Solution
\[ y^{\prime \prime }-2xy^{\prime }+2ny=0\qquad -\infty <x<\infty \]
\[ g\left ( x,t\right ) =\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}\] Differentiating w.r.t \(x\), and assuming term by term differentiation is allowed, gives\[ \frac{\partial g\left ( x,t\right ) }{\partial x}=\sum _{n=0}^{\infty }H_{n}^{\prime }\left ( x\right ) \frac{t^{n}}{n!}\] Using \(H_{n}^{\prime }\left ( x\right ) =2nH_{n-1}\left ( x\right ) \) in the above results in\[ \frac{\partial g\left ( x,t\right ) }{\partial x}=\sum _{n=0}^{\infty }2nH_{n-1}\left ( x\right ) \frac{t^{n}}{n!}\] But for \(n=0\), the first term is zero, so the sum can start from \(1\) and give the same result\[ \frac{\partial g\left ( x,t\right ) }{\partial x}=\sum _{n=1}^{\infty }2nH_{n-1}\left ( x\right ) \frac{t^{n}}{n!}\] Now, decreasing the summation index by \(1\) and increasing the \(n\) inside the sum by \(1\) gives\begin{align*} \frac{\partial g\left ( x,t\right ) }{\partial x} & =\sum _{n=0}^{\infty }2\left ( n+1\right ) H_{n}\left ( x\right ) \frac{t^{n+1}}{\left ( n+1\right ) !}\\ & =\sum _{n=0}^{\infty }2\left ( n+1\right ) H_{n}\left ( x\right ) \frac{t^{n+1}}{\left ( n+1\right ) n!}\\ & =\sum _{n=0}^{\infty }2H_{n}\left ( x\right ) \frac{t^{n+1}}{n!}\\ & =\sum _{n=0}^{\infty }2t\left ( H_{n}\left ( x\right ) \frac{t^{n}}{n!}\right ) \\ & =2t\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!} \end{align*}
But \(\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}=g\left ( x,t\right ) \) and the above reduces to\[ \frac{\partial g\left ( x,t\right ) }{\partial x}=2tg\left ( x,t\right ) \] The problem says it is supposed to be a first order differential equation and not a first order partial differential equation. Therefore, by assuming \(x\) to be a fixed parameter instead of another independent variable, the above can now be written as\[ \frac{d}{dx}g\left ( x,t\right ) -2tg\left ( x,t\right ) =0 \]
From the solution found in part (1)\begin{align*} \frac{\frac{d}{dx}g\left ( x,t\right ) }{g\left ( x,t\right ) } & =2t\\ \frac{dg\left ( x,t\right ) }{g\left ( x,t\right ) } & =2tdx \end{align*}
Integrating both sides gives\begin{align*} \int \frac{dg\left ( x,t\right ) }{g\left ( x,t\right ) } & =\int 2tdx\\ \ln \left \vert g\left ( x,t\right ) \right \vert & =2tx+C\\ g\left ( x,t\right ) & =e^{2tx+C}\\ g\left ( x,t\right ) & =C_{1}e^{2tx} \end{align*}
Where \(C_{1}=e^{C}\) a new constant. Let \(g\left ( 0,t\right ) =g_{0}\) then the above shows that \(C_{1}=g_{0}\) and the above can now be written as\[ g\left ( x,t\right ) =g\left ( 0,t\right ) e^{2tx}\]
Using the given definition of \(g\left ( x,t\right ) =\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}\) and when \(x=0\) then\begin{align*} g\left ( 0,t\right ) & =\sum _{n=0}^{\infty }H_{n}\left ( 0\right ) \frac{t^{n}}{n!}\\ & =H_{0}\left ( 0\right ) +H_{1}\left ( 0\right ) +\sum _{n=2}^{\infty }H_{n}\left ( 0\right ) \frac{t^{n}}{n!} \end{align*}
