BVP4c Solving two equations simultaneously

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 comentario

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.

Iniciar sesión para comentar.

Respuestas (1)

Torsten
Torsten el 25 de En. de 2019

0 votos

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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Preguntada:

el 24 de En. de 2019

Respondida:

el 25 de En. de 2019

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