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3.4 HW4

  3.4.1 problem 10.5 (page 540)
  3.4.2 problem 10.6
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3.4.1 problem 10.5 (page 540)

problem Use WKB to obtain solution to \begin{equation} \varepsilon y^{\prime \prime }+a\left ( x\right ) y^{\prime }+b\left ( x\right ) y=0 \tag{1} \end{equation} with a\left ( x\right ) >0,y\left ( 0\right ) =A,y\left ( 1\right ) =B correct to order \varepsilon .

solution

Assuming 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\begin{align*} y^{\prime }\left ( x\right ) & \thicksim \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) \exp \left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}\left ( x\right ) \right ) \\ y^{\prime \prime }\left ( x\right ) & \thicksim \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime \prime }\left ( x\right ) \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 ) \end{align*}

Substituting the above into (1) and simplifying gives (writing = instead of \thicksim for simplicity for now)\begin{align*} \varepsilon \left [ \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime \prime }\left ( x\right ) +\left ( \frac{1}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) \right ) ^{2}\right ] +\frac{a}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) +b & =0\\ \frac{\varepsilon }{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime \prime }\left ( x\right ) +\frac{\varepsilon }{\delta ^{2}}\left ( \sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) \sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) \right ) +\frac{a}{\delta }\sum _{n=0}^{\infty }\delta ^{n}S_{n}^{\prime }\left ( x\right ) +b & =0 \end{align*}

Expanding gives\begin{multline*} \frac{\varepsilon }{\delta }\left ( S_{0}^{\prime \prime }+\delta S_{1}^{\prime \prime }+\delta ^{2}S_{2}^{\prime \prime }+\cdots \right ) \\ +\frac{\varepsilon }{\delta ^{2}}\left ( \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 ) \right ) \\ +\frac{a}{\delta }\left ( S_{0}^{\prime }+\delta S_{1}^{\prime }+\delta ^{2}S_{2}^{\prime }+\cdots \right ) +b=0 \end{multline*} Simplifying\begin{multline} \left ( \frac{\varepsilon }{\delta }S_{0}^{\prime \prime }+\varepsilon S_{1}^{\prime \prime }+\varepsilon \delta S_{2}^{\prime \prime }+\cdots \right ) \nonumber \\ +\left ( \frac{\varepsilon }{\delta ^{2}}\left ( S_{0}^{\prime }\right ) ^{2}+\frac{2\varepsilon }{\delta }\left ( S_{0}^{\prime }S_{1}^{\prime }\right ) +\varepsilon \left ( 2S_{0}^{\prime }S_{2}^{\prime }+\left ( S_{1}^{\prime }\right ) ^{2}\right ) +\cdots \right ) \nonumber \\ +\left ( \frac{a}{\delta }S_{0}^{\prime }+aS_{1}^{\prime }+a\delta S_{2}^{\prime }+\cdots \right ) +b=0\tag{1A} \end{multline} The largest terms in the left are \frac{\varepsilon }{\delta ^{2}}\left ( S_{0}^{\prime }\right ) ^{2} and \frac{a}{\delta }S_{0}^{\prime }. By dominant balance these must be equal in magnitude. Hence \frac{\varepsilon }{\delta ^{2}}=O\left ( \frac{1}{\delta }\right ) or \frac{\varepsilon }{\delta }=O\left ( 1\right ) . Therefore \delta is proportional to \varepsilon and for simplicity \varepsilon is taken as equal to \delta , hence (1A) becomes\begin{multline*} \left ( S_{0}^{\prime \prime }+\varepsilon S_{1}^{\prime \prime }+\varepsilon ^{2}S_{2}^{\prime \prime }+\cdots \right ) \\ +\left ( \varepsilon ^{-1}\left ( S_{0}^{\prime }\right ) ^{2}+2S_{0}^{\prime }S_{1}^{\prime }+\varepsilon \left ( 2S_{0}^{\prime }S_{2}^{\prime }+\left ( S_{1}^{\prime }\right ) ^{2}\right ) +\cdots \right ) \\ +\left ( a\varepsilon ^{-1}S_{0}^{\prime }+aS_{1}^{\prime }+a\varepsilon S_{2}^{\prime }+\cdots \right ) +b=0 \end{multline*} Terms of O\left ( \varepsilon ^{-1}\right ) give\begin{equation} \left ( S_{0}^{\prime }\right ) ^{2}+aS_{0}^{\prime }=0\tag{2} \end{equation} And terms of O\left ( 1\right ) give\begin{equation} S_{0}^{\prime \prime }+2S_{0}^{\prime }S_{1}^{\prime }+aS_{1}^{\prime }+b=0\tag{3} \end{equation} And terms of O\left ( \varepsilon \right ) give\begin{align} 2S_{0}^{\prime }S_{2}^{\prime }+aS_{2}^{\prime }+\left ( S_{1}^{\prime }\right ) ^{2}+S_{1}^{\prime \prime } & =0\nonumber \\ S_{2}^{\prime } & =-\frac{\left ( S_{1}^{\prime }\right ) ^{2}+S_{1}^{\prime \prime }}{\left ( a+2S_{0}^{\prime }\right ) }\tag{4} \end{align}

