This code for an ODE function with boundary condition, and it can only given an answer at selected time,in this coding is at t=1 t. I need to add in a loop with a small time increment, from t=0 to t=20 at increment of t=0.01. Can anyone please help me ? I need to solve this problem.
solinit = bvpinit(linspace(0,1,10),[0,0,]);
sol = bvp4c(@ex1ode,@ex1bc,solinit);
y = linspace(0,1);
u = deval(sol, y);
plot(u(1,:),y)
title ('beta = 1');
xlabel ('u*');
ylabel ('y*');
legend ('t=1');
function dudy = ex1ode(y,u)
t=1;n=1;U = 1; pho = 1; eta = 1;
uini = 0;
dudy = [ u(2)^(1/n)
(pho/eta) *((u(1)-0)/ t)];
end
function res = ex1bc (ua,ub)
t=1;U=1; omega = 1;
res=[ua(1)- U*sin(omega*t)
ub(1)];
end

 Réponse acceptée

Torsten
Torsten le 16 Mai 2018
t=0.01:0.01:20;
for i=1:numel(t)
solinit = bvpinit(linspace(0,1,10),[0,0]);
sol{i} = bvp4c(@(y,u)ex1ode(y,u,t(i)),@(ua,ub)ex1bc(ua,ub,t(i)),solinit);
end
function dudy = ex1ode(y,u,t)
n=1;U = 1; pho = 1; eta = 1;
uini = 0;
dudy = [ u(2)^(1/n)
(pho/eta) *((u(1)-0)/ t)];
end
function res = ex1bc (ua,ub,t)
U=1; omega = 1;
res=[ua(1)- U*sin(omega*t)
ub(1)];
end
Best wishes
Torsten.

1 commentaire

Chee Hao Hor
Chee Hao Hor le 16 Mai 2018
I had found my problem, thanks anyway. btw, your code seem like not able to run.

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