Solution
\begin{align} X^{\prime }\left ( x\right ) +xX^{\prime \prime }\left ( x\right ) +\frac{\lambda }{x}X\left ( x\right ) & =0\nonumber \\ x^{2}X^{\prime \prime }\left ( x\right ) +xX^{\prime }\left ( x\right ) +\lambda X\left ( x\right ) & =0 \tag{1} \end{align}
To transform the above to \(X^{\prime \prime }\left ( s\right ) +\lambda X\left ( s\right ) =0\,\), let \(x=e^{s}\). Therefore \(\frac{dx}{ds}=e^{s}\) or \(\frac{ds}{dx}=e^{-s}\). Now\begin{align} \frac{dX}{dx} & =\frac{dX}{ds}\frac{ds}{dx}\nonumber \\ & =\frac{dX}{ds}e^{-s} \tag{2} \end{align}
And\begin{align*} \frac{d^{2}X}{dx^{2}} & =\frac{d}{dx}\left ( \frac{dX}{dx}\right ) \\ & =\frac{d}{dx}\left ( \frac{dX}{ds}e^{-s}\right ) \end{align*}
Hence, by product rule\begin{align} \frac{d^{2}X}{dx^{2}} & =\frac{d^{2}X}{ds^{2}}\frac{ds}{dx}e^{-s}+\frac{dX}{ds}\frac{d}{dx}\left ( e^{-s}\right ) \nonumber \\ & =\frac{d^{2}X}{ds^{2}}e^{-s}e^{-s}+\frac{dX}{ds}\frac{d}{ds}\left ( e^{-s}\right ) \frac{ds}{dx}\nonumber \\ & =\frac{d^{2}X}{ds^{2}}e^{-2s}+\frac{dX}{ds}\left ( -e^{-s}\right ) \left ( e^{-s}\right ) \nonumber \\ & =e^{-2s}\frac{d^{2}X}{ds^{2}}-e^{-2s}\frac{dX}{ds} \tag{3} \end{align}
Substituting (2,3) back into (1) gives\[ x^{2}\left ( e^{-2s}\frac{d^{2}X}{ds^{2}}-e^{-2s}\frac{dX}{ds}\right ) +x\left ( \frac{dX}{ds}e^{-s}\right ) +\lambda X=0 \] But \(x=e^{s}\) and the above simplifies to\begin{align*} e^{2s}\left ( e^{-2s}\frac{d^{2}X}{ds^{2}}-e^{-2s}\frac{dX}{ds}\right ) +e^{s}\left ( \frac{dX}{ds}e^{-s}\right ) +\lambda X & =0\\ \frac{d^{2}X}{ds^{2}}-\frac{dX}{ds}+\frac{dX}{ds}+\lambda X & =0\\ \frac{d^{2}X\left ( s\right ) }{ds^{2}}+\lambda X\left ( s\right ) & =0 \end{align*}
When \(X\left ( 1\right ) =0\), which means when \(x=1\), and since \(x=e^{s}\), then when \(s=0\). Hence \(X\left ( 1\right ) =0\) becomes \(X\left ( 0\right ) =0\). And when \(x=b\), then \(s=\ln \left ( b\right ) \). Hence the second condition becomes \(X\left ( \ln \left ( b\right ) \right ) =0\). Therefore the new B.C. are\begin{align*} X\left ( 0\right ) & =0\\ X\left ( \ln \left ( b\right ) \right ) & =0 \end{align*}
By referring to problem (4) in section 35 we see that the eigenvalues are\[ \lambda _{n}=\left ( \frac{n\pi }{c}\right ) ^{2}\] Where here \(c=\ln \left ( b\right ) \). Hence\begin{align*} \lambda _{n} & =\left ( \frac{n\pi }{\ln \left ( b\right ) }\right ) ^{2}\qquad n=1,2,3,\cdots \\ & =\alpha _{n}^{2} \end{align*}
Where \(\alpha _{n}=\frac{n\pi }{\ln \left ( b\right ) }\). And the eigenfunctions are, per section 35\[ X_{n}\left ( s\right ) =\sin \left ( \alpha _{n}s\right ) \] In terms of \(x\), the eigenfunctions become\[ X_{n}\left ( s\right ) =\sin \left ( \alpha _{n}\ln x\right ) \]