But \(H_{0}\left ( x\right ) =1\), hence \(H_{0}\left ( 0\right ) =1\) and \(H_{1}\left ( x\right ) =2x\), hence \(H_{1}\left ( 0\right ) =0\) and the above becomes\[ g\left ( 0,t\right ) =1+\sum _{n=2}^{\infty }H_{n}\left ( 0\right ) \frac{t^{n}}{n!}\] For the remaining series, it can be written as sum of even and odd terms\[ g\left ( 0,t\right ) =1+\sum _{n=2,4,6,\cdots }^{\infty }H_{n}\left ( 0\right ) \frac{t^{n}}{n!}+\sum _{n=3,5,7,\cdots }^{\infty }H_{n}\left ( 0\right ) \frac{t^{n}}{n!}\] Or, equivalently\[ g\left ( 0,t\right ) =1+\sum _{n=1,2,3,\cdots }^{\infty }H_{2n}\left ( 0\right ) \frac{t^{2n}}{\left ( 2n\right ) !}+\sum _{n=1,2,3,\cdots }^{\infty }H_{2n+1}\left ( 0\right ) \frac{t^{2n+1}}{\left ( 2n+1\right ) !}\] But using the hint given that \(H_{2n+1}\left ( 0\right ) =0\) and \(H_{2n}\left ( 0\right ) =\frac{\left ( -1\right ) ^{n}\left ( 2n\right ) !}{n!}\) the above simplifies to \begin{align*} g\left ( 0,t\right ) & =1+\sum _{n=1,2,3,\cdots }^{\infty }\frac{\left ( -1\right ) ^{n}\left ( 2n\right ) !}{n!}\frac{t^{2n}}{\left ( 2n\right ) !}\\ & =1+\sum _{n=1,2,3,\cdots }^{\infty }\left ( -1\right ) ^{n}\frac{t^{2n}}{n!} \end{align*}
But since \(\left ( -1\right ) ^{n}\frac{t^{2n}}{n!}=1\) when \(n=0\), then the above sum can be made to start as zero and it simplifies to\[ g\left ( 0,t\right ) =\sum _{n=0}^{\infty }\left ( -1\right ) ^{n}\frac{t^{2n}}{n!}\] Therefore the solution \(g\left ( x,t\right ) =g\left ( 0,t\right ) e^{tx}\) found in part (2) becomes\begin{equation} g\left ( x,t\right ) =\left ( \sum _{n=0}^{\infty }\left ( -1\right ) ^{n}\frac{t^{2n}}{n!}\right ) e^{2tx} \tag{1} \end{equation} Now the sum \(\sum _{n=0}^{\infty }\left ( -1\right ) ^{n}\frac{t^{2n}}{n!}=1-t^{2}+\frac{t^{4}}{2!}-\frac{t^{6}}{3!}+\cdots \) and comparing this sum to standard series of \(e^{z}=1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\cdots \), then this shows that when \(z=-t^{2}\) and series for \(e^{-t^{2}}\) becomes\begin{align*} e^{-t^{2}} & =1+\left ( -t^{2}\right ) +\frac{\left ( -t^{2}\right ) ^{2}}{2!}+\frac{\left ( -t^{2}\right ) ^{3}}{3!}+\frac{\left ( -t^{2}\right ) ^{4}}{4!}\cdots \\ & =1-t^{2}+\frac{t^{4}}{2!}-\frac{t^{6}}{3!}+\frac{t^{8}}{4!}\cdots \end{align*}
Hence \[ \sum _{n=0}^{\infty }\left ( -1\right ) ^{n}\frac{t^{2n}}{n!}=e^{-t^{2}}\] Substituting this into (1) gives\begin{align*} g\left ( x,t\right ) & =e^{-t^{2}}e^{2tx}\\ & =e^{2tx-t^{2}} \end{align*}
Since \(g\left ( x,t\right ) =e^{2tx-t^{2}}\) from part (3), then\begin{align*} \frac{\partial }{\partial t}g\left ( x,t\right ) & =\left ( 2x-2t\right ) e^{2tx-t^{2}}\\ & =\left ( 2x-2t\right ) g\left ( x,t\right ) \end{align*}
But \(g\left ( x,t\right ) =\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}\), therefore the above can be written as\begin{align} \frac{\partial }{\partial t}g\left ( x,t\right ) & =\left ( 2x-2t\right ) \sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}\nonumber \\ & =2x\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}-2t\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}\nonumber \\ & =2x\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}-2\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n+1}}{n!}\nonumber \\ & =2x\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}-2\sum _{n=1}^{\infty }H_{n-1}\left ( x\right ) \frac{t^{n}}{\left ( n-1\right ) !}\nonumber \\ & =2x\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}-2\sum _{n=1}^{\infty }nH_{n-1}\left ( x\right ) \frac{t^{n}}{n\left ( n-1\right ) !}\nonumber \\ & =2x\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}-2\sum _{n=1}^{\infty }nH_{n-1}\left ( x\right ) \frac{t^{n}}{n!} \tag{1} \end{align}
