Mixing
BVP4c Solving two equations simultaneously
0
$begingroup$
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?
ordinary-differential-equations numerical-methods matlab boundary-value-problem
asked Jan 24 at 1:07
tenicholstenichols
83
$endgroup$
add a comment |
0
$begingroup$
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?
ordinary-differential-equations numerical-methods matlab boundary-value-problem
asked Jan 24 at 1:07
tenicholstenichols
83
$endgroup$
$begingroup$
You can simply add a fifth variable y(5) which takes the role of V. From the ODE you obtain one additional entry in the vector rhs and from the boundary conditions you obtain two additional entries in the vector bc.
$endgroup$
– Christoph
Jan 24 at 5:21
2
$begingroup$
Is the equation supposed to be $y'''' = Ay'' + y$?
$endgroup$
– Dylan
Jan 24 at 7:03
$begingroup$
No, that is not possible. The number of boundary conditions has always to be equal to the dimension of the state. If you make the parameter $A$ variable, it adds a state dimension, so that you could have these 6 BC.
$endgroup$
– LutzL
Jan 24 at 7:59
add a comment |
0
0
0
$begingroup$
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?
ordinary-differential-equations numerical-methods matlab boundary-value-problem
asked Jan 24 at 1:07
tenicholstenichols
83
$endgroup$
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?
ordinary-differential-equations numerical-methods matlab boundary-value-problem
ordinary-differential-equations numerical-methods matlab boundary-value-problem
asked Jan 24 at 1:07
tenicholstenichols
83
asked Jan 24 at 1:07
tenicholstenichols
83
asked Jan 24 at 1:07
tenicholstenichols
83
asked Jan 24 at 1:07
tenicholstenichols
83
asked Jan 24 at 1:07
tenicholstenichols
83
83
$begingroup$
You can simply add a fifth variable y(5) which takes the role of V. From the ODE you obtain one additional entry in the vector rhs and from the boundary conditions you obtain two additional entries in the vector bc.
$endgroup$
– Christoph
Jan 24 at 5:21
2
$begingroup$
Is the equation supposed to be $y'''' = Ay'' + y$?
$endgroup$
– Dylan
Jan 24 at 7:03
$begingroup$
No, that is not possible. The number of boundary conditions has always to be equal to the dimension of the state. If you make the parameter $A$ variable, it adds a state dimension, so that you could have these 6 BC.
$endgroup$
– LutzL
Jan 24 at 7:59
add a comment |
$begingroup$
You can simply add a fifth variable y(5) which takes the role of V. From the ODE you obtain one additional entry in the vector rhs and from the boundary conditions you obtain two additional entries in the vector bc.
$endgroup$
– Christoph
Jan 24 at 5:21
2
$begingroup$
Is the equation supposed to be $y'''' = Ay'' + y$?
$endgroup$
– Dylan
Jan 24 at 7:03
$begingroup$
No, that is not possible. The number of boundary conditions has always to be equal to the dimension of the state. If you make the parameter $A$ variable, it adds a state dimension, so that you could have these 6 BC.
$endgroup$
– LutzL
Jan 24 at 7:59
$begingroup$
You can simply add a fifth variable y(5) which takes the role of V. From the ODE you obtain one additional entry in the vector rhs and from the boundary conditions you obtain two additional entries in the vector bc.
$endgroup$
– Christoph
Jan 24 at 5:21
$begingroup$
You can simply add a fifth variable y(5) which takes the role of V. From the ODE you obtain one additional entry in the vector rhs and from the boundary conditions you obtain two additional entries in the vector bc.
$endgroup$
– Christoph
Jan 24 at 5:21
2
2
$begingroup$
Is the equation supposed to be $y'''' = Ay'' + y$?
$endgroup$
– Dylan
Jan 24 at 7:03
$begingroup$
Is the equation supposed to be $y'''' = Ay'' + y$?
$endgroup$
– Dylan
Jan 24 at 7:03
$begingroup$
No, that is not possible. The number of boundary conditions has always to be equal to the dimension of the state. If you make the parameter $A$ variable, it adds a state dimension, so that you could have these 6 BC.
$endgroup$
– LutzL
Jan 24 at 7:59
$begingroup$
No, that is not possible. The number of boundary conditions has always to be equal to the dimension of the state. If you make the parameter $A$ variable, it adds a state dimension, so that you could have these 6 BC.
$endgroup$
– LutzL
Jan 24 at 7:59
add a comment |
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$begingroup$
You can simply add a fifth variable y(5) which takes the role of V. From the ODE you obtain one additional entry in the vector rhs and from the boundary conditions you obtain two additional entries in the vector bc.
$endgroup$
– Christoph
Jan 24 at 5:21
2
$begingroup$
Is the equation supposed to be $y'''' = Ay'' + y$?
$endgroup$
– Dylan
Jan 24 at 7:03
$begingroup$
No, that is not possible. The number of boundary conditions has always to be equal to the dimension of the state. If you make the parameter $A$ variable, it adds a state dimension, so that you could have these 6 BC.
$endgroup$
– LutzL
Jan 24 at 7:59