\[ \left \langle X_{n}\left ( x\right ) ,X_{m}\left ( x\right ) \right \rangle =\int _{1}^{b}\sin \left ( \alpha _{n}\ln x\right ) \sin \left ( \alpha _{m}\ln x\right ) p\left ( x\right ) dx \] But from \(\left ( xX^{\prime }\left ( x\right ) \right ) ^{\prime }+\frac{\lambda }{x}X\left ( x\right ) =0\) and comparing this to \(\left ( rX^{\prime }\right ) ^{\prime }+\left ( \lambda p+q\right ) X=0\), we see that \(r\left ( x\right ) =x\) and \(q=0\) and \(p=\frac{1}{x}\). Hence the above integral becomes\[ \left \langle X_{n}\left ( x\right ) ,X_{m}\left ( x\right ) \right \rangle =\int _{1}^{b}\frac{1}{x}\sin \left ( \alpha _{n}\ln x\right ) \sin \left ( \alpha _{m}\ln x\right ) dx \] Let \(s=\frac{\ln x}{\ln b}\pi \). Then \(\frac{ds}{dx}=\frac{1}{x}\frac{\pi }{\ln b}\) or \(dx=\frac{x}{\pi }\ln \left ( b\right ) ds\). When \(x=1\) then \(s=0\) and when \(x=b\) then \(s=\pi \). Hence the above integral becomes\begin{align*} \left \langle X_{n}\left ( x\right ) ,X_{m}\left ( x\right ) \right \rangle & =\int _{s=0}^{s=\pi }\frac{1}{x}\sin \left ( \alpha _{n}\frac{s\ln b}{\pi }\right ) \sin \left ( \alpha _{m}\frac{s\ln b}{\pi }\right ) \left ( \frac{x}{\pi }\ln \left ( b\right ) ds\right ) \\ & =\frac{1}{\pi }\ln \left ( b\right ) \int _{0}^{\pi }\sin \left ( \alpha _{n}\frac{s\ln b}{\pi }\right ) \sin \left ( \alpha _{m}\frac{s\ln b}{\pi }\right ) ds \end{align*}
But \(\alpha _{n}=\frac{n\pi }{\ln \left ( b\right ) }\) and \(\alpha _{m}=\frac{m\pi }{\ln \left ( b\right ) }\), therefore the above becomes\begin{align} \left \langle X_{n}\left ( x\right ) ,X_{m}\left ( x\right ) \right \rangle & =\frac{1}{\pi }\ln \left ( b\right ) \int _{0}^{\pi }\sin \left ( \frac{n\pi }{\ln \left ( b\right ) }\frac{s\ln b}{\pi }\right ) \sin \left ( \frac{m\pi }{\ln \left ( b\right ) }\frac{s\ln b}{\pi }\right ) ds\nonumber \\ & =\frac{1}{\pi }\ln \left ( b\right ) \int _{0}^{\pi }\sin \left ( ns\right ) \sin \left ( ms\right ) ds \tag{1} \end{align}
Referring to Problem 9., section 5 which says that \[ \int _{0}^{\pi }\sin \left ( nx\right ) \sin \left ( mx\right ) dx=\left \{ \begin{array} [c]{ccc}0 & & n\neq m\\ \frac{\pi }{2} & & n=0 \end{array} \right . \] Applying this to (1) shows that \[ \left \langle X_{n}\left ( x\right ) ,X_{m}\left ( x\right ) \right \rangle =\left \{ \begin{array} [c]{ccc}0 & & n\neq m\\ \frac{\pi }{2} & & n=0 \end{array} \right . \] Hence \(X_{n}\left ( x\right ) \) and \(X_{m}\left ( x\right ) \) are orthogonal, since this is the definition of orthogonality.
