### 5.1 Exam 1

5.1.3 problem 3
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#### 5.1.1 problem 3.26 (page 139)

Problem Perform local analysis solution to $$\left ( x-1\right ) y^{\prime \prime }-xy^{\prime }+y=0$$ at $$x=1$$. Use the result of this analysis to prove that a Taylor series expansion of any solution about $$x=0$$ has an inﬁnite radius of convergence. Find the exact solution by summing the series.

solution

Writing the ODE in standard form\begin{align} y^{\prime \prime }\left ( x\right ) +a\left ( x\right ) y^{\prime }\left ( x\right ) +b\left ( x\right ) y\left ( x\right ) & =0\tag{1}\\ y^{\prime \prime }-\frac{x}{\left ( x-1\right ) }y^{\prime }+\frac{1}{\left ( x-1\right ) }y & =0 \tag{2} \end{align}

Where $$a\left ( x\right ) =\frac{-x}{\left ( x-1\right ) },b\left ( x\right ) =\frac{1}{\left ( x-1\right ) }$$. The above shows that $$x=1$$ is singular point for both $$a\left ( x\right )$$ and $$b\left ( x\right )$$. The next step is to classify the type of the singular point. Is it regular singular point or irregular singular point?\begin{align*} \lim _{x\rightarrow 1}\left ( x-1\right ) a\left ( x\right ) & =\lim _{x\rightarrow 1}\left ( x-1\right ) \frac{-x}{\left ( x-1\right ) }\\ & =-1 \end{align*}

And\begin{align*} \lim _{x\rightarrow 1}\left ( x-1\right ) ^{2}b\left ( x\right ) & =\lim _{x\rightarrow 1}\left ( x-1\right ) ^{2}\frac{1}{\left ( x-1\right ) }\\ & =0 \end{align*}

Because the limit exist, then $$x=1$$ is a regular singular point. Therefore solution is assumed to be a Frobenius power series given by$y\left ( x\right ) =\sum _{n=0}^{\infty }a_{n}\left ( x-1\right ) ^{n+r}$ Substituting this in the original ODE $$\left ( x-1\right ) y^{\prime \prime }-xy^{\prime }+y=0$$ gives\begin{align*} y^{\prime }\left ( x\right ) & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}\left ( x-1\right ) ^{n+r-1}\\ y^{\prime \prime }\left ( x\right ) & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}\left ( x-1\right ) ^{n+r-2} \end{align*}

In order to move the $$\left ( x-1\right )$$ inside the summation, the original ODE $$\left ( x-1\right ) y^{\prime \prime }-xy^{\prime }+y=0$$ is ﬁrst rewritten as$$\left ( x-1\right ) y^{\prime \prime }-\left ( x-1\right ) y^{\prime }-y^{\prime }+y=0 \tag{3}$$ Substituting the Frobenius series into the above gives

Or

Adjusting all powers of $$\left ( x-1\right )$$ to be the same by rewriting exponents and summation indices gives
Collecting terms with same powers in $$\left ( x-1\right )$$ simpliﬁes the above to$$\sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) -\left ( n+r\right ) \right ) a_{n}\left ( x-1\right ) ^{n+r-1}-\sum _{n=1}^{\infty }\left ( n+r-2\right ) a_{n-1}\left ( x-1\right ) ^{n+r-1}=0\tag{4}$$ Setting $$n=0$$ gives the indicial equation\begin{align*} \left ( \left ( n+r\right ) \left ( n+r-1\right ) -\left ( n+r\right ) \right ) a_{0} & =0\\ \left ( \left ( r\right ) \left ( r-1\right ) -r\right ) a_{0} & =0 \end{align*}

Since $$a_{0}\neq 0$$ then the indicial equation is\begin{align*} \left ( r\right ) \left ( r-1\right ) -r & =0\\ r^{2}-2r & =0\\ r\left ( r-2\right ) & =0 \end{align*}

The roots of the indicial equation are therefore\begin{align*} r_{1} & =2\\ r_{2} & =0 \end{align*}

Each one of these roots generates a solution to the ODE. The next step is to ﬁnd the solution $$y_{1}\left ( x\right )$$ associated with $$r=2$$.  (The largest root is used ﬁrst). Using $$r=2$$ in equation (4) gives\begin{align} \sum _{n=0}^{\infty }\left ( \left ( n+2\right ) \left ( n+1\right ) -\left ( n+2\right ) \right ) a_{n}\left ( x-1\right ) ^{n+1}-\sum _{n=1}^{\infty }na_{n-1}\left ( x-1\right ) ^{n+1} & =0\nonumber \\ \sum _{n=0}^{\infty }n\left ( n+2\right ) a_{n}\left ( x-1\right ) ^{n+1}-\sum _{n=1}^{\infty }na_{n-1}\left ( x-1\right ) ^{n+1} & =0\tag{5} \end{align}

At $$n\geq 1$$, the recursive relation is found and used to generate the coeﬃcients of the Frobenius power series\begin{align*} n\left ( n+2\right ) a_{n}-na_{n-1} & =0\\ a_{n} & =\frac{n}{n\left ( n+2\right ) }a_{n-1} \end{align*}

Few terms are now generated to see the pattern of the series and to determine the closed form. For $$n=1$$$a_{1}=\frac{1}{3}a_{0}$ For $$n=2$$$a_{2}=\frac{2}{2\left ( 2+2\right ) }a_{1}=\frac{2}{8}\frac{1}{3}a_{0}=\frac{1}{12}a_{0}$ For $$n=3$$$a_{3}=\frac{3}{3\left ( 3+2\right ) }a_{2}=\frac{3}{15}\frac{1}{12}a_{0}=\frac{1}{60}a_{0}$ For $$n=4$$$a_{4}=\frac{4}{4\left ( 4+2\right ) }a_{3}=\frac{1}{6}\frac{1}{60}a_{0}=\frac{1}{360}a_{0}$ And so on. From the above, the ﬁrst solution becomes\begin{align} y_{1}\left ( x\right ) & =\sum _{n=0}^{\infty }a_{n}\left ( x-1\right ) ^{n+2}\nonumber \\ & =a_{0}\left ( x-1\right ) ^{2}+a_{1}\left ( x-1\right ) ^{3}+a_{2}\left ( x-1\right ) ^{4}+a_{3}\left ( x-1\right ) ^{4}+a_{4}\left ( x-1\right ) ^{5}+\cdots \nonumber \\ & =\left ( x-1\right ) ^{2}\left ( a_{0}+a_{1}\left ( x-1\right ) +a_{2}\left ( x-1\right ) ^{2}+a_{3}\left ( x-1\right ) ^{3}+a_{4}\left ( x-1\right ) ^{4}+\cdots \right ) \nonumber \\ & =\left ( x-1\right ) ^{2}\left ( a_{0}+\frac{1}{3}a_{0}\left ( x-1\right ) +\frac{1}{12}a_{0}\left ( x-1\right ) ^{2}+\frac{1}{60}a_{0}\left ( x-1\right ) ^{3}+\frac{1}{360}a_{0}\left ( x-1\right ) ^{4}+\cdots \right ) \nonumber \\ & =a_{0}\left ( x-1\right ) ^{2}\left ( 1+\frac{1}{3}\left ( x-1\right ) +\frac{1}{12}\left ( x-1\right ) ^{2}+\frac{1}{60}\left ( x-1\right ) ^{3}+\frac{1}{360}\left ( x-1\right ) ^{4}+\cdots \right ) \tag{6} \end{align}

To ﬁnd closed form solution to $$y_{1}\left ( x\right )$$, Taylor series expansion of $$e^{x}$$ around $$x=1$$ is found ﬁrst\begin{align*} e^{x} & \approx e+e\left ( x-1\right ) +\frac{e}{2}\left ( x-1\right ) ^{2}+\frac{e}{3!}\left ( x-1\right ) ^{3}+\frac{e}{4!}\left ( x-1\right ) ^{4}+\frac{e}{5!}\left ( x-1\right ) ^{5}+\cdots \\ & \approx e+e\left ( x-1\right ) +\frac{e}{2}\left ( x-1\right ) ^{2}+\frac{e}{6}\left ( x-1\right ) ^{3}+\frac{e}{24}\left ( x-1\right ) ^{4}+\frac{e}{120}\left ( x-1\right ) ^{5}+\cdots \\ & \approx e\left ( 1+\left ( x-1\right ) +\frac{1}{2}\left ( x-1\right ) ^{2}+\frac{1}{6}\left ( x-1\right ) ^{3}+\frac{1}{24}\left ( x-1\right ) ^{4}+\frac{1}{120}\left ( x-1\right ) ^{5}+\cdots \right ) \end{align*}

