Singular problem for bvp4c
2 次查看(过去 30 天)
显示 更早的评论
Hi All,
I encounter some singular problems regarding bvp4c, please see my code for my equations and implementation. I know that when x = -1 and 1, my derivative goes to infinity. And it seems like that MATLAB complaints about this. I wonder is there any workaround for this problem? I tried on ode113, and instead of evaluating my initial point at x = -1, I just do it at x = -1 + 1e-7 to avoid this singular points. And it is solvable. I would like to use bvp4c because I have other equations and boundary condition to coupled together. I wonder is there similar strategy for bvp4c? Here are my code
solinit = bvpinit(linspace(-1,1,50),0);
sol = bvp4c(@height_ode,@height_bc,solinit);
y = deval(sol,linspace(-1,1,50));
plot(linspace(-1,1,50),y(1,:))
and defintion for height_ode and height_bc are following:
function dhdx = height_ode(x,h)
w = 0.7;
q = 1/2*w*(1-x^2);
dqdx = -w*x;
kappa = 1.469;
dhdx = (560*h(1)^0.5*q - 64*h(1).^4 + 6*w.^(7/8)*(72*kappa - 77)*h(1)*q*dqdx)/(3*w.^(7/8)*(96*kappa - 77)*q.^2);
end
function res = height_bc(yleft, yright)
res = [yleft];
3 个评论
回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Boundary Value Problems 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!