Mixing
stability analysis of an ODE
$begingroup$
I need help on how to linearize the following ODE equation so that I am able to do Stability analysis for the equation. Thanks for the help.
$frac{dQ}{dz} = 2aM^{1/2}$
$frac{dM}{dz} = frac{QF}{M}$
$frac{dF}{dz} = bQ$
Where a and b are constants. Initial points are $Q = M = 0, F = 1$ at z = 0.
Non-dimensional form of the above equation is:
$frac{d hat{Q}}{dhat{z}} = hat{M}^{1/2}$
$hat{M}frac{d hat{M}}{dhat{z}} = hat{F}hat{Q}$
$frac{d hat{F}}{dhat{z}} = -hat{Q}$
Case 1:
begin{equation}
begin{pmatrix}
frac{dQ}{dz} \
frac{dM}{dz} \
frac{dF}{dz}
end{pmatrix}=begin{pmatrix}
0 & frac{1}{M^{1/2}} & 0 \
0 & 0 & frac{Q}{M} \
-1 & 0 & 0
end{pmatrix}
begin{pmatrix}
Q \
M \
F
end{pmatrix}
end{equation}
Case 2:
begin{equation}
begin{pmatrix}
frac{dQ}{dz} \
frac{dM}{dz} \
frac{dF}{dz}
end{pmatrix}=begin{pmatrix}
0 & frac{1}{M^{1/2}} & 0 \
frac{F}{M} & 0 & 0 \
-1 & 0 & 0
end{pmatrix}
begin{pmatrix}
Q \
M \
F
end{pmatrix}
end{equation}
stability-in-odes
asked Feb 3 at 12:45
DerejeDereje
44
$endgroup$
|
show 5 more comments
$begingroup$
I need help on how to linearize the following ODE equation so that I am able to do Stability analysis for the equation. Thanks for the help.
$frac{dQ}{dz} = 2aM^{1/2}$
$frac{dM}{dz} = frac{QF}{M}$
$frac{dF}{dz} = bQ$
Where a and b are constants. Initial points are $Q = M = 0, F = 1$ at z = 0.
Non-dimensional form of the above equation is:
$frac{d hat{Q}}{dhat{z}} = hat{M}^{1/2}$
$hat{M}frac{d hat{M}}{dhat{z}} = hat{F}hat{Q}$
$frac{d hat{F}}{dhat{z}} = -hat{Q}$
Case 1:
begin{equation}
begin{pmatrix}
frac{dQ}{dz} \
frac{dM}{dz} \
frac{dF}{dz}
end{pmatrix}=begin{pmatrix}
0 & frac{1}{M^{1/2}} & 0 \
0 & 0 & frac{Q}{M} \
-1 & 0 & 0
end{pmatrix}
begin{pmatrix}
Q \
M \
F
end{pmatrix}
end{equation}
Case 2:
begin{equation}
begin{pmatrix}
frac{dQ}{dz} \
frac{dM}{dz} \
frac{dF}{dz}
end{pmatrix}=begin{pmatrix}
0 & frac{1}{M^{1/2}} & 0 \
frac{F}{M} & 0 & 0 \
-1 & 0 & 0
end{pmatrix}
begin{pmatrix}
Q \
M \
F
end{pmatrix}
end{equation}
stability-in-odes
asked Feb 3 at 12:45
DerejeDereje
44
$endgroup$
$begingroup$
Welcome to MSE. Are you sure that you put everything right? For your initial values, in the second equation you have indeterminate $0/0$.
$endgroup$
– user539887
Feb 8 at 10:30
$begingroup$
Yes actually M can not be zero. But it could be assumed that very small value.
$endgroup$
– Dereje
Feb 9 at 11:54
$begingroup$
I still do not understand. Do you mean linearization around an equilibrium? But there are no equilibria, since by the first equation $M$ must be zero, which is incompatible with the second equation.
$endgroup$
– user539887
Feb 9 at 13:02
$begingroup$
Yes you are right, may be if try to convert the equations into non-dimensional equations, the indeterminate might disappear.
$endgroup$
– Dereje
Feb 9 at 20:39
$begingroup$
I've done the non-dimensional form of the above equation(I wrote it above in my original question) but it doesn't make as such a difference.
