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Let $0<\alpha<1$ and $N\in\mathbb{N}$. I would like to solve the following difference equation

$$n=0: s_0 = 1+\alpha s_1 $$ $$n\in\{1,\dots,N-2\}: s_n = 1+\alpha s_{n+1}+(1-\alpha)s_n$$ $$n=N-1:s_{N-1}= 1+\alpha s_0 + (1-\alpha)s_{N-1}$$

and the boundary condition $s_N=0$. I was able to calculate the homogeneous solution, i.e. the solution to the equations:

$$n=0: s_0 = 1+\alpha s_1 $$ $$n\in\{1,\dots,N-2\}: s_n = \alpha s_{n+1}+(1-\alpha)s_n$$ $$n=N-1:s_{N-1}= 1+\alpha s_0 + (1-\alpha)s_{N-1}$$

to be $s_n = \frac{1}{\alpha},0<n<N$, $s_0=0$. Is this correct and how can I now solve this for a non-homogneous casse? To get the homogeneous solution I used that for $0<n<N$ we see that $s_n$ are all equal. Then I sued the equation for $n-1$ and $0$ to find the particular values for $s_0$ and all other values

swissy
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1 Answers1

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$$n\in\{1,\dots,N-2\}: s_n = 1+\alpha s_{n+1}+(1-\alpha)s_n \\ \implies -1 = \alpha (s_{n+1}-s_n). $$

Therefore $s_1, s_2, \ldots, s_{N-1}$ is an arithmetic sequence.

Can you end it now?

Neat Math
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  • thanks for your hint, does this mean we have $s_n = s_1 - \frac{n-1}{\alpha}$ for $n\in{1,\dots,N-1}$? – swissy Nov 21 '20 at 15:25
  • It should be $s_n = s_1-\frac{n-1}{\alpha}$. – Neat Math Nov 21 '20 at 15:27
  • yes indeed! just edited while you were typing.. Is this now the general solution, or just for the non-homogeneous part? I.e. do I need to construct the solution via the sum of the solution of the homogeneous and the non homogeneous one? I'm a bit confused which to take – swissy Nov 21 '20 at 15:31
  • This is general, but you still have to go back to the other two equations and solve for $s_1$. – Neat Math Nov 21 '20 at 15:33