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

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Amanda Liu 2021-6-16

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/857290-how-do-i-supply-different-constants-into-a-function-of-a-loop

编辑： Amanda Liu 2021-6-16

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 2021-6-16

But you are already doing it. Your question is not clear.

Amanda Liu 2021-6-16

编辑：Amanda Liu 2021-6-16

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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Answer 1

Walter Roberson 2021-6-16

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/857290-how-do-i-supply-different-constants-into-a-function-of-a-loop#answer_725795

Start with the code I posted in response to another one of your Questions, https://www.mathworks.com/matlabcentral/answers/857275-loop-never-stops#answer_725790

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 2021-6-16

编辑：Amanda Liu 2021-6-16

在 MATLAB Online 中打开

I've transformed both of my constants of A and of B into Matrix so that at different position will have different values of k1 and k2. I changed my n to 20 but my final Tn is of size 400x400?

tic
n = 20;
T=sym(zeros(n,n));
T(1,:)   = 410; % Top
T(end,:) = 420; % Bottom
T(:,1)   = 400; % Left
T(:,end) = 400; % Right
T_old = T; 
for k = 1:20 %time steps  
    for j = 2: (n-1)
        for i = 2: (n-1)
           T(i,j) = expand(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
toc
% Constant A
c1 = 0.02;
c2 = 0.33;
% ConstantB at (x<=a & y<=b) & (x>=L-d & y<=c)
c3 = 0.04;
c4 = 0.67;
% Dimension A
L = 20;
H = 5;
% Dimension B
a = 1;
b = 3;
c = 1;
d = 3;
% Matrix for k1 
k1 = c1*ones(n,n); %constant A (c1)
k1(ceil(n*b/H : n) , ceil(1 : n*a/L)) = c3; %(x<=a & y<=b)
k1(ceil(n*(H-c)/H : n) , ceil(n*(L-d)/L : n)) = c3; %(x>=L-d & y<=c)
% Matrix for k2
k2 = c2*ones(n,n); %constant A (c2)
k2(ceil(n*b/H : n) , ceil(1 : n*a/L)) = c4; %(x<=a & y<=b)
k2(ceil(n*(H-c)/H) : n , ceil(n*(L-d)/L : n)) = c4; %(x>=L-d & y<=c)
Tn = subs(T);
vpa(Tn,5)

Walter Roberson 2021-6-16

在 MATLAB Online 中打开

H is not defined here, and you forgot to

syms k1 k2

Also

k1 = c1*ones(n,n); %constant A (c1)
k2 = c2*ones(n,n); %constant A (c2)

those are each 20 x 20

Tn = subs(T);

T is 20 x 20, and involves symbolic scalar variables k1 and k2.

When you subs() an array (k1, k2 are 20 x 20 each) in place of symbolic variables, then in every array element in T, the 20 x 20 numeric arrays k1 and k2 will be substituted for the symbolic variables k1, k2. So what was T(3,7) for example that involved symbolic scalars k1 and k2, will become a 20 x 20 array. But the same is true for all 400 (20 x 200) elements of T, so you end up with an array that is 400 x 400.

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.

Instead of getting a 400 x 400 array, with a little work you could get a 20 x 20 x 20 x 20 array -- like

T4(:,:,K,L) = subs(T, {sym('k1'), sym('k2')}, {k1(K,L), k2(K,L)})

However, if what you wanted to do was have

Tn(I,J) = subs(T(I,J), {sym('k1'), sym('k2')}, {k1(K,L), k2(K,L)})

which is to say that you want to substitute in corresponding k1, k2, then you need to do a bit more work. Take subsets of T and subs() scalar values into the subsets.

%do not construct numeric k1, k2 matrices.
rb1 = ceil(n*b/H);
cb1 = floor(n*a/L);
rb2 = ceil(N*(H-c)/H);
cb2 = ceil(N*(L-d)/L);
Tn = subs(T, {k1,k2}, {c1,c2});
Tn(rb1:n, cb1:n) = subs(T(rb1:n, cb1:n), {k1,k2}, {c3, c4});
Tn(rb2:n, cb2:n) = subs(T(rb2:n, cb2:n), {k1,k2}, {c3, c4});

Amanda Liu 2021-6-16

编辑：Amanda Liu 2021-6-16

Finally, i got everything to work! I've spent 1 week on this. Thank you so much!

Amanda Liu 2021-6-16

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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