Problem Show that \[ \int _{0}^{\infty }\frac{1}{x}J_{m}\left ( x\right ) J_{n}\left ( x\right ) dx=\frac{2}{\pi }\frac{\sin \left ( \left ( m-n\right ) \frac{\pi }{2}\right ) }{m^{2}-n^{2}}\] Solution\begin{align*} x^{2}J_{n}^{\prime \prime }\left ( x\right ) +xJ_{n}^{\prime }\left ( x\right ) +\left ( x^{2}-n^{2}\right ) J_{n}\left ( x\right ) & =0\\ x^{2}J_{m}^{\prime \prime }\left ( x\right ) +xJ_{m}^{\prime }\left ( x\right ) +\left ( x^{2}-m^{2}\right ) J_{m}\left ( x\right ) & =0 \end{align*}
Dividing both equations by \(x^{2}\) gives\begin{align*} J_{n}^{\prime \prime }\left ( x\right ) +\frac{1}{x}J_{n}^{\prime }\left ( x\right ) +\left ( 1-\frac{n^{2}}{x^{2}}\right ) J_{n}\left ( x\right ) & =0\\ J_{m}^{\prime \prime }\left ( x\right ) +\frac{1}{x}J_{m}^{\prime }\left ( x\right ) +\left ( 1-\frac{m^{2}}{x^{2}}\right ) J_{m}\left ( x\right ) & =0 \end{align*}
Multiplying the first ODE by \(xJ_{m}\left ( x\right ) \) and the second by \(xJ_{n}\left ( x\right ) \) gives (multiplying by just \(x\) did not lead to a result that I could use).\begin{align*} xJ_{m}J_{n}^{\prime \prime }+J_{m}J_{n}^{\prime }+x\left ( 1-\frac{n^{2}}{x^{2}}\right ) J_{m}J_{n} & =0\\ xJ_{n}J_{m}^{\prime \prime }+J_{n}J_{m}^{\prime }+x\left ( 1-\frac{m^{2}}{x^{2}}\right ) J_{n}J_{m} & =0 \end{align*}
Subtracting gives\begin{align*} \left ( xJ_{m}J_{n}^{\prime \prime }+J_{m}J_{n}^{\prime }+x\left ( 1-\frac{n^{2}}{x^{2}}\right ) J_{m}J_{n}\right ) -\left ( xJ_{n}J_{m}^{\prime \prime }+J_{n}J_{m}^{\prime }+x\left ( 1-\frac{m^{2}}{x^{2}}\right ) J_{n}J_{m}\right ) & =0\\ x\left ( J_{m}J_{n}^{\prime \prime }-J_{n}J_{m}^{\prime \prime }\right ) +J_{m}J_{n}^{\prime }-J_{n}J_{m}^{\prime }-xJ_{m}J_{n}\left ( \left ( 1-\frac{n^{2}}{x^{2}}\right ) -\left ( 1-\frac{m^{2}}{x^{2}}\right ) \right ) & =0 \end{align*}
Or\begin{equation} x\left ( J_{m}J_{n}^{\prime \prime }-J_{n}J_{m}^{\prime \prime }\right ) +J_{m}J_{n}^{\prime }-J_{n}J_{m}^{\prime }=xJ_{m}J_{n}\left ( \left ( 1-\frac{m^{2}}{x^{2}}\right ) -\left ( 1-\frac{n^{2}}{x^{2}}\right ) \right ) \tag{1} \end{equation} But the LHS above is complete differential3 \begin{equation} x\left ( J_{m}J_{n}^{\prime \prime }-J_{n}J_{m}^{\prime \prime }\right ) +J_{m}J_{n}^{\prime }-J_{n}J_{m}^{\prime }=\left ( x\left ( J_{m}J_{n}^{\prime }-J_{n}J_{m}^{\prime }\right ) \right ) ^{\prime } \tag{2} \end{equation} Hence using (2) in (1), then (1) simplifies to\begin{align*} \left ( x\left ( J_{m}J_{n}^{\prime }-J_{n}J_{m}^{\prime }\right ) \right ) ^{\prime } & =xJ_{m}J_{n}\left ( \left ( 1-\frac{m^{2}}{x^{2}}\right ) -\left ( 1-\frac{n^{2}}{x^{2}}\right ) \right ) \\ & =xJ_{m}J_{n}\left ( \frac{n^{2}}{x^{2}}-\frac{m^{2}}{x^{2}}\right ) \\ & =\frac{J_{m}J_{n}}{x}\left ( n^{2}-m^{2}\right ) \end{align*}
