Matrix Population from formula

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Alexander Brooks
Alexander Brooks 2024-11-20，17:54
编辑： Saurabh 2024-11-21，4:08
I have a i x j matrix that follows the formula....
[(u(i-1,j)-2*u(i,j)+u(i+1,j)/(.2^2)]-[(u(i,j)-u(i,j-1))/(.2)]=0
How do I populate a 6x6 matrix using this as the formula
I have the following boundary conditions: if i=1 u=0, if i=6 u=100, if j=1 u=0, if j=6 u=0.
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William Rose
William Rose 2024-11-20，18:15
@Alexander Brooks,
The equation ofr u(i,j) looks like a second derivative with respect to i, minus a first derivative with respect to j, where the delta value equals 0.2 for the i and j directions.
Start with matrix u having the boundary conditons along the boundaries, and zero everywhere else. Or let the initial value be a linear gradienmt from 0 to 100 in the u direction. Then apply the equaitons for u(ij) inthe interior, once. See what you have. Then do it again, and again, and again. See if it is evloving twoard a constant value, or getting more oscillatory, or something else.
By the way, your boundary contions are problematic: Does u(6,6)=0 or 100?
Alexander Brooks
Alexander Brooks 2024-11-20，18:32
编辑：Alexander Brooks 2024-11-20，18:44
That is what this is. What would a code for this look like? u(6,6)= 100

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Saurabh
Saurabh 2024-11-20，18:08
编辑：Saurabh 2024-11-21，4:08
Hi @Alexander Brooks,
I understand that you want to populate a 6x6 matrix using the given formula. In MATLAB, you can use the 'zeros' function to create a matrix of any size with initial values set to zero. You can then iterate over the matrix and apply the formula to each element.
And then explicitly assign the boundary condition later.
For more information on the 'zeros' function referred to the following documentation:
https://www.mathworks.com/help/matlab/ref/zeros.html
I hope this helps.
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John D'Errico
John D'Errico 2024-11-20，20:50
编辑：John D'Errico 2024-11-20，20:53
@Alexander Brooks
Just use a loop! Surely you can use a nested loops, over i and j? TRY IT! You will get better and more help if you make an effort.
Walter Roberson
Walter Roberson 2024-11-20，20:59
It seems to me that you would likely have to loop several times to get the matrix to settle down.
On the other hand, it looks like a matrix that is all zero satisfies the conditions.

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