Multiplying the above by $$2$$ gives$2e^{x}\approx e\left ( 2+2\left ( x-1\right ) +\left ( x-1\right ) ^{2}+\frac{1}{3}\left ( x-1\right ) ^{3}+\frac{1}{12}\left ( x-1\right ) ^{4}+\frac{1}{60}\left ( x-1\right ) ^{5}+\cdots \right )$ Factoring $$\left ( x-1\right ) ^{2}$$ from the RHS results in$$2e^{x}\approx e\left ( 2+2\left ( x-1\right ) +\left ( x-1\right ) ^{2}\left ( 1+\frac{1}{3}\left ( x-1\right ) +\frac{1}{12}\left ( x-1\right ) ^{2}+\frac{1}{60}\left ( x-1\right ) ^{3}+\cdots \right ) \right ) \tag{6A}$$ Comparing the above result with the solution $$y_{1}\left ( x\right )$$ in (6), shows that the (6A) can be written in terms of $$y_{1}\left ( x\right )$$ as$2e^{x}=e\left ( 2+2\left ( x-1\right ) +\left ( x-1\right ) ^{2}\left ( \frac{y_{1}\left ( x\right ) }{a_{0}\left ( x-1\right ) ^{2}}\right ) \right )$ Therefore\begin{align*} 2e^{x} & =e\left ( 2+2\left ( x-1\right ) +\frac{y_{1}\left ( x\right ) }{a_{0}}\right ) \\ 2e^{x-1} & =2+2\left ( x-1\right ) +\frac{y_{1}\left ( x\right ) }{a_{0}}\\ 2e^{x-1}-2-2\left ( x-1\right ) & =\frac{y_{1}\left ( x\right ) }{a_{0}} \end{align*}

Solving for $$y_{1}\left ( x\right )$$\begin{align*} y_{1}\left ( x\right ) & =a_{0}\left ( 2e^{x-1}-2-2\left ( x-1\right ) \right ) \\ & =a_{0}\left ( 2e^{x-1}-2-2x+2\right ) \\ & =a_{0}\left ( 2e^{x-1}-2x\right ) \\ & =\frac{2a_{0}}{e}e^{x}-2a_{0}x \end{align*}

Let $$\frac{2a_{0}}{e}=C_{1}$$ and $$-2a_{0}=C_{2}$$, then the above solution can be written as$y_{1}\left ( x\right ) =C_{1}e^{x}+C_{2}x$ Now that $$y_{1}\left ( x\right )$$ is found, which is the solution associated with $$r=2$$, the next step is to ﬁnd the second solution $$y_{2}\left ( x\right )$$ associated with $$r=0$$. Since $$r_{2}-r_{1}=2$$ is an integer, the solution can be either case $$II(b)\left ( i\right )$$ or case $$II\left ( b\right ) \left ( ii\right )$$ as given in the text book at page 72.

From equation (3.3.9) at page 72 of the text, using $$N=2$$ since $$N=r_{2}-r_{1}$$ and where $$p\left ( x\right ) =-x$$ and $$q\left ( x\right ) =1$$ in this problem by comparing our ODE with the standard ODE in (3.3.2) at page 70 given by$y^{\prime \prime }+\frac{p\left ( x\right ) }{\left ( x-x_{0}\right ) }y^{\prime }+\frac{q\left ( x\right ) }{\left ( x-x_{0}\right ) ^{2}}y=0$ Expanding $$p\left ( x\right ) ,q\left ( x\right )$$ in Taylor series\begin{align*} p\left ( x\right ) & =\sum _{n=0}^{\infty }p_{n}\left ( x-1\right ) ^{n}\\ q\left ( x\right ) & =\sum _{n=0}^{\infty }q_{n}\left ( x-1\right ) ^{n} \end{align*}

Since $$p\left ( x\right ) =-x$$ in our ODE, then $$p_{0}=-1$$ and $$p_{1}=-1$$ and all other terms are zero. For $$q\left ( x\right )$$, which is just $$1$$ in our ODE, then $$q_{0}=1$$ and all other terms are zero. Hence

\begin{align*} p_{0} & =-1\\ p_{1} & =-1\\ q_{0} & =1\\ N & =2\\ r & =0 \end{align*}

The above values are now used to evaluate RHS of 3.3.9 in order to ﬁnd which case it is. (book uses $$\alpha$$ for $$r$$)$$0a_{N}=-\sum _{k=0}^{N-1}\left [ \left ( r+k\right ) p_{N-k}+q_{N-k}\right ] a_{k}\tag{3.3.9}$$ Since $$N=2$$ the above becomes$0a_{2}=-\sum _{k=0}^{1}\left [ \left ( r+k\right ) p_{2-k}+q_{2-k}\right ] a_{k}$ Using $$r=0$$, since this is the second root, gives\begin{align*} 0a_{2} & =-\sum _{k=0}^{1}\left ( kp_{2-k}+q_{2-k}\right ) a_{k}\\ & =-\left ( \left ( 0p_{2-0}+q_{2-0}\right ) a_{0}+\left ( p_{2-1}+q_{2-1}\right ) a_{1}\right ) \\ & =-\left ( \left ( 0p_{2}+q_{2}\right ) a_{0}+\left ( p_{1}+q_{1}\right ) a_{1}\right ) \\ & =-\left ( 0+q_{2}\right ) a_{0}-\left ( p_{1}+q_{1}\right ) a_{1} \end{align*}

Since $$q_{2}=0,p_{1}=-1,q_{1}=1$$, therefore\begin{align*} 0a_{2} & =-\left ( 0+0\right ) a_{0}-\left ( -1+1\right ) a_{1}\\ & =0 \end{align*}

The above shows that this is case $$II\left ( b\right ) \left ( ii\right )$$, because the right side of $$3.3.9$$ is zero. This means the second solution $$y_{2}\left ( x\right )$$ is also a Fronbenius series. If the above was not zero, the method of reduction of order would be used to ﬁnd second solution.

Assuming $$y_{2}\left ( x\right ) =\sum b_{n}\left ( x-1\right ) ^{n+r}$$, and since $$r=0$$, therefore$y_{2}\left ( x\right ) =\sum _{n=0}^{\infty }b_{n}\left ( x-1\right ) ^{n}$ Following the same method used to ﬁnd the ﬁrst solution, this series is now used in the ODE to determine $$b_{n}$$. \begin{align*} y_{2}^{\prime }\left ( x\right ) & =\sum _{n=0}^{\infty }nb_{n}\left ( x-1\right ) ^{n-1}=\sum _{n=1}^{\infty }nb_{n}\left ( x-1\right ) ^{n-1}=\sum _{n=0}^{\infty }\left ( n+1\right ) b_{n+1}\left ( x-1\right ) ^{n}\\ y_{2}^{\prime \prime }\left ( x\right ) & =\sum _{n=0}^{\infty }n\left ( n+1\right ) b_{n+1}\left ( x-1\right ) ^{n-1}=\sum _{n=1}^{\infty }n\left ( n+1\right ) b_{n+1}\left ( x-1\right ) ^{n-1}=\sum _{n=0}^{\infty }\left ( n+1\right ) \left ( n+2\right ) b_{n+2}\left ( x-1\right ) ^{n} \end{align*}

The ODE $$\left ( x-1\right ) y^{\prime \prime }-\left ( x-1\right ) y^{\prime }-y^{\prime }+y=0$$ now becomes

Or
Hence$\sum _{n=1}^{\infty }\left ( n\right ) \left ( n+1\right ) b_{n+1}\left ( x-1\right ) ^{n}-\sum _{n=1}^{\infty }nb_{n}\left ( x-1\right ) ^{n}-\sum _{n=0}^{\infty }\left ( n+1\right ) b_{n+1}\left ( x-1\right ) ^{n}+\sum _{n=0}^{\infty }b_{n}\left ( x-1\right ) ^{n}=0$ $$n=0$$ gives\begin{align*} -\left ( n+1\right ) b_{n+1}+b_{n} & =0\\ -b_{1}+b_{0} & =0\\ b_{1} & =b_{0} \end{align*}

$$n\geq 1$$ generates the recursive relation to ﬁnd all remaining $$b_{n}$$ coeﬃcients\begin{align*} \left ( n\right ) \left ( n+1\right ) b_{n+1}-nb_{n}-\left ( n+1\right ) b_{n+1}+b_{n} & =0\\ \left ( n\right ) \left ( n+1\right ) b_{n+1}-\left ( n+1\right ) b_{n+1} & =nb_{n}-b_{n}\\ b_{n+1}\left ( \left ( n\right ) \left ( n+1\right ) -\left ( n+1\right ) \right ) & =b_{n}\left ( n-1\right ) \\ b_{n+1} & =b_{n}\frac{\left ( n-1\right ) }{\left ( n\right ) \left ( n+1\right ) -\left ( n+1\right ) } \end{align*}