Integrating both sides above gives\[ \left [ x\left ( J_{m}J_{n}^{\prime }-J_{n}J_{m}^{\prime }\right ) \right ] _{0}^{\infty }=\left ( n^{2}-m^{2}\right ) \int _{0}^{\infty }\frac{J_{m}J_{n}}{x}dx \] Therefore\begin{equation} \int _{0}^{\infty }\frac{J_{m}\left ( x\right ) J_{n}\left ( x\right ) }{x}dx=\frac{1}{\left ( m^{2}-n^{2}\right ) }\left [ x\left ( J_{n}\left ( x\right ) J_{m}^{\prime }\left ( x\right ) -J_{m}\left ( x\right ) J_{n}^{\prime }\left ( x\right ) \right ) \right ] _{0}^{\infty } \tag{3} \end{equation} At \(x=0\) the expression \(x\left ( J_{n}\left ( x\right ) J_{m}^{\prime }\left ( x\right ) -J_{m}\left ( x\right ) J_{n}^{\prime }\left ( x\right ) \right ) =0\). And at \(x=\infty \) we can use the asymptotic approximation given by\begin{align*} J_{n}\left ( x\right ) & =\sqrt{\frac{2}{\pi x}}\cos \left ( x-\frac{n\pi }{2}-\frac{\pi }{4}\right ) \\ J_{n}^{\prime }\left ( x\right ) & =-\sqrt{\frac{2}{\pi x}}\sin \left ( x-\frac{n\pi }{2}-\frac{\pi }{4}\right ) -\frac{1}{\sqrt{2\pi }}\left ( \frac{1}{x}\right ) ^{\frac{3}{2}}\cos \left ( x-\frac{n\pi }{2}-\frac{\pi }{4}\right ) \end{align*}
And similarly for \(J_{m}\left ( x\right ) \)\begin{align*} J_{m}\left ( x\right ) & =\sqrt{\frac{2}{\pi x}}\cos \left ( x-\frac{m\pi }{2}-\frac{\pi }{4}\right ) \\ J_{m}^{\prime }\left ( x\right ) & =-\sqrt{\frac{2}{\pi x}}\sin \left ( x-\frac{m\pi }{2}-\frac{\pi }{4}\right ) -\frac{1}{\sqrt{2\pi }}\left ( \frac{1}{x}\right ) ^{\frac{3}{2}}\cos \left ( x-\frac{m\pi }{2}-\frac{\pi }{4}\right ) \end{align*}
Therefore\[ J_{n}\left ( x\right ) J_{m}^{\prime }\left ( x\right ) =\sqrt{\frac{2}{\pi x}}\cos \left ( x-\frac{n\pi }{2}-\frac{\pi }{4}\right ) \left ( -\sqrt{\frac{2}{\pi x}}\sin \left ( x-\frac{m\pi }{2}-\frac{\pi }{4}\right ) -\frac{1}{\sqrt{2\pi }}\left ( \frac{1}{x}\right ) ^{\frac{3}{2}}\cos \left ( x-\frac{m\pi }{2}-\frac{\pi }{4}\right ) \right ) \] Let \(x-\frac{n\pi }{2}-\frac{\pi }{4}=\alpha \), and let \(x-\frac{m\pi }{2}-\frac{\pi }{4}=\beta \), then the above becomes\begin{align} J_{n}\left ( x\right ) J_{m}^{\prime }\left ( x\right ) & =\sqrt{\frac{2}{\pi x}}\cos \left ( \alpha \right ) \left ( -\sqrt{\frac{2}{\pi x}}\sin \left ( \beta \right ) -\frac{1}{\sqrt{2\pi }}\left ( \frac{1}{x}\right ) ^{\frac{3}{2}}\cos \left ( \beta \right ) \right ) \nonumber \\ & =-\frac{2}{\pi x}\cos \left ( \alpha \right ) \sin \left ( \beta \right ) -\sqrt{\frac{2}{\pi x}}\sqrt{\frac{1}{2\pi }}\left ( \frac{1}{x}\right ) ^{\frac{3}{2}}\cos \left ( \alpha \right ) \cos \left ( \beta \right ) \nonumber \\ & =-\frac{2}{\pi x}\cos \left ( \alpha \right ) \sin \left ( \beta \right ) -\frac{1}{\pi }\left ( \frac{1}{x}\right ) ^{\frac{1}{2}}\left ( \frac{1}{x}\right ) ^{\frac{3}{2}}\cos \left ( \alpha \right ) \cos \left ( \beta \right ) \nonumber \\ & =-\frac{2}{\pi x}\cos \left ( \alpha \right ) \sin \left ( \beta \right ) -\frac{1}{\pi }\left ( \frac{1}{x}\right ) ^{2}\cos \left ( \alpha \right ) \cos \left ( \beta \right ) \tag{4} \end{align}
Similarly\begin{equation} J_{m}\left ( x\right ) J_{n}^{\prime }\left ( x\right ) =-\frac{2}{\pi x}\cos \left ( \beta \right ) \sin \left ( \alpha \right ) -\frac{1}{\pi }\left ( \frac{1}{x}\right ) ^{2}\cos \left ( \beta \right ) \cos \left ( \alpha \right ) \tag{5} \end{equation} Substituting (4,5) into (3) gives (only the term as \(x\rightarrow \infty \) remains)
