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}