On the other hand, \begin{align*} \frac{\partial }{\partial t}g\left ( x,t\right ) & =\frac{\partial }{\partial t}\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}\\ & =\sum _{n=0}^{\infty }nH_{n}\left ( x\right ) \frac{t^{n-1}}{n!} \end{align*}
Since at \(n=0\) the sum is zero, then it can be started from \(n=1\) without changing the result\begin{align} \frac{\partial }{\partial t}g\left ( x,t\right ) & =\sum _{n=1}^{\infty }nH_{n}\left ( x\right ) \frac{t^{n-1}}{n!}\nonumber \\ & =\sum _{n=0}^{\infty }\left ( n+1\right ) H_{n+1}\left ( x\right ) \frac{t^{n}}{\left ( n+1\right ) !}\nonumber \\ & =\sum _{n=0}^{\infty }\left ( n+1\right ) H_{n+1}\left ( x\right ) \frac{t^{n}}{\left ( n+1\right ) n!}\nonumber \\ & =\sum _{n=0}^{\infty }H_{n+1}\left ( x\right ) \frac{t^{n}}{n!} \tag{2} \end{align}
Equating (1) and (2) gives\[ \sum _{n=0}^{\infty }H_{n+1}\left ( x\right ) \frac{t^{n}}{n!}=2x\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}-2\sum _{n=1}^{\infty }nH_{n-1}\left ( x\right ) \frac{t^{n}}{n!}\] But \(\sum _{n=1}^{\infty }nH_{n-1}\left ( x\right ) \frac{t^{n}}{n!}=\sum _{n=0}^{\infty }nH_{n-1}\left ( x\right ) \frac{t^{n}}{n!}\) because at \(n=0\) it is zero, so it does not affect the result to start the sum from zero, and now the above can be written as\[ \sum _{n=0}^{\infty }H_{n+1}\left ( x\right ) \frac{t^{n}}{n!}=2x\sum _{n=0}^{\infty }H_{n}\left ( x\right ) \frac{t^{n}}{n!}-2\sum _{n=0}^{\infty }nH_{n-1}\left ( x\right ) \frac{t^{n}}{n!}\] Now since all the sums start from \(n=0\) then the above means the same as\[ H_{n+1}\left ( x\right ) \frac{t^{n}}{n!}=2xH_{n}\left ( x\right ) \frac{t^{n}}{n!}-2nH_{n-1}\left ( x\right ) \frac{t^{n}}{n!}\] Canceling \(\frac{t^{n}}{n!}\) from each term gives\[ H_{n+1}\left ( x\right ) =2xH_{n}\left ( x\right ) -2nH_{n-1}\left ( x\right ) \] Which is the result required to show.
The problem is asking to show that \[ \int _{-\infty }^{\infty }e^{-x^{2}}H_{m}\left ( x\right ) H_{n}\left ( x\right ) dx=\left \{ \begin{array} [c]{cc}0 & n\neq m\\ 2^{n}n!\sqrt{\pi } & n=m \end{array} \right . \] The first part below will show the case for \(n\neq m\) and the second part part will show the case for \(n=m\)
case \(n\neq m\) This is shown by using the differential equation directly. I found this method easier and more direct. Before starting, the ODE \(y^{\prime \prime }-2xy^{\prime }+2ny=0\) is rewritten as\begin{equation} e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}y^{\prime }\right ) +2ny=0 \tag{1} \end{equation} The above form is exactly the same as the original ODE as can be seen by expanding it. Now, Let \(H_{n}\left ( x\right ) \) be one solution to (1) and let \(H_{m}\left ( x\right ) \) be another solution to (1) which results in the following two ODE’s\begin{align} e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }\right ) +2nH_{n} & =0\tag{1A}\\ e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }\right ) +2mH_{m} & =0 \tag{2A} \end{align}
