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