Therefore the recursive relation is$b_{n+1}=\frac{b_{n}}{n+1}$ Few terms are generated to see the pattern and to ﬁnd the closed form solution for $$y_{2}\left ( x\right )$$. For $$n=1$$$b_{2}=b_{1}\frac{1}{2}=\frac{1}{2}b_{0}$ For $$n=2$$$b_{3}=\frac{b_{2}}{3}=\frac{1}{3}\frac{1}{2}b_{0}=\frac{1}{6}b_{0}$ For $$n=3$$$b_{4}=\frac{b_{3}}{3+1}=\frac{1}{4}\frac{1}{6}b_{0}=\frac{1}{24}b_{0}$ For $$n=4$$$b_{5}=\frac{b_{4}}{4+1}=\frac{1}{5}\frac{1}{24}b_{0}=\frac{1}{120}b_{0}$ And so on. Therefore, the second solution is\begin{align} y_{2}\left ( x\right ) & =\sum _{n=0}^{\infty }b_{n}\left ( x-1\right ) ^{n}\nonumber \\ & =b_{0}+b_{1}\left ( x-1\right ) +b_{2}\left ( x-1\right ) ^{2}+\cdots \nonumber \\ & =b_{0}+b_{0}\left ( x-1\right ) +\frac{1}{2}b_{0}\left ( x-1\right ) ^{2}+\frac{1}{6}b_{0}\left ( x-1\right ) ^{3}+\frac{1}{24}b_{0}\left ( x-1\right ) ^{4}+\frac{1}{120}b_{0}\left ( x-1\right ) ^{5}+\cdots \nonumber \\ & =b_{0}\left ( 1+\left ( x-1\right ) +\frac{1}{2}\left ( x-1\right ) ^{2}+\frac{1}{6}\left ( x-1\right ) ^{3}+\frac{1}{24}\left ( x-1\right ) ^{4}+\frac{1}{120}\left ( x-1\right ) ^{5}+\cdots \right ) \tag{7A} \end{align}

The Taylor series for $$e^{x}$$ around $$x=1$$ is \begin{align} e^{x} & \approx e+e\left ( x-1\right ) +\frac{e}{2}\left ( x-1\right ) ^{2}+\frac{e}{6}\left ( x-1\right ) ^{3}+\frac{e}{24}\left ( x-1\right ) ^{4}+\frac{e}{120}\left ( x-1\right ) ^{5}+\cdots \nonumber \\ & \approx e\left ( 1+\left ( x-1\right ) +\frac{1}{2}\left ( x-1\right ) ^{2}+\frac{1}{6}\left ( x-1\right ) ^{3}+\frac{1}{24}\left ( x-1\right ) ^{4}+\frac{1}{120}\left ( x-1\right ) ^{5}+\cdots \right ) \tag{7B} \end{align}

Comparing (7A) with (7B) shows that the second solution closed form is$y_{2}\left ( x\right ) =b_{0}\frac{e^{x}}{e}$ Let $$\frac{b_{0}}{e}$$ be some constant, say $$C_{3}$$, the second solution above becomes$y_{2}\left ( x\right ) =C_{3}e^{x}$ Both solutions $$y_{1}\left ( x\right ) ,y_{2}\left ( x\right )$$ have now been found. The ﬁnal solution is\begin{align*} y\left ( x\right ) & =y_{1}\left ( x\right ) +y_{2}\left ( x\right ) \\ & =\overset{y_{1}\left ( x\right ) }{\overbrace{C_{1}e^{x}+C_{2}x}}+\overset{y_{2}\left ( x\right ) }{\overbrace{C_{3}e^{x}}}\\ & =C_{4}e^{x}+C_{2}x \end{align*}

Hence, the exact solution is$$y\left ( x\right ) =Ae^{x}+Bx\tag{7}$$ Where $$A,B\,$$ are constants to be found from initial conditions if given. Above solution is now veriﬁed by substituting it back to original ODE\begin{align*} y^{\prime } & =Ae^{x}+B\\ y^{\prime \prime } & =Ae^{x} \end{align*}

Substituting these into $$\left ( x-1\right ) y^{\prime \prime }-xy^{\prime }+y=0$$ gives\begin{align*} \left ( x-1\right ) Ae^{x}-x\left ( Ae^{x}+B\right ) +Ae^{x}+Bx & =0\\ xAe^{x}-Ae^{x}-xAe^{x}-xB+Ae^{x}+Bx & =0\\ -Ae^{x}-xB+Ae^{x}+Bx & =0\\ 0 & =0 \end{align*}

To answer the ﬁnal part of the question, the above solution (7) is analytic around $$x=0$$ with inﬁnite radius of convergence since $$\exp \left ( \cdot \right )$$ is analytic everywhere. Writing the solution as$y\left ( x\right ) =\left ( A\sum _{n=0}^{\infty }\frac{x^{n}}{n!}\right ) +Bx$ The function $$x$$ have inﬁnite radius of convergence, since it is its own series. And the exponential function has inﬁnite radius of convergence as known, veriﬁed by using standard ratio test $A\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert =A\lim _{n\rightarrow \infty }\left \vert \frac{x^{n+1}n!}{\left ( n+1\right ) !x^{n}}\right \vert =A\lim _{n\rightarrow \infty }\left \vert \frac{xn!}{\left ( n+1\right ) !}\right \vert =A\lim _{n\rightarrow \infty }\left \vert \frac{x}{n+1}\right \vert =0$ For any $$x$$. Since the ratio is less than $$1$$, then the solution $$y\left ( x\right )$$ expanded around $$x=0$$ has an inﬁnite radius of convergence.

#### 5.1.2 problem 9.8 (page 480)

Problem Use boundary layer to ﬁnd uniform approximation with error of order $$O\left ( \varepsilon ^{2}\right )$$ for the problem $$\varepsilon y^{\prime \prime }+y^{\prime }+y=0$$ with $$y\left ( 0\right ) =e,y\left ( 1\right ) =1$$. Compare your solution to exact solution. Plot the solution for some values of $$\varepsilon$$.

solution

$$\varepsilon y^{\prime \prime }+y^{\prime }+y=0\tag{1}$$ Since $$a\left ( x\right ) =1>0$$, then a boundary layer is expected at the left side, near $$x=0$$. Matching will fail if this was not the case. Starting with the outer solution near $$x=1$$. Let $y^{out}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}\left ( x\right )$ Substituting this into (1) gives$$\varepsilon \left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\varepsilon ^{2}y_{2}^{\prime \prime }+\cdots \right ) +\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\varepsilon ^{2}y_{2}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots \right ) =0\tag{2}$$ Collecting powers of $$O\left ( \varepsilon ^{0}\right )$$ results in the ODE\begin{align} y_{0}^{\prime } & \thicksim -y_{0}\nonumber \\ \frac{dy_{0}}{y_{0}} & \thicksim -dx\nonumber \\ \ln \left \vert y_{0}\right \vert & \thicksim -x+C_{1}\nonumber \\ y_{0}^{out}\left ( x\right ) & \thicksim C_{1}e^{-x}+O\left ( \varepsilon \right ) \tag{3} \end{align}

$$C_{1}$$ is found from boundary conditions $$y\left ( 1\right ) =1$$. Equation (3) gives\begin{align*} 1 & =C_{1}e^{-1}\\ C_{1} & =e \end{align*}

Hence solution (3) becomes$y_{0}^{out}\left ( x\right ) \thicksim e^{1-x}$ $$y_{1}^{out}\left ( x\right )$$ is now found. Using (2) and collecting terms of $$O\left ( \varepsilon ^{1}\right )$$ gives the ODE$$y_{1}^{\prime }+y_{1}\thicksim -y_{0}^{\prime \prime }\tag{4}$$ But \begin{align*} y_{0}^{\prime }\left ( x\right ) & =-e^{1-x}\\ y_{0}^{\prime \prime }\left ( x\right ) & =e^{1-x} \end{align*}

Using the above in the RHS of (4) gives$y_{1}^{\prime }+y_{1}\thicksim -e^{1-x}$ The integrating factor is $$e^{x}$$, hence the above becomes\begin{align*} \frac{d}{dx}\left ( y_{1}e^{x}\right ) & \thicksim -e^{x}e^{1-x}\\ \frac{d}{dx}\left ( y_{1}e^{x}\right ) & \thicksim -e \end{align*}

Integrating both sides gives\begin{align} y_{1}e^{x} & \thicksim -ex+C_{2}\nonumber \\ y_{1}^{out}\left ( x\right ) & \thicksim -xe^{1-x}+C_{2}e^{-x}\tag{5} \end{align}

Applying boundary conditions $$y\left ( 1\right ) =0$$ to the above gives\begin{align*} 0 & =-1+C_{2}e^{-1}\\ C_{2} & =e \end{align*}

Hence the solution in (5) becomes\begin{align*} y_{1}^{out}\left ( x\right ) & \thicksim -xe^{1-x}+e^{1-x}\\ & \thicksim \left ( 1-x\right ) e^{1-x} \end{align*}

Therefore the outer solution is\begin{align} y^{out}\left ( x\right ) & =y_{0}+\varepsilon y_{1}\nonumber \\ & =e^{1-x}+\varepsilon \left ( 1-x\right ) e^{1-x}\tag{6} \end{align}

