align*}
Now we find y_{in}. First we introduce interval variable \xi =\frac{x}{\varepsilon ^{p}} And transform the ODE. Since \frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx} then \frac{dy}{dx}=\frac{dy}{d\xi }\varepsilon ^{-p}. Hence \frac{d}{dx}\equiv \varepsilon ^{-p}\frac{d}{d\xi }\begin{align*} \frac{d^{2}}{dx^{2}} & =\frac{d}{dx}\frac{d}{dx}\\ & =\left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \\ & =\varepsilon ^{-2p}\frac{d^{2}}{d\xi ^{2}} \end{align*}
Therefore \frac{d^{2}y}{dx^{2}}=\varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}} and the ODE becomes\begin{align*} \varepsilon \varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}}+a\left ( \xi \right ) \frac{dy}{d\xi }\varepsilon ^{-p}+b\left ( \xi \right ) y & =0\\ \varepsilon ^{1-2p}y^{\prime \prime }+a\varepsilon ^{-p}y^{\prime }+by & =0 \end{align*}
Balancing 1-2p with -p shows that \fbox{$p=1$} Hence \varepsilon ^{-1}y^{\prime \prime }+a\varepsilon ^{-1}y^{\prime }+by=0 Substituting y_{in}=\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots in the above gives \varepsilon ^{-1}\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\varepsilon ^{2}y_{2}^{\prime \prime }+\cdots \right ) +a\varepsilon ^{-1}\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) +b\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =0 Collecting terms with order O\left ( \varepsilon ^{-1}\right ) gives y_{0}^{\prime \prime }\thicksim -ay_{0}^{\prime } Assuming z=y_{0}^{\prime } then the above becomes z^{\prime }\thicksim -az or \frac{dz}{d\xi }\thicksim -az. This is separable. The solution is \frac{dz}{z}\thicksim -ad\xi or \begin{align*} \ln \left \vert z\right \vert & \thicksim -\int _{0}^{\xi }a\left ( s\right ) ds+C_{1}\\ z & \thicksim C_{1}e^{-\int _{0}^{\xi }a\left ( s\right ) ds} \end{align*}
Hence \begin{align*} \frac{dy_{0}}{d\xi } & \thicksim C_{1}e^{-\int _{0}^{\xi }a\left ( s\right ) ds}\\ dy_{0} & \thicksim \left ( C_{1}e^{-\int _{0}^{\xi }a\left ( s\right ) ds}\right ) d\xi \end{align*}
Integrating again y_{0}^{in}\thicksim \int _{0}^{\xi }\left ( C_{1}e^{-\int _{0}^{\eta }a\left ( s\right ) ds}\right ) d\eta +C_{2} Applying initial conditions at y\left ( 0\right ) since this is where the y_{in} exist. Using y_{in}\left ( 0\right ) =1 then the above becomes 1=C_{2} Hence the solution becomes y_{0}^{in}\thicksim \int _{0}^{\xi }\left ( C_{1}e^{-\int _{0}^{\eta }a\left ( s\right ) ds}\right ) d\eta +1 To apply the second initial condition, which is y^{\prime }\left ( 0\right ) =1, we first take derivative of the above w.r.t. \xi y_{0}^{\prime }\thicksim C_{1}e^{-\int _{0}^{\xi }a\left ( s\right ) ds} Applying y_{0}^{\prime }\left ( 0\right ) =1 gives 1=C_{1} Hence y_{0}^{in}\thicksim 1+\int _{0}^{\xi }e^{-\int _{0}^{\eta }a\left ( s\right ) ds}d\eta Now to find constant of integration for y^{out} from earlier, we need to do matching.\begin{align*} \lim _{\xi \rightarrow \infty }y_{0}^{in} & \thicksim \lim _{x\rightarrow 0}y_{0}^{out}\\ \lim _{\xi \rightarrow \infty }1+\int _{0}^{\xi }e^{-\int _{0}^{\eta }a\left ( s\right ) ds}d\eta & \thicksim \lim _{x\rightarrow 0}Ce^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} \end{align*}
On the LHS the integral \int _{0}^{\xi }e^{-\int _{0}^{\eta }a\left ( s\right ) ds}d\eta since a>0 and negative power on the exponential. So as \xi \rightarrow \infty the integral value is zero. So we have now 1\thicksim \lim _{x\rightarrow 0}Ce^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} Let \lim _{x\rightarrow 0}e^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}\rightarrow E, where E is the value of the definite integral Ce^{-\int _{1}^{0}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}. Another constant, which if we know a\left ( x\right ) ,b\left ( x\right ) we can evaluate. Hence the above gives the value of C as C=\frac{1}{E} The uniform solution can now be written as\begin{align} y_{\text{uniform}} & =y_{in}+y_{out}-y_{\text{match}}\nonumber \\ & =1+\int _{0}^{\xi }e^{-\int _{0}^{\eta }a\left ( s\right ) ds}d\eta +\frac{1}{E}e^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}-1\nonumber \\ & =\int _{0}^{\xi }e^{-\int _{0}^{\eta }a\left ( s\right ) ds}d\eta +\frac{1}{E}e^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} \tag{2} \end{align}
Finally, we need to show that y_{\text{uniform}}\left ( 0\right ) =1 but not y_{\text{uniform}}^{\prime }\left ( 0\right ) =1. From (2), at x=0\, which also means \xi =0, since boundary layer at left side, equation (2) becomes y_{\text{uniform}}\left ( 0\right ) =0+\frac{1}{E}\lim _{x\rightarrow 0}e^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} But we said that \lim _{x\rightarrow 0}e^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}=E, therefore y_{\text{uniform}}\left ( 0\right ) =1 Now we take derivative of (2) w.r.t. x and obtain\begin{align*} y_{\text{uniform}}^{\prime }\left ( x\right ) & =\frac{d}{dx}\left ( \int _{0}^{\frac{x}{\varepsilon }}e^{-\int _{0}^{\eta }a\left ( s\right ) ds}d\eta \right ) +\frac{1}{E}\frac{d}{dx}\left ( e^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds}\right ) \\ & =e^{-\int _{0}^{\frac{x}{\varepsilon }}a\left ( s\right ) ds}-\frac{1}{E}\frac{b\left ( x\right ) }{a\left ( x\right ) }e^{-\int _{1}^{x}\frac{b\left ( s\right ) }{a\left ( s\right ) }ds} \end{align*}
And at x=0 the above becomes y_{\text{uniform}}^{\prime }\left ( 0\right ) =1-\frac{1}{E}\frac{b\left ( 0\right ) }{a\left ( 0\right ) } The above is zero only if b\left ( 0\right ) =0 (since we know a\left ( 0\right ) >0). Therefore, we see that y_{\text{uniform}}^{\prime }\left ( 0\right ) \neq 1 for any arbitrary b\left ( x\right ) . Which is what we are asked to show.