\begin{align} \int _{0}^{\infty }\frac{J_{m}\left ( x\right ) J_{n}\left ( x\right ) }{x}dx & =\frac{1}{\left ( m^{2}-n^{2}\right ) }\left [ x\left ( J_{n}\left ( x\right ) J_{m}^{\prime }\left ( x\right ) -J_{m}\left ( x\right ) J_{n}^{\prime }\left ( x\right ) \right ) \right ] _{0}^{\infty }\nonumber \\ & =\frac{x}{\left ( m^{2}-n^{2}\right ) }\left ( \left ( -\frac{2}{\pi x}\cos \left ( \alpha \right ) \sin \left ( \beta \right ) -\frac{1}{\pi }\left ( \frac{1}{x}\right ) ^{2}\cos \left ( \alpha \right ) \cos \left ( \beta \right ) \right ) -\left ( -\frac{2}{\pi x}\cos \left ( \beta \right ) \sin \left ( \alpha \right ) -\frac{1}{\pi }\left ( \frac{1}{x}\right ) ^{2}\cos \left ( \beta \right ) \cos \left ( \alpha \right ) \right ) \right ) \nonumber \\ & =\frac{x}{\left ( m^{2}-n^{2}\right ) }\left ( -\frac{2}{\pi x}\cos \left ( \alpha \right ) \sin \left ( \beta \right ) -\frac{1}{\pi }\left ( \frac{1}{x}\right ) ^{2}\cos \left ( \alpha \right ) \cos \left ( \beta \right ) +\frac{2}{\pi x}\cos \left ( \beta \right ) \sin \left ( \alpha \right ) +\frac{1}{\pi }\left ( \frac{1}{x}\right ) ^{2}\cos \left ( \beta \right ) \cos \left ( \alpha \right ) \right ) \nonumber \\ & =\frac{x}{\left ( m^{2}-n^{2}\right ) }\left ( -\frac{2}{\pi x}\cos \left ( \alpha \right ) \sin \left ( \beta \right ) +\frac{2}{\pi x}\cos \left ( \beta \right ) \sin \left ( \alpha \right ) \right ) \nonumber \\ & =\frac{2}{\pi }\frac{1}{\left ( m^{2}-n^{2}\right ) }\left ( \sin \left ( \alpha \right ) \cos \left ( \beta \right ) -\cos \left ( \alpha \right ) \sin \left ( \beta \right ) \right ) \tag{6} \end{align}
But \begin{align*} \sin \left ( \alpha \right ) \cos \left ( \beta \right ) -\cos \left ( \alpha \right ) \sin \left ( \beta \right ) & =\sin \left ( \alpha -\beta \right ) \\ & =\sin \left ( \left ( x-\frac{n\pi }{2}-\frac{\pi }{4}\right ) -\left ( x-\frac{m\pi }{2}-\frac{\pi }{4}\right ) \right ) \\ & =\sin \left ( x-\frac{n\pi }{2}-\frac{\pi }{4}-x+\frac{m\pi }{2}+\frac{\pi }{4}\right ) \\ & =\sin \left ( \frac{m\pi }{2}-\frac{n\pi }{2}\right ) \\ & =\sin \left ( \left ( m-n\right ) \frac{\pi }{2}\right ) \end{align*}
Using the above in (6) gives\[ \int _{0}^{\infty }\frac{J_{m}\left ( x\right ) J_{n}\left ( x\right ) }{x}dx=\frac{2}{\pi }\frac{\sin \left ( \left ( m-n\right ) \frac{\pi }{2}\right ) }{\left ( m^{2}-n^{2}\right ) }\] Which is the result required to show. QED.
Problem What linear second order ODE does the function \(x^{m}J_{n}\left ( ax^{k}\right ) \) solves? Are there any required relationships among \(m,n,k\)? Use this to solve \(y^{\prime \prime }+x^{2}y=0\)
Solution
We know that the Bessel ODE \begin{equation} t^{2}z^{\prime \prime }\left ( t\right ) +tz^{\prime }\left ( t\right ) +\left ( t^{2}-\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) z\left ( t\right ) =0 \tag{1} \end{equation} I am using the order as \(\frac{\alpha }{\beta }\) instead of \(n\) to make it more general. At the end, \(\frac{\alpha }{\beta }\) can always be replaced back by \(n\).