Multiplying (1A) by \(H_{m}\) and (2A) by \(H_{n}\) and subtracting gives\begin{align} H_{m}\left ( e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }\right ) +2nH_{n}\right ) -H_{n}\left ( e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }\right ) +2mH_{m}\right ) & =0\nonumber \\ \left ( H_{m}e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }\right ) +2nH_{n}H_{m}\right ) -\left ( H_{n}e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }\right ) +2mH_{n}H_{m}\right ) & =0\nonumber \\ H_{m}e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }\right ) -H_{n}e^{x^{2}}\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }\right ) +2\left ( n-m\right ) H_{n}H_{m} & =0\nonumber \\ H_{m}\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }\right ) -H_{n}\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }\right ) +2\left ( n-m\right ) H_{n}H_{m}e^{-x^{2}} & =0 \tag{3} \end{align}
But \[ H_{m}\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }\right ) =\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }H_{m}\right ) -e^{-x^{2}}H_{n}^{\prime }H_{m}^{\prime }\] And\[ H_{n}\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }\right ) =\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }H_{n}\right ) -e^{-x^{2}}H_{m}^{\prime }H_{n}^{\prime }\] Therefore\begin{align*} H_{m}\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }\right ) -H_{n}\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }\right ) & =\left ( \frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }H_{m}\right ) -e^{-x^{2}}H_{n}^{\prime }H_{m}^{\prime }\right ) -\left ( \frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }H_{n}\right ) -e^{-x^{2}}H_{m}^{\prime }H_{n}^{\prime }\right ) \\ & =\frac{d}{dx}\left ( e^{-x^{2}}H_{n}^{\prime }H_{m}\right ) -\frac{d}{dx}\left ( e^{-x^{2}}H_{m}^{\prime }H_{n}\right ) \\ & =\frac{d}{dx}\left ( e^{-x^{2}}\left ( H_{n}^{\prime }H_{m}-H_{m}^{\prime }H_{n}\right ) \right ) \end{align*}
Substituting the above relation back into (3) gives\[ \frac{d}{dx}\left ( e^{-x^{2}}\left ( H_{n}^{\prime }H_{m}-H_{m}^{\prime }H_{n}\right ) \right ) +2\left ( n-m\right ) H_{n}H_{m}e^{-x^{2}}=0 \] Integrating gives\begin{align*} \int _{-\infty }^{\infty }\frac{d}{dx}\left ( e^{-x^{2}}\left ( H_{n}^{\prime }H_{m}-H_{m}^{\prime }H_{n}\right ) \right ) dx+\int _{-\infty }^{\infty }2\left ( n-m\right ) H_{n}H_{m}e^{-x^{2}}dx & =0\\ \int _{-\infty }^{\infty }d\left ( e^{-x^{2}}\left ( H_{n}^{\prime }H_{m}-H_{m}^{\prime }H_{n}\right ) \right ) +2\left ( n-m\right ) \int _{-\infty }^{\infty }H_{n}H_{m}e^{-x^{2}}dx & =0\\ \left [ e^{-x^{2}}\left ( H_{n}^{\prime }H_{m}-H_{m}^{\prime }H_{n}\right ) \right ] _{-\infty }^{\infty }+2\left ( n-m\right ) \int _{-\infty }^{\infty }H_{n}H_{m}e^{-x^{2}}dx & =0 \end{align*}
But \(\lim _{x\rightarrow \pm \infty }e^{-x^{2}}\rightarrow 0\) so the first term above vanishes and the above becomes\[ 2\left ( n-m\right ) \int _{-\infty }^{\infty }H_{n}H_{m}e^{-x^{2}}dx=0 \] Since this is the case where \(n\neq m\) then the above shows that \[ \int _{-\infty }^{\infty }H_{n}H_{m}e^{-x^{2}}dx=0\qquad n\neq m \] Now the case \(n=m\) is proofed. When \(H_{n}=H_{m}\) then the integral becomes \(\int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx\). Using the known Rodrigues formula for Hermite polynomials, given by \[ H_{n}\left ( x\right ) =\left ( -1\right ) ^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}\] Then applying the above the above to one of the \(H_{n}\left ( x\right ) \) in the integral \(\int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx\), gives\begin{align*} \int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx & =\int _{-\infty }^{\infty }\left ( \left ( -1\right ) ^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}e^{-x^{2}}\right ) H_{n}e^{-x^{2}}dx\\ & =\left ( -1\right ) ^{n}\int _{-\infty }^{\infty }\left ( \frac{d^{n}}{dx^{n}}e^{-x^{2}}\right ) H_{n}dx \end{align*}