Will now solve the whole problem again, when a,b are constants.\begin{align} \varepsilon y^{\prime \prime }+ay^{\prime }+by & =0\tag{1A}\\ y\left ( 0\right ) & =1\nonumber \\ y^{\prime }\left ( 0\right ) & =1\nonumber \end{align}
And a>0. And compare leading order uniform approximation with the exact solution to the problem when a\left ( x\right ) ,b\left ( x\right ) are constants. Since a>0 then boundary layer will occur at x=0. We start by finding y_{out}\left ( x\right ) . y_{out}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots Substituting this into (1) gives\begin{equation} \varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\varepsilon ^{2}y_{2}^{\prime \prime }+\cdots \right ) +a\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) +b\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =0\nonumber \end{equation} Collecting terms with O\left ( 1\right ) results in\begin{align*} ay_{0}^{\prime } & \thicksim -by_{0}\\ \frac{dy_{0}}{dx} & \thicksim -\frac{b}{a}y_{0} \end{align*}
This is separable\begin{align*} \int \frac{dy_{0}}{y_{0}} & \thicksim -\frac{b}{a}dx\\ \ln \left \vert y_{0}\right \vert & \thicksim -\frac{b}{a}x+C\\ y_{0}^{out} & \thicksim C_{1}e^{-\frac{b}{a}x} \end{align*}
Now we find y_{in}. First we introduce internal variable \xi =\frac{x}{\varepsilon ^{p}} and transform the ODE as we did above. This results in \varepsilon ^{-1}\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\varepsilon ^{2}y_{2}^{\prime \prime }+\cdots \right ) +a\varepsilon ^{-1}\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) +b\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =0 Collecting terms with order O\left ( \varepsilon ^{-1}\right ) gives y_{0}^{\prime \prime }\thicksim -ay_{0}^{\prime } Assuming z=y_{0}^{\prime } then the above becomes z^{\prime }\thicksim -az or \frac{dz}{d\xi }\thicksim -az. This is separable. The solution is \frac{dz}{z}\thicksim -ad\xi or \begin{align*} \ln \left \vert z\right \vert & \thicksim -a\xi +E_{1}\\ z & \thicksim E_{1}e^{-a\xi } \end{align*}
Hence \begin{align*} \frac{dy_{0}}{d\xi } & \thicksim E_{1}e^{-a\xi }\\ dy_{0} & \thicksim E_{1}e^{-a\xi }d\xi \end{align*}
Integrating again y_{0}^{in}\thicksim E_{1}\left ( \frac{-1}{a}\right ) e^{-a\xi }+E_{2} Applying initial conditions at y\left ( 0\right ) since this is where the y_{in} exist. Using y_{in}\left ( 0\right ) =1 then the above becomes\begin{align*} 1 & =E_{1}\left ( \frac{-1}{a}\right ) +E_{2}\\ a\left ( E_{2}-1\right ) & =E_{1} \end{align*}
Hence the solution becomes\begin{equation} y_{0}^{in}\thicksim \left ( 1-E_{2}\right ) e^{-a\xi }+E_{2} \tag{1B} \end{equation}
To apply the second initial condition, which is y^{\prime }\left ( 0\right ) =1, we first take derivative of the above w.r.t. \xi
y_{0}^{\prime }\thicksim -a\left ( 1-E_{2}\right ) e^{-a\xi }
Hence y^{\prime }\left ( 0\right ) =1 gives
\begin{align*} 1 & =-a\left ( 1-E_{2}\right ) \\ 1 & =-a+aE_{2}\\ E_{2} & =\frac{1+a}{a} \end{align*}
And the solution y_{in} in (1B) becomes \begin{align*} y_{0}^{in} & \thicksim \left ( 1-\frac{1+a}{a}\right ) e^{-a\xi }+\frac{1+a}{a}\\ & \thicksim \left ( \frac{-1}{a}\right ) e^{-a\xi }+\frac{1+a}{a}\\ & \thicksim \frac{\left ( 1+a\right ) -e^{-a\xi }}{a} \end{align*}
Now to find constant of integration for y^{out}\left ( x\right ) from earlier, we need to do matching.\begin{align*} \lim _{\xi \rightarrow \infty }y_{0}^{in} & \thicksim \lim _{x\rightarrow 0}y_{0}^{out}\\ \lim _{\xi \rightarrow \infty }\frac{\left ( 1+a\right ) -e^{-a\xi }}{a} & \thicksim \lim _{x\rightarrow 0}C_{1}e^{-\frac{b}{a}x}\\ \frac{1+a}{a} & \thicksim C_{1} \end{align*}