Now the boundary layer (inner) solution $$y^{in}\left ( x\right )$$ near $$x=0$$ is found. Let $$\xi =\frac{x}{\varepsilon ^{p}}$$ be the inner variable. The original ODE is expressed using this new variable, and $$p$$ is found. Since $$\frac{dy}{dx}=\frac{dy}{d\xi }\frac{d\xi }{dx}$$ then $$\frac{dy}{dx}=\frac{dy}{d\xi }\varepsilon ^{-p}$$. The diﬀerential operator is $$\frac{d}{dx}\equiv \varepsilon ^{-p}\frac{d}{d\xi }$$ therefore\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*}

Hence $$\frac{d^{2}y}{dx^{2}}=\varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}}$$ and $$\varepsilon y^{\prime \prime }+y^{\prime }+y=0$$ becomes\begin{align} \varepsilon \left ( \varepsilon ^{-2p}\frac{d^{2}y}{d\xi ^{2}}\right ) +\varepsilon ^{-p}\frac{dy}{d\xi }+y & =0\nonumber \\ \varepsilon ^{1-2p}y^{\prime \prime }+\varepsilon ^{-p}y^{\prime }+y & =0\tag{7A} \end{align}

The largest terms are $$\left \{ \varepsilon ^{1-2p},\varepsilon ^{-p}\right \}$$, balance gives $$1-2p=-p$$ or $\fbox{p=1}$ The ODE (7A) becomes$$\varepsilon ^{-1}y^{\prime \prime }+\varepsilon ^{-1}y^{\prime }+y=0\tag{7}$$ Assuming that solution is$y_{in}\left ( x\right ) =\sum _{n=0}^{\infty }\varepsilon ^{n}y_{n}=y_{0}+\varepsilon y_{1}+\varepsilon ^{2}y_{2}+\cdots$ Substituting the above into (7) gives$$\varepsilon ^{-1}\left ( y_{0}^{\prime \prime }+\varepsilon y_{1}^{\prime \prime }+\cdots \right ) +\varepsilon ^{-1}\left ( y_{0}^{\prime }+\varepsilon y_{1}^{\prime }+\cdots \right ) +\left ( y_{0}+\varepsilon y_{1}+\cdots \right ) =0\tag{8}$$ Collecting terms with $$O\left ( \varepsilon ^{-1}\right )$$ gives the ﬁrst order ODE to solve$y_{0}^{\prime \prime }\thicksim -y_{0}^{\prime }$ Let $$z=y_{0}^{\prime }$$, the above becomes\begin{align*} z^{\prime } & \thicksim -z\\ \frac{dz}{z} & \thicksim -d\xi \\ \ln \left \vert z\right \vert & \thicksim -\xi +C_{4}\\ z & \thicksim C_{4}e^{-\xi } \end{align*}

Hence$y_{0}^{\prime }\thicksim C_{4}e^{-\xi }$ Integrating \begin{align} y_{0}^{in}\left ( \xi \right ) & \thicksim C_{4}\int e^{-\xi }d\xi +C_{5}\nonumber \\ & \thicksim -C_{4}e^{-\xi }+C_{5}\tag{9} \end{align}

Applying boundary conditions $$y\left ( 0\right ) =e$$ gives\begin{align*} e & =-C_{4}+C_{5}\\ C_{5} & =e+C_{4} \end{align*}

Equation (9) becomes\begin{align} y_{0}^{in}\left ( \xi \right ) & \thicksim -C_{4}e^{-\xi }+e+C_{4}\nonumber \\ & \thicksim C_{4}\left ( 1-e^{-\xi }\right ) +e\tag{10} \end{align}

The next leading order $$y_{1}^{in}\left ( \xi \right )$$ is found from (8) by collecting terms in $$O\left ( \varepsilon ^{0}\right )$$, which results in the ODE$y_{1}^{\prime \prime }+y_{1}^{\prime }\thicksim -y_{0}$ Since $$y_{0}^{in}\left ( \xi \right ) \thicksim C_{4}\left ( 1-e^{-\xi }\right ) +e$$, therefore $$y_{0}^{\prime }\thicksim C_{4}e^{-\xi }$$ and the above becomes$y_{1}^{\prime \prime }+y_{1}^{\prime }\thicksim -C_{4}e^{-\xi }$ The homogenous solution is found ﬁrst, then method of undetermined coeﬃcients is used to ﬁnd particular solution. The homogenous ODE is $y_{1,h}^{\prime \prime }\thicksim y_{1,h}^{\prime }$ This was solved above for $$y_{0}^{in}$$, and the solution is$y_{1,h}\thicksim -C_{5}e^{-\xi }+C_{6}$ To ﬁnd the particular solution, let $$y_{1,p}\thicksim A\xi e^{-\xi }$$, where $$\xi$$ was added since $$e^{-\xi }$$ shows up in the homogenous solution. Hence\begin{align*} y_{1,p}^{\prime } & \thicksim Ae^{-\xi }-A\xi e^{-\xi }\\ y_{1,p}^{\prime \prime } & \thicksim -Ae^{-\xi }-\left ( Ae^{-\xi }-A\xi e^{-\xi }\right ) \\ & \thicksim -2Ae^{-\xi }+A\xi e^{-\xi } \end{align*}

Substituting these in the ODE $$y_{1,p}^{\prime \prime }+y_{1,p}^{\prime }\thicksim -C_{4}e^{-\xi }$$ results in\begin{align*} -2Ae^{-\xi }+A\xi e^{-\xi }+Ae^{-\xi }-A\xi e^{-\xi } & \thicksim -C_{4}e^{-\xi }\\ -A & =-C_{4}\\ A & =C_{4} \end{align*}

Therefore the particular solution is $y_{1,p}\thicksim C_{4}\xi e^{-\xi }$ And therefore the complete solution is \begin{align*} y_{1}^{in}\left ( \xi \right ) & \thicksim y_{1,h}+y_{1,p}\\ & \thicksim -C_{5}e^{-\xi }+C_{6}+C_{4}\xi e^{-\xi } \end{align*}

Applying boundary conditions $$y\left ( 0\right ) =0$$ to the above gives\begin{align*} 0 & =-C_{5}+C_{6}\\ C_{6} & =C_{5} \end{align*}

Hence the solution becomes\begin{align} y_{1}^{in}\left ( \xi \right ) & \thicksim -C_{5}e^{-\xi }+C_{5}+C_{4}\xi e^{-\xi }\nonumber \\ & \thicksim C_{5}\left ( 1-e^{-\xi }\right ) +C_{4}\xi e^{-\xi }\tag{11} \end{align}

The complete inner solution now becomes\begin{align} y^{in}\left ( \xi \right ) & \thicksim y_{0}^{in}+\varepsilon y_{1}^{in}\nonumber \\ & \thicksim C_{4}\left ( 1-e^{-\xi }\right ) +e+\varepsilon \left ( C_{5}\left ( 1-e^{-\xi }\right ) +C_{4}\xi e^{-\xi }\right ) \tag{12} \end{align}

There are two constants that need to be determined in the above from matching with the outer solution.\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 }C_{4}\left ( 1-e^{-\xi }\right ) +e+\varepsilon \left ( C_{5}\left ( 1-e^{-\xi }\right ) +C_{4}\xi e^{-\xi }\right ) & \thicksim \lim _{x\rightarrow 0}e^{1-x}+\varepsilon \left ( 1-x\right ) e^{1-x}\\ C_{4}+e+\varepsilon C_{5} & \thicksim e+\varepsilon e \end{align*}

The above shows that\begin{align*} C_{5} & =e\\ C_{4}+e & =e\\ C_{4} & =0 \end{align*}

This gives the boundary layer solution $$y^{in}\left ( \xi \right )$$ as \begin{align*} y^{in}\left ( \xi \right ) & \thicksim e+\varepsilon e\left ( 1-e^{-\xi }\right ) \\ & \thicksim e\left ( 1+\varepsilon \left ( 1-e^{-\xi }\right ) \right ) \end{align*}

In terms of $$x$$, since $$\xi =\frac{x}{\varepsilon },$$ the above can be written as$y^{in}\left ( x\right ) \thicksim e\left ( 1+\varepsilon \left ( 1-e^{-\frac{x}{\varepsilon }}\right ) \right )$ The uniform solution is therefore\begin{align*} y_{\text{uniform}}\left ( x\right ) & \thicksim y^{in}\left ( x\right ) +y^{out}\left ( x\right ) -y_{match}\\ & \thicksim \overset{y^{in}}{\overbrace{e\left ( 1+\varepsilon \left ( 1-e^{-\frac{x}{\varepsilon }}\right ) \right ) }}+\overset{y^{out}}{\overbrace{e^{1-x}+\varepsilon \left ( 1-x\right ) e^{1-x}}}-\left ( e+\varepsilon e\right ) \\ & \thicksim e+e\varepsilon \left ( 1-e^{-\frac{x}{\varepsilon }}\right ) +e^{1-x}+\left ( \varepsilon -\varepsilon x\right ) e^{1-x}-\left ( e+\varepsilon e\right ) \\ & \thicksim e+e\varepsilon -\varepsilon e^{1-\frac{x}{\varepsilon }}+e^{1-x}+\varepsilon e^{1-x}-\varepsilon xe^{1-x}-e-\varepsilon e\\ & \thicksim -\varepsilon e^{1-\frac{x}{\varepsilon }}+e^{1-x}+\varepsilon e^{1-x}-\varepsilon xe^{1-x}\\ & \thicksim e^{1-x}\left ( -\varepsilon e^{-\frac{1}{\varepsilon }}+1+\varepsilon -\varepsilon x\right ) \end{align*}

Or$y_{\text{uniform}}\left ( x\right ) \thicksim e^{1-x}\left ( 1+\varepsilon \left ( 1-x-e^{-\frac{1}{\varepsilon }}\right ) \right )$ With error $$O\left ( \varepsilon ^{2}\right )$$.