Now integration by parts is carried out. \(\int udv=uv-\int vdu\). Let \(u=H_{n}\) and let \(dv=\frac{d^{n}}{dx^{n}}e^{-x^{2}}\), therefore \(du=H_{n}^{\prime }\left ( x\right ) =2nH_{n-1}\left ( x\right ) \) and \(v=\frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}\), therefore\[ \int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx=\left ( -1\right ) ^{n}\left ( \left [ H_{n}\left ( x\right ) \frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}\right ] _{-\infty }^{\infty }-\int _{-\infty }^{\infty }\left ( \frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}\right ) 2nH_{n-1}\left ( x\right ) dx\right ) \] But \(\left [ H_{n}\left ( x\right ) \frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}\right ] _{-\infty }^{\infty }\rightarrow 0\) as \(x\rightarrow \pm \infty \) because each derivative of \(\frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}\) produces a term with \(e^{-x^{2}}\) which vanishes at both ends of the real line. Hence the above integral now becomes\[ \int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx=\left ( -1\right ) ^{n}\left ( -2n\int _{-\infty }^{\infty }\left ( \frac{d^{n-1}}{dx^{n-1}}e^{-x^{2}}\right ) H_{n-1}\left ( x\right ) dx\right ) \] Now the process is repeated, doing one more integration by parts. This results in\[ \int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx=\left ( -1\right ) ^{n}\left ( -2n\left ( -2\left ( n-1\right ) \int _{-\infty }^{\infty }\left ( \frac{d^{n-2}}{dx^{n-2}}e^{-x^{2}}\right ) H_{n-2}\left ( x\right ) dx\right ) \right ) \] And again\[ \int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx=\left ( -1\right ) ^{n}\left ( -2n\left ( -2\left ( n-1\right ) \left ( -2\left ( n-2\right ) \int _{-\infty }^{\infty }\left ( \frac{d^{n-3}}{dx^{n-3}}e^{-x^{2}}\right ) H_{n-3}\left ( x\right ) dx\right ) \right ) \right ) \] This process continues \(n\) times. After \(n\) integrations by parts, the above becomes\begin{align*} \int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx & =\left ( -1\right ) ^{n}\left ( -2n\left ( -2\left ( n-1\right ) \left ( -2\left ( n-2\right ) \left ( \cdots \left ( \int _{-\infty }^{\infty }e^{-x^{2}}H_{0}\left ( x\right ) dx\right ) \right ) \right ) \right ) \right ) \\ & =\left ( -1\right ) ^{n}\left ( -2\right ) ^{n}n!\int _{-\infty }^{\infty }e^{-x^{2}}H_{0}\left ( x\right ) dx\\ & =2^{n}n!\int _{-\infty }^{\infty }e^{-x^{2}}H_{0}\left ( x\right ) dx \end{align*}
But \(H_{0}\left ( x\right ) =1\), therefore the above becomes\[ \int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx=2^{n}n!\int _{-\infty }^{\infty }e^{-x^{2}}dx \] But \begin{align*} \int _{-\infty }^{\infty }e^{-x^{2}} & =2\int _{0}^{\infty }e^{-x^{2}}\\ & =2\frac{\sqrt{\pi }}{2}\\ & =\sqrt{\pi } \end{align*}
