Caught in loop
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My code is as follows:
alpha = 0.5;
beta = 0.1;
a_0 = 1;
dr = 0.1;
dt = input('Value of dt:');
h = 0.001;
c_inf = 0;
c_a = 1;
c_s = 2;
mesh_points = 3;
rho = dt/(dr)^2;
R= 30;
N_i = R/dr;
t_diss = 5.8;
N_t = t_diss/dt;
if rho >= 0.5;
disp('Invalid dt value');
return;
end
for i=1:N_i;
r(i)=(i-1)*dr;
if r(i)< a_0
c(i,1)= c_s;
else
c(i,1)= c_inf;
end
end
a(1) = a_0;
for j = 1:N_t -1;
b_1 = floor (a(j)/dr) + 2;
b_2 = b_1 + 1;
c_ah = (a(j)+h-b_1)*(a(j)+h-b_2)*(c_a)/((a(j)-b_1)*(a(j)-b_2))+h*(a(j)+h-b_2)*c(b_1,j)/((b_1-a(j))*(b_1-b_2)) + h*(a(j)+h-b_1)*c(b_2,j)/((b_2-a(j))*(b_2-b_1));
dc_dr = ((c_ah)-c_a)/h;
a(j+1) = a(j) +dt*beta*dc_dr;
for i=2:N_i - 1;
if r(i) > a(j+1);
d = rho - (rho/(i-1)) - (a(j)^2)*(alpha-1)*(a(j+1)-a(j))/(2*(i-1)^2*(dr)^3);
e = 1-2*rho;
f = rho + (rho/(i-1)) + (a(j)^2)*(alpha-1)*(a(j+1)-a(j))/(2*(i-1)^2*(dr)^3);
c(i, j+1) = d*c(i-1,j) + e*c(i,j) + f*c(i+1, j);
elseif r(i) == a(j+1);
c(i,j+1) = c_a;
else c(i,j+1) = c_s;
end
disp(c(i,j));
end
c(1, j+1) = c(2, j+1);
end
It basically calculates concentration near a dissolving sphere but when it gets to disp(c(i,j)) I get caught in loop of displaying numbers that change periodically. Any ideas?
0 Comments
Accepted Answer
Jan
on 7 May 2012
And some cleanup:
...
% Vectorization of the "for i=1:N_i;" loop:
r = (0:N_i) * dr;
c = zeros(N_i, N_t - 1); % Pre-allocate!
c(:, 1) = c_inf;
c(r < a_0, 1) = c_s;
...
dr3 = dr ^ 3;
...
% Reduce repeated calculations inside the "for j = 1:N_t -1" loop:
ajp1 = a(j) +dt*beta*dc_dr;
a(j+1) = ajp1;
tmp1 = (a(j)^2) * (alpha-1) * (ajp1-a(j)) / (2 * dr3);
for i = 2:N_i - 1
if r(i) > ajp1
tmp2 = rho / (i-1) + tmp1 / ((i-1)^2);
d = rho - tmp2;
e = 1 - 2 * rho;
f = rho + tmp2;
c(i, j+1) = d*c(i-1,j) + e*c(i,j) + f*c(i+1, j);
elseif r(i) == ajp1
c(i, j+1) = c_a;
else
c(i, j+1) = c_s;
end
% disp(c(i,j));
end
5 Comments
More Answers (1)
Richard Brown
on 7 May 2012
Your code works fine. If you remove the disp statement (which is causing it to slow down heaps), you'll find that it runs completely in around 10 seconds or so.
2 Comments
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