The ODE above has solution \[ z\left ( t\right ) =J_{\frac{\alpha }{\beta }}\left ( t\right ) \] Hence using the transformation \begin{equation} t=ax^{k} \tag{2} \end{equation} The solution \(y\left ( x\right ) \equiv z\left ( ax^{k}\right ) \) will becomes\[ y\left ( x\right ) =J_{\frac{\alpha }{\beta }}\left ( ax^{k}\right ) \] Therefore the question now is, how does ODE (1) transforms under (2)? From (2) \[ x=\left ( \frac{t}{a}\right ) ^{\frac{1}{k}}\] Hence\begin{align} \frac{dx}{dt} & =\frac{1}{k}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\nonumber \\ & =\frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1} \tag{3} \end{align}
Now\begin{align} \frac{dz}{dt} & =\frac{dz}{dx}\frac{dx}{dt}\nonumber \\ & =\frac{dz}{dx}\frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1} \tag{5} \end{align}
And\begin{align} \frac{d^{2}z}{dt^{2}} & =\frac{d}{dt}\left ( \frac{dz}{dt}\right ) \nonumber \\ & =\frac{d}{dt}\left ( \frac{dz}{dx}\frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\right ) \nonumber \\ & =\frac{d^{2}z}{dx^{2}}\frac{dx}{dt}\left ( \frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\right ) +\frac{dz}{dx}\frac{d}{dt}\left ( \frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\right ) \nonumber \\ & =\frac{d^{2}z}{dx^{2}}\left ( \frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\right ) ^{2}+\frac{dz}{dx}\left ( \frac{1}{a^{2}k}\left ( \frac{1}{k}-1\right ) \left ( \frac{t}{a}\right ) ^{\frac{1}{k}-2}\right ) \tag{6} \end{align}
Using (5,6) then ODE (1) becomes
\[ t^{2}\left ( z^{\prime \prime }\left ( ax^{k}\right ) \left ( \frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\right ) ^{2}+z^{\prime }\left ( ax^{k}\right ) \frac{1}{a^{2}k}\left ( \frac{1}{k}-1\right ) \left ( \frac{t}{a}\right ) ^{\frac{1}{k}-2}\right ) +t\left ( z^{\prime }\left ( ax^{k}\right ) \frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\right ) +\left ( t^{2}-\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) z\left ( ax^{k}\right ) =0 \]
Writing \(y\left ( x\right ) \equiv z\left ( ax^{k}\right ) \) so we do not have to keep writing \(z\left ( ax^{k}\right ) \), the above becomes\[ t^{2}\left ( y^{\prime \prime }\left ( x\right ) \left ( \frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\right ) ^{2}+y^{\prime }\left ( x\right ) \frac{1}{a^{2}k}\left ( \frac{1}{k}-1\right ) \left ( \frac{t}{a}\right ) ^{\frac{1}{k}-2}\right ) +t\left ( y^{\prime }\left ( x\right ) \frac{1}{ak}\left ( \frac{t}{a}\right ) ^{\frac{1}{k}-1}\right ) +\left ( t^{2}-\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) y\left ( x\right ) =0 \] But \(t=ax^{k}\) and the above becomes
\[ a^{2}x^{2k}\left ( y^{\prime \prime }\left ( x\right ) \left ( \frac{1}{ak}\left ( \frac{ax^{k}}{a}\right ) ^{\frac{1}{k}-1}\right ) ^{2}+y^{\prime }\left ( x\right ) \frac{1}{a^{2}k}\left ( \frac{1}{k}-1\right ) \left ( \frac{ax^{k}}{a}\right ) ^{\frac{1}{k}-2}\right ) +ax^{k}\left ( y^{\prime }\left ( x\right ) \frac{1}{ak}\left ( \frac{ax^{k}}{a}\right ) ^{\frac{1}{k}-1}\right ) +\left ( a^{2}x^{2k}-\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) y\left ( x\right ) =0 \]
Which is simplified more as follows\begin{align} a^{2}x^{2k}\left ( y^{\prime \prime }\left ( x\right ) \left ( \frac{1}{ak}\left ( \frac{x}{x^{k}}\right ) \right ) ^{2}+y^{\prime }\left ( x\right ) \frac{1}{a^{2}k}\left ( \frac{1}{k}-1\right ) \frac{x}{x^{2k}}\right ) +ax^{k}\left ( y^{\prime }\left ( x\right ) \frac{1}{ak}\frac{x}{x^{k}}\right ) +\left ( a^{2}x^{2k}-\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) y\left ( x\right ) & =0\nonumber \\ a^{2}x^{2k}\left ( y^{\prime \prime }\left ( x\right ) \frac{1}{a^{2}k^{2}}\left ( \frac{x^{2}}{x^{2k}}\right ) +y^{\prime }\left ( x\right ) \frac{1}{a^{2}k}\left ( \frac{1}{k}-1\right ) \frac{x}{x^{2k}}\right ) +ax^{k}\left ( y^{\prime }\left ( x\right ) \frac{1}{ak}\frac{x}{x^{k}}\right ) +\left ( a^{2}x^{2k}-\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) y\left ( x\right ) & =0\nonumber \\ x^{2k}\left ( y^{\prime \prime }\left ( x\right ) \frac{1}{k^{2}}\left ( \frac{x^{2}}{x^{2k}}\right ) +y^{\prime }\left ( x\right ) \frac{1}{k}\left ( \frac{1}{k}-1\right ) \frac{x}{x^{2k}}\right ) +y^{\prime }\left ( x\right ) \frac{x}{k}+\left ( a^{2}x^{2k}-\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) y\left ( x\right ) & =0\nonumber \\ \frac{x^{2}}{k^{2}}y^{\prime \prime }\left ( x\right ) +y^{\prime }\left ( x\right ) \frac{1}{k}\left ( \frac{1}{k}-1\right ) x+y^{\prime }\left ( x\right ) \frac{x}{k}+\left ( a^{2}x^{2k}-\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) y\left ( x\right ) & =0\nonumber \\ x^{2}y^{\prime \prime }\left ( x\right ) +y^{\prime }\left ( x\right ) k\left ( \frac{1}{k}-1\right ) x+y^{\prime }\left ( x\right ) kx+\left ( k^{2}a^{2}x^{2k}-k^{2}\left ( \frac{\alpha }{\beta }\right ) ^{2}\right ) y\left ( x\right ) & =0\nonumber \\ x^{2}y^{\prime \prime }\left ( x\right ) +xy^{\prime }\left ( x\right ) +\left ( k^{2}a^{2}x^{2k}-\frac{k^{2}\alpha ^{2}}{\beta ^{2}}\right ) y\left ( x\right ) & =0 \tag{7} \end{align}
We know that the above ODE has one solution as \(y\left ( x\right ) =J_{\frac{\alpha }{\beta }}\left ( ax^{k}\right ) \) because this is how the above was constructed. Now assuming that\begin{align*} w\left ( x\right ) & =x^{m}y\left ( x\right ) \\ & =x^{m}J_{\frac{\alpha }{\beta }}\left ( ax^{k}\right ) \end{align*}
Then \(w\left ( x\right ) \) is the solution we want. This means we need to express (7) in terms of \(w\left ( x\right ) \) instead of \(y\left ( x\right ) \) in order to find the ODE whose solution is \(x^{m}J_{\frac{\alpha }{\beta }}\left ( ax^{k}\right ) \).
Since \(y\left ( x\right ) =w\left ( x\right ) x^{-m}\) then\begin{align*} y^{\prime }\left ( x\right ) & =\frac{d}{dx}\left ( x^{-m}w\right ) \\ & =-mx^{-m-1}w+x^{-m}w^{\prime } \end{align*}
And\begin{align*} y^{\prime \prime }\left ( x\right ) & =\frac{d}{dx}\left ( -mx^{-m-1}w+x^{-m}w^{\prime }\right ) \\ & =-m\left ( -m-1\right ) x^{-m-2}w-mx^{-m-1}w^{\prime }-mx^{-m-1}w^{\prime }+x^{-m}w^{\prime \prime }\\ & =m\left ( m+1\right ) x^{-m-2}w-2w^{\prime }mx^{-m-1}+x^{-m}w^{\prime \prime } \end{align*}
Substituting the above results back into (7) gives\[ x^{2}\left ( m\left ( m+1\right ) x^{-m-2}w-2w^{\prime }mx^{-m-1}+x^{-m}w^{\prime \prime }\right ) +x\left ( -mx^{-m-1}w+x^{-m}w^{\prime }\right ) +\left ( k^{2}a^{2}x^{2k}-\frac{k^{2}\alpha ^{2}}{\beta ^{2}}\right ) wx^{-m}=0 \] Dividing by \(x^{-m}\)\begin{align} x^{2}\left ( m\left ( m+1\right ) x^{-2}w-2w^{\prime }mx^{-1}+w^{\prime \prime }\right ) +x\left ( -mx^{-1}w+w^{\prime }\right ) +\left ( k^{2}a^{2}x^{2k}-\frac{k^{2}\alpha ^{2}}{\beta ^{2}}\right ) w & =0\nonumber \\ m\left ( m+1\right ) w-2xw^{\prime }m+x^{2}w^{\prime \prime }-mw+xw^{\prime }+\left ( k^{2}a^{2}x^{2k}-\frac{k^{2}\alpha ^{2}}{\beta ^{2}}\right ) w & =0\nonumber \\ x^{2}w^{\prime \prime }+w^{\prime }\left ( -2xm+x\right ) +\left ( k^{2}a^{2}x^{2k}+m\left ( m+1\right ) -m-\frac{k^{2}\alpha ^{2}}{\beta ^{2}}\right ) w & =0\nonumber \\ x^{2}w^{\prime \prime }+\left ( 1-2m\right ) xw^{\prime }+\left ( k^{2}a^{2}x^{2k}+m^{2}-\frac{k^{2}\alpha ^{2}}{\beta ^{2}}\right ) w & =0 \tag{8} \end{align}
Hence the above ODE (8) will have the solution \(x^{m}J_{\frac{\alpha }{\beta }}\left ( ax^{k}\right ) \). We can now let \(n=\frac{\alpha }{\beta }\) and the above ODE becomes\begin{equation} x^{2}w^{\prime \prime }+\left ( 1-2m\right ) xw^{\prime }+\left ( k^{2}a^{2}x^{2k}+m^{2}-k^{2}n^{2}\right ) w=0 \tag{9} \end{equation} Has the required solution \(x^{m}J_{n}\left ( ax^{k}\right ) \).