Solution
Solve for eigenvalues and normalized eigenfunctions.\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X^{\prime }\left ( 0\right ) & =0\\ X\left ( c\right ) & =0 \end{align*}
Writing the boundary conditions in SL standard form\begin{align*} a_{1}X\left ( 0\right ) +a_{2}X^{\prime }\left ( 0\right ) & =0\\ b_{1}X\left ( c\right ) +b_{2}X^{\prime }\left ( c\right ) & =0 \end{align*}
Shows that \(a_{1}=0,a_{2}=1\) and \(b_{1}=1,b_{2}=0\,\). Therefore \(a_{1}a_{2}=0\) and \(b_{1}b_{2}=0\). But we know that if \(a_{1}a_{2}\geq 0\) and \(b_{1}b_{2}\geq 0\), then \(\lambda >0\) is only possible eigenvalues. Let \(\lambda _{n}=\alpha _{n}^{2}\). \(\alpha >0\). Hence the solution to the ODE is\begin{align*} X_{n}\left ( x\right ) & =A\cos \left ( \alpha _{n}x\right ) +B\sin \left ( \alpha _{n}x\right ) \\ X_{n}^{\prime }\left ( x\right ) & =-A\alpha _{n}\sin \left ( \alpha _{n}x\right ) +B\alpha _{n}\cos \left ( \alpha _{n}x\right ) \end{align*}
First B.C \(X^{\prime }\left ( 0\right ) =0\) gives\[ 0=B\alpha _{n}\] Which implies \(B=0\). Hence the solution now becomes \(X_{n}\left ( x\right ) =A\cos \left ( \alpha _{n}x\right ) \). For the second BC\begin{align*} 0 & =A\cos \left ( \alpha _{n}c\right ) \\ 0 & =\cos \left ( \alpha _{n}c\right ) \end{align*}
Which implies\begin{align*} \alpha _{n}c & =\frac{\pi }{2},3\frac{\pi }{2},5\frac{\pi }{2},\cdots \\ & =\left ( 2n-1\right ) \frac{\pi }{2}\qquad n=1,2,3,\cdots \end{align*}
Hence\[ \alpha _{n}=\frac{\left ( 2n-1\right ) }{c}\frac{\pi }{2}\qquad n=1,2,3,\cdots \] And the corresponding eigenfunctions are\begin{align*} X_{n}\left ( x\right ) & =\cos \left ( \alpha _{n}x\right ) \\ & =\cos \left ( \frac{\left ( 2n-1\right ) }{c}\frac{\pi }{2}x\right ) \end{align*}
To find the normalized \(X_{n}\left ( x\right ) \) which we call it \(\phi _{n}\left ( x\right ) \), then by definition \[ \phi _{n}\left ( x\right ) =\frac{X_{n}\left ( x\right ) }{\left \Vert X_{n}\left ( x\right ) \right \Vert }\] But \[ \left \Vert X_{n}\left ( x\right ) \right \Vert ^{2}=\int _{0}^{c}p\left ( x\right ) X_{n}^{2}\left ( x\right ) dx \] Comparing the ODE \(X^{\prime \prime }+\lambda X=0\) to \(\left ( rX^{\prime }\right ) ^{\prime }+\left ( \lambda p+q\right ) X=0\), we see that \(r\left ( x\right ) =1\) and \(q=0\) and \(p=1\). Hence the above becomes\begin{align*} \left \Vert X_{n}\left ( x\right ) \right \Vert ^{2} & =\int _{0}^{c}\cos ^{2}\left ( \alpha _{n}x\right ) dx\\ & =\frac{c}{2} \end{align*}
Therefore \(\left \Vert X_{n}\left ( x\right ) \right \Vert =\sqrt{\frac{c}{2}}\) which shows that\begin{align*} \phi _{n}\left ( x\right ) & =\frac{X_{n}\left ( x\right ) }{\sqrt{\frac{c}{2}}}\\ & =\sqrt{\frac{2}{c}}\cos \left ( \alpha _{n}x\right ) \end{align*}
where \[ \alpha _{n}=\frac{\left ( 2n-1\right ) }{c}\frac{\pi }{2}\qquad n=1,2,3,\cdots \] Which is what required to show.
Solution
\begin{align*} X_{n}\left ( x\right ) & =\sin \left ( \alpha _{n}\ln x\right ) \\ \alpha _{n} & =\frac{n\pi }{\ln b}\qquad n=1,2,3,\cdots \end{align*}
The normalized eigenfunction is given by \[ \phi _{n}\left ( x\right ) =\frac{X_{n}\left ( x\right ) }{\left \Vert X_{n}\left ( x\right ) \right \Vert }\] But \[ \left \Vert X_{n}\left ( x\right ) \right \Vert ^{2}=\int _{1}^{b}p\left ( x\right ) X_{n}^{2}\left ( x\right ) dx \] Comparing the ODE \(\left ( xX^{\prime }\right ) ^{\prime }+\frac{\lambda }{x}X=0\) to \(\left ( rX^{\prime }\right ) ^{\prime }+\left ( \lambda p+q\right ) X=0\), we see that \(r\left ( x\right ) =x\) and \(q=0\) and \(p=\frac{1}{x}\). Hence the above becomes
