matrix from a matrix matlab

I am trying to get the function to output an array T that has each value inside the fixed outer rows and columns, averaged with itself and the 4 numbers surrounding it. I made X to recieve all 9 of the values from my larger array, S to select only the ones I wanted and A to use when averaging, yet it will not work, I believe the problem lies in the X(ii,jj) = T((ii-1):(ii+1), (jj-1):(jj+1)). Any help much appreciated

function T = tempsim(rows, cols, top开发者_C百科NsideTemp, bottomTemp, tol)
T = zeros(rows,cols);
T(1,:) = topNsideTemp;
T(:,1) = topNsideTemp;
T(:,rows) = topNsideTemp;
T(rows,:) = bottomTemp;
S = [0 1 0; 1 1 1; 0 1 0]; 
X = zeros(3,3);
A = zeros(3,3);
for ii = 2:(cols-1);
    jj = 2:(rows-1);
    X(ii,jj) = T((ii-1):(ii+1), (jj-1):(jj+1))
    A = X.*S;
    T = (sum(sum(A)))/5

---

What you are doing looks like a convolution as Jouni points out. So using that knowledge, I came up with following code:

function T = tempsim(rows, cols, topNsideTemp, bottomTemp, tol)
sz = [rows,cols];

topEdge    = sub2ind(sz, ones(1,cols)     , 1:cols);
bottomEdge = sub2ind(sz, ones(1,cols)*rows, 1:cols);
leftEdge   = sub2ind(sz, 1:rows           , ones(1,rows));
rightEdge  = sub2ind(sz, 1:rows           , ones(1,rows)*cols);

otherEdges = [topEdge leftEdge rightEdge];
edges      = [bottomEdge otherEdges];

%% set initial grid
T0             = zeros(sz);
T0(otherEdges) = topNsideTemp;
T0(bottomEdge) = bottomTemp;

%% average filter
F = [0 1 0 
     1 1 1 
     0 1 0]; 
F = F/sum(F(:));

%% simulation
T = T0; % initial condition
T = conv2(T, F, 'same');
T(edges) = T0(edges); % this keeps the edges set to the initial values

If you run this, you will get following results:

T = tempsim(10,10,100,-100)
T0 =

   100   100   100   100   100   100   100   100   100   100
   100     0     0     0     0     0     0     0     0   100
   100     0     0     0     0     0     0     0     0   100
   100     0     0     0     0     0     0     0     0   100
   100     0     0     0     0     0     0     0     0   100
   100     0     0     0     0     0     0     0     0   100
   100     0     0     0     0     0     0     0     0   100
   100     0     0     0     0     0     0     0     0   100
   100     0     0     0     0     0     0     0     0   100
   100  -100  -100  -100  -100  -100  -100  -100  -100   100


T =

   100   100   100   100   100   100   100   100   100   100
   100    40    20    20    20    20    20    20    40   100
   100    20     0     0     0     0     0     0    20   100
   100    20     0     0     0     0     0     0    20   100
   100    20     0     0     0     0     0     0    20   100
   100    20     0     0     0     0     0     0    20   100
   100    20     0     0     0     0     0     0    20   100
   100    20     0     0     0     0     0     0    20   100
   100     0   -20   -20   -20   -20   -20   -20     0   100
   100  -100  -100  -100  -100  -100  -100  -100  -100   100

I also showed T0 for clarity as you can see that T(2,2) == 40, which is equal to (100 + 100 + 0 + 0 + 0)/5 from the same position in T0.

From the context, I guess you'll be studying the convergence of this problem. If that's the case, you will have to repeat the last 2 lines until it converges.

But depending on your actual problem, I think you can improve the initial conditions to speed up convergence by initializing the grid to a temperature different from 0. In the current code your boundary conditions will heat up the complete grid, which takes some time. If you just provide a proper guess for the bulk temperature (in lieu of 0), this can speed up the convergence considerably. In my example I need about 40 steps for convergence up to a certain tolerance, with a proper guess (50 in my case) this can be reduced to about 20 steps for the same tolerance level. For larger grid, I expect to see even larger gains in efficiency.

This converges to the following values (and the mirror image for the other values):

      100          100          100          100          100
      100       96.502       93.464       91.254       90.097
      100       92.989       86.925       82.533       80.245
      100       89.229       79.995       73.386       69.974
      100       84.579       71.615       62.556       57.963
      100        77.78        59.86       47.904       42.037
      100       66.515       41.786       26.614       19.565
      100       45.939       13.075      -4.3143       -11.72
      100       3.4985      -32.392      -46.997      -52.455
      100         -100         -100         -100         -100

You can verify that this solution is an approximate fixpoint by verifying that for each element in the bulk it is equal to the calculated average within a certain tolerance.