Hence now the uniform solution can be written as\begin{align} y_{\text{uniform}}\left ( x\right ) & \thicksim y_{in}+y_{out}-y_{\text{match}}\nonumber \\ & \thicksim \overset{\,y_{in}}{\overbrace{\frac{\left ( 1+a\right ) -e^{-a\frac{x}{\varepsilon }}}{a}}}+\overset{y_{out}}{\overbrace{\frac{1+a}{a}e^{-\frac{b}{a}x}}}-\frac{1+a}{a}\nonumber \\ & \thicksim \frac{\left ( 1+a\right ) }{a}-\frac{e^{-a\frac{x}{\varepsilon }}}{a}+\frac{1+a}{a}e^{-\frac{b}{a}x}-\frac{1+a}{a}\nonumber \\ & \thicksim -\frac{e^{-a\frac{x}{\varepsilon }}}{a}+\frac{1+a}{a}e^{-\frac{b}{a}x}\nonumber \\ & \thicksim \frac{1}{a}\left ( \left ( 1+a\right ) e^{-\frac{b}{a}x}-e^{-a\frac{x}{\varepsilon }}\right ) \tag{2A} \end{align}
Now we compare the above, which is the leading order uniform approximation, to the exact solution. Since now a,b are constants, then the exact solution is\begin{equation} y_{exact}\left ( x\right ) =Ae^{\lambda _{1}x}+Be^{\lambda _{2}x} \tag{3} \end{equation} Where \lambda _{1,2} are roots of characteristic equation of \varepsilon y^{\prime \prime }+ay^{\prime }+by=0. These are \lambda =\frac{-a}{2\varepsilon }\pm \frac{1}{2\varepsilon }\sqrt{a^{2}-4\varepsilon b}. Hence\begin{align*} \lambda _{1} & =\frac{-a}{2}+\frac{1}{2}\sqrt{a^{2}-4\varepsilon b}\\ \lambda _{2} & =\frac{-a}{2}-\frac{1}{2}\sqrt{a^{2}-4\varepsilon b} \end{align*}
Applying initial conditions to (3). y\left ( 0\right ) =1 gives\begin{align*} 1 & =A+B\\ B & =1-A \end{align*}
And solution becomes y_{exact}\left ( x\right ) =Ae^{\lambda _{1}x}+\left ( 1-A\right ) e^{\lambda _{2}x}. Taking derivatives gives y_{exact}^{\prime }\left ( x\right ) =A\lambda _{1}e^{\lambda _{1}x}+\left ( 1-A\right ) \lambda _{2}e^{\lambda _{2}x} Using y^{\prime }\left ( 0\right ) =1 gives\begin{align*} 1 & =A\lambda _{1}+\left ( 1-A\right ) \lambda _{2}\\ 1 & =A\left ( \lambda _{1}-\lambda _{2}\right ) +\lambda _{2}\\ A & =\frac{1-\lambda _{2}}{\lambda _{1}-\lambda _{2}} \end{align*}
Therefore, B=1-\frac{1-\lambda _{2}}{\lambda _{1}-\lambda _{2}} and the exact solution becomes\begin{align} y_{exact}\left ( x\right ) & =\frac{1-\lambda _{2}}{\lambda _{1}-\lambda _{2}}e^{\lambda _{1}x}+\left ( 1-\frac{1-\lambda _{2}}{\lambda _{1}-\lambda _{2}}\right ) e^{\lambda _{2}x}\nonumber \\ & =\frac{1-\lambda _{2}}{\lambda _{1}-\lambda _{2}}e^{\lambda _{1}x}+\left ( \frac{\left ( \lambda _{1}-\lambda _{2}\right ) -\left ( 1-\lambda _{2}\right ) }{\lambda _{1}-\lambda _{2}}\right ) e^{\lambda _{2}x}\nonumber \\ & =\frac{1-\lambda _{2}}{\lambda _{1}-\lambda _{2}}e^{\lambda _{1}x}+\left ( \frac{\lambda _{1}-1}{\lambda _{1}-\lambda _{2}}\right ) e^{\lambda _{2}x} \tag{4} \end{align}
While the uniform solution above was found to be \frac{1}{a}\left ( \left ( 1+a\right ) e^{-\frac{b}{a}x}-e^{-a\frac{x}{\varepsilon }}\right ) . Here is a plot of the exact solution above, for \varepsilon =\left \{ 1/10,1/50,1/100\right \} , and for some values for a,b such as a=1,b=10 in order to compare with the uniform solution. Note that the uniform solution is O\left ( \varepsilon \right ) . As \varepsilon becomes smaller, the leading order uniform solution will better approximate the exact solution. At \varepsilon =0.01 the uniform approximation gives very good approximation. This is using only leading term approximation.