The above solution is now compared to the exact solution of $$\varepsilon y^{\prime \prime }+y^{\prime }+y=0$$ with $$y\left ( 0\right ) =e,y\left ( 1\right ) =1$$. Since this is a homogenous second order ODE with constant coeﬃcient, it is easily solved using characteristic equation.$\varepsilon \lambda ^{2}+\lambda +1=0$ The roots are\begin{align*} \lambda & =\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}\\ & =\frac{-1}{2\varepsilon }\pm \frac{\sqrt{1-4\varepsilon }}{2\varepsilon } \end{align*}

Therefore the solution is\begin{align*} y\left ( x\right ) & =Ae^{\lambda _{1}x}+Be^{\lambda _{2}x}\\ & =Ae^{\left ( \frac{-1}{2\varepsilon }+\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }\right ) x}+Be^{\left ( \frac{-1}{2\varepsilon }-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }\right ) x}\\ & =Ae^{\frac{-x}{2\varepsilon }}e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}+Be^{\frac{-x}{2\varepsilon }}e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }x}\\ & =e^{\frac{-x}{2\varepsilon }}\left ( Ae^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}+Be^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right ) \end{align*}

Applying ﬁrst boundary conditions $$y\left ( 0\right ) =e$$ to the above gives\begin{align*} e & =A+B\\ B & =e-A \end{align*}

Hence the solution becomes\begin{align} y\left ( x\right ) & =e^{\frac{-x}{2\varepsilon }}\left ( Ae^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}+\left ( e-A\right ) e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right ) \nonumber \\ & =e^{\frac{-x}{2\varepsilon }}\left ( Ae^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}+e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}-Ae^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right ) \nonumber \\ & =e^{\frac{-x}{2\varepsilon }}\left ( A\left ( e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right ) +e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right ) \tag{13} \end{align}

Applying second boundary conditions $$y\left ( 1\right ) =1$$ gives\begin{align} 1 & =e^{\frac{-1}{2\varepsilon }}\left ( A\left ( e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }}\right ) +e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}\right ) \nonumber \\ e^{\frac{1}{2\varepsilon }} & =A\left ( e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }}\right ) +e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}\nonumber \\ A & =\frac{e^{\frac{1}{2\varepsilon }}-e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}}{e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }}}\tag{14} \end{align}

Substituting this into (13) results in$y^{\text{exact}}\left ( x\right ) =e^{\frac{-x}{2\varepsilon }}\left ( A\left ( e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right ) +e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right )$ Where $$A$$ is given in (14). hence$y^{\text{exact}}\left ( x\right ) =e^{\frac{-x}{2\varepsilon }}\left ( \left ( \frac{e^{\frac{1}{2\varepsilon }}-e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}}{e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }}}\right ) \left ( e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right ) +e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right )$ In summary

 exact solution asymptotic solution $$e^{\frac{-x}{2\varepsilon }}\left ( \left ( \frac{e^{\frac{1}{2\varepsilon }}-e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}}{e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }}}\right ) \left ( e^{\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}-e^{\frac{-\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right ) +e^{1-\frac{\sqrt{1-4\varepsilon }}{2\varepsilon }x}\right )$$ $$e^{1-x}+\varepsilon e^{1-x}\left ( 1-x-e^{-\frac{1}{\varepsilon }}\right ) +O\left ( \varepsilon ^{2}\right )$$

The following plot compares the exact solution with the asymptotic solution for $$\varepsilon =0.1$$

The following plot compares the exact solution with the asymptotic solution for $$\varepsilon =0.01$$. The diﬀerence was too small to notice in this case, the plot below is zoomed to be near $$x=0$$

At $$\varepsilon =0.001$$, the diﬀerence between the exact and the asymptotic solution was not noticeable. Therefore, to better compare the solutions, the following plot shows the relative percentage error given by $100\left \vert \frac{y^{\text{exact}}-y^{\text{uniform}}}{y^{\text{exact}}}\right \vert \%$ For diﬀerent $$\varepsilon$$.

Some observations: The above plot shows more clearly how the diﬀerence between the exact solution and the asymptotic solution became smaller as $$\varepsilon$$ became smaller. The plot also shows that the boundary layer near $$x=0$$ is becoming more narrow are $$\varepsilon$$ becomes smaller as expected.  It also shows that the relative error is smaller in the outer region than in the boundary layer region. For example, for $$\varepsilon =0.05$$, the largest percentage error in the outer region was less than $$1\%$$, while in the boundary layer, very near $$x=0$$, the error grows to about $$5\%.$$ Another observation is that at the matching location, the relative error goes down to zero. One also notices that the matching location drifts towards $$x=0$$ as $$\varepsilon$$ becomes smaller because the boundary layer is becoming more narrow. The following table summarizes these observations.

 $$\varepsilon$$ $$\%$$ error near $$x=0$$ apparent width of boundary layer $$0.1$$ $$10$$ $$0.2$$ $$0.05$$ $$5$$ $$0.12$$ $$0.01$$ $$1$$ $$0.02$$

#### 5.1.3 problem 3

5.1.3.1 Part a
5.1.3.2 Part b
5.1.3.3 Part c

Problem (a) Find physical optics approximation to the eigenvalue and eigenfunctions of the Sturm-Liouville problem are $$\lambda \rightarrow \infty$$\begin{align*} -y^{\prime \prime } & =\lambda \left ( \sin \left ( x\right ) +1\right ) ^{2}y\\ y\left ( 0\right ) & =0\\ y\left ( \pi \right ) & =0 \end{align*}

(b) What is the integral relation necessary to make the eigenfunctions orthonormal? For some reasonable choice of scaling coeﬃcient (give the value), plot the eigenfunctions for $$n=5,n=20.$$

(c) Estimate how large $$\lambda$$ should be for the relative error of less than $$0.1\%$$

solution

##### 5.1.3.1 Part a

Writing the ODE as$y^{\prime \prime }+\lambda \left ( \sin \left ( x\right ) +1\right ) ^{2}y=0$ Let1 $\fbox{\lambda =\frac{1}{\varepsilon ^2}}$ Then the given ODE becomes$$\varepsilon ^{2}y^{\prime \prime }\left ( x\right ) +\left ( \sin \left ( x\right ) +1\right ) ^{2}y\left ( x\right ) =0\tag{1}$$ Physical optics approximation is obtained when $$\lambda \rightarrow \infty$$ which implies $$\varepsilon \rightarrow 0^{+}$$. Since the ODE is linear and the highest derivative is now multiplied by a very small parameter $$\varepsilon$$, WKB can therefore be used to solve it. WKB starts by assuming that the solution has the form$y\left ( x\right ) \thicksim \exp \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}\left ( x\right ) \right ) \qquad \delta \rightarrow 0$ Therefore, taking derivatives and substituting back in the ODE results in\begin{align*} y^{\prime }\left ( x\right ) & \thicksim \exp \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}\left ( x\right ) \right ) \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) \right ) \\ y^{\prime \prime }\left ( x\right ) & \thicksim \exp \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}\left ( x\right ) \right ) \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) \right ) ^{2}+\exp \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}\left ( x\right ) \right ) \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime \prime }\left ( x\right ) \right ) \end{align*}

Substituting these into (1) and canceling the exponential terms gives\begin{align} \varepsilon ^{2}\left ( \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) \right ) ^{2}+\frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime \prime }\left ( x\right ) \right ) & \thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2}\nonumber \\ \frac{\varepsilon ^{2}}{\delta ^{2}}\left ( S_{0}^{\prime }+\delta S_{1}^{\prime }+\cdots \right ) \left ( S_{0}^{\prime }+\delta S_{1}^{\prime }+\cdots \right ) +\frac{\varepsilon ^{2}}{\delta }\left ( S_{0}^{\prime \prime }+\delta S_{1}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2}\nonumber \\ \frac{\varepsilon ^{2}}{\delta ^{2}}\left ( \left ( S_{0}^{\prime }\right ) ^{2}+\delta \left ( 2S_{1}^{\prime }S_{0}^{\prime }\right ) +\cdots \right ) +\frac{\varepsilon ^{2}}{\delta }\left ( S_{0}^{\prime \prime }+\delta S_{1}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2}\nonumber \\ \left ( \frac{\varepsilon ^{2}}{\delta ^{2}}\left ( S_{0}^{\prime }\right ) ^{2}+\frac{2\varepsilon ^{2}}{\delta }S_{1}^{\prime }S_{0}^{\prime }+\cdots \right ) +\left ( \frac{\varepsilon ^{2}}{\delta }S_{0}^{\prime \prime }+\varepsilon ^{2}S_{1}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2}\tag{2} \end{align}

