Solving a linear difference equation subject to boundary conditions

Hi everyone,

I read this on a book, which says we have a linear difference equation:

p_{k}= 1/2 (p_{k+1} + p_{K-1}) if 0<k<N,

s.t. boundary conditions: p_{0}= 1, p_{N}= 0.

The solution is also provided however I don't get it, it says:

Put b_{k} = p_{k} - p_{k-1}, to obtain b_{k} = b_{k-1} and hance b_{k} = b_{1} for all k

[How do we get b_{k} = b_{k-1} ?]

Thus, p_{k} = b_{1} + p_{k-1} = 2b_{1} + p_{k-2} =...= kb_{1} + p_{0}

Boundary conditions imply p_{0} = 1, b_{1} = -1/N

so we have p_{k} = 1 - (k/N)

At first glance everything looks fine...but when I read it again I don't quite understand how does b_{k} = b_{k-1}? Could anyone explain why it is the case?

Thank you very much for your help :)

Re: Solving a linear difference equation subject to boundary conditions

alrite seems no one wants to answer this Q probably because it is too easy...:(

after read it a few more times I think I see it

so I just answer myself:

to see this: split LHS p_{k} in to 1/2(p_{k} + p_{k}) and the first equation becomes 1/2(p_{k} + p_{k})= 1/2 (p_{k+1} + p_{K-1})

rewrite above in to: 1/2(p_{k+1}-p_{k}) = 1/2(p_{k}-p_{K-1}) which is: 1/2b_{k+1} = 1/2b_{k} gives us b_{k} = b_{k-1}