Problem Find first order uniform approximation valid as \varepsilon \rightarrow 0^{+} for 0\leq x\leq 1\begin{align} \varepsilon y^{\prime \prime }+\left ( x^{2}+1\right ) y^{\prime }-x^{3}y & =0\tag{1}\\ y\left ( 0\right ) & =1\nonumber \\ y\left ( 1\right ) & =1\nonumber \end{align}
Solution
Since a\left ( x\right ) =\left ( x^{2}+1\right ) \, is positive for 0\leq x\leq 1, therefore we expect the boundary layer to be on the left side at x=0. Assuming this is the case for now (if it is not, then we expect not to be able to do the matching). We start by finding y_{out}\left ( x\right ) . y_{out}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots Substituting this into (1) gives\begin{equation} \varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\varepsilon ^{2}y_{2}^{\prime \prime }+\cdots \right ) +\left ( x^{2}+1\right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) -x^{3}\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =0 \tag{2} \end{equation} Collecting terms with O\left ( 1\right ) results in\begin{align*} \left ( x^{2}+1\right ) y_{0}^{\prime } & \thicksim x^{3}y_{0}\\ \frac{dy_{0}}{dx} & \thicksim \frac{x^{3}}{\left ( x^{2}+1\right ) }y_{0} \end{align*}
This is separable.\begin{align*} \int \frac{dy_{0}}{y_{0}} & \thicksim \int \frac{x^{3}}{\left ( x^{2}+1\right ) }dx\\ \ln \left \vert y_{0}\right \vert & \thicksim \int x-\frac{x}{1+x^{2}}dx\\ & \thicksim \frac{x^{2}}{2}-\frac{1}{2}\ln \left ( 1+x^{2}\right ) +C \end{align*}
Hence\begin{align*} y_{0} & \thicksim e^{\frac{x^{2}}{2}-\frac{1}{2}\ln \left ( 1+x^{2}\right ) +C}\\ & \thicksim \frac{Ce^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}} \end{align*}
Applying y_{0}^{out}\left ( 1\right ) =1 to the above (since this is where the outer solution is), we solve for C\begin{align*} 1 & \thicksim \frac{Ce^{\frac{1}{2}}}{\sqrt{2}}\\ C & \thicksim \sqrt{2}e^{\frac{-1}{2}} \end{align*}
Therefore\begin{align*} y_{0}^{out} & \thicksim \frac{\sqrt{2}e^{\frac{-1}{2}}e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\\ & \thicksim \sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}} \end{align*}
Now we need to find y_{1}^{out}. From (2), but now collecting terms in O\left ( \varepsilon \right ) gives\begin{equation} y_{0}^{\prime \prime }+\left ( x^{2}+1\right ) y_{1}^{\prime }\backsim x^{3}y_{1} \tag{3} \end{equation} In the above y_{0}^{\prime \prime } is known.\begin{align*} y_{0}^{\prime }\left ( x\right ) & =\sqrt{\frac{2}{e}}\frac{d}{dx}\left ( \frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\right ) \\ & =\sqrt{\frac{2}{e}}\frac{x^{3}e^{\frac{x^{2}}{2}}}{\left ( 1+x^{2}\right ) ^{\frac{3}{2}}} \end{align*}
And y_{0}^{\prime \prime }\left ( x\right ) =\sqrt{\frac{2}{e}}\frac{x^{2}e^{\frac{x^{2}}{2}}\left ( x^{4}+x^{2}+3\right ) }{\left ( 1+x^{2}\right ) ^{\frac{5}{2}}} Hence (3) becomes\begin{align*} \left ( x^{2}+1\right ) y_{1}^{\prime } & \backsim x^{3}y_{1}-y_{0}^{\prime \prime }\\ \left ( x^{2}+1\right ) y_{1}^{\prime } & \backsim x^{3}y_{1}-\sqrt{\frac{2}{e}}\frac{x^{2}e^{\frac{x^{2}}{2}}\left ( x^{4}+x^{2}+3\right ) }{\left ( 1+x^{2}\right ) ^{\frac{5}{2}}}\\ y_{1}^{\prime }-\frac{x^{3}}{\left ( x^{2}+1\right ) }y_{1} & \backsim -\sqrt{\frac{2}{e}}\frac{x^{2}e^{\frac{x^{2}}{2}}\left ( x^{4}+x^{2}+3\right ) }{\left ( 1+x^{2}\right ) ^{\frac{7}{2}}} \end{align*}
Integrating factor is \mu =e^{-\int \frac{x^{3}}{\left ( x^{2}+1\right ) }dx}=e^{-\frac{x^{2}}{2}+\frac{1}{2}\ln \left ( 1+x^{2}\right ) }=\left ( 1+x^{2}\right ) ^{\frac{1}{2}}e^{\frac{-x^{2}}{2}}, hence the above becomes\begin{align*} \frac{d}{dx}\left ( \left ( 1+x^{2}\right ) ^{\frac{1}{2}}e^{\frac{-x^{2}}{2}}y_{1}\right ) & \backsim -\sqrt{\frac{2}{e}}\left ( 1+x^{2}\right ) ^{\frac{1}{2}}e^{\frac{-x^{2}}{2}}\frac{x^{2}e^{\frac{x^{2}}{2}}\left ( x^{4}+x^{2}+3\right ) }{\left ( 1+x^{2}\right ) ^{\frac{7}{2}}}\\ & \backsim -\sqrt{\frac{2}{e}}\frac{x^{2}\left ( x^{4}+x^{2}+3\right ) }{\left ( 1+x^{2}\right ) ^{3}} \end{align*}
Integrating gives (with help from CAS)\begin{align*} \left ( 1+x^{2}\right ) ^{\frac{1}{2}}e^{\frac{-x^{2}}{2}}y_{1}\left ( x\right ) & \backsim -\sqrt{\frac{2}{e}}\int \frac{x^{2}\left ( x^{4}+x^{2}+3\right ) }{\left ( 1+x^{2}\right ) ^{3}}dx\\ & \backsim -\sqrt{\frac{2}{e}}\int 1-\frac{3}{\left ( 1+x^{2}\right ) ^{3}}+\frac{4}{\left ( 1+x^{2}\right ) ^{2}}-\frac{2}{1+x^{2}}dx\\ & \backsim -\sqrt{\frac{2}{e}}\left ( x-\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}+\frac{7x}{8\left ( 1+x^{2}\right ) }-9\frac{\arctan \left ( x\right ) }{8}\right ) +C_{1} \end{align*}
Hence y_{1}^{out}\left ( x\right ) \backsim -\sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\left ( 1+x^{2}\right ) ^{\frac{1}{2}}}\left ( x-\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}+\frac{7x}{8\left ( 1+x^{2}\right ) }-9\frac{\arctan \left ( x\right ) }{8}\right ) +C_{1}\frac{e^{\frac{x^{2}}{2}}}{\left ( 1+x^{2}\right ) ^{\frac{1}{2}}} Now we find C_{1} from boundary conditions y_{1}\left ( 1\right ) =0. (notice the BC now is y_{1}\left ( 1\right ) =0 and not y_{1}\left ( 1\right ) =1, since we used y_{1}\left ( 1\right ) =1 already).\begin{align*} \sqrt{\frac{2}{e}}\frac{e^{\frac{1}{2}}}{\left ( 1+1\right ) ^{\frac{1}{2}}}\left ( 1-\frac{3}{4\left ( 1+1\right ) ^{2}}+\frac{7}{8\left ( 1+1\right ) }-9\frac{\arctan \left ( 1\right ) }{8}\right ) & =C_{1}\frac{e^{\frac{1}{2}}}{\left ( 1+1\right ) ^{\frac{1}{2}}}\\ \sqrt{\frac{2}{e}}\frac{e^{\frac{1}{2}}}{\sqrt{2}}\left ( 1-\frac{3}{16}+\frac{7}{16}-\frac{9}{8}\arctan \left ( 1\right ) \right ) & =C_{1}\frac{e^{\frac{1}{2}}}{\sqrt{2}} \end{align*}
