Solving third order nonlinear boundary value problem
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I have a boundary value problem in this form:
where alpha is a nonzero constant and boudary conditions are:
f(-1) = 0
f(0) = 1
f(1) = 0
How can I solve this problem using bvp5c?
5 个评论
David Goodmanson
2020-12-22
编辑：David Goodmanson
2020-12-22
Hi parham,
This looks doable if you constuct an f vector of length 6, where the first three elements are f,f',f'' in the left hand region -1<=x<=0 and the second three elements are f,f',f'' in the right hand region 0<=x<=1. Then the differential equation part will be
dfdt(1) = f(2)
dfdt(2) = f(3)
dfdt(3) = a*f(1)*f(2) + 4*a^2*f(2)
and the same for f variables (4,5,6) since the differential equation is the same. At the boundary x = 0, the first and second derivatives of f have to be continuous across the boundary. So for the boundary part in terms of fa and fb, remembering that the a,b boundaries are -1,0 for f (1,2,3) and 0,1 for f(4,5,6) ,the follwing values are zero:
fa(1)
fb(1)-1
fa(4)-1
fb(4)
fb(2)-fa(5) % continuous 1st derivative
fb(3)-fa(6) % continuous 2nd derivative
PRITESH
2023-9-27
I have the exact same equation, and I understood everything about your code, except for one thing.
Instead of dfdt(3) = a*f(1)*f(2) + 4*a^2*f(2), shouldn't it be dfdt(3) = -a*f(1)*f(2) - 4*a^2*f(2) ?
Torsten
2023-9-27
Instead of dfdt(3) = a*f(1)*f(2) + 4*a^2*f(2), shouldn't it be dfdt(3) = -a*f(1)*f(2) - 4*a^2*f(2) ?
Yes.
I am trying to use bvp4c and am getting singular Jacobian values.
I don't:
alpha = 1;
xc = 0;
xmesh = [linspace(-1,xc,1000),linspace(xc,1,1000)];
solinit = bvpinit(xmesh, [0 0 0]);
sol = bvp5c(@(x,y,r)f(x,y,r,alpha),@bc,solinit);
plot(sol.x,sol.y(1,:))
function dydx = f(x,y,region,alpha)
dydx = [y(2);y(3);-alpha*y(1)*y(2)-4*alpha^2*y(2)];
end
function res = bc(yl,yr)
res = [yl(1,1);yr(1,1)-1;yr(1,1)-yl(1,2);yr(2,1)-yl(2,2);yr(3,1)-yl(3,2);yr(1,2)];
end
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2020-12-21
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