Starting with (2) S_{0}^{\prime }\left ( S_{0}^{\prime }+a\right ) =0 There are two cases to consider.

case 1 S_{0}^{\prime }=0. This means that S_{0}\left ( x\right ) =c_{0}. A constant. Using this result in (3) gives an ODE to solve for S_{1}\left ( x\right ) \begin{align*} aS_{1}^{\prime }+b & =0\\ S_{1}^{\prime } & =-\frac{b\left ( x\right ) }{a\left ( x\right ) }\\ S_{1} & \thicksim -\int _{0}^{x}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt+c_{1} \end{align*}

Using this result in (4) gives an ODE to solve for S_{2}\left ( x\right ) \begin{align*} S_{2}^{\prime } & =-\frac{\left ( -\frac{b\left ( x\right ) }{a\left ( x\right ) }\right ) ^{2}+\left ( -\frac{b\left ( x\right ) }{a\left ( x\right ) }\right ) ^{\prime }}{a\left ( x\right ) }\\ & =-\frac{\frac{b^{2}\left ( x\right ) }{a^{2}\left ( x\right ) }-\left ( \frac{b^{\prime }\left ( x\right ) }{a\left ( x\right ) }-\frac{b\left ( x\right ) a^{\prime }\left ( x\right ) }{a^{2}\left ( x\right ) }\right ) }{a\left ( x\right ) }\\ & =-\frac{\frac{b^{2}\left ( x\right ) }{a^{2}\left ( x\right ) }-\frac{a\left ( x\right ) b^{\prime }\left ( x\right ) }{a^{2}\left ( x\right ) }+\frac{a^{\prime }\left ( x\right ) b\left ( x\right ) }{a^{2}\left ( x\right ) }}{a\left ( x\right ) }\\ & =\frac{a\left ( x\right ) b^{\prime }\left ( x\right ) -b^{2}\left ( x\right ) -a^{\prime }\left ( x\right ) b\left ( x\right ) }{a^{3}\left ( x\right ) } \end{align*}

Therefore S_{2}=\int _{0}^{x}\frac{a\left ( t\right ) b^{\prime }\left ( t\right ) -b^{2}\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt+c_{2} For case one, the solution becomes\begin{align} y_{1}\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}{\varepsilon }\left ( S_{0}\left ( x\right ) +\varepsilon S_{1}\left ( x\right ) +\varepsilon ^{2}S_{2}\left ( x\right ) \right ) \right ) \qquad \varepsilon \rightarrow 0^{+}\nonumber \\ & \thicksim \exp \left ( \frac{1}{\varepsilon }S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +\varepsilon S_{2}\left ( x\right ) \right ) \nonumber \\ & \thicksim \exp \left ( \frac{1}{\varepsilon }c_{0}-\int _{0}^{x}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt+c_{1}+\varepsilon \int _{0}^{x}\frac{a\left ( t\right ) b^{\prime }\left ( t\right ) -b^{2}\left ( t\right ) -b\left ( t\right ) }{a^{3}\left ( t\right ) }dt+c_{2}\right ) \nonumber \\ & \thicksim C_{1}\exp \left ( -\int _{0}^{x}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt+\varepsilon \int _{0}^{x}\frac{a\left ( t\right ) b^{\prime }\left ( t\right ) -b^{2}\left ( t\right ) -b\left ( t\right ) }{a^{3}\left ( t\right ) }dt\right ) \tag{5} \end{align}