Simplifying\begin{align*} 1-\frac{3}{16}+\frac{7}{16}-\frac{9}{8}\arctan \left ( 1\right ) & =C_{1}\frac{e^{\frac{1}{2}}}{\sqrt{2}}\\ C_{1} & =\sqrt{\frac{2}{e}}\left ( \frac{5}{4}-\frac{9}{8}\arctan \left ( 1\right ) \right ) \\ C_{1} & =\sqrt{\frac{2}{e}}\left ( \frac{5}{4}-\frac{9}{32}\pi \right ) \\ & =0.31431 \end{align*}
Hence \begin{align*} y_{1}^{out} & \thicksim -\sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{\left ( 1+x^{2}\right ) }}\left ( x-\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}+\frac{7x}{8\left ( 1+x^{2}\right ) }-\frac{9}{8}\arctan \left ( x\right ) \right ) +\sqrt{\frac{2}{e}}\left ( \frac{5}{4}-\frac{9}{32}\pi \right ) \frac{e^{\frac{x^{2}}{2}}}{\sqrt{\left ( 1+x^{2}\right ) }}\\ & \thicksim \sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{\left ( 1+x^{2}\right ) }}\left ( \left ( \frac{5}{4}-\frac{9}{32}\pi \right ) -\left ( x-\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}+\frac{7x}{8\left ( 1+x^{2}\right ) }-\frac{9}{8}\arctan \left ( x\right ) \right ) \right ) \\ & \thicksim \sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{\left ( 1+x^{2}\right ) }}\left ( \frac{5}{4}-\frac{9}{32}\pi -x+\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}-\frac{7x}{8\left ( 1+x^{2}\right ) }+\frac{9}{8}\arctan \left ( x\right ) \right ) \end{align*}
Hence\begin{align} y^{out}\left ( x\right ) & \thicksim y_{0}^{out}+\varepsilon y_{1}^{out}\nonumber \\ & \thicksim \sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}+\varepsilon \sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{\left ( 1+x^{2}\right ) }}\left ( \frac{5}{4}-\frac{9}{32}\pi -x+\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}-\frac{7x}{8\left ( 1+x^{2}\right ) }+\frac{9}{8}\arctan \left ( x\right ) \right ) +O\left ( \varepsilon ^{2}\right ) \nonumber \\ & \thicksim \sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\left ( 1+\varepsilon \left ( \frac{5}{4}-\frac{9}{32}\pi -x+\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}-\frac{7x}{8\left ( 1+x^{2}\right ) }+\frac{9}{8}\arctan \left ( x\right ) \right ) \right ) +O\left ( \varepsilon ^{2}\right ) \tag{3A} \end{align}
Now that we found y^{out}\left ( x\right ) , we need to find y^{in}\left ( x\right ) and then do the matching and the find uniform approximation. Since the boundary layer at x=0, we introduce inner variable \xi =\frac{x}{\varepsilon ^{p}} and then express the original ODE using this new variable. We also need to determine p in the above expression. Since \frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx} then \frac{dy}{dx}=\frac{dy}{d\xi }\varepsilon ^{-p}. Hence \frac{d}{dx}\equiv \varepsilon ^{-p}\frac{d}{d\xi }\begin{align*} \frac{d^{2}}{dx^{2}} & =\frac{d}{dx}\frac{d}{dx}\\ & =\left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \\ & =\varepsilon ^{-2p}\frac{d^{2}}{d\xi ^{2}} \end{align*}
Therefore \frac{d^{2}y}{dx^{2}}=\varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}} and the ODE\ \varepsilon y^{\prime \prime }+\left ( x^{2}+1\right ) y^{\prime }-x^{3}y=0 now becomes\begin{align*} \varepsilon \varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}}+\left ( \left ( \xi \varepsilon ^{p}\right ) ^{2}+1\right ) \varepsilon ^{-p}\frac{dy}{d\xi }-\left ( \xi \varepsilon ^{p}\right ) ^{3}y & =0\\ \varepsilon ^{1-2p}\frac{d^{2}y}{d\xi ^{2}}+\left ( \xi ^{2}\varepsilon ^{p}+\varepsilon ^{-p}\right ) \frac{dy}{d\xi }-\xi ^{3}\varepsilon ^{3p}y & =0 \end{align*}
The largest terms are \left \{ \varepsilon ^{1-2p},\varepsilon ^{-p}\right \} , therefore matching them gives p=1. The ODE now becomes\begin{equation} \varepsilon ^{-1}\frac{d^{2}y}{d\xi ^{2}}+\left ( \xi ^{2}\varepsilon +\varepsilon ^{-1}\right ) \frac{dy}{d\xi }-\xi ^{3}\varepsilon ^{3}y=0 \tag{4} \end{equation} Assuming that y_{in}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots And substituting the above into (4) gives\begin{equation} \varepsilon ^{-1}\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\left ( \xi ^{2}\varepsilon +\varepsilon ^{-1}\right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) -\xi ^{3}\varepsilon ^{3}\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0 \tag{4A} \end{equation} Collecting terms in O\left ( \varepsilon ^{-1}\right ) gives y_{0}^{\prime \prime }\backsim -y_{0}^{\prime } Letting z=y_{0}^{\prime }, the above becomes\begin{align*} \frac{dz}{d\xi } & \backsim -z\\ \frac{dz}{z} & \backsim -d\xi \\ \ln \left \vert z\right \vert & \backsim -\xi +C_{1}\\ z & \backsim C_{1}e^{-\xi } \end{align*}
Hence\begin{align} \frac{dy_{0}}{d\xi } & \backsim C_{1}e^{-\xi }\nonumber \\ y_{0} & \backsim C_{1}\int e^{-\xi }d\xi +C_{2}\nonumber \\ & \backsim -C_{1}e^{-\xi }+C_{2} \tag{5} \end{align}