The largest term in the left side is $$\frac{\varepsilon ^{2}}{\delta ^{2}}\left ( S_{0}^{\prime }\right ) ^{2}$$. By dominant balance, this term has the same order of magnitude as the right side $$-\left ( \sin \left ( x\right ) +1\right ) ^{2}$$. This implies that $$\delta ^{2}$$ is proportional to $$\varepsilon ^{2}.$$ For simplicity (following the book) $$\delta$$ can be taken as equal to $$\varepsilon$$$\delta =\varepsilon$ Using the above in equation (2) results in$\left ( \left ( S_{0}^{\prime }\right ) ^{2}+2\varepsilon S_{1}^{\prime }S_{0}^{\prime }+\cdots \right ) +\left ( \varepsilon S_{0}^{\prime \prime }+\varepsilon ^{2}S_{1}^{\prime \prime }+\cdots \right ) \thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2}$ Balance of $$O\left ( 1\right )$$ gives$$\left ( S_{0}^{\prime }\right ) ^{2}\thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2}\tag{3}$$ Balance of $$O\left ( \varepsilon \right )$$ gives$$2S_{1}^{\prime }S_{0}^{\prime }\thicksim -S_{0}^{\prime \prime }\tag{4}$$ Equation (3) is solved ﬁrst in order to ﬁnd $$S_{0}\left ( x\right )$$. $S_{0}^{\prime }\thicksim \pm i\left ( \sin \left ( x\right ) +1\right )$ Hence\begin{align} S_{0}\left ( x\right ) & \thicksim \pm i\int _{0}^{x}\left ( \sin \left ( t\right ) +1\right ) dt+C^{\pm }\nonumber \\ & \thicksim \pm i\left ( t-\cos \left ( t\right ) \right ) _{0}^{x}+C^{\pm }\nonumber \\ & \thicksim \pm i\left ( 1+x-\cos \left ( x\right ) \right ) +C^{\pm }\tag{5} \end{align}

$$S_{1}\left ( x\right )$$ is now found from (4) and using $$S_{0}^{\prime \prime }=\pm i\cos \left ( x\right )$$ gives\begin{align*} S_{1}^{\prime } & \thicksim -\frac{1}{2}\frac{S_{0}^{\prime \prime }}{S_{0}^{\prime }}\\ & \thicksim -\frac{1}{2}\frac{\pm i\cos \left ( x\right ) }{\pm i\left ( \sin \left ( x\right ) +1\right ) }\\ & \thicksim -\frac{1}{2}\frac{\cos \left ( x\right ) }{\left ( \sin \left ( x\right ) +1\right ) } \end{align*}

Hence the solution is$$S_{1}\left ( x\right ) \thicksim -\frac{1}{2}\ln \left ( 1+\sin \left ( x\right ) \right ) \tag{6}$$ Having found $$S_{0}\left ( x\right )$$ and $$S_{1}\left ( x\right )$$, the leading behavior is now obtained from\begin{align*} y\left ( x\right ) & \thicksim \exp \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}\left ( x\right ) \right ) \\ & \thicksim \exp \left ( \frac{1}{\varepsilon }\left ( S_{0}\left ( x\right ) +\varepsilon S_{1}\left ( x\right ) \right ) +\cdots \right ) \\ & \thicksim \exp \left ( \frac{1}{\varepsilon }S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +\cdots \right ) \end{align*}

The leading behavior is only the ﬁrst two terms (called physical optics approximation in WKB), therefore\begin{align*} y\left ( x\right ) & \thicksim \exp \left ( \frac{1}{\varepsilon }S_{0}\left ( x\right ) +S_{1}\left ( x\right ) \right ) \\ & \thicksim \exp \left ( \pm \frac{i}{\varepsilon }\left ( 1+x-\cos \left ( x\right ) \right ) +C^{\pm }-\frac{1}{2}\ln \left ( 1+\sin \left ( x\right ) \right ) \right ) \\ & \thicksim \frac{1}{\sqrt{1+\sin x}}\exp \left ( \pm \frac{i}{\varepsilon }\left ( 1+x-\cos \left ( x\right ) \right ) +C^{\pm }\right ) \end{align*}

Which can be written as$y\left ( x\right ) \thicksim \frac{C}{\sqrt{1+\sin x}}\exp \left ( \frac{i}{\varepsilon }\left ( 1+x-\cos \left ( x\right ) \right ) \right ) -\frac{C}{\sqrt{1+\sin x}}\exp \left ( \frac{-i}{\varepsilon }\left ( 1+x-\cos \left ( x\right ) \right ) \right )$ In terms of $$\sin$$ and $$\cos$$ the above becomes (using the standard Euler relation simpliﬁcations)$y\left ( x\right ) \thicksim \frac{A}{\sqrt{1+\sin x}}\cos \left ( \frac{1}{\varepsilon }\left ( 1+x-\cos \left ( x\right ) \right ) \right ) +\frac{B}{\sqrt{1+\sin x}}\sin \left ( \frac{1}{\varepsilon }\left ( 1+x-\cos \left ( x\right ) \right ) \right )$ Where $$A,B$$ are the new constants. But $$\lambda =\frac{1}{\varepsilon ^{2}}$$, and the above becomes$$y\left ( x\right ) \thicksim \frac{A}{\sqrt{1+\sin x}}\cos \left ( \sqrt{\lambda }\left ( 1+x-\cos \left ( x\right ) \right ) \right ) +\frac{B}{\sqrt{1+\sin x}}\sin \left ( \sqrt{\lambda }\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \tag{7}$$ Boundary conditions are now applied to determine $$A,B$$.\begin{align*} y\left ( 0\right ) & =0\\ y\left ( \pi \right ) & =0 \end{align*}

First B.C. applied to (7) gives (where now $$\thicksim$$ is replaced by $$=$$ for notation simplicity)\begin{align*} 0 & =A\cos \left ( \sqrt{\lambda }\left ( 1-\cos \left ( 0\right ) \right ) \right ) +B\sin \left ( \sqrt{\lambda }\left ( 1-\cos \left ( 0\right ) \right ) \right ) \\ 0 & =A\cos \left ( 0\right ) +B\sin \left ( 0\right ) \\ 0 & =A \end{align*}

Hence solution (7) becomes$y\left ( x\right ) \thicksim \frac{B}{\sqrt{1+\sin x}}\sin \left ( \sqrt{\lambda }\left ( 1+x-\cos \left ( x\right ) \right ) \right )$ Applying the second B.C. $$y\left ( \pi \right ) =0$$ to the above results in\begin{align*} 0 & =\frac{B}{\sqrt{1+\sin \pi }}\sin \left ( \sqrt{\lambda }\left ( 1+\pi -\cos \left ( \pi \right ) \right ) \right ) \\ 0 & =B\sin \left ( \sqrt{\lambda }\left ( 1+\pi +1\right ) \right ) \\ & =B\sin \left ( \left ( 2+\pi \right ) \sqrt{\lambda }\right ) \end{align*}

Hence, non-trivial solution implies that\begin{align*} \left ( 2+\pi \right ) \sqrt{\lambda _{n}} & =n\pi \qquad n=1,2,3,\cdots \\ \sqrt{\lambda _{n}} & =\frac{n\pi }{2+\pi } \end{align*}

The eigenvalues are$\lambda _{n}=\frac{n^{2}\pi ^{2}}{\left ( 2+\pi \right ) ^{2}}\qquad n=1,2,3,\cdots$ Hence $$\lambda _{n}\approx n^{2}$$ for large $$n$$. The eigenfunctions are$y_{n}\left ( x\right ) \thicksim \frac{B_{n}}{\sqrt{1+\sin x}}\sin \left ( \sqrt{\lambda _{n}}\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \qquad n=1,2,3,\cdots$ The solution is therefore a linear combination of the eigenfunctions\begin{align} y\left ( x\right ) & \thicksim \sum _{n=1}^{\infty }y_{n}\left ( x\right ) \nonumber \\ & \thicksim \sum _{n=1}^{\infty }\frac{B_{n}}{\sqrt{1+\sin x}}\sin \left ( \sqrt{\lambda _{n}}\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \tag{7A} \end{align}

This solution becomes more accurate for large $$\lambda$$ or large $$n$$.