Where C_{1} is a constant that combines all e^{\frac{1}{\varepsilon }c_{0}+c_{1}+c_{2}} constants into one. Equation (5) gives the first WKB solution of order O\left ( \varepsilon \right ) for case one. Case 2 now is considered.

case 2 In this case S_{0}^{\prime }=-a, therefore S_{0}=-\int _{0}^{x}a\left ( t\right ) dt+c_{0} Equation (3) now gives

\begin{align*} S_{0}^{\prime \prime }+2S_{0}^{\prime }S_{1}^{\prime }+aS_{1}^{\prime }+b & =0\\ -a^{\prime }-2aS_{1}^{\prime }+aS_{1}^{\prime }+b & =0\\ -aS_{1}^{\prime } & =a^{\prime }-b\\ S_{1}^{\prime } & =\frac{b-a^{\prime }}{a}\\ S_{1}^{\prime } & =\frac{b}{a}-\frac{a^{\prime }}{a} \end{align*}

Integrating the above results in S_{1}=\int _{0}^{x}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt-\ln \left ( a\right ) +c_{1} S_{2}\left ( x\right ) is now found from (4)\begin{align*} S_{2}^{\prime } & =-\frac{\left ( S_{1}^{\prime }\right ) ^{2}+S_{1}^{\prime \prime }}{\left ( a+2S_{0}^{\prime }\right ) }\\ & =-\frac{\left ( \frac{b-a^{\prime }}{a}\right ) ^{2}+\left ( \frac{b-a^{\prime }}{a}\right ) ^{\prime }}{\left ( a+2\left ( -a\right ) \right ) }\\ & =-\frac{\frac{b^{2}+\left ( a^{\prime }\right ) ^{2}-2ba^{\prime }}{a^{2}}+\frac{b^{\prime }-a^{\prime \prime }}{a}-\frac{a^{\prime }b-\left ( a^{\prime }\right ) ^{2}}{a^{2}}}{-a}\\ & =\frac{b^{2}+\left ( a^{\prime }\right ) ^{2}-2ba^{\prime }+ab^{\prime }-aa^{\prime \prime }-a^{\prime }b-\left ( a^{\prime }\right ) ^{2}}{a^{3}}\\ & =\frac{b^{2}-2ba^{\prime }+ab^{\prime }-aa^{\prime \prime }-a^{\prime }b}{a^{3}} \end{align*}

Hence S_{2}=\int _{0}^{x}\frac{b^{2}\left ( t\right ) -2b\left ( t\right ) a^{\prime }\left ( t\right ) +a\left ( t\right ) b^{\prime }\left ( t\right ) -a\left ( t\right ) a^{\prime \prime }\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt+c_{2} Therefore for this case the solution becomes\begin{align} y_{2}\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}{\varepsilon }\left ( S_{0}\left ( x\right ) +\varepsilon S_{1}\left ( x\right ) +\varepsilon ^{2}S_{2}\left ( x\right ) \right ) \right ) \qquad \varepsilon \rightarrow 0^{+}\nonumber \\ & \thicksim \exp \left ( \frac{1}{\varepsilon }S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +\varepsilon S_{2}\left ( x\right ) \right ) \nonumber \end{align}