Applying boundary conditions y_{0}^{in}\left ( 0\right ) =1 gives\begin{align*} 1 & =-C_{1}+C_{2}\\ C_{2} & =1+C_{1} \end{align*}
And (5) becomes\begin{align} y_{0}^{in}\left ( \xi \right ) & \backsim -C_{1}e^{-\xi }+\left ( 1+C_{1}\right ) \nonumber \\ & \backsim 1+C_{1}\left ( 1-e^{-\xi }\right ) \tag{6} \end{align}
We now find y_{1}^{in}. Going back to (4) and collecting terms in O\left ( 1\right ) gives the ODE y_{1}^{\prime \prime }\backsim y_{1}^{\prime } This is the same ODE we solved above. But it will have different B.C. Hence y_{1}^{in}\backsim -C_{3}e^{-\xi }+C_{4} Applying boundary conditions y_{1}^{in}\left ( 0\right ) =0 gives\begin{align*} 0 & =-C_{3}+C_{4}\\ C_{3} & =C_{4} \end{align*}
Therefore\begin{align*} y_{1}^{in} & \backsim -C_{3}e^{-\xi }+C_{3}\\ & \backsim C_{3}\left ( 1-e^{-\xi }\right ) \end{align*}
Now we have the leading order y^{in}\begin{align} y^{in}\left ( \xi \right ) & =y_{0}^{in}+\varepsilon y_{1}^{in}\nonumber \\ & =1+C_{1}\left ( 1-e^{-\xi }\right ) +\varepsilon C_{3}\left ( 1-e^{-\xi }\right ) +O\left ( \varepsilon ^{2}\right ) \tag{7} \end{align}
Now we are ready to do the matching between (7) and (3A)\begin{multline*} \lim _{\xi \rightarrow \infty }1+C_{1}\left ( 1-e^{-\xi }\right ) +\varepsilon C_{3}\left ( 1-e^{-\xi }\right ) \thicksim \\ \lim _{x\rightarrow 0}\sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\left ( 1+\varepsilon \left ( \frac{5}{4}-\frac{9}{32}\pi -x+\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}-\frac{7x}{8\left ( 1+x^{2}\right ) }+\frac{9}{8}\arctan \left ( x\right ) \right ) \right ) \end{multline*} Or 1+C_{1}+\varepsilon C_{3}\thicksim \sqrt{\frac{2}{e}}\lim _{x\rightarrow 0}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}+\sqrt{\frac{2}{e}}\varepsilon \lim _{x\rightarrow 0}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\left ( \frac{5}{4}-\frac{9}{32}\pi -x+\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}-\frac{7x}{8\left ( 1+x^{2}\right ) }+\frac{9}{8}\arctan \left ( x\right ) \right ) But \lim _{x\rightarrow 0}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\rightarrow 1, \lim _{x\rightarrow 0}\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}\rightarrow 0 \lim _{x\rightarrow 0}\frac{7x}{8\left ( 1+x^{2}\right ) }\rightarrow 0 therefore the above becomes 1+C_{1}+\varepsilon C_{3}\thicksim \sqrt{\frac{2}{e}}+\sqrt{\frac{2}{e}}\varepsilon \left ( \frac{5}{4}-\frac{9}{32}\pi \right ) Hence \begin{align*} 1+C_{1} & =\sqrt{\frac{2}{e}}\\ C_{1} & =\sqrt{\frac{2}{e}}-1\\ C_{3} & =\sqrt{\frac{2}{e}}\left ( \frac{5}{4}-\frac{9}{32}\pi \right ) \end{align*}
This means that \begin{align*} y^{in}\left ( \xi \right ) & \thicksim 1+C_{1}\left ( 1-e^{-\xi }\right ) +\varepsilon C_{3}\left ( 1-e^{-\xi }\right ) \\ & \thicksim 1+\left ( \sqrt{\frac{2}{e}}-1\right ) \left ( 1-e^{-\xi }\right ) +\varepsilon \sqrt{\frac{2}{e}}\left ( \frac{5}{4}-\frac{9}{32}\pi \right ) \left ( 1-e^{-\xi }\right ) \end{align*}
Therefore\begin{align*} y_{\text{uniform}\left ( x\right ) } & \thicksim y^{in}\left ( \xi \right ) +y^{out}\left ( x\right ) -y_{match}\\ & \thicksim 1+\left ( \sqrt{\frac{2}{e}}-1\right ) \left ( 1-e^{-\xi }\right ) +\varepsilon \sqrt{\frac{2}{e}}\left ( \frac{5}{4}-\frac{9}{32}\pi \right ) \left ( 1-e^{-\xi }\right ) \\ & +\sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\left ( 1+\varepsilon \left ( \frac{5}{4}-\frac{9}{32}\pi -x+\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}-\frac{7x}{8\left ( 1+x^{2}\right ) }+\frac{9}{8}\arctan \left ( x\right ) \right ) \right ) \\ & -\left ( \sqrt{\frac{2}{e}}+\sqrt{\frac{2}{e}}\varepsilon \left ( \frac{5}{4}-\frac{9}{32}\pi \right ) \right ) \end{align*}
Or (replacing \xi by \frac{x}{\varepsilon } and simplifying)\begin{align*} y_{\text{uniform}\left ( x\right ) } & \thicksim 1+\left ( \sqrt{\frac{2}{e}}-1\right ) \left ( 1-e^{-\xi }\right ) -e^{-\xi }\varepsilon \sqrt{\frac{2}{e}}\left ( \frac{5}{4}-\frac{9}{32}\pi \right ) \\ & +\sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\left ( 1+\varepsilon \left ( \frac{5}{4}-\frac{9}{32}\pi -x+\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}-\frac{7x}{8\left ( 1+x^{2}\right ) }+\frac{9}{8}\arctan \left ( x\right ) \right ) \right ) \\ & -\sqrt{\frac{2}{e}} \end{align*}
Or\begin{multline*} y_{\text{uniform}\left ( x\right ) }\thicksim -\sqrt{\frac{2}{e}}e^{-\xi }+e^{-\xi }-e^{-\xi }\varepsilon \sqrt{\frac{2}{e}}\left ( \frac{5}{4}-\frac{9}{32}\pi \right ) \\ +\sqrt{\frac{2}{e}}\frac{e^{\frac{x^{2}}{2}}}{\sqrt{1+x^{2}}}\left ( 1+\varepsilon \left ( \frac{5}{4}-\frac{9}{32}\pi -x+\frac{3x}{4\left ( 1+x^{2}\right ) ^{2}}-\frac{7x}{8\left ( 1+x^{2}\right ) }+\frac{9}{8}\arctan \left ( x\right ) \right ) \right ) \end{multline*} To check validity of the above solution, the approximate solution is plotted against the numerical solution for different values of \varepsilon =\left \{ 0.1,0.05,0.01\right \} . This shows very good agreement with the numerical solution. At \varepsilon =0.01 the solutions are almost the same.
