Solving the Laguerre Equation: 0 as Regular Singular Point

[evinda]

Hello! (Wave)

The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given.

• Show that the equation has $0$ as its singular regular point .
• Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (m \in \mathbb{R})$
• Show that if $a=n$, where $n \in \mathbb{N}$ then there is a polynomial solution of degree $n$.
• Let $L_n$ the polynomial $L_n(x)=e^x \frac{d^n}{dx^n} (x^n \cdot e^{-x})$ (show that it is a polynomial), $n=1,2,3, \dots$. Show that $L_n$ satisfies the equation Laguerre if $a=n(n=1,2, \dots)$

That's what I have tried:

• For $x \neq 0$ the differential equation can be written as: $y''+ \frac{1-x}{x}y'+ \frac{a}{x}y=0$.
  $p(x)= \frac{1-x}{x}, q(x)= \frac{a}{x}$
  The functions $x \cdot p(x)= 1-x, \ x^2q(x)=ax$ can be written as power series in a region of $0$.
  Thus, $0$ is a singular regular point.
• We suppose that there is a solution of the form $y(x)=x^m \sum_{n=0}^{\infty} a_n x^n= \sum_{n=0}^{\infty} a_n x^{n+m}$.
  Then $y'(x)= \sum_{n=0}^{\infty} a_n(n+m) x^{n+m-1} \Rightarrow -xy'(x)= \sum_{n=0}^{\infty} -a_n(n+m) x^{n+m}$ and $y''(x)= \sum_{n=0}^{\infty} a_n(n+m)(n+m-1) x^{n+m-2} \Rightarrow xy''(x)= \sum_{n=0}^{\infty} a_n(n+m)(n+m-1) x^{n+m-1}$So we have $\sum_{n=0}^{\infty} a_n (n+m)(n+m-1) x^{n+m-1}+ \sum_{n=0}^{\infty} a_n (n+m) x^{n+m-1} + \sum_{n=1}^{\infty} -a_{n-1} (n+m-1) x^{n+m-1}+ \sum_{n=1}^{\infty} a a_{n-1} x^{n+m-1}=0 \\ \Rightarrow a_0 m (m-1) x^{m-1}+ a_0 m x^{m-1}+ \sum_{n=1}^{\infty} \left[ a_n (n+m) (n+m-1)+ a_n(n+m)-a_{n-1}(n+m-1)+ a a_{n-1} \right] x^{n+m-1}=0$
  
  It has to hold:
  
  $$a_0 m^2=0 \overset{a_0 \neq 0}{ \Rightarrow } m=0$$
  
  $$a_n (n+m) (n+m-1)+ a_n (n+m)- a_{n-1} (n+m-1)+ a a_{n-1}=0$$
  
  For $m=0$: $a_{n} n (n-1)+ a_n n-a_{n-1} (n-1)+ a a_{n-1}=0 \Rightarrow a_n n^2+ a_{n-1} (a-n+1)=0 \Rightarrow a_n=- \frac{a_{n-1}(a-n+1)}{n^2}$
  
  For $n=1: \ a_1=-aa_0$
  For $n=2: \ a_2= \frac{aa_0(a-1)}{2^2} $
  For $n=3: \ a_3=- \frac{aa_0(a-1) (a-2)}{2^2 3^2} $
  For $n=4: \ a_4= \frac{aa_0(a-1)(a-2)(a-3)}{2^2 3^2 4^2} $
  For $n=5: \ a_5= -\frac{aa_0(a-1)(a-2)(a-3)(a-4)}{2^2 3^2 4^2 5^2} $
  
  We see that $a_n=(-1)^n a_0 \frac{\prod_{i=0}^{n-1} (a-i)}{\prod_{i=2}^n i^2}$$$\frac{a_{n+1}}{a_n}=(-1) \frac{a-n}{(n+1)^2} \to 0$$
  
  So the series $\sum_{n=0}^{\infty} a_n x^n$ converges and its radius of convergence is equal to $+\infty$.

Is it right so far? Could you give me a hint what we could do in order to answer the other two questions?

 

[jvicens]

Yes, your solution for the first part is correct. For the second part, we can use the fact that $L_n$ is a polynomial to show that it satisfies the differential equation. We have:

$$L_n(x) = e^x \frac{d^n}{dx^n}(x^n e^{-x}) = e^x \left(\frac{d^n}{dx^n}x^n \right)e^{-x} + x^n \frac{d^n}{dx^n}(e^{-x})$$

Using the product rule and the fact that $\frac{d}{dx}e^{-x}=-e^{-x}$, we can simplify this to:

$$L_n(x) = e^x \left(n(n-1)(n-2)...(n-n+1)x^{n-1} \right)e^{-x} + (-1)^n x^n e^{-x}$$

Now, we can see that this is equivalent to:

$$L_n(x) = n(n-1)(n-2)...(n-n+1)x^{n-1} + (-1)^n x^n$$

Using the definition of $p(x)$ and $q(x)$ from the first part, we can rewrite this as:

$$L_n(x) = n(n-1)p(x)p(x-1)...p(x-n+1)x^{n-1} + (-1)^n x^n$$

From this, we can see that $L_n(x)$ satisfies the differential equation Laguerre with $a=n$.

For the third part, we can use this result to show that if $a=n$, where $n \in \mathbb{N}$, then there is a polynomial solution of degree $n$. We have already shown that $L_n(x)$ satisfies the differential equation, so we just need to show that it is a polynomial of degree $n$. From the definition of $L_n(x)$, we can see that the highest power of $x$ in $L_n(x)$ is $n$. Therefore, $L_n(x)$ is a polynomial of degree $n$ and satisfies the differential equation.