To answer the final part about the relation between \(n,m,k\). One restriction is that \(m=\frac{1}{2}\). One relation between the order \(n\) and \(k\) is that \(m^{2}-k^{2}n^{2}\) being a rational number. This means\[ m^{2}-k^{2}n^{2}=\frac{N}{M}\] Where \(N,M\) are integers.
\begin{equation} y^{\prime \prime }\left ( x\right ) +x^{2}y\left ( x\right ) =0\tag{1} \end{equation} Comparing this ODE to one found in part (a), written below again, now using \(y\left ( x\right ) \) to make it easier to compare\begin{align} x^{2}y^{\prime \prime }\left ( x\right ) +\left ( 1-2m\right ) xy^{\prime }\left ( x\right ) +\left ( k^{2}a^{2}x^{2k}+m^{2}-k^{2}n^{2}\right ) y\left ( x\right ) & =0\nonumber \\ y^{\prime \prime }\left ( x\right ) +\frac{\left ( 1-2m\right ) }{x}y^{\prime }\left ( x\right ) +\frac{1}{x^{2}}\left ( k^{2}a^{2}x^{2k}+m^{2}-k^{2}n^{2}\right ) y\left ( x\right ) & =0\tag{2} \end{align}
To make (2) same as (1), we want \(\left ( 1-2m\right ) =0\) or \(m=\frac{1}{2}\). Also need \(2k=4\) or \(k=2\). Using these the above reduces to\[ y^{\prime \prime }\left ( x\right ) +\left ( 4a^{2}x^{2}+\frac{\frac{1}{4}-4n^{2}}{x^{2}}\right ) y\left ( x\right ) =0 \] Therefore, we need also that \(n^{2}=\frac{1}{16}\) in order to cancel extra term above. Hence \(n=\frac{1}{4}\). Now the above becomes\[ y^{\prime \prime }\left ( x\right ) +4a^{2}x^{2}y\left ( x\right ) =0 \] Finally, if we let \(a^{2}=\frac{1}{4}\) or \(a=\frac{1}{2}\), then the above becomes\[ y^{\prime \prime }\left ( x\right ) +x^{2}y\left ( x\right ) =0 \] Therefore, we found that \begin{align*} n & =\frac{1}{4}\\ a & =\frac{1}{2}\\ k & =2\\ m & =\frac{1}{2} \end{align*}
Hence the following solves the ODE\begin{align*} y\left ( x\right ) & =x^{m}J_{n}\left ( ax^{k}\right ) \\ & =\sqrt{x}J_{\frac{1}{4}}\left ( \frac{1}{2}x^{2}\right ) \end{align*}
To verify the above result, it is solved again directly. We first need to convert this ODE to Bessel ODE. Let \[ y=x^{\frac{1}{2}}z\left ( x\right ) \] Then\begin{align*} \frac{dy}{dx} & =\frac{1}{2}x^{-\frac{1}{2}}z+x^{\frac{1}{2}}z^{\prime }\\ \frac{d^{2}y}{dx^{2}} & =-\frac{1}{4}x^{-\frac{3}{2}}z+\frac{1}{2}x^{-\frac{1}{2}}z^{\prime }+\frac{1}{2}x^{-\frac{1}{2}}z^{\prime }+x^{\frac{1}{2}}z^{\prime \prime }\\ & =-\frac{1}{4}x^{-\frac{3}{2}}z+x^{-\frac{1}{2}}z^{\prime }+x^{\frac{1}{2}}z^{\prime \prime } \end{align*}
Substituting the above into (1) gives\begin{align*} \left ( -\frac{1}{4}x^{-\frac{3}{2}}z+x^{-\frac{1}{2}}z^{\prime }+x^{\frac{1}{2}}z^{\prime \prime }\right ) +x^{2}x^{\frac{1}{2}}z & =0\\ x^{\frac{1}{2}}z^{\prime \prime }+x^{-\frac{1}{2}}z^{\prime }+\left ( x^{\frac{5}{2}}-\frac{1}{4}x^{-\frac{3}{2}}\right ) z & =0 \end{align*}
Multiplying both sides by \(x^{\frac{3}{2}}\) gives\begin{equation} x^{2}z^{\prime \prime }+xz^{\prime }+\left ( x^{4}-\frac{1}{4}\right ) z=0\tag{2} \end{equation} Where the derivatives above is with respect to \(x\). Now let \(t=\frac{x^{2}}{2}\). Then \[ \frac{dz}{dx}=\frac{dz}{dt}\frac{dt}{dx}=x\frac{dz}{dt}\] And\begin{align*} \frac{d^{2}z}{dx^{2}} & =\frac{d^{2}z}{dt^{2}}\left ( \frac{dt}{dx}\right ) \left ( x\right ) +\frac{dz}{dt}\\ & =\frac{d^{2}z}{dt^{2}}x^{2}+\frac{dz}{dt} \end{align*}