Or

Which simplifies to\begin{equation} y_{2}\left ( x\right ) \thicksim \frac{C_{2}}{a}\exp \left ( \frac{-1}{\varepsilon }\int _{0}^{x}a\left ( t\right ) dt+\int _{0}^{x}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt+\varepsilon \int _{0}^{x}\frac{b^{2}\left ( t\right ) -2b\left ( t\right ) a^{\prime }\left ( t\right ) +a\left ( t\right ) b^{\prime }\left ( t\right ) -a\left ( t\right ) a^{\prime \prime }\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt\right ) \tag{6} \end{equation} Where C_{2} is new constant that combines c_{0},c_{1},c_{2} constants. The general solution is linear combinations of y_{1},y_{2} y\left ( x\right ) \thicksim Ay_{1}\left ( x\right ) +By_{2}\left ( x\right ) Or
Now boundary conditions are applied to find C_{1},C_{2}. Using y\left ( 0\right ) =A in the above gives\begin{equation} A=C_{1}+\frac{C_{2}}{a\left ( 0\right ) } \tag{7} \end{equation} And using y\left ( 1\right ) =B gives\begin{multline*} B=C_{1}\exp \left ( -\int _{0}^{1}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt+\varepsilon \int _{0}^{1}\frac{a\left ( t\right ) b^{\prime }\left ( t\right ) -b^{2}\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt\right ) \\ +\frac{C_{2}}{a\left ( 1\right ) }\exp \left ( \frac{-1}{\varepsilon }\int _{0}^{1}a\left ( t\right ) dt+\int _{0}^{1}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt+\varepsilon \int _{0}^{1}\frac{b^{2}\left ( t\right ) -2b\left ( t\right ) a^{\prime }\left ( t\right ) +a\left ( t\right ) b^{\prime }\left ( t\right ) -a\left ( t\right ) a^{\prime \prime }\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt\right ) \end{multline*} Neglecting exponentially small terms involving e^{\frac{-1}{\varepsilon }} the above becomes\begin{multline} B=C_{1}\exp \left ( -\int _{0}^{1}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt\right ) \exp \left ( \varepsilon \int _{0}^{1}\frac{a\left ( t\right ) b^{\prime }\left ( t\right ) -b^{2}\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt\right ) \nonumber \\ +\frac{C_{2}}{a\left ( 1\right ) }\exp \left ( \int _{0}^{1}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt\right ) \exp \left ( \varepsilon \int _{0}^{1}\frac{b^{2}\left ( t\right ) -2b\left ( t\right ) a^{\prime }\left ( t\right ) +a\left ( t\right ) b^{\prime }\left ( t\right ) -a\left ( t\right ) a^{\prime \prime }\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt\right ) \tag{8} \end{multline} To simplify the rest of the solution which finds C_{1},C_{2}, let \begin{align*} z_{1} & =\int _{0}^{1}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt\\ z_{2} & =\int _{0}^{1}\frac{a\left ( t\right ) b^{\prime }\left ( t\right ) -b^{2}\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt\\ z_{3} & =\int _{0}^{1}\frac{b^{2}\left ( t\right ) -2b\left ( t\right ) a^{\prime }\left ( t\right ) +a\left ( t\right ) b^{\prime }\left ( t\right ) -a\left ( t\right ) a^{\prime \prime }\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt \end{align*}

Hence (8) becomes\begin{equation} B=C_{1}e^{-z_{1}}e^{\varepsilon z_{2}}+\frac{C_{2}}{a\left ( 1\right ) }e^{z_{1}}e^{\varepsilon z_{3}} \tag{8A} \end{equation} From (7) C_{2}=a\left ( 0\right ) \left ( A-C_{1}\right ) . Substituting this in (8A) gives\begin{align} B & =C_{1}e^{-z_{1}}e^{\varepsilon z_{2}}+\frac{a\left ( 0\right ) \left ( A-C_{1}\right ) }{a\left ( 1\right ) }e^{z_{1}}e^{\varepsilon z_{3}}\nonumber \\ & =C_{1}e^{-z_{1}}e^{\varepsilon z_{2}}+\frac{a\left ( 0\right ) }{a\left ( 1\right ) }Ae^{z_{1}}e^{\varepsilon z_{3}}-\frac{a\left ( 0\right ) }{a\left ( 1\right ) }C_{1}e^{z_{1}}e^{\varepsilon z_{3}}\nonumber \\ B & =C_{1}\left ( e^{-z_{1}}e^{\varepsilon z_{2}}-\frac{a\left ( 0\right ) }{a\left ( 1\right ) }e^{z_{1}}e^{\varepsilon z_{3}}\right ) +A\frac{a\left ( 0\right ) }{a\left ( 1\right ) }e^{z_{1}}e^{\varepsilon z_{3}}\nonumber \\ C_{1} & =\frac{B-A\frac{a\left ( 0\right ) }{a\left ( 1\right ) }e^{z_{1}}e^{\varepsilon z_{3}}}{e^{-z_{1}}e^{\varepsilon z_{2}}-\frac{a\left ( 0\right ) }{a\left ( 1\right ) }e^{z_{1}}e^{\varepsilon z_{3}}}\nonumber \\ & =\frac{a\left ( 1\right ) B-Aa\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}{a\left ( 1\right ) e^{-z_{1}}e^{\varepsilon z_{2}}-a\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}} \tag{9} \end{align}