Problem Find lowest order uniform approximation to boundary value problem \begin{align*} \varepsilon y^{\prime \prime }+\left ( \sin x\right ) y^{\prime }+y\sin \left ( 2x\right ) & =0\\ y\left ( 0\right ) & =\pi \\ y\left ( \pi \right ) & =0 \end{align*}
Solution
We expect a boundary layer at left end at x=0. Therefore, we need to find y^{in}\left ( \xi \right ) ,y^{out}\left ( x\right ) , where \xi is an inner variable defined by \xi =\frac{x}{\varepsilon ^{p}}.
Finding y^{in}\left ( \xi \right )
At x=0, we introduce inner variable \xi =\frac{x}{\varepsilon ^{p}} and then express the original ODE using this new variable. We also need to determine p in the above expression. Since \frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx} then \frac{dy}{dx}=\frac{dy}{d\xi }\varepsilon ^{-p}. Hence \frac{d}{dx}\equiv \varepsilon ^{-p}\frac{d}{d\xi }\begin{align*} \frac{d^{2}}{dx^{2}} & =\frac{d}{dx}\frac{d}{dx}\\ & =\left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \left ( \varepsilon ^{-p}\frac{d}{d\xi }\right ) \\ & =\varepsilon ^{-2p}\frac{d^{2}}{d\xi ^{2}} \end{align*}
Therefore \frac{d^{2}y}{dx^{2}}=\varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}} and the ODE\ \varepsilon y^{\prime \prime }+\left ( \sin x\right ) y^{\prime }+y\sin \left ( 2x\right ) =0 now becomes\begin{align*} \varepsilon \varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}}+\left ( \sin \left ( \xi \varepsilon ^{p}\right ) \right ) \varepsilon ^{-p}\frac{dy}{d\xi }+\sin \left ( 2\xi \varepsilon ^{p}\right ) y & =0\\ \varepsilon ^{1-2p}\frac{d^{2}y}{d\xi ^{2}}+\left ( \sin \left ( \xi \varepsilon ^{p}\right ) \right ) \varepsilon ^{-p}\frac{dy}{d\xi }+\sin \left ( 2\xi \varepsilon ^{p}\right ) y & =0 \end{align*}
Expanding the \sin terms in the above, in Taylor series around zero, \sin \left ( x\right ) =x-\frac{x^{3}}{3!}+\cdots gives\begin{align*} \varepsilon ^{1-2p}\frac{d^{2}y}{d\xi ^{2}}+\left ( \xi \varepsilon ^{p}-\frac{\left ( \xi \varepsilon ^{p}\right ) ^{3}}{3!}+\cdots \right ) \varepsilon ^{-p}\frac{dy}{d\xi }+\left ( 2\xi \varepsilon ^{p}-\frac{\left ( 2\xi \varepsilon ^{p}\right ) ^{3}}{3!}+\cdots \right ) y & =0\\ \varepsilon ^{1-2p}\frac{d^{2}y}{d\xi ^{2}}+\left ( \xi -\frac{\xi ^{3}\varepsilon ^{2p}}{3!}+\cdots \right ) \frac{dy}{d\xi }+\left ( 2\xi \varepsilon ^{p}-\frac{\left ( 2\xi \varepsilon ^{p}\right ) ^{3}}{3!}+\cdots \right ) y & =0 \end{align*}
Then the largest terms are \left \{ \varepsilon ^{1-2p},1\right \} , therefore 1-2p=0 or \fbox{$p=\frac{1}{2}$} The ODE now becomes\begin{equation} y^{\prime \prime }+\left ( \xi -\frac{\xi ^{3}\varepsilon }{3!}+\cdots \right ) y^{\prime }+\left ( 2\xi \sqrt{\varepsilon }-\frac{\left ( 2\xi \sqrt{\varepsilon }\right ) ^{3}}{3!}+\cdots \right ) y=0 \tag{1} \end{equation} Assuming that y^{left}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots Then (1) becomes \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\left ( \xi -\frac{\xi ^{3}\varepsilon }{3!}+\cdots \right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( 2\xi \sqrt{\varepsilon }-\frac{\left ( 2\xi \sqrt{\varepsilon }\right ) ^{3}}{3!}+\cdots \right ) \left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0 Collecting terms in O\left ( 1\right ) gives the balance\begin{align*} y_{0}^{\prime \prime }\left ( \xi \right ) & \thicksim -\xi y_{0}^{\prime }\left ( \xi \right ) \\ y_{0}\left ( 0\right ) & =\pi \end{align*}
Assuming z=y_{0}^{\prime }, then\begin{align*} z^{\prime } & \thicksim -\xi z\\ \frac{dz}{z} & \thicksim -\xi \\ \ln \left \vert z\right \vert & \thicksim -\frac{\xi ^{2}}{2}+C_{1}\\ z & \thicksim C_{1}e^{\frac{-\xi ^{2}}{2}} \end{align*}