##### 5.1.3.2 Part b

For normalization, the requirement is that$\int _{0}^{\pi }y_{n}^{2}\left ( x\right ) \overset{\text{weight}}{\overbrace{\left ( \sin \left ( x\right ) +1\right ) ^{2}}}dx=1$ Substituting the eigenfunction $$y_{n}\left ( x\right )$$ solution obtained in ﬁrst part in the above results in$\int _{0}^{\pi }\left ( \frac{B_{n}}{\sqrt{1+\sin x}}\sin \left ( \sqrt{\lambda _{n}}\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \right ) ^{2}\left ( \sin \left ( x\right ) +1\right ) ^{2}dx\thicksim 1$ The above is now solved for constant $$B_{n}$$. The constant $$B_{n}$$ will the same for each $$n$$ for normalization. Therefore any $$n$$ can be used for the purpose of ﬁnding the scaling constant. Selecting $$n=1$$ in the above gives\begin{align} \int _{0}^{\pi }\left ( \frac{B}{\sqrt{1+\sin x}}\sin \left ( \frac{\pi }{2+\pi }\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \right ) ^{2}\left ( \sin \left ( x\right ) +1\right ) ^{2}dx & \thicksim 1\nonumber \\ B^{2}\int _{0}^{\pi }\frac{1}{1+\sin x}\sin ^{2}\left ( \frac{\pi }{2+\pi }\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \left ( \sin \left ( x\right ) +1\right ) ^{2}dx & \thicksim 1\nonumber \\ B^{2}\int _{0}^{\pi }\sin ^{2}\left ( \frac{\pi }{2+\pi }\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \left ( \sin \left ( x\right ) +1\right ) dx & \thicksim 1\tag{8} \end{align}

Letting $$u=\frac{\pi }{2+\pi }\left ( 1+x-\cos \left ( x\right ) \right )$$, then$\frac{du}{dx}=\frac{\pi }{2+\pi }\left ( 1+\sin \left ( x\right ) \right )$ When $$x=0$$, then $$u=\frac{\pi }{2+\pi }\left ( 1+0-\cos \left ( 0\right ) \right ) =0$$ and when $$x=\pi$$ then $$u=\frac{\pi }{2+\pi }\left ( 1+\pi -\cos \left ( \pi \right ) \right ) =\frac{\pi }{2+\pi }\left ( 2+\pi \right ) =\pi$$, hence (8) becomes\begin{align*} B^{2}\int _{0}^{\pi }\sin ^{2}\left ( u\right ) \frac{2+\pi }{\pi }\frac{du}{dx}dx & =1\\ \frac{2+\pi }{\pi }B^{2}\int _{0}^{\pi }\sin ^{2}\left ( u\right ) du & =1 \end{align*}

But $$\sin ^{2}\left ( u\right ) =\frac{1}{2}-\frac{1}{2}\cos 2u$$, therefore the above becomes\begin{align*} \frac{2+\pi }{\pi }B^{2}\int _{0}^{\pi }\left ( \frac{1}{2}-\frac{1}{2}\cos 2u\right ) du & =1\\ \frac{1}{2}\frac{2+\pi }{\pi }B^{2}\left ( u-\frac{\sin 2u}{2}\right ) _{0}^{\pi } & =1\\ \frac{2+\pi }{2\pi }B^{2}\left ( \left ( \pi -\frac{\sin 2\pi }{2}\right ) -\left ( 0-\frac{\sin 0}{2}\right ) \right ) & =1\\ \frac{2+\pi }{2\pi }B^{2}\pi & =1\\ B^{2} & =\frac{2}{2+\pi } \end{align*}

Therefore\begin{align*} B & =\sqrt{\frac{2}{\pi +2}}\\ & =0.62369 \end{align*}

Using the above for each $$B_{n}$$ in the solution obtained for the eigenfunctions in (7A), and pulling this scaling constant out of the sum results in$$y^{\text{normalized}}\thicksim \sqrt{\frac{2}{\pi +2}}\sum _{n=1}^{\infty }\frac{1}{\sqrt{1+\sin x}}\sin \left ( \sqrt{\lambda _{n}}\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \tag{9}$$ Where $\sqrt{\lambda _{n}}=\frac{n\pi }{2+\pi }\qquad n=1,2,3,\cdots$ The following are plots for the normalized $$y_{n}\left ( x\right )$$ for $$n$$ values it asks to show.

The following shows  the $$y\left ( x\right )$$ as more eigenfunctions are added up to $$55$$.

##### 5.1.3.3 Part c

Since approximate solution is\begin{align} y\left ( x\right ) & \thicksim \exp \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}\left ( x\right ) \right ) \qquad \delta \rightarrow 0\nonumber \\ & \thicksim \exp \left ( \frac{1}{\delta }S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +\delta S_{2}\left ( x\right ) +\cdots \right ) \tag{1} \end{align}

And the physical optics approximation includes the ﬁrst two terms in the series above, then the relative error between physical optics and exact solution is given by $$\delta S_{2}\left ( x\right )$$. But $$\delta =\varepsilon$$. Hence (1) becomes$y\left ( x\right ) \thicksim \exp \left ( \frac{1}{\varepsilon }S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +\varepsilon S_{2}\left ( x\right ) +\cdots \right )$ Hence the relative error must be such that$$\left \vert \varepsilon S_{2}\left ( x\right ) \right \vert _{\max }\leq 0.001\tag{1A}$$

Now $$S_{2}\left ( x\right )$$ is found. From (2) in part(a)\begin{align*} \frac{\varepsilon ^{2}}{\delta ^{2}}\left ( S_{0}^{\prime }+\delta S_{1}^{\prime }+\delta ^{2}S_{2}^{\prime }+\cdots \right ) \left ( S_{0}^{\prime }+\delta S_{1}^{\prime }+\delta ^{2}S_{2}^{\prime }+\cdots \right ) +\frac{\varepsilon ^{2}}{\delta }\left ( S_{0}^{\prime \prime }+\delta S_{1}^{\prime \prime }+\delta ^{2}S_{2}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2}\\ \frac{\varepsilon ^{2}}{\delta ^{2}}\left ( \left ( S_{0}^{\prime }\right ) ^{2}+\delta \left ( 2S_{1}^{\prime }S_{0}^{\prime }\right ) +\delta ^{2}\left ( 2S_{0}^{\prime }S_{2}^{\prime }+\left ( S_{1}^{\prime }\right ) ^{2}\right ) +\cdots \right ) +\frac{\varepsilon ^{2}}{\delta }\left ( S_{0}^{\prime \prime }+\delta S_{1}^{\prime \prime }+\delta ^{2}S_{2}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2}\\ \left ( \left ( S_{0}^{\prime }\right ) ^{2}+\varepsilon \left ( 2S_{1}^{\prime }S_{0}^{\prime }\right ) +\varepsilon ^{2}\left ( 2S_{0}^{\prime }S_{2}^{\prime }+\left ( S_{1}^{\prime }\right ) ^{2}\right ) +\cdots \right ) +\left ( \varepsilon S_{0}^{\prime \prime }+\varepsilon ^{2}S_{1}^{\prime \prime }+\varepsilon ^{3}S_{2}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( \sin \left ( x\right ) +1\right ) ^{2} \end{align*}

A balance on $$O\left ( \varepsilon ^{2}\right ) ~$$ gives the ODE to solve to ﬁnd $$S_{2}$$$$2S_{0}^{\prime }S_{2}^{\prime }\thicksim -\left ( S_{1}^{\prime }\right ) ^{2}-S_{1}^{\prime \prime }\tag{2}$$ But \begin{align*} S_{0}^{\prime } & \thicksim \pm i\left ( 1+\sin \left ( x\right ) \right ) \\ \left ( S_{1}^{\prime }\right ) ^{2} & \thicksim \left ( -\frac{1}{2}\frac{\cos \left ( x\right ) }{\left ( \sin \left ( x\right ) +1\right ) }\right ) ^{2}\\ & \thicksim \frac{1}{4}\frac{\cos ^{2}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{2}}\\ S_{1}^{\prime \prime } & \thicksim -\frac{1}{2}\frac{d}{dx}\left ( \frac{\cos \left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) }\right ) \\ & \thicksim \frac{1}{2}\left ( \frac{1}{1+\sin \left ( x\right ) }\right ) \end{align*}

