how to change the code below with while loop to for loop?

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Nilly
Nilly el 17 de En. de 2018
Respondida: Mihir el 18 de Jun. de 2023
I would like to change this while loop to for loop and is there any way to re-code it without mod(c,2500)?
% code
dx = 1;
x = 0:dx:1000;
x(end) = [];
u = zeros(size(x));
u(x<=220) = 1;
t = 0;
dt = 450;
uinit = u;
k = 0.001;
s = 0.22;
tend = 4000000;
v = s*k;
c = 1;
vmax = 0;
unew(1,:) = uinit;
front1 = 0.75*min(uinit)+0.25*max(uinit);
front2 = 0.25*min(uinit)+0.75*max(uinit);
front = length(uinit([uinit>front1 & uinit<front2]));
figure(1)
plot(x,uinit)
hold on
while t < tend
D = k*u;
D = (D([2:end,1])+D)/2;
dudx = (u([2:end,1])-u)/dx;
q = -D.*dudx+u*v;
u = u - dt * (q-q([end,1:end-1]))/dx;
if mod(c,2500)==0
plot(x,u)
unew(end+1,:) = u;
end
front1 = 0.75*min(u)+0.25*max(u);
front2 = 0.25*min(u)+0.75*max(u);
front(end+1) = length(u([u>front1 & u< front2]));
vmax(end+1) = max(k*(s+dudx));
t = t+dt;
c = c+1;
end

Respuestas (1)

Mihir
Mihir el 18 de Jun. de 2023
Hi, the replacement of the outer while loop by the for loop can be made by firstly declaring the upperbound of the for loop (num_plots) in the below code and for replacing the inner if condition, you can write one more for loop the runs for 2500 times. So basically nested for loops will meet the requirements.
num_plots = floor(tend / dt / 2500);
for plot_index = 1:num_plots
for c = 1:2500
Perform the operations here
end
plot(x,u)
unew(end+1,:) = u;
end

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