Using (7), now C_{2} is found\begin{align} A & =C_{1}+\frac{C_{2}}{a\left ( 0\right ) }\nonumber \\ A & =\frac{a\left ( 1\right ) B-Aa\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}{a\left ( 1\right ) e^{-z_{1}}e^{\varepsilon z_{2}}-a\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}+\frac{C_{2}}{a\left ( 0\right ) }\nonumber \\ C_{2} & =a\left ( 0\right ) \left ( A-\frac{a\left ( 1\right ) B-Aa\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}{a\left ( 1\right ) e^{-z_{1}}e^{\varepsilon z_{2}}-a\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}\right ) \tag{10} \end{align}

The constants C_{1},C_{2}, are now found, hence the solution is now complete.

Summary of solution

Where\begin{align*} C_{1} & =\frac{a\left ( 1\right ) B-Aa\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}{a\left ( 1\right ) e^{-z_{1}}e^{\varepsilon z_{2}}-a\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}\\ C_{2} & =a\left ( 0\right ) \left ( A-\frac{a\left ( 1\right ) B-Aa\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}{a\left ( 1\right ) e^{-z_{1}}e^{\varepsilon z_{2}}-a\left ( 0\right ) e^{z_{1}}e^{\varepsilon z_{3}}}\right ) \end{align*}

And\begin{align*} z_{1} & =\int _{0}^{1}\frac{b\left ( t\right ) }{a\left ( t\right ) }dt\\ z_{2} & =\int _{0}^{1}\frac{a\left ( t\right ) b^{\prime }\left ( t\right ) -b^{2}\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt\\ z_{3} & =\int _{0}^{1}\frac{b^{2}\left ( t\right ) -2b\left ( t\right ) a^{\prime }\left ( t\right ) +a\left ( t\right ) b^{\prime }\left ( t\right ) -a\left ( t\right ) a^{\prime \prime }\left ( t\right ) -a^{\prime }\left ( t\right ) b\left ( t\right ) }{a^{3}\left ( t\right ) }dt \end{align*}

3.4.2 problem 10.6

problem Use second order WKB to derive formula which is more accurate than (10.1.31) for the n^{th} eigenvalue of the Sturm-Liouville problem in 10.1.27. Let Q\left ( x\right ) =\left ( x+\pi \right ) ^{4} and compare your formula with value of E_{n} in table 10.1