Substituting the above into (2) gives\[ x^{2}\left ( x^{2}z^{\prime \prime }+z^{\prime }\right ) +x\left ( xz^{\prime }\right ) +\left ( x^{4}-\frac{1}{4}\right ) z=0 \] Where the derivatives above is with respect to \(t\) now. This simplifies to\[ x^{4}z^{\prime \prime }+2x^{2}z^{\prime }+\left ( x^{4}-\frac{1}{4}\right ) z=0 \] But \(t=\frac{x^{2}}{2}\), hence the above becomes\begin{align*} 4t^{2}z^{\prime \prime }+4tz^{\prime }+\left ( 4t^{2}-\frac{1}{4}\right ) z & =0\\ t^{2}z^{\prime \prime }+tz^{\prime }+\left ( t^{2}-\frac{1}{16}\right ) z & =0 \end{align*}
This now in the form of Bessel ODE\[ t^{2}z^{\prime \prime }+tz^{\prime }+\left ( t^{2}-n^{2}\right ) z=0 \] Where \(n=\frac{1}{4}\). Hence one solution is \begin{align*} z\left ( t\right ) & =J_{n}\left ( t\right ) \\ & =J_{\frac{1}{4}}\left ( t\right ) \end{align*}
But \(y\left ( x\right ) =\sqrt{x}z\left ( x\right ) \) and \(t=\frac{x^{2}}{2}\), therefore the above becomes\begin{equation} y\left ( x\right ) =\sqrt{x}J_{\frac{1}{4}}\left ( \frac{x^{2}}{2}\right ) \tag{3} \end{equation}
Which is the same as found in part (b)
Problem Prove that \(\left \vert J_{n}\left ( x\right ) \right \vert \leq 1\) for all integers \(n\)
Solution
From the integral representation of \(J_{n}\left ( x\right ) \) for integer \(n\)\[ J_{n}\left ( x\right ) =\frac{1}{\pi }\int _{0}^{\pi }\cos \left ( n\theta -x\sin \theta \right ) d\theta \] Then\begin{align*} \left \vert J_{n}\left ( x\right ) \right \vert & \leq \frac{1}{\pi }\left \vert \int _{0}^{\pi }\cos \left ( n\theta -x\sin \theta \right ) d\theta \right \vert _{\max }\\ & \leq \frac{1}{\pi }\int _{0}^{\pi }\left \vert \cos \left ( n\theta -x\sin \theta \right ) \right \vert _{\max }d\theta \\ & =\frac{1}{\pi }\left \vert M\right \vert _{\max }\int _{0}^{\pi }d\theta \\ & =\frac{1}{\pi }\left \vert M\right \vert _{\max }\pi \\ & =\left \vert M\right \vert _{\max } \end{align*}
Where \(\left \vert M\right \vert _{\max }=\left \vert \cos \left ( n\theta -x\sin \theta \right ) \right \vert _{\max }\) over \(\theta =0\cdots \pi \). But this is \(1\) for the cosine function. Hence\[ \left \vert J_{n}\left ( x\right ) \right \vert \leq 1 \]
Problem Starting with the integral formula for hypergeometric function, express the following in terms of elementary functions \(_{2}F_{1}\left ( 1,1,2;x\right ) \) and \(_{2}F_{1}\left ( a,1,1;x\right ) \)
Solution
\begin{align} _{2}F_{1}\left ( a,b,c;x\right ) & =\frac{\Gamma \left ( c\right ) }{\Gamma \left ( b\right ) \Gamma \left ( c-b\right ) }\int _{0}^{1}t^{b-1}\left ( 1-t\right ) ^{c-b-1}\left ( 1-tx\right ) ^{-a}dt\tag{1}\\ & =\frac{\Gamma \left ( c\right ) }{\Gamma \left ( a\right ) \Gamma \left ( b\right ) }\sum _{n=0}^{\infty }\frac{\Gamma \left ( a+n\right ) \Gamma \left ( b+n\right ) }{\Gamma \left ( c+n\right ) }\frac{x^{n}}{n!} \tag{2} \end{align}
Here \(a=1,b=1,c=2\). Therefore, using (1) representation gives\begin{align*} _{2}F_{1}\left ( 1,1,2;x\right ) & =\frac{\Gamma \left ( 2\right ) }{\Gamma \left ( 1\right ) \Gamma \left ( 2-1\right ) }\int _{0}^{1}t^{1-1}\left ( 1-t\right ) ^{2-1-1}\left ( 1-tx\right ) ^{-1}dt\\ & =\frac{\Gamma \left ( 2\right ) }{\Gamma \left ( 1\right ) \Gamma \left ( 1\right ) }\int _{0}^{1}\frac{dt}{1-tx} \end{align*}
But \(\Gamma \left ( 2\right ) =1,\Gamma \left ( 1\right ) =0\), therefore the above becomes\begin{align*} _{2}F_{1}\left ( 1,1,2;x\right ) & =\int _{0}^{1}\frac{dt}{1-tx}\\ & =\left [ \frac{-\ln \left ( 1-tx\right ) }{x}\right ] _{0}^{1}\\ & =-\left ( \frac{\ln \left ( 1-x\right ) }{x}-\frac{-\ln \left ( 1-0\right ) }{x}\right ) \\ & =-\frac{\ln \left ( 1-x\right ) }{x} \end{align*}
