BVP4c Solving two equations simultaneously
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Say for example I have an 4th order ODE
y''' = A*y''+y
on the boundary [0 1] with the BC
y(0) = 1; y'(0) = 0; y(1) = 2; y'(1) = 0
I have my code setup like this.
init = bvpinit(linspace(0,1,10),[0,0,0,0]);
sol = bvp4c(@rhs_bvp, @bc_bvp, init);
x1 = linspace(0,1,100);
BS = deval(sol, x1);
function [ rhs ] = rhs_bvp( x, y )
A = 10;
rhs = [y(2);
y(3);
y(4);
A*y(3)+y(1)];
end
function [ bc ] = bc_bvp( yl, yr)
bc = [yl(1) - hi;
yl(2);
yr(1) - ho;
yr(2)];
end
Now Say I want to add another equation to solve simultaneously
V' = y
on the same boundary with BC
V(0) = 0; V(1) = 1
How would I go about including this new equation into this solver?
1 个评论
David Goodmanson
2019-1-25
Hi Taylor,
If you start with (your original expression was a typo)
y'''' = A*y''+ y
and plug in y = V', you obtain a fifth order differential equation. But now you have six boundary conditions,which appears to be one too many.
回答（1 个）
Torsten
2019-1-25
0 个投票
init = bvpinit(linspace(0,1,10),[0,0,0,0,0]);
sol = bvp4c(@rhs_bvp, @bc_bvp, init);
x1 = linspace(0,1,100);
BS = deval(sol, x1);
function [ rhs ] = rhs_bvp( x, y )
A = 10;
rhs = [y(2);
y(3);
y(4);
A*y(3)+y(1);
y(1)];
end
function [ bc ] = bc_bvp( yl, yr)
bc = [yl(1) - hi;
yl(2);
yr(1) - ho;
yr(2)
yl(5)];
%or yr(5)-1.0];
end
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2019-1-24
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