Hence (2) becomes\begin{align*} 2S_{0}^{\prime }S_{2}^{\prime } & \thicksim -\left ( S_{1}^{\prime }\right ) ^{2}-S_{1}^{\prime \prime }\\ S_{2}^{\prime } & \thicksim -\frac{\left ( \left ( S_{1}^{\prime }\right ) ^{2}+S_{1}^{\prime \prime }\right ) }{2S_{0}^{\prime }}\\ & \thicksim -\frac{\left ( \frac{1}{4}\frac{\cos ^{2}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{2}}+\frac{1}{2}\left ( \frac{1}{1+\sin \left ( x\right ) }\right ) \right ) }{\pm 2i\left ( 1+\sin \left ( x\right ) \right ) }\\ & \thicksim \pm \frac{i\left ( \frac{1}{4}\frac{\cos ^{2}\left ( x\right ) }{\left ( \sin \left ( x\right ) +1\right ) ^{2}}+\frac{1}{2}\left ( \frac{1}{1+\sin \left ( x\right ) }\right ) \right ) }{2\left ( \sin \left ( x\right ) +1\right ) }\\ & \thicksim \pm \frac{i\frac{1}{4}\left ( \frac{\cos ^{2}\left ( x\right ) +2\left ( 1+\sin \left ( x\right ) \right ) }{\left ( \sin \left ( x\right ) +1\right ) ^{2}}\right ) }{2\left ( \sin \left ( x\right ) +1\right ) }\\ & \thicksim \pm \frac{i}{8}\frac{\cos ^{2}\left ( x\right ) +2\left ( 1+\sin \left ( x\right ) \right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{3}} \end{align*}

Therefore\begin{align} S_{2}\left ( x\right ) & \thicksim \pm \frac{i}{8}\int _{0}^{x}\frac{\cos ^{2}\left ( t\right ) +2\left ( 1+\sin \left ( t\right ) \right ) }{\left ( 1+\sin \left ( t\right ) \right ) ^{3}}dt\nonumber \\ & \thicksim \pm \frac{i}{8}\left ( \int _{0}^{x}\frac{\cos ^{2}\left ( t\right ) }{\left ( 1+\sin \left ( t\right ) \right ) ^{3}}dt+2\int _{0}^{x}\frac{1}{\left ( 1+\sin \left ( t\right ) \right ) ^{2}}dt\right ) \tag{3} \end{align}

To do $$\int _{0}^{x}\frac{\cos ^{2}\left ( t\right ) }{\left ( 1+\sin \left ( t\right ) \right ) ^{3}}dt$$, I used a lookup integration rule from tables which says $$\int \cos ^{p}\left ( t\right ) \left ( a+\sin t\right ) ^{m}dt=\frac{1}{\left ( a\right ) \left ( m\right ) }\cos ^{p+1}\left ( t\right ) \left ( a+\sin t\right ) ^{m}$$, therefore using this rule the integral becomes, where now $$m=-3,p=2,a=1$$,\begin{align*} \int _{0}^{x}\frac{\cos ^{2}t}{\left ( 1+\sin t\right ) ^{3}}dt & =\frac{1}{-3}\left ( \frac{\cos ^{3}t}{\left ( 1+\sin t\right ) ^{3}}\right ) _{0}^{x}\\ & =\frac{1}{-3}\left ( \frac{\cos ^{3}x}{\left ( 1+\sin x\right ) ^{3}}-1\right ) \\ & =\frac{1}{3}\left ( 1-\frac{\cos ^{3}x}{\left ( 1+\sin x\right ) ^{3}}\right ) \end{align*}

And for $$\int \frac{1}{\left ( 1+\sin \left ( x\right ) \right ) ^{2}}dx$$, half angle substitution can be used. I do not know what other substitution to use. Using CAS for little help on this, I get \begin{align*} \int _{0}^{x}\frac{1}{\left ( 1+\sin t\right ) ^{2}}dt & =\left ( -\frac{\cos t}{3\left ( 1+\sin t\right ) ^{2}}-\frac{1}{3}\frac{\cos t}{1+\sin t}\right ) _{0}^{x}\\ & =\left ( -\frac{\cos x}{3\left ( 1+\sin x\right ) ^{2}}-\frac{1}{3}\frac{\cos x}{1+\sin x}\right ) -\left ( -\frac{1}{3}-\frac{1}{3}\right ) \\ & =\frac{2}{3}-\frac{\cos x}{3\left ( 1+\sin x\right ) ^{2}}-\frac{1}{3}\frac{\cos x}{1+\sin x} \end{align*}

Hence from (3)\begin{align*} S_{2}\left ( x\right ) & \thicksim \pm \frac{i}{8}\left ( \frac{1}{3}\left ( 1-\frac{\cos ^{3}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{3}}\right ) +2\left ( \frac{2}{3}-\frac{\cos x}{3\left ( 1+\sin x\right ) ^{2}}-\frac{1}{3}\frac{\cos x}{1+\sin x}\right ) \right ) \\ & \thicksim \pm \frac{i}{8}\left ( \frac{1}{3}-\frac{1}{3}\frac{\cos ^{3}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{3}}+\frac{4}{3}-\frac{2\cos x}{3\left ( 1+\sin x\right ) ^{2}}-\frac{2}{3}\frac{\cos x}{1+\sin x}\right ) \\ & \thicksim \pm \frac{i}{24}\left ( 1-\frac{\cos ^{3}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{3}}+4-\frac{2\cos x}{\left ( 1+\sin x\right ) ^{2}}-\frac{2\cos x}{1+\sin x}\right ) \end{align*}

Therefore, from (1A)\begin{align} \left \vert \varepsilon S_{2}\left ( x\right ) \right \vert _{\max } & \leq 0.001\nonumber \\ \left \vert \varepsilon \frac{i}{24}\left ( 1-\frac{\cos ^{3}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{3}}+4-\frac{2\cos x}{\left ( 1+\sin x\right ) ^{2}}-\frac{2\cos x}{1+\sin x}\right ) \right \vert _{\max } & \leq 0.001\nonumber \\ \frac{1}{24}\left \vert \varepsilon \left ( 1-\frac{\cos ^{3}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{3}}+4-\frac{2\cos x}{\left ( 1+\sin x\right ) ^{2}}-\frac{2\cos x}{1+\sin x}\right ) \right \vert _{\max } & \leq 0.001\nonumber \\ \left \vert \varepsilon \left ( 1-\frac{\cos ^{3}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{3}}+4-\frac{2\cos x}{\left ( 1+\sin x\right ) ^{2}}-\frac{2\cos x}{1+\sin x}\right ) \right \vert _{\max } & \leq 0.024\tag{2} \end{align}

The maximum value of $$\left ( 1-\frac{\cos ^{3}\left ( x\right ) }{\left ( 1+\sin \left ( x\right ) \right ) ^{3}}+4-\frac{2\cos x}{\left ( 1+\sin x\right ) ^{2}}-\frac{2\cos x}{1+\sin x}\right )$$ between $$x=0$$ and $$x=\pi$$ is now found and used to ﬁnd $$\varepsilon$$. A plot of the above shows the maximum is maximum at the end, at $$x=\pi$$ (Taking the derivative and setting it to zero to determine where the maximum is can also be used).

Therefore, at $$x=\pi$$ \begin{align*} \left ( 1-\frac{\cos ^{3}x}{\left ( 1+\sin x\right ) ^{3}}+4-\frac{2\cos x}{\left ( 1+\sin x\right ) ^{2}}-\frac{2\cos x}{1+\sin x}\right ) _{x=\pi } & =\left ( 1-\frac{\cos ^{3}\left ( \pi \right ) }{\left ( 1+\sin \pi \right ) ^{3}}+4-\frac{2\cos \pi }{\left ( 1+\sin \pi \right ) ^{2}}-\frac{2\cos \pi }{1+\sin \pi }\right ) \\ & =10 \end{align*}

Hence (2) becomes\begin{align*} 10\varepsilon & \leq 0.024\\ \varepsilon & \leq 0.0024 \end{align*}

But since $$\lambda =\frac{1}{\varepsilon ^{2}}$$ the above becomes\begin{align*} \frac{1}{\sqrt{\lambda }} & \leq 0.0024\\ \sqrt{\lambda } & \geq \frac{1}{0.0024}\\ \sqrt{\lambda } & \geq 416.67 \end{align*}

Hence $\fbox{\lambda \geq 17351.1}$ To ﬁnd which mode this corresponds to, since $$\lambda _{n}=\frac{n^{2}\pi ^{2}}{\left ( 2+\pi \right ) ^{2}}$$, then need to solve for $$n$$\begin{align*} 17351.1 & =\frac{n^{2}\pi ^{2}}{\left ( 2+\pi \right ) ^{2}}\\ n^{2}\pi ^{2} & =\left ( 17351.1\right ) \left ( 2+\pi \right ) ^{2}\\ n & =\sqrt{\frac{\left ( 17351.1\right ) \left ( 2+\pi \right ) ^{2}}{\pi ^{2}}}\\ & =215.58 \end{align*}

Hence the next largest integer is used$\fbox{n=216}$ To have relative error less than $$0.1\%$$ compared to exact solution. Therefore using the result obtained in (9) in part (b) the normalized solution needed is$$y^{\text{normalized}}\thicksim \sqrt{\frac{2}{\pi +2}}\sum _{n=1}^{216}\frac{1}{\sqrt{1+\sin x}}\sin \left ( \frac{n\pi }{2+\pi }\left ( 1+x-\cos \left ( x\right ) \right ) \right ) \nonumber$$

The following is a plot of the above solution adding all the ﬁrst $$216$$ modes for illustration.