solution

Problem 10.1.27 is

y^{\prime \prime }+EQ\left ( x\right ) y=0 With Q\left ( x\right ) =\left ( x+\pi \right ) ^{4} and boundary conditions y\left ( 0\right ) =0,y\left ( \pi \right ) =0. Letting E=\frac{1}{\varepsilon } Then the ODE becomes\begin{equation} \varepsilon y^{\prime \prime }\left ( x\right ) +\left ( x+\pi \right ) ^{4}y\left ( x\right ) =0\tag{1} \end{equation} Physical optics approximation is obtained when \lambda \rightarrow \infty or \varepsilon \rightarrow 0^{+}. Since the ODE is linear, and the highest derivative is now multiplied by a very small parameter \varepsilon , WKB can be used to solve it. Assuming the solution is 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 Then\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 \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 ( x+\pi \right ) ^{4}\nonumber \\ \frac{\varepsilon }{\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 }{\delta }\left ( S_{0}^{\prime \prime }+\delta S_{1}^{\prime \prime }+\delta ^{2}S_{2}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( x+\pi \right ) ^{4}\nonumber \\ \frac{\varepsilon }{\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_{0}^{\prime }\right ) ^{2}\right ) +\cdots \right ) +\frac{\varepsilon }{\delta }\left ( S_{0}^{\prime \prime }+\delta S_{1}^{\prime \prime }+\delta ^{2}S_{2}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( x+\pi \right ) ^{4}\nonumber \\ \left ( \frac{\varepsilon }{\delta ^{2}}\left ( S_{0}^{\prime }\right ) ^{2}+\frac{\varepsilon }{\delta }\left ( 2S_{1}^{\prime }S_{0}^{\prime }\right ) +\varepsilon \left ( 2S_{0}^{\prime }S_{2}^{\prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) +\cdots \right ) +\left ( \frac{\varepsilon }{\delta }S_{0}^{\prime \prime }+\varepsilon ^{2}S_{1}^{\prime \prime }+\cdots \right ) & \thicksim -\left ( x+\pi \right ) ^{4}\tag{2} \end{align}

The largest term in the left side is \frac{\varepsilon }{\delta ^{2}}\left ( S_{0}^{\prime }\right ) ^{2}. By dominant balance, this term has the same order of magnitude as right side -\left ( x+\pi \right ) ^{4}. Hence \delta ^{2} is proportional to \varepsilon and for simplicity, \delta can be taken equal to \sqrt{\varepsilon } or \delta =\sqrt{\varepsilon } Equation (2) becomes \left ( \left ( S_{0}^{\prime }\right ) ^{2}+\sqrt{\varepsilon }\left ( 2S_{1}^{\prime }S_{0}^{\prime }\right ) +\varepsilon \left ( 2S_{0}^{\prime }S_{2}^{\prime }+\left ( S_{0}^{\prime }\right ) ^{2}\right ) +\cdots \right ) +\left ( \sqrt{\varepsilon }S_{0}^{\prime \prime }+\varepsilon S_{1}^{\prime \prime }+\cdots \right ) \thicksim -\left ( x+\pi \right ) ^{4} Balance of O\left ( 1\right ) gives\begin{equation} \left ( S_{0}^{\prime }\right ) ^{2}\thicksim -\left ( x+\pi \right ) ^{4}\tag{3} \end{equation} And Balance of O\left ( \sqrt{\varepsilon }\right ) gives\begin{equation} 2S_{1}^{\prime }S_{0}^{\prime }\thicksim -S_{0}^{\prime \prime }\tag{4} \end{equation} Balance of O\left ( \varepsilon \right ) gives\begin{equation} 2S_{0}^{\prime }S_{2}^{\prime }+\left ( S_{1}^{\prime }\right ) ^{2}\thicksim -S_{1}^{\prime \prime }\tag{5} \end{equation} Equation (3) is solved first in order to find S_{0}\left ( x\right ) . Therefore S_{0}^{\prime }\thicksim \pm i\left ( x+\pi \right ) ^{2} Hence\begin{align} S_{0}\left ( x\right ) & \thicksim \pm i\int _{0}^{x}\left ( t+\pi \right ) ^{2}dt+C^{\pm }\nonumber \\ & \thicksim \pm i\left ( \frac{t^{3}}{3}+\pi t^{2}+\pi ^{2}t\right ) _{0}^{x}+C^{\pm }\nonumber \\ & \thicksim \pm i\left ( \frac{x^{3}}{3}+\pi x^{2}+\pi ^{2}x\right ) +C^{\pm }\tag{6} \end{align}

S_{1}\left ( x\right ) is now found from (4), and since S_{0}^{\prime \prime }=\pm 2i\left ( x+\pi \right ) therefore\begin{align*} S_{1}^{\prime } & \thicksim -\frac{1}{2}\frac{S_{0}^{\prime \prime }}{S_{0}^{\prime }}\\ & \thicksim -\frac{1}{2}\frac{\pm 2i\left ( x+\pi \right ) }{\pm i\left ( x+\pi \right ) ^{2}}\\ & \thicksim -\frac{1}{x+\pi } \end{align*}