Therefore\[ \int _{-\infty }^{\infty }H_{n}H_{n}e^{-x^{2}}dx=2^{n}n!\sqrt{\pi }\] This completes the case for \(n=m.\) Hence\[ \int _{-\infty }^{\infty }e^{-x^{2}}H_{m}\left ( x\right ) H_{n}\left ( x\right ) dx=\left \{ \begin{array} [c]{cc}0 & n\neq m\\ 2^{n}n!\sqrt{\pi } & n=m \end{array} \right . \] Which is what the problem asked to show.
Solution
\[ y^{\prime \prime }\left ( r\right ) +\frac{1}{r}y^{\prime }\left ( r\right ) -\frac{n^{2}}{r^{2}}y\left ( r\right ) =0\qquad 0<r<\infty \] Or\[ r^{2}y^{\prime \prime }\left ( r\right ) +ry^{\prime }\left ( r\right ) -n^{2}y\left ( r\right ) =0 \] case \(n=0\)
The ode becomes \(r^{2}y^{\prime \prime }\left ( r\right ) +ry^{\prime }\left ( r\right ) =0\). Let \(z=y^{\prime }\) and it becomes \(r^{2}z^{\prime }\left ( r\right ) +rz\left ( r\right ) =0\) or \(z^{\prime }\left ( r\right ) +\frac{1}{r}z\left ( r\right ) =0\). This is linear in \(z\left ( r\right ) \). Integrating factor is \(I=e^{\int \frac{1}{r}dr}=r\). Multiplying the ode by \(I\) it becomes exact differential \(\frac{d}{dr}\left ( zr\right ) =0\) or \(d\left ( zr\right ) =0\), hence \(z=\frac{c_{1}}{r}\) where \(c_{1}\) is constant of integration. Therefore\[ y^{\prime }\left ( r\right ) =\frac{c_{1}}{r}\] Integrating again gives \[ y\left ( r\right ) =\frac{c_{1}}{\ln r}+c_{2}\] Since \(\lim _{r\rightarrow 0}\) the solution is bounded, then \(c_{1}\) must be zero. Therefore \(0=c_{2}\) and this implies \(c_{2}=0\) also. Therefore when \(n=0\) the solution is \[ y\left ( r\right ) =0 \]
Case \(n\neq 0\)
Since powers of \(r\) is the same as order of derivative in each term, this is an Euler ODE. It is solved by assuming \(y=r^{\alpha }\). Hence \(y^{\prime }=\alpha r^{\alpha -1},y^{\prime \prime }=\alpha \left ( \alpha -1\right ) r^{\alpha -2}\). Substituting these into the above ODE gives\begin{align*} r^{2}\alpha \left ( \alpha -1\right ) r^{\alpha -2}+r\alpha r^{\alpha -1}-n^{2}r^{\alpha } & =0\\ \alpha \left ( \alpha -1\right ) r^{\alpha }+\alpha r^{\alpha }-n^{2}r^{\alpha } & =0\\ r^{\alpha }\left ( \alpha \left ( \alpha -1\right ) +\alpha -n^{2}\right ) & =0 \end{align*}
Assuming non-trivial solution \(r^{\alpha }\neq 0\), then the indicial equation is\begin{align*} \alpha \left ( \alpha -1\right ) +\alpha -n^{2} & =0\\ \alpha ^{2} & =n^{2}\\ \alpha & =\pm n \end{align*}
Hence one solution is \[ y_{1}\left ( r\right ) =r^{n}\] And second solution is \[ y_{2}\left ( r\right ) =r^{-n}\] And the general solution is linear combination of these solutions\[ y\left ( r\right ) =c_{1}r^{n}+c_{2}r^{-n}\] The above shows that \(\lim _{r\rightarrow 0}y_{1}\left ( r\right ) =0\) and \(\lim _{r\rightarrow \infty }y_{2}\left ( r\right ) =0\).
Short version of the solution
To simplify the notations, \(r_{0}\) is used instead of \(r^{\prime }\) in all the following.\[ y^{\prime \prime }\left ( r\right ) +\frac{1}{r}y^{\prime }\left ( r\right ) -\frac{n^{2}}{r^{2}}y\left ( r\right ) =\frac{1}{r}\delta \left ( r-r_{0}\right ) \qquad 0<r<\infty \] Multiplying both sides by \(r\) the above becomes\begin{equation} ry^{\prime \prime }\left ( r\right ) +y^{\prime }\left ( r\right ) -\frac{n^{2}}{r}y\left ( r\right ) =\delta \left ( r-r_{0}\right ) \tag{1} \end{equation} But the two solutions2 to the homogeneous ODE \(ry^{\prime \prime }\left ( r\right ) +y^{\prime }\left ( r\right ) -\frac{n^{2}}{r}y\left ( r\right ) =0\) were found in part (a). These are\begin{align} y_{1}\left ( r\right ) & =r^{n}\tag{1A}\\ y_{2}\left ( r\right ) & =r^{-n}\nonumber \end{align}