Here \(a=a,b=1,c=1\). Therefore (2) representation gives\begin{align*} _{2}F_{1}\left ( a,1,1;x\right ) & =\frac{\Gamma \left ( c\right ) }{\Gamma \left ( a\right ) \Gamma \left ( b\right ) }\sum _{n=0}^{\infty }\frac{\Gamma \left ( a+n\right ) \Gamma \left ( b+n\right ) }{\Gamma \left ( c+n\right ) }\frac{x^{n}}{n!}\\ & =\frac{1}{\Gamma \left ( a\right ) }\sum _{n=0}^{\infty }\frac{\Gamma \left ( a+n\right ) \Gamma \left ( 1+n\right ) }{\Gamma \left ( 1+n\right ) }\frac{x^{n}}{n!}\\ & =\sum _{n=0}^{\infty }\frac{\Gamma \left ( a+n\right ) }{\Gamma \left ( a\right ) }\frac{x^{n}}{n!} \end{align*}
Looking at few values
| \(n\) | \(_{2}F_{1}\left ( a,1,1;x\right ) \) |
| \(0\) | \(\frac{\Gamma \left ( a\right ) }{\Gamma \left ( a\right ) }=1\) |
| \(1\) | \(\frac{\Gamma \left ( a+1\right ) }{\Gamma \left ( a\right ) }x\) |
| \(2\) | \(\frac{\Gamma \left ( a+2\right ) }{\Gamma \left ( a\right ) }\frac{x^{2}}{2!}\) |
| \(3\) | \(\frac{\Gamma \left ( a+3\right ) }{\Gamma \left ( a\right ) }\frac{x^{3}}{3!}\) |
| \(\vdots \) | \(\vdots \) |
Using the recursive relation \(\Gamma \left ( a+1\right ) =a\Gamma \left ( a\right ) \), which works for integer and non integer \(a\), then we see that\[ \Gamma \left ( a+1\right ) =a\Gamma \left ( a\right ) \] And\begin{align*} \Gamma \left ( a+2\right ) & =\Gamma \left ( \left ( a+1\right ) +1\right ) \\ & =\left ( a+1\right ) \Gamma \left ( a+1\right ) \\ & =\left ( a+1\right ) a\Gamma \left ( a\right ) \end{align*}
And\begin{align*} \Gamma \left ( a+3\right ) & =\Gamma \left ( \left ( a+2\right ) +1\right ) \\ & =\left ( a+2\right ) \Gamma \left ( \left ( a+2\right ) \right ) \\ & =\left ( a+2\right ) \left ( a+1\right ) a\Gamma \left ( a\right ) \end{align*}
And so on. Hence the above now becomes
| \(n\) | \(_{2}F_{1}\left ( a,1,1;x\right ) \) |
| \(0\) | \(1\) |
| \(1\) | \(\frac{a\Gamma \left ( a\right ) }{\Gamma \left ( a\right ) }x=ax\) |
| \(2\) | \(\frac{\left ( a+1\right ) a\Gamma \left ( a\right ) }{\Gamma \left ( a\right ) }\frac{x^{2}}{2!}=a\left ( a+1\right ) \frac{x^{2}}{2!}\) |
| \(3\) | \(\frac{\left ( a+2\right ) \left ( a+1\right ) a\Gamma \left ( a\right ) }{\Gamma \left ( a\right ) }\frac{x^{3}}{3!}=a\left ( a+1\right ) \left ( a+2\right ) \frac{x^{3}}{3!}\) |
| \(\vdots \) | \(\vdots \) |
We see from the above the pattern of the sequence is as follows\begin{equation} _{2}F_{1}\left ( a,1,1;x\right ) =1+ax+a\left ( a+1\right ) \frac{x^{2}}{2!}+a\left ( a+1\right ) \left ( a+2\right ) \frac{x^{3}}{3!}+\cdots \tag{1} \end{equation} Comparing the above to the Binomial expansion given by\begin{equation} \left ( 1+z\right ) ^{n}=1+nz+n\left ( n-1\right ) \frac{z^{2}}{2!}+n\left ( n-1\right ) \left ( n-2\right ) \frac{z^{3}}{3!}+\cdots \tag{2} \end{equation} By replacing \(z\rightarrow -x\) and \(n\rightarrow -a\), the above becomes\begin{align*} \left ( 1-x\right ) ^{-a} & =1+\left ( -a\right ) \left ( -x\right ) +\left ( -a\right ) \left ( \left ( -a\right ) -1\right ) \frac{\left ( -x\right ) ^{2}}{2!}+\left ( -a\right ) \left ( \left ( -a\right ) -1\right ) \left ( \left ( -a\right ) -2\right ) \frac{\left ( -x\right ) ^{3}}{3!}+\cdots \\ & =1+ax+\left ( a\right ) \left ( a+1\right ) \frac{x^{2}}{2!}+\left ( a\right ) \left ( a+1\right ) \left ( a+2\right ) \frac{x^{3}}{3!}+\cdots \end{align*}
Comparing the above to (1) shows it is the same series. Hence\[ _{2}F_{1}\left ( a,1,1;x\right ) =\left ( 1-x\right ) ^{-a}\]