# general solution to linear second order difference equation

Is there a general solution to difference equations of the form: $$u(n+2) + a(n)u(n+1) + b(n)u(n) = 0$$ Thank you in advance

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I attended a talk from Antonio Duran once where he solved something like this. I do not recall everything, but he was looking at polynomials solutions $\{p_n\}$ of the the difference equation (all over $\mathbb{C}$) $$a_n \bigtriangleup \bigtriangledown p_n + b_n \bigtriangledown p_n = \lambda_n p_n$$ where $\bigtriangleup(f) = f(x+1) - f(x)$ and $\bigtriangledown(f) = f(x) - f(x-1)$. All these solutions where completely solved by Lancaster in 1941. The classes of solutions are Charlier polynomials, Meixner polynomials, Krawtchouk polynomials and Hahn polynomials.

I think that is is possible to rewrite this difference equation to a difference equation similar to yours. It may not solve you problem completely, but you have some solutions for special cases.

You could have a look at: Orthogonal polynomials defined by difference equations from Lancaster and Orthogonal Polynomials Satisfying Higher-Order Difference Equations from Duran.

I hope this helps, good luck!

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You mean there is no known general solution. – Amr May 30 '12 at 15:45
is it impossible to find one written using the summation operator – Amr May 30 '12 at 15:46
That's right. For some special, restricted forms of $a(n)$ and $b(n)$ there might be solutions, but there won't be a solution in such generality. – Gerry Myerson May 31 '12 at 10:01
When both $a(n)$ and $b(n)$ are polynomials, using 'integral kernel method' can lead to become an ODE problem. – doraemonpaul Jun 1 '12 at 3:15
@doraemonpaul, and is there a solution to the ODE problem? – Gerry Myerson Jun 1 '12 at 3:40

Besides linear ODEs (http://en.wikipedia.org/wiki/Abel%27s_identity#Proof_using_Liouville.27s_formula)(Does this ODE question have closed form solution?), linear difference equations also have similar tricks:

In fact $u(n+2)+a(n)u(n+1)+b(n)u(n)=0$ can rewrite to the matrix form

$\begin{pmatrix}u(n+1)\\u(n+2)\end{pmatrix}=\begin{pmatrix}0&1\\-b(n)&-a(n)\end{pmatrix}\begin{pmatrix}u(n)\\u(n+1)\end{pmatrix}$

$\begin{pmatrix}u(n)\\u(n+1)\end{pmatrix}=\begin{pmatrix}0&1\\-b(n-1)&-a(n-1)\end{pmatrix}\begin{pmatrix}u(n-1)\\u(n)\end{pmatrix}$

This becomes to a matrix difference equation.

For the discrete version $\begin{pmatrix}u_n\\u_{n+1}\end{pmatrix}=\begin{pmatrix}0&1\\-b(n-1)&-a(n-1)\end{pmatrix}\begin{pmatrix}u_{n-1}\\u_n\end{pmatrix}$ ,

$\begin{pmatrix}u_n\\u_{n+1}\end{pmatrix}=\prod_{k=n-1}^c\begin{pmatrix}0&1\\-b(k)&-a(k)\end{pmatrix}\begin{pmatrix}u_c\\u_{c+1}\end{pmatrix}$

$\therefore u_n=C_1f(n)+C_2g(n)$ , where $C_1$ and $C_2$ are arbitrary constants, $f(n)$ and $g(n)$ are the top-left and top-right elements of the simplified matrix expression $\prod_{k=n-1}^c\begin{pmatrix}0&1\\-b(k)&-a(k)\end{pmatrix}$ respectively.

Now for the continuous version $\begin{pmatrix}u(n)\\u(n+1)\end{pmatrix}=\begin{pmatrix}0&1\\-b(n-1)&-a(n-1)\end{pmatrix}\begin{pmatrix}u(n-1)\\u(n)\end{pmatrix}$

$\therefore u(n)=T_1(n)f(n)+T_2(n)g(n)$ , where $T_1(n)$ and $T_2(n)$ are arbitrary periodic functions with period $1$, $f(n)$ and $g(n)$ are the top-left and top-right elements of the simplified and continuous extended matrix expression $\prod_{k=n-1}^c\begin{pmatrix}0&1\\-b(k)&-a(k)\end{pmatrix}$ respectively.

But the simplification process and the continuous extension process of $\prod_{k=n-1}^c\begin{pmatrix}0&1\\-b(k)&-a(k)\end{pmatrix}$ are super complicated.

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