# How do I supply different constants into a function of a loop?

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Amanda Liu el 16 de Jun. de 2021
Editada: Amanda Liu el 16 de Jun. de 2021
My code is currently supplying 2 constants of A only. How do I supply another 2 constants of B into the same function?
n=30;
% Initialization
T=zeros(n,n);
T(1,:) = 410; % Top
T(end,:) = 420; % Bottom
T(:,1) = 400; % Left
T(:,end) = 400; % Right
T_old = T;
% Constants of A
k1 = 0.02;
k2 = 0.33;
% Constants of B
% k1 = 0.04;
% k2 = 0.67;
for k = 1:20 %time steps
for j = 2: (n-1)
for i = 2: (n-1)
T(i,j) = T_old(i,j)*(1-2*k1-2*k2)+k1*(T_old(i-1,j)+T_old(i+1,j))+k2*(T_old(i,j-1)+T_old(i,j+1));
end
end
T_old = T;
end
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KSSV el 16 de Jun. de 2021
Amanda Liu el 16 de Jun. de 2021
Editada: Amanda Liu el 16 de Jun. de 2021
I'm so sorry that I'm not good in English and I didn't explain it clearly enough.
What I meant was I have 2 values of k1 and 2 values of k2. But the function in my loop only takes 1 value of k1 and 1 value of k2 from A.
There are 2 values of k1 and of k2 due to different x and y intervals.

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Walter Roberson el 16 de Jun. de 2021
and subs() different k1, k2 values as desired.
If you were careful enough about how you proceeded, you could create 2D array of combinations of k1, k2 values, move those into the 3rd and 4th dimension, and subs() once, to get a single 4D array.... though visualizing the result could be tricky!
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Amanda Liu el 16 de Jun. de 2021
Editada: Amanda Liu el 16 de Jun. de 2021
Finally, i got everything to work! I've spent 1 week on this. Thank you so much!
Amanda Liu el 16 de Jun. de 2021
I've also tried the numeric matrices approach by turning the scalar k into vector k(i,j). It is much faster compared with symbolic method. But anyway, thank you!

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