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Can behaviour of subsystem of a coupled system be analyzed using the partial Jacobian?
Clash Royale CLAN TAG#URR8PPP
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Consider the following coupled dynamical system,
$$dot x = f(x,y)$$
$$dot y = h(x,y)$$
For example, $dot x$ could be the dynamics of the lung, $dot y$ could be the dynamics of the heart.
Suppose we assign state $z_1 = x$, $z_2 = y$, then we have a new system given by $dot z = [dot z_1, dot z_2]^T = [dot x, dot y]^T = mathbff$.
Linearizing this system, yields the Jacobian matrix:
$$J[mathbff] = beginbmatrix dfracpartial fpartial x & dfracpartial fpartial y \ dfracpartial hpartial x & dfracpartial hpartial y endbmatrix$$
Normally, in order to analyze the behaviour system around an equilibrium point, we solve the Jacobian around an equilibrium point $(bar x, bar y)$, then we find the eigenvalues associated with the Jacobian, and deduce the behaviour of the nonlinear system in a local neighborhood of that equilibrium point.
However, let's say, I am only interested in the dynamics of the lung, i.e., $dot x$ and I know the point $(bar x, bar y)$ which I want to linearize around.
It is possible to understand the convergence behaviour of the solution of $dot x = f(x,y)$ only by looking at $dfracpartial fpartial x|_(bar x, bar y)$ alone? Suppose that $dfracpartial fpartial x|_(bar x, bar y) = -1$, then can we say that $x(t)$ will converge towards the equilibrium point $bar x$ in a small neighborhood around $bar x$?
In other words: can we analyze a subsystem of a coupled system by only looking at the sub-matrix of the Jacobian, and not the entire Jacobian?
Note: this is trivially true if the system is decoupled, e.g.,
$$dot x= - x$$ $$dot y = y$$
I guess a refinement of the question would be, what types of coupled system can we recover this result for decoupled system? When does the coupling terms not matter in the overall analysis? Or, can the coupling be "weak" enough that do not affect the result provided using linearization?
differential-equations analysis dynamical-systems nonlinear-system linearization
asked Jul 21 at 21:27
Enlightened One
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Consider the following coupled dynamical system,
$$dot x = f(x,y)$$
$$dot y = h(x,y)$$
For example, $dot x$ could be the dynamics of the lung, $dot y$ could be the dynamics of the heart.
Suppose we assign state $z_1 = x$, $z_2 = y$, then we have a new system given by $dot z = [dot z_1, dot z_2]^T = [dot x, dot y]^T = mathbff$.
Linearizing this system, yields the Jacobian matrix:
$$J[mathbff] = beginbmatrix dfracpartial fpartial x & dfracpartial fpartial y \ dfracpartial hpartial x & dfracpartial hpartial y endbmatrix$$
Normally, in order to analyze the behaviour system around an equilibrium point, we solve the Jacobian around an equilibrium point $(bar x, bar y)$, then we find the eigenvalues associated with the Jacobian, and deduce the behaviour of the nonlinear system in a local neighborhood of that equilibrium point.
However, let's say, I am only interested in the dynamics of the lung, i.e., $dot x$ and I know the point $(bar x, bar y)$ which I want to linearize around.
It is possible to understand the convergence behaviour of the solution of $dot x = f(x,y)$ only by looking at $dfracpartial fpartial x|_(bar x, bar y)$ alone? Suppose that $dfracpartial fpartial x|_(bar x, bar y) = -1$, then can we say that $x(t)$ will converge towards the equilibrium point $bar x$ in a small neighborhood around $bar x$?
In other words: can we analyze a subsystem of a coupled system by only looking at the sub-matrix of the Jacobian, and not the entire Jacobian?
Note: this is trivially true if the system is decoupled, e.g.,
$$dot x= - x$$ $$dot y = y$$
I guess a refinement of the question would be, what types of coupled system can we recover this result for decoupled system? When does the coupling terms not matter in the overall analysis? Or, can the coupling be "weak" enough that do not affect the result provided using linearization?
differential-equations analysis dynamical-systems nonlinear-system linearization
asked Jul 21 at 21:27
Enlightened One
4,60231244
add a comment |Â
up vote
0
down vote
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up vote
0
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Consider the following coupled dynamical system,
$$dot x = f(x,y)$$
$$dot y = h(x,y)$$
For example, $dot x$ could be the dynamics of the lung, $dot y$ could be the dynamics of the heart.
Suppose we assign state $z_1 = x$, $z_2 = y$, then we have a new system given by $dot z = [dot z_1, dot z_2]^T = [dot x, dot y]^T = mathbff$.
Linearizing this system, yields the Jacobian matrix:
$$J[mathbff] = beginbmatrix dfracpartial fpartial x & dfracpartial fpartial y \ dfracpartial hpartial x & dfracpartial hpartial y endbmatrix$$
Normally, in order to analyze the behaviour system around an equilibrium point, we solve the Jacobian around an equilibrium point $(bar x, bar y)$, then we find the eigenvalues associated with the Jacobian, and deduce the behaviour of the nonlinear system in a local neighborhood of that equilibrium point.
However, let's say, I am only interested in the dynamics of the lung, i.e., $dot x$ and I know the point $(bar x, bar y)$ which I want to linearize around.
It is possible to understand the convergence behaviour of the solution of $dot x = f(x,y)$ only by looking at $dfracpartial fpartial x|_(bar x, bar y)$ alone? Suppose that $dfracpartial fpartial x|_(bar x, bar y) = -1$, then can we say that $x(t)$ will converge towards the equilibrium point $bar x$ in a small neighborhood around $bar x$?
In other words: can we analyze a subsystem of a coupled system by only looking at the sub-matrix of the Jacobian, and not the entire Jacobian?
Note: this is trivially true if the system is decoupled, e.g.,
$$dot x= - x$$ $$dot y = y$$
I guess a refinement of the question would be, what types of coupled system can we recover this result for decoupled system? When does the coupling terms not matter in the overall analysis? Or, can the coupling be "weak" enough that do not affect the result provided using linearization?
differential-equations analysis dynamical-systems nonlinear