Difference equation with one definition for odd $n$ and another for $n$ even.

Question

Let us define the difference equation:

$u_{n+1} - au_n + u_{n-1} = 0 $, if $n$ is odd,

$u_{n+1} - bu_n + u_{n-1} = 0$, if $n$ is even.

I've been struggling to solve this equation for the last few days and I'm very stuck. I know that if either of the equations above was standalone $\forall n$ it would be a simple characteristic equation, but I'm not sure what to do here with how to split into an odd or even cases.

I thought about writing $u_n$ in a piecewise way:

$u_n=v_n$ for $n$ odd and $u_n = w_n$ for $n$ even. And treating it as a system:

$v_{n+1} - av_n + v_{n-1} = 0$

$w_{n+1} - bw_n + w_{n-1} = 0$$

But I didn't get very far with this approach. Thank you for your help.

Answer 1

Call $x_n=u_{2n}$ and $y_n=u_{2n+1}$. Then $$\begin{pmatrix}x_{n+1}\\ y_{n+1}\end{pmatrix}=\begin{pmatrix}au_{2n+1}-u_{2n}\\ bu_{2n+2}-u_{2n+1}\end{pmatrix}=\begin{pmatrix}ay_n-x_n\\ bx_{n+1}-y_n\end{pmatrix}=\begin{pmatrix}ay_n-x_n\\ bay_n-bx_n-y_n\end{pmatrix}=\\=\begin{pmatrix}-1&a\\ -b& ba-1\end{pmatrix}\begin{pmatrix}x_n\\ y_n\end{pmatrix}$$

Therefore $\begin{pmatrix}u_{2n}\\ u_{2n+1}\end{pmatrix}=\begin{pmatrix}-1 &a\\ -b& ba-1\end{pmatrix}^n\begin{pmatrix}u_0\\ u_1\end{pmatrix}$, which may be calculated with standard methods via the Jordan normal form of the matrix.

Answer 2

Okay lets look:

Odd case: $$u_{n+1} - au_n + u_{n-1} = 0\tag{1}$$ Even case: $$u_{n+1} - bu_n + u_{n-1} = 0\tag{2}$$ Rearranged Even case: $$u_{n+1}=bu_n - u_{n-1}\tag{2a}$$ Index edited Even case: $$u_{n}=bu_{n-1} - u_{n-2}\tag{2b}$$ Substitution Odd case: $$u_{n+1} - a(bu_{n-1} - u_{n-2}) + u_{n-1} = 0\tag{1a}$$ Disribution Odd case: $$u_{n+1} - abu_{n-1} - au_{n-2} + u_{n-1} = 0\tag{1b}$$ Odd case: $$u_{n+1} - au_n + u_{n-1} = 0\tag{1}$$ Even case: $$u_{n+1} - bu_n + u_{n-1} = 0\tag{2}$$ Rearranged Odd case: $$u_{n+1}=au_n - u_{n-1}\tag{1c}$$ Index edited Odd case: $$u_{n}=au_{n-1} - u_{n-2}\tag{1d}$$ Substitution Even case: $$u_{n+1} - b(au_{n-1} - u_{n-2}) + u_{n-1} = 0\tag{2c}$$ Disribution Even case: $$u_{n+1} - abu_{n-1} - bu_{n-2} + u_{n-1} = 0\tag{2d}$$

Answer 3

Here is an approach by rearranging the recurrence relations and separating them in even and odd parts.

We start with the given relations consisting of even part, odd part and initial conditions. \begin{align*} &u_{2n+2}-au_{2n+1}+u_{2n}=0\qquad\qquad n\geq0\\ &u_{2n+1}-bu_{2n}+u_{2n-1}=0\\ &u_0,u_1 \end{align*}

We obtain for $n\geq0$: \begin{align*} u_{2n+1}=\frac{1}{a}\left(u_{2n+2}+u_{2n}\right),\qquad\qquad u_{2n}=\frac{1}{b}\left(u_{2n+1}+u_{2n-1}\right) \end{align*} and we obtain recurrence relations where even and odd terms are separated. \begin{align*} &u_{2n+2}+(2-ab)u_{2n}+u_{2n-2}=0\qquad\qquad u_0,u_2=au_1-u_0\tag{1}\\ &u_{2n+3}+(2-ab)u_{2n+1}+u_{2n-1}=0\qquad\quad u_1,u_3=(ab-1)u_1-bu_0\tag{2} \end{align*} The recurrence relations (1) and (2) can be solved independently and can then be combined to give a final result.

Using a generating function approach $$U(z)=\sum_{n=0}^\infty u_nz^n$$ with even function $U_{E}(z)=\sum_{n=0}^\infty u_{2n}z^{2n}$ and odd function $U_{O}(z)=\sum_{n=0}^\infty u_{2n+1}z^{2n+1}$,

we can calculate $U_{E}(z)$:

\begin{align*} \sum_{n=2}^\infty u_{2n}&=\sum_{n=0}^\infty u_{2n+4}z^{2n+4}\\ &=\sum_{n=0}^\infty (ab-2)u_{2n+2}z^{2n+4}-\sum_{n=0}^\infty u_{2n}z^{2n+4}\\ &=(ab-2)z^2\sum_{n=1}^\infty u_{2n}z^{2n}-z^4\sum_{n=0}^\infty u_{2n}z^{2n}\\ &=(ab-2)z^2\left(U_{E}-u_0\right)-z^4U_{E}(z)\\ U_{E}(z)-u_0-u_2z^2&=(ab-2)z^2\left(U_{E}-u_0\right)-z^4U_{E}(z)\\ \color{blue}{U_{E}(z)}&\color{blue}{=\frac{u_0+\left((1-ab)u_0+au_1\right)z^2}{1+(2-ab)z^2+z^4}}\tag{3} \end{align*}

and similarly we obtain \begin{align*} \color{blue}{U_{O}(z)}&\color{blue}{=\frac{u_1z+\left(u_1-bu_0\right)z^3}{1+(2-ab)z^2+z^4}}\tag{4} \end{align*}

Note the equal denominator in (3) and (4) showing that $U_{E}(z)$ and $U_{O}(z)$ have essentially the same structure. This is also indicated by the recurrence relations (1) and (2). Combining (3) and (4) we find a generating function \begin{align*} U(z)&=U_{E}(z)+U_{O}(z)\\ &=\frac{u_0+u_1z+\left(au_1+(1-ab)u_0\right)z^2+(u_1-bu_0)z^3}{1+(2-ab)z^2+z^4}\tag{5}\\ &=u_0+u_1z+(au_1-u_0)z^2+\left((ab-1)u_1-bu_0\right)z^3\\ &\qquad+\left(\left(a^2b-2a\right)u_1+\left(1-ab\right)u_0\right)z^4\\ &\qquad+\left(\left(a^2b^2-3ab+1\right)u_1-\left(ab^2-2b\right)u_0\right)z^5+\cdots \end{align*} The expansion of (5) was calculated with some help of WolframAlpha. The manual way to derive a representation of $u_n$ from $U(z)$ is usually done by partial fraction decomposition of (5) followed by a geometric series expansion of the partial fractions.