$endgroup$
– Dereje
Feb 9 at 22:15
|
show 5 more comments
$begingroup$
I need help on how to linearize the following ODE equation so that I am able to do Stability analysis for the equation. Thanks for the help.
$frac{dQ}{dz} = 2aM^{1/2}$
$frac{dM}{dz} = frac{QF}{M}$
$frac{dF}{dz} = bQ$
Where a and b are constants. Initial points are $Q = M = 0, F = 1$ at z = 0.
Non-dimensional form of the above equation is:
$frac{d hat{Q}}{dhat{z}} = hat{M}^{1/2}$
$hat{M}frac{d hat{M}}{dhat{z}} = hat{F}hat{Q}$
$frac{d hat{F}}{dhat{z}} = -hat{Q}$
Case 1:
begin{equation}
begin{pmatrix}
frac{dQ}{dz} \
frac{dM}{dz} \
frac{dF}{dz}
end{pmatrix}=begin{pmatrix}
0 & frac{1}{M^{1/2}} & 0 \
0 & 0 & frac{Q}{M} \
-1 & 0 & 0
end{pmatrix}
begin{pmatrix}
Q \
M \
F
end{pmatrix}
end{equation}
Case 2:
begin{equation}
begin{pmatrix}
frac{dQ}{dz} \
frac{dM}{dz} \
frac{dF}{dz}
end{pmatrix}=begin{pmatrix}
0 & frac{1}{M^{1/2}} & 0 \
frac{F}{M} & 0 & 0 \
-1 & 0 & 0
end{pmatrix}
begin{pmatrix}
Q \
M \
F
end{pmatrix}
end{equation}
stability-in-odes
asked Feb 3 at 12:45
DerejeDereje
44
$endgroup$
I need help on how to linearize the following ODE equation so that I am able to do Stability analysis for the equation. Thanks for the help.
$frac{dQ}{dz} = 2aM^{1/2}$
$frac{dM}{dz} = frac{QF}{M}$
$frac{dF}{dz} = bQ$
Where a and b are constants. Initial points are $Q = M = 0, F = 1$ at z = 0.
Non-dimensional form of the above equation is:
$frac{d hat{Q}}{dhat{z}} = hat{M}^{1/2}$
$hat{M}frac{d hat{M}}{dhat{z}} = hat{F}hat{Q}$
$frac{d hat{F}}{dhat{z}} = -hat{Q}$
Case 1:
begin{equation}
begin{pmatrix}
frac{dQ}{dz} \
frac{dM}{dz} \
frac{dF}{dz}
end{pmatrix}=begin{pmatrix}
0 & frac{1}{M^{1/2}} & 0 \
0 & 0 & frac{Q}{M} \
-1 & 0 & 0
end{pmatrix}
begin{pmatrix}
Q \
M \
F
end{pmatrix}
end{equation}
Case 2:
begin{equation}
begin{pmatrix}
frac{dQ}{dz} \
frac{dM}{dz} \
frac{dF}{dz}
end{pmatrix}=begin{pmatrix}
0 & frac{1}{M^{1/2}} & 0 \
frac{F}{M} & 0 & 0 \
-1 & 0 & 0
end{pmatrix}
begin{pmatrix}
Q \
M \
F
end{pmatrix}
end{equation}
stability-in-odes
stability-in-odes
asked Feb 3 at 12:45
DerejeDereje
44
asked Feb 3 at 12:45
DerejeDereje
44
edited Feb 11 at 0:38
Dereje
asked Feb 3 at 12:45
DerejeDereje
44
asked Feb 3 at 12:45
DerejeDereje
44
asked Feb 3 at 12:45
DerejeDereje
44
44
$begingroup$
Welcome to MSE. Are you sure that you put everything right? For your initial values, in the second equation you have indeterminate $0/0$.
$endgroup$
– user539887
Feb 8 at 10:30
$begingroup$
Yes actually M can not be zero. But it could be assumed that very small value.
$endgroup$
– Dereje
Feb 9 at 11:54
$begingroup$
I still do not understand. Do you mean linearization around an equilibrium? But there are no equilibria, since by the first equation $M$ must be zero, which is incompatible with the second equation.