Hence\begin{align} S_{1}\left ( x\right ) & \thicksim -\int _{0}^{x}\frac{1}{t+\pi }dt\nonumber \\ & \thicksim -\ln \left ( \frac{\pi +x}{\pi }\right ) \tag{7} \end{align}

S_{2}\left ( x\right ) is now solved from from (5)\begin{align*} 2S_{0}^{\prime }S_{2}^{\prime }+\left ( S_{1}^{\prime }\right ) ^{2} & \thicksim -S_{1}^{\prime \prime }\\ S_{2}^{\prime } & \thicksim \frac{-S_{1}^{\prime \prime }-\left ( S_{1}^{\prime }\right ) ^{2}}{2S_{0}^{\prime }} \end{align*}

Since S_{1}^{\prime }\thicksim -\frac{1}{x+\pi }, then S_{1}^{\prime \prime }\thicksim \frac{1}{\left ( x+\pi \right ) ^{2}} and the above becomes\begin{align*} S_{2}^{\prime } & \thicksim \frac{-\frac{1}{\left ( x+\pi \right ) ^{2}}-\left ( -\frac{1}{x+\pi }\right ) ^{2}}{2\left ( \pm i\left ( x+\pi \right ) ^{2}\right ) }\\ & \thicksim \frac{-\frac{1}{\left ( x+\pi \right ) ^{2}}-\frac{1}{\left ( x+\pi \right ) ^{2}}}{\pm 2i\left ( x+\pi \right ) ^{2}}\\ & \thicksim \pm i\frac{1}{\left ( x+\pi \right ) ^{4}} \end{align*}

Hence\begin{align*} S_{2} & \thicksim \pm i\left ( \int _{0}^{x}\frac{1}{\left ( t+\pi \right ) ^{4}}dt\right ) +k^{\pm }\\ & \thicksim \pm \frac{i}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\left ( \pi +x\right ) ^{3}}\right ) +k^{\pm } \end{align*}

Therefore the solution becomes\begin{align*} y\left ( x\right ) & \thicksim \exp \left ( \frac{1}{\sqrt{\varepsilon }}\left ( S_{0}\left ( x\right ) +\sqrt{\varepsilon }S_{1}\left ( x\right ) +\varepsilon S_{2}\left ( x\right ) \right ) \right ) \\ & \thicksim \exp \left ( \frac{1}{\sqrt{\varepsilon }}S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +\sqrt{\varepsilon }S_{2}\left ( x\right ) \right ) \end{align*}

But E=\frac{1}{\varepsilon }, hence \sqrt{\varepsilon }=\frac{1}{\sqrt{E}}, and the above becomes y\left ( x\right ) \thicksim \exp \left ( \sqrt{E}S_{0}\left ( x\right ) +S_{1}\left ( x\right ) +\frac{1}{\sqrt{E}}S_{2}\left ( x\right ) \right ) Therefore y\left ( x\right ) \thicksim \exp \left ( \pm i\sqrt{E}\left ( \frac{x^{3}}{3}+\pi x^{2}+\pi ^{2}x\right ) +C^{\pm }-\ln \left ( \frac{\pi +x}{\pi }\right ) \pm i\frac{1}{\sqrt{E}}\frac{1}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\left ( \pi +x\right ) ^{3}}\right ) +k^{\pm }\right ) Which can be written as\begin{multline*} y\left ( x\right ) \thicksim \left ( \frac{\pi +x}{\pi }\right ) ^{-1}C\exp \left ( i\left ( \sqrt{E}\left ( \frac{x^{3}}{3}+\pi x^{2}+\pi ^{2}x\right ) +\frac{1}{\sqrt{E}}\frac{1}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\left ( \pi +x\right ) ^{3}}\right ) \right ) \right ) \\ -C\left ( \frac{\pi +x}{\pi }\right ) ^{-1}\exp \left ( -i\left ( \sqrt{E}\left ( \frac{x^{3}}{3}+\pi x^{2}+\pi ^{2}x\right ) +\frac{1}{\sqrt{E}}\frac{1}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\left ( \pi +x\right ) ^{3}}\right ) \right ) \right ) \end{multline*} Where all constants combined into \pm C.  In terms of \sin /\cos the above becomes\begin{align} y\left ( x\right ) & \thicksim \frac{\pi A}{\pi +x}\cos \left ( \sqrt{E}\left ( \frac{x^{3}}{3}+\pi x^{2}+\pi ^{2}x\right ) +\frac{1}{\sqrt{E}}\frac{1}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\left ( \pi +x\right ) ^{3}}\right ) \right ) \nonumber \\ & +\frac{\pi B}{\pi +x}\sin \left ( \sqrt{E}\left ( \frac{x^{3}}{3}+\pi x^{2}+\pi ^{2}x\right ) +\frac{1}{\sqrt{E}}\frac{1}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\left ( \pi +x\right ) ^{3}}\right ) \right ) \tag{8} \end{align}

