I have a homogeneous recurrence equation.

$x_i = (1-p)x_{i-1} + px_{i+1}, p \in (0,1)$

Using the classical methods (solving the characteristic polynomial) I can show that

if $p=1/2$, then the sequences described by $x_i = c_1+c_2i$ are solutions $\forall c_1,c_2 \in \mathbb R$.

if $p \ne 1/2$, then the sequences described by $x_i = c_1+c_2(\frac{1-p}{p})^i$ are solutions $\forall c_1,c_2 \in \mathbb R$

According to a theorem here (point 2) there is no other solutions.

I am now asked to explain why there are no other solutions. Is the only explanation a reference to the theorem/ its proof or there is a straightforward argument why in my case there could be no other solutions?


$c_1$ and $c_2$ are determined by two given $x_i$ and $x_k$ (for different indices). Suppose now that there is another solution $y$ not identical to the initial solution $x$ but which also satisfies the given xs. The difference of this other solution and your initial solution, write it as $d_l=x_l-y_l$, solves the difference equation as well with $d_i=d_k=0$. If $d_{i\pm 1}=0$ then the difference equation gives, successively, that all $d_l$s are zero, contradicting that $x$ and $y$ are not identical. So the $d_{i\pm 1}$ elements have to be different from zero. But then this non zero $d$ is then propagated by the difference equation and we get $d_k\neq 0$, a contradiction again.

  • $\begingroup$ could you please write it a bit more formally. Currently I do not see why "It then follows that the difference is the zero sequence", the difference sequence is indeed $0$ in points $i,k$, but in all other (infinite number of) points it can differ from $0$, can not it? $\endgroup$ – Sergey Zykov Nov 2 '15 at 13:22
  • $\begingroup$ I clarified my answer. Hope its better now. $\endgroup$ – Elsa Nov 2 '15 at 14:28

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