$endgroup$
– user539887
Feb 9 at 13:02
$begingroup$
Yes you are right, may be if try to convert the equations into non-dimensional equations, the indeterminate might disappear.
$endgroup$
– Dereje
Feb 9 at 20:39
$begingroup$
I've done the non-dimensional form of the above equation(I wrote it above in my original question) but it doesn't make as such a difference.
$endgroup$
– Dereje
Feb 9 at 22:15
|
show 5 more comments
$begingroup$
Welcome to MSE. Are you sure that you put everything right? For your initial values, in the second equation you have indeterminate $0/0$.
$endgroup$
– user539887
Feb 8 at 10:30
$begingroup$
Yes actually M can not be zero. But it could be assumed that very small value.
$endgroup$
– Dereje
Feb 9 at 11:54
$begingroup$
I still do not understand. Do you mean linearization around an equilibrium? But there are no equilibria, since by the first equation $M$ must be zero, which is incompatible with the second equation.
$endgroup$
– user539887
Feb 9 at 13:02
$begingroup$
Yes you are right, may be if try to convert the equations into non-dimensional equations, the indeterminate might disappear.
$endgroup$
– Dereje
Feb 9 at 20:39
$begingroup$
I've done the non-dimensional form of the above equation(I wrote it above in my original question) but it doesn't make as such a difference.
$endgroup$
– Dereje
Feb 9 at 22:15
$begingroup$
Welcome to MSE. Are you sure that you put everything right? For your initial values, in the second equation you have indeterminate $0/0$.
$endgroup$
– user539887
Feb 8 at 10:30
$begingroup$
Welcome to MSE. Are you sure that you put everything right? For your initial values, in the second equation you have indeterminate $0/0$.
$endgroup$
– user539887
Feb 8 at 10:30
$begingroup$
Yes actually M can not be zero. But it could be assumed that very small value.
$endgroup$
– Dereje
Feb 9 at 11:54
$begingroup$
Yes actually M can not be zero. But it could be assumed that very small value.
$endgroup$
– Dereje
Feb 9 at 11:54
$begingroup$
I still do not understand. Do you mean linearization around an equilibrium? But there are no equilibria, since by the first equation $M$ must be zero, which is incompatible with the second equation.
$endgroup$
– user539887
Feb 9 at 13:02
$begingroup$
I still do not understand. Do you mean linearization around an equilibrium? But there are no equilibria, since by the first equation $M$ must be zero, which is incompatible with the second equation.
$endgroup$
– user539887
Feb 9 at 13:02
$begingroup$
Yes you are right, may be if try to convert the equations into non-dimensional equations, the indeterminate might disappear.
$endgroup$
– Dereje
Feb 9 at 20:39
$begingroup$
Yes you are right, may be if try to convert the equations into non-dimensional equations, the indeterminate might disappear.
$endgroup$
– Dereje
Feb 9 at 20:39
$begingroup$
I've done the non-dimensional form of the above equation(I wrote it above in my original question) but it doesn't make as such a difference.
$endgroup$
– Dereje
Feb 9 at 22:15
$begingroup$
I've done the non-dimensional form of the above equation(I wrote it above in my original question) but it doesn't make as such a difference.
$endgroup$
– Dereje
Feb 9 at 22:15
|
show 5 more comments
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$begingroup$
Welcome to MSE. Are you sure that you put everything right? For your initial values, in the second equation you have indeterminate $0/0$.
$endgroup$
– user539887
Feb 8 at 10:30
$begingroup$
Yes actually M can not be zero. But it could be assumed that very small value.
$endgroup$
– Dereje
Feb 9 at 11:54
$begingroup$
I still do not understand. Do you mean linearization around an equilibrium? But there are no equilibria, since by the first equation $M$ must be zero, which is incompatible with the second equation.
$endgroup$
– user539887
Feb 9 at 13:02
$begingroup$
Yes you are right, may be if try to convert the equations into non-dimensional equations, the indeterminate might disappear.
$endgroup$
– Dereje
Feb 9 at 20:39
$begingroup$
I've done the non-dimensional form of the above equation(I wrote it above in my original question) but it doesn't make as such a difference.
$endgroup$
– Dereje
Feb 9 at 22:15