\[ \left \Vert X_{n}\left ( x\right ) \right \Vert ^{2}=\int _{1}^{b}\frac{1}{x}\sin ^{2}\left ( \alpha _{n}\ln x\right ) dx \] Let \(s=\frac{\ln x}{\ln b}\pi \). Then \(\frac{ds}{dx}=\frac{1}{x}\frac{\pi }{\ln b}\) or \(dx=\frac{x}{\pi }\ln \left ( b\right ) ds\). When \(x=1\) then \(s=0\) and when \(x=b\) then \(s=\pi \). Hence the above integral becomes\begin{align*} \left \Vert X_{n}\left ( x\right ) \right \Vert ^{2} & =\int _{s=0}^{s=\pi }\frac{1}{x}\sin ^{2}\left ( \alpha _{n}\frac{s\ln b}{\pi }\right ) \left ( \frac{x}{\pi }\ln \left ( b\right ) ds\right ) \\ & =\frac{1}{\pi }\ln \left ( b\right ) \int _{0}^{\pi }\sin ^{2}\left ( \alpha _{n}\frac{s\ln b}{\pi }\right ) ds \end{align*}
But \(\alpha _{n}=\frac{n\pi }{\ln \left ( b\right ) }\) therefore the above becomes\begin{align*} \left \Vert X_{n}\left ( x\right ) \right \Vert ^{2} & =\frac{1}{\pi }\ln \left ( b\right ) \int _{0}^{\pi }\sin ^{2}\left ( \frac{n\pi }{\ln \left ( b\right ) }\frac{s\ln b}{\pi }\right ) ds\\ & =\frac{1}{\pi }\ln \left ( b\right ) \int _{0}^{\pi }\sin ^{2}\left ( ns\right ) ds\\ & =\frac{1}{\pi }\ln \left ( b\right ) \int _{0}^{\pi }\frac{1}{2}-\frac{1}{2}\cos \left ( 2ns\right ) ds\\ & =\frac{1}{\pi }\ln \left ( b\right ) \left ( \frac{\pi }{2}-\frac{1}{2}\sin \left ( \frac{2ns}{2n}\right ) _{0}^{\pi }\right ) \\ & =\frac{1}{\pi }\ln \left ( b\right ) \left ( \frac{\pi }{2}-\frac{1}{2}\sin \left ( s\right ) _{0}^{\pi }\right ) \\ & =\frac{1}{2}\ln \left ( b\right ) \end{align*}
Hence\begin{align*} \phi _{n}\left ( x\right ) & =\frac{\sin \left ( \alpha _{n}\ln x\right ) }{\sqrt{\frac{1}{2}\ln \left ( b\right ) }}\\ & =\sqrt{\frac{2}{\ln \left ( b\right ) }}\sin \left ( \alpha _{n}\ln x\right ) \end{align*}
Which is what required to show.
solution
From problem section 69 problem 1, we know that \(\left ( xX^{\prime }\left ( x\right ) \right ) ^{\prime }+\frac{\lambda }{x}X\left ( x\right ) =0\) can be transformed to \(X^{\prime \prime }\left ( s\right ) +\lambda X\left ( s\right ) =0\) using \(x=e^{s}\). With boundary conditions in \(s\) found as follows. When \(x=1\) then \(s=0\) and when \(x=b\) then \(s=\ln b\). Hence we obtain the SL problem\begin{align} X^{\prime \prime }\left ( s\right ) +\lambda X\left ( s\right ) & =0\tag{1}\\ X^{\prime }\left ( 0\right ) & =0\nonumber \\ X\left ( \ln b\right ) & =0\nonumber \end{align}
But problem 3 is\begin{align} X^{\prime \prime }+\lambda X & =0\tag{2}\\ X^{\prime }\left ( 0\right ) & =0\nonumber \\ X\left ( c\right ) & =0\nonumber \end{align}
And it had the solution\[ \phi _{n}\left ( x\right ) =\sqrt{\frac{2}{c}}\cos \left ( \alpha _{n}x\right ) \] where \[ \alpha _{n}=\frac{\left ( 2n-1\right ) }{c}\frac{\pi }{2}\qquad n=1,2,3,\cdots \] By comparing (2) and (1) we see it is the same problem, except \(c\rightarrow \ln b\). Hence the solution to (2) is the same as the solution in (1) but with \(c\) replaced by \(\ln b\). Hence the solution is\begin{align*} \phi _{n}\left ( s\right ) & =\sqrt{\frac{2}{\ln b}}\cos \left ( \alpha _{n}s\right ) \\ \alpha _{n} & =\frac{\left ( 2n-1\right ) }{\ln b}\frac{\pi }{2}\qquad n=1,2,3,\cdots \end{align*}
But \(s=\ln x\), hence the above becomes\begin{align*} \phi _{n}\left ( x\right ) & =\sqrt{\frac{2}{\ln b}}\cos \left ( \alpha _{n}\ln x\right ) \\ \alpha _{n} & =\frac{\left ( 2n-1\right ) }{\ln b}\frac{\pi }{2}\qquad n=1,2,3,\cdots \end{align*}
Which is what required to show.