initial guessing - for mixed boundary condition
2 次查看(过去 30 天)
显示 更早的评论
f'''+(1/2)*f*f''+lamda*theta
theta''+pr*(1/2*f*theta'+f'*theta
with B.C f(eta) = 0,f'(eta) = delta*f''(eta) , theata(eta)=1+beta*theta'(eta) at eta = zero
f'(eta) = 1 ,theta(eta) = 0 as eta = infinity
close all
solinit = bvpinit(linspace(0,2),@exlinit );
options = bvpset('RelTol',1e-6);
sol = bvp4c(@convec,@bvp,solinit,options);
figure(1)
plot(sol.x,sol.y(4,:),'linewidth',1.5);
%plot(sol.x,sol.y(5,:),'--','linewidth',1.5);
hold on
function dy = convec(t,y)
lamda = 1;
Pr = 6.2;
dy(1) = y(2);
dy(2) = y(3);
dy(3) = -(1/2)*y(1)*y(3)+lamda*y(4);
dy(4) = y(5);
dy(5) = -(Pr)*(1/2*y(1)*y(5)+y(2)*y(4));
dy =[dy(1);dy(2);dy(3);dy(4);dy(5)];
end
function res = bvp(ya,yb)
delta = 0.5;
beta = 1;
res = [ya(1)-0
ya(2)-delta*ya(3)
yb(2)-1;
ya(4)-1-beta*ya(5);
yb(4)-0];
end
function v = exlinit(t)
v = [0
1
1-exp(-t)
1
exp(-t)];
end
the program is running but not getting correct one for example temperature profile ie y(4)
kindly give a suggestion for guessing a initial condition for the problem
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1117110/image.jpeg)
1 个评论
Torsten
2022-9-5
From your equations, it should be
dy(3) = -(1/2)*y(1)*y(3)-lamda*y(4);
instead of
dy(3) = -(1/2)*y(1)*y(3)+lamda*y(4);
回答(1 个)
Sarthak
2023-9-13
Hi N Nithya,
I understand you are looking for a way of initial guessing, but I found something else in the code which might help.
The equation for dy(3) should be:
dy(3) = -(1/2)*y(1)*y(3) - lamda*y(4)
The original equation, dy(3) = -(1/2)y(1)y(3) + lamday(4), is incorrect because the term lamday(4) should be negative.
The negative sign in front of the term lamda*y(4) is important because it ensures that the derivative of dy(3) is always negative. This is necessary because dy(3) represents the rate of change of y(3), and the rate of change of y(3) should always be negative when y(3) is increasing.
Hope this solves your query.
0 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Function Creation 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!