Boundary conditions y\left ( 0\right ) =0 gives\begin{align*} 0 & \thicksim A\cos \left ( 0+\frac{1}{\sqrt{E}}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\pi ^{3}}\right ) \right ) +B\sin \left ( 0+\frac{1}{\sqrt{E}}\frac{1}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\pi ^{3}}\right ) \right ) \\ & \thicksim A \end{align*}

Hence solution in (8) reduces to y\left ( x\right ) \thicksim \frac{\pi B}{\pi +x}\sin \left ( \sqrt{E}\left ( \frac{x^{3}}{3}+\pi x^{2}+\pi ^{2}x\right ) +\frac{1}{\sqrt{E}}\frac{1}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\left ( \pi +x\right ) ^{3}}\right ) \right ) Applying B.C. y\left ( \pi \right ) =0 gives\begin{align*} 0 & \thicksim \frac{\pi B}{\pi +\pi }\sin \left ( \sqrt{E}\left ( \frac{\pi ^{3}}{3}+\pi \pi ^{2}+\pi ^{2}\pi \right ) +\frac{1}{\sqrt{E}}\frac{1}{3}\left ( \frac{1}{\pi ^{3}}-\frac{1}{\left ( \pi +\pi \right ) ^{3}}\right ) \right ) \\ & \thicksim \frac{B}{2}\sin \left ( \sqrt{E}\left ( \frac{7}{3}\pi ^{3}\right ) +\frac{1}{\sqrt{E}}\left ( \frac{7}{24\pi ^{3}}\right ) \right ) \end{align*}

For non trivial solution, therefore \sqrt{E_{n}}\left ( \frac{7}{3}\pi ^{3}\right ) +\frac{1}{\sqrt{E_{n}}}\left ( \frac{7}{24\pi ^{3}}\right ) =n\pi \qquad n=1,2,3,\cdots Solving for \sqrt{E_{n}}. Let \sqrt{E_{n}}=x, then the above becomes\begin{align*} x^{2}\left ( \frac{7}{3}\pi ^{3}\right ) +\left ( \frac{7}{24\pi ^{6}}\right ) & =xn\pi \\ x^{2}-\frac{3}{7\pi ^{2}}xn+\frac{1}{8\pi ^{6}} & =0 \end{align*}

Solving using quadratic formula and taking the positive root, since E_{n}>0\,\, gives\begin{align*} x & =\frac{1}{28\pi ^{3}}\left ( \sqrt{2}\sqrt{18\pi ^{2}n^{2}-49}+6\pi n\right ) \qquad n=1,2,3,\cdots \\ \sqrt{E_{n}} & =\frac{1}{28\pi ^{3}}\left ( \sqrt{2}\sqrt{18\pi ^{2}n^{2}-49}+6\pi n\right ) \\ E_{n} & =\left ( \sqrt{2}\sqrt{18\pi ^{2}n^{2}-49}+6\pi n\right ) ^{2} \end{align*}

Table 10.1 is now reproduced to compare the above more accurate E_{n}. The following table shows the actual E_{n} values obtained this more accurate method. Values computed from above formula are in column 3.

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The following table shows the relative error in place of the actual values of E_{n} to better compare how more accurate the result obtained in this solution is compared to the book result

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The above shows clearly that adding one more term in the WKB series resulted in more accurate eigenvalue estimate.