Matrix Population from formula

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Alexander Brooks
Alexander Brooks am 20 Nov. 2024 um 17:54
Bearbeitet: Saurabh am 21 Nov. 2024 um 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.
  3 Kommentare
William Rose
William Rose am 20 Nov. 2024 um 18:15
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 am 20 Nov. 2024 um 18:32
Bearbeitet: Alexander Brooks am 20 Nov. 2024 um 18:44
That is what this is. What would a code for this look like? u(6,6)= 100

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Antworten (1)

Saurabh
Saurabh am 20 Nov. 2024 um 18:08
Bearbeitet: Saurabh am 21 Nov. 2024 um 4:08
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:
I hope this helps.
  3 Kommentare
John D'Errico
John D'Errico am 20 Nov. 2024 um 20:50
Bearbeitet: John D'Errico am 20 Nov. 2024 um 20:53
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 am 20 Nov. 2024 um 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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