The Green function is the solution to\begin{align} rG\left ( r,r_{0}\right ) +G\left ( r,r_{0}\right ) -\frac{n^{2}}{r}G\left ( r,r_{0}\right ) & =\delta \left ( r-r_{0}\right ) \tag{1B}\\ \lim _{r\rightarrow 0}G\left ( r,r_{0}\right ) & =0\nonumber \\ \lim _{r\rightarrow \infty }G\left ( r,r_{0}\right ) & =0\nonumber \end{align}
Which is given by (Using class notes, Lecture December 5, 2018) as\begin{equation} G\left ( r,r_{0}\right ) =\frac{1}{C}\left \{ \begin{array} [c]{ccc}y_{1}\left ( r\right ) y_{2}\left ( r_{0}\right ) & & 0<r<r_{0}\\ y_{1}\left ( r_{0}\right ) y_{2}\left ( r\right ) & & r_{0}<r<\infty \end{array} \right . \tag{2} \end{equation} Note, I used \(\frac{+1}{C}\) and not \(\frac{-1}{C}\) as in class notes, since I am using \(L=-\left ( \left ( py^{\prime }\right ) ^{\prime }-qy\right ) \) as the operator and not \(L=+\left ( \left ( py^{\prime }\right ) ^{\prime }+qy\right ) \). Now \(C\) is given by \[ C=p\left ( r_{0}\right ) \left ( y_{1}\left ( r_{0}\right ) y_{2}^{\prime }\left ( r_{0}\right ) -y_{1}^{\prime }\left ( r_{0}\right ) y_{2}\left ( r_{0}\right ) \right ) \] Where from (1A) we see that \begin{align*} y_{1}\left ( r_{0}\right ) & =r_{0}^{n}\\ y_{2}^{\prime }\left ( r_{0}\right ) & =-nr_{0}^{-n-1}\\ y_{1}^{\prime }\left ( r_{0}\right ) & =nr_{0}^{n-1}\\ y_{2}\left ( r_{0}\right ) & =r_{0}^{-n} \end{align*}
Therefore \(C\) becomes\begin{align*} C & =p\left ( r_{0}\right ) \left ( -nr_{0}^{-n-1}r_{0}^{n}-nr_{0}^{n-1}r_{0}^{-n}\right ) \\ & =2nr_{0}^{-1}p\left ( r_{0}\right ) \end{align*}
We just need now to find \(p\left ( r_{0}\right ) \). This comes from Sturm Liouville form. We need to convert the ODE \(r^{2}y^{\prime \prime }\left ( r\right ) +ry^{\prime }\left ( r\right ) -n^{2}y\left ( r\right ) =0\) to Sturm Liouville. Writing this ODE as \(ay^{\prime \prime }+by^{\prime }+\left ( c+\lambda \right ) y=0\) where \(a=r^{2},b=r,c=0,\lambda =-n^{2}\), therefore\begin{align*} p & =e^{\int \frac{b}{a}dr}=e^{\int \frac{r}{r^{2}}dr}=r\\ q & =-p\frac{c}{a}=0\\ \rho & =\frac{p}{a}=\frac{r}{r^{2}}=\frac{1}{r} \end{align*}
Hence the SL form is \(\left ( py^{\prime }\right ) ^{\prime }-qy+\lambda \rho y=0\). Hence the SL form is \(\left ( py^{\prime }\right ) ^{\prime }-qy+\lambda \rho y=0\) or\begin{equation} \left ( ry^{\prime }\right ) ^{\prime }-\frac{1}{r}n^{2}y=0\tag{2A} \end{equation} Hence the operator is \(L\left [ y\right ] =-\left ( \frac{d}{dr}\left ( r\frac{d}{dr}\right ) \right ) \left [ y\right ] \) and in standard form it becomes \(L\left [ y\right ] +\frac{1}{r}n^{2}y=0\).
The above shows that \(p\left ( r_{0}\right ) =r_{0}\). Therefore \[ C=2n \] Hence Green function is now found from (2) as, for \(n\neq 0\)\[ G\left ( r,r_{0}\right ) =\frac{1}{2n}\left \{ \begin{array} [c]{ccc}r^{n}r_{0}^{-n} & & 0<r<r_{0}\\ r_{0}^{n}r^{-n} & & r_{0}<r<\infty \end{array} \right . \] Since \(f\left ( r\right ) \) in the original ODE is zero, there is nothing to convolve with. i.e. \(y\left ( r\right ) =\int _{0}^{\infty }G\left ( r,r_{0}\right ) f\left ( r_{0}\right ) dr_{0}\) here is not needed since there is no \(f\left ( r\right ) \). Therefore the above is the final solution.
