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Suppose we have the following data:

  1. $ r=\frac{vk}{h^2}$

  2. $u[i,j]$ is a function of two variables with $i,j$ integers.

  3. $\frac{u[i,j+1]-u[i,j]}{k}=v\frac{u[i+1,j]-2u[i,j]+u[i-1,j]}{h^2}$

Then I would like to show using Mathematica that the following equation is true:

$u[i,j+1]=(1-2r)u[i,j]+r(u[i+1,j]+u[i-1,j])$

This is trivial to do by hand but when it comes to prove it with Reduce, it get nowhere...more precisly I give the following commands:

r=v k/h^2;

u[i_,j_];

(u[i,j+1]-u[i,j])/k==v (u[i+1,j]-2 u[i,j]+u[i-1,j])/h^2;

Reduce[u[i,j+1]=(1-2 r) u[i,j]+r (u[i+1,j]+u[i-1,j])]

And what I get are strange messages....Can someone tell me what I am doing wrong or propose a better way to do that? Thanks.

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Perhaps,

ClearAll[r, i, j, u, v, k, h]
Simplify[u[i, j + 1] == (1 - 2 r) u[i, j] + r (u[i + 1, j] + u[i - 1, j]),
 {r == v k/h^2, (u[i, j + 1] - u[i, j])/k == v (u[i + 1, j] - 2 u[i, j] + u[i - 1, j])/h^2}
]

True

| improve this answer | |
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  • $\begingroup$ Thanks for the answer, but it does not seem to work in the notebook I open in the Wolfram Cloud... I get as a result:u[i,1+j]==(1-2 r) u[i,j]+r (u[-1+i,j]+u[1+i,j]) $\endgroup$ – dmtri Jan 25 at 10:27

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