Therefore y_{0}^{\prime }\thicksim C_{1}e^{\frac{-\xi ^{2}}{2}}. Hence y_{0}\left ( \xi \right ) \thicksim C_{1}\int _{0}^{\xi }e^{\frac{-s^{2}}{2}}ds+C_{2} With boundary conditions y\left ( 0\right ) =\pi . Hence \pi =C_{2} And the solution becomes\begin{equation} y_{0}^{in}\left ( \xi \right ) \thicksim C_{1}\int _{0}^{\xi }e^{\frac{-s^{2}}{2}}ds+\pi \tag{2} \end{equation} Now we need to find y^{out}\left ( x\right ) . Assuming that y^{out}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots Then \varepsilon y^{\prime \prime }+\left ( \sin x\right ) y^{\prime }+y\sin \left ( 2x\right ) =0 becomes \varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\sin \left ( x\right ) \left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\sin \left ( 2x\right ) \left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0 Collecting terms in O\left ( 1\right ) gives the balance\begin{align*} \sin \left ( x\right ) y_{0}^{\prime }\left ( x\right ) & \thicksim -\sin \left ( 2x\right ) y_{0}\left ( x\right ) \\ \frac{dy_{0}}{y_{0}} & \thicksim -\frac{\sin \left ( 2x\right ) }{\sin \left ( x\right ) }dx\\ \ln \left \vert y_{0}\right \vert & \thicksim -\int \frac{\sin \left ( 2x\right ) }{\sin \left ( x\right ) }dx\\ & \thicksim -\int \frac{2\sin x\cos x}{\sin \left ( x\right ) }dx\\ & \thicksim -\int 2\cos xdx\\ & \thicksim -2\sin x+C_{5} \end{align*}
Hence\begin{align*} y_{0}^{out}\left ( x\right ) & \thicksim Ae^{-2\sin x}\\ y_{0}\left ( \pi \right ) & =0 \end{align*}
Therefore A=0 and y_{0}^{out}\left ( x\right ) =0.Now that we found all solutions, we can do the matching. The matching on the left side gives\begin{align} \lim _{\xi \rightarrow \infty }y^{in}\left ( \xi \right ) & =\lim _{x\rightarrow 0}y^{out}\left ( x\right ) \nonumber \\ \lim _{\xi \rightarrow \infty }C_{1}\int _{0}^{\xi }e^{\frac{-s^{2}}{2}}ds+\pi & =\lim _{x\rightarrow 0}C_{5}e^{-2\sin x}\nonumber \\ \lim _{\xi \rightarrow \infty }C_{1}\int _{0}^{\xi }e^{\frac{-s^{2}}{2}}ds+\pi & =0 \tag{3} \end{align}
But \int _{0}^{\xi }e^{\frac{-s^{2}}{2}}ds=\sqrt{\frac{\pi }{2}}\operatorname{erf}\left ( \frac{\xi }{\sqrt{2}}\right ) And \lim _{\xi \rightarrow \infty }\operatorname{erf}\left ( \frac{\xi }{\sqrt{2}}\right ) =1, hence (3) becomes\begin{align} C_{1}\sqrt{\frac{\pi }{2}}+\pi & =0\nonumber \\ C_{1} & =-\pi \sqrt{\frac{2}{\pi }}\nonumber \\ & =-\sqrt{2\pi } \tag{4} \end{align}
Therefore from (2)\begin{equation} y_{0}^{in}\left ( \xi \right ) \thicksim -\sqrt{2\pi }\int _{0}^{\xi }e^{\frac{-s^{2}}{2}}ds+\pi \tag{5} \end{equation} Near x=\pi , using \eta =\frac{\pi -x}{\varepsilon ^{p}}. Expansion y^{in}\left ( \eta \right ) \thicksim y_{0}\left ( \eta \right ) +\varepsilon y_{1}\left ( \eta \right ) +O\left ( \varepsilon ^{2}\right ) gives p=\frac{1}{2}. Hence O\left ( 1\right ) terms gives\begin{align*} y_{0}^{\prime \prime }\left ( \eta \right ) & \thicksim \eta y_{0}^{\prime }\left ( \eta \right ) \\ y_{0}^{in}\left ( 0\right ) & =0\\ y_{0}^{in}\left ( \eta \right ) & \thicksim D\int _{0}^{\eta }e^{\frac{-s^{2}}{2}}ds \end{align*}
And matching on the right side gives\begin{align} \lim _{\eta \rightarrow \infty }y^{in}\left ( \eta \right ) & =\lim _{x\rightarrow \pi }y^{out}\left ( x\right ) \nonumber \\ \lim _{\eta \rightarrow \infty }D\int _{0}^{\eta }e^{\frac{-s^{2}}{2}}ds & =0\nonumber \\ D & =0 \tag{6} \end{align}
Therefore the solution is\begin{align} y\left ( x\right ) & \thicksim y^{in}\left ( \xi \right ) +y^{in}\left ( \eta \right ) +y^{out}\left ( x\right ) -y^{match}\nonumber \\ & \thicksim -\sqrt{2\pi }\int _{0}^{\xi }e^{\frac{-s^{2}}{2}}ds+\pi +0\nonumber \\ & \thicksim -\sqrt{2\pi }\sqrt{\frac{\pi }{2}}\operatorname{erf}\left ( \frac{\xi }{\sqrt{2}}\right ) +\pi \nonumber \\ & \thicksim -\sqrt{2\pi }\sqrt{\frac{\pi }{2}}\operatorname{erf}\left ( \frac{x}{\sqrt{2\varepsilon }}\right ) +\pi \nonumber \\ & \thicksim \pi -\pi \operatorname{erf}\left ( \frac{x}{\sqrt{2\varepsilon }}\right ) \tag{7} \end{align}
The following plot compares exact solution with (7) for \varepsilon =0.1,0.05. We see from these results, that as \varepsilon decreased, the approximation solution improved.
