Loading web-font TeX/Size2/Regular

2.10 HW 10

2.10.1 HW 10 questions

PDF (letter size)

PDF (legal size)

2.10.2 Problem 1

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.

2.10.3 Problem 2

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

2.10.3.1 Part (a)

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.

2.10.3.2 Part (b)

$$\begin{equation} y^{\prime \prime }\left ( x\right ) +x^{2}y\left ( x\right ) =0\tag{1} \end{equation}$$ 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

$$\begin{align*} y\left ( x\right ) & =x^{m}J_{n}\left ( ax^{k}\right ) \\ & =\sqrt{x}J_{\frac{1}{4}}\left ( \frac{1}{2}x^{2}\right ) \end{align*}$$

2.10.3.3 Appendix

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)

2.10.4 Problem 3

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

$$\left \vert J_{n}\left ( x\right ) \right \vert \leq 1$$

2.10.5 Problem 4

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

$$\begin{align} _{2}F_{1}\left ( a,b,c;x\right ) & =\frac{\Gamma \left ( c\right ) }{\Gamma \left ( b\right ) \Gamma \left ( c-b\right ) }\int _{0}^{1}t^{b-1}\left ( 1-t\right ) ^{c-b-1}\left ( 1-tx\right ) ^{-a}dt\tag{1}\\ & =\frac{\Gamma \left ( c\right ) }{\Gamma \left ( a\right ) \Gamma \left ( b\right ) }\sum _{n=0}^{\infty }\frac{\Gamma \left ( a+n\right ) \Gamma \left ( b+n\right ) }{\Gamma \left ( c+n\right ) }\frac{x^{n}}{n!} \tag{2} \end{align}$$

2.10.5.1 Part (a)

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

$$\begin{align*} _{2}F_{1}\left ( 1,1,2;x\right ) & =\int _{0}^{1}\frac{dt}{1-tx}\\ & =\left [ \frac{-\ln \left ( 1-tx\right ) }{x}\right ] _{0}^{1}\\ & =-\left ( \frac{\ln \left ( 1-x\right ) }{x}-\frac{-\ln \left ( 1-0\right ) }{x}\right ) \\ & =-\frac{\ln \left ( 1-x\right ) }{x} \end{align*}$$

2.10.5.2 Part (b)

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

$$_{2}F_{1}\left ( a,1,1;x\right ) =\left ( 1-x\right ) ^{-a}$$

2.10.6 Key solution for HW 10

PDF