Extended solution
This solution shows derivation of (2) above. It can be considered as an appendix. The Green function is the solution to\begin{align} rG\left ( r,r_{0}\right ) +G\left ( r,r_{0}\right ) -\frac{n^{2}}{r}G\left ( r,r_{0}\right ) & =\delta \left ( r-r_{0}\right ) \tag{1B}\\ \lim _{r\rightarrow 0}G\left ( r,r_{0}\right ) & =0\nonumber \\ \lim _{r\rightarrow \infty }G\left ( r,r_{0}\right ) & =0\nonumber \end{align}
In (1B), \(r_{0}\) is the location of the impulse and \(r\) is the location of the observed response due to this impulse. The solution to the above ODE is now broken to two regions\begin{equation} G\left ( r,r_{0}\right ) =\left \{ \begin{array} [c]{ccc}A_{1}y_{1}\left ( r\right ) +A_{2}y_{2}\left ( r\right ) & & 0<r<r_{0}\\ B_{1}y_{1}\left ( r\right ) +B_{1}y_{2}\left ( r\right ) & & r_{0}<r<\infty \end{array} \right . \tag{2} \end{equation} Where \(y_{1}\left ( r\right ) ,y_{2}\left ( r\right ) \) are the solution to \(ry^{\prime \prime }\left ( r\right ) +y^{\prime }\left ( r\right ) -\frac{n^{2}}{r}y\left ( r\right ) =0\) and these were found in part (a) to be \(y_{1}\left ( r\right ) =r^{n},y_{2}\left ( r\right ) =r^{-n}\) and \(A_{1},A_{2},B_{1},B_{2}\) needs to be determined. Hence (2) becomes\begin{equation} G\left ( r,r_{0}\right ) =\left \{ \begin{array} [c]{ccc}A_{1}r^{n}+A_{2}r^{-n} & & 0<r<r_{0}\\ B_{1}r^{n}+B_{2}r^{-n} & & r_{0}<r<\infty \end{array} \right . \tag{3} \end{equation} The left boundary condition \(\lim _{r\rightarrow 0}\) \(G\left ( r,r_{0}\right ) =0\) implies \(A_{2}=0\) and the right boundary condition \(\lim _{r\rightarrow \infty }\) \(G\left ( r,r_{0}\right ) =0\) implies \(B_{1}=0\). This is needed to keep the solution bounded. Hence (3) simplifies to\begin{equation} G\left ( r,r_{0}\right ) =\left \{ \begin{array} [c]{ccc}A_{1}r^{n} & & 0<r<r_{0}\\ B_{2}r^{-n} & & r_{0}<r<\infty \end{array} \right . \tag{4} \end{equation} To determine the remaining two constants \(A_{1},B_{2}\), two additional conditions are needed. The first is that \(G\left ( r,r_{0}\right ) \) is continuous at \(r=r_{0}\) which implies\begin{equation} A_{1}r_{0}^{n}=B_{2}r_{0}^{-n}\tag{5} \end{equation} The second condition is the jump in the derivative of \(G\left ( r,r_{0}\right ) \) given by\[ \left . \frac{d}{dr}G\left ( r,r_{0}\right ) \right \vert _{r>r_{0}}-\left . \frac{d}{dr}G\left ( r,r_{0}\right ) \right \vert _{r<r_{0}}=\frac{-1}{p\left ( r_{0}\right ) }\] Where \(p\left ( r_{0}\right ) \) comes from the Sturm Liouville form of the homogeneous ODE. This was found above as \(p\left ( r_{0}\right ) =r_{0}\). Hence the above condition becomes\[ \left . \frac{d}{dr}G\left ( r,r_{0}\right ) \right \vert _{r>r_{0}}-\left . \frac{d}{dr}G\left ( r,r_{0}\right ) \right \vert _{r<r_{0}}=\frac{-1}{r_{0}}\] Equation (4) shows that \(\left . \frac{d}{dr}G\left ( r,r_{0}\right ) \right \vert _{r>r_{0}}=-nB_{2}r_{0}^{-n-1}\) and that \(\left . \frac{d}{dr}G\left ( r,r_{0}\right ) \right \vert _{r<r_{0}}=nA_{1}r_{0}^{n-1}\). Using these in the above gives the second equation needed\begin{equation} -nB_{2}r_{0}^{-n-1}-nA_{1}r_{0}^{n-1}=\frac{-1}{r_{0}}\tag{6} \end{equation} Solving (5,6) for \(A_{1},B_{1}\): From (5) \(A_{1}=B_{2}r_{0}^{-2n}\). Substituting this in (6) gives\begin{align*} -nB_{2}r_{0}^{-n-1}-n\left ( B_{2}r_{0}^{-2n}\right ) r^{n-1} & =\frac{-1}{r_{0}}\\ -nB_{2}r^{-n-1}-nB_{2}r^{-n-1} & =\frac{-1}{r_{0}}\\ -2nB_{2}r_{0}^{-n-1} & =-r_{0}^{-1}\\ B_{2} & =\frac{-r_{0}^{-1}}{-2nr_{0}^{-n-1}}\\ & =\frac{1}{2n}r_{0}^{n} \end{align*}
But since \(A_{1}=B_{2}r_{0}^{-2n}\), then \begin{align*} A_{1} & =\frac{1}{2n}r_{0}^{n}r_{0}^{-2n}\\ & =\frac{1}{2n}r_{0}^{-n} \end{align*}
Therefore the solution (4), which is the Green function, becomes, for \(n\neq 0\)\begin{equation} G\left ( r,r_{0}\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2n}r_{0}^{-n}r^{n} & & 0<r<r_{0}\\ \frac{1}{2n}r_{0}^{n}r^{-n} & & r_{0}<r<\infty \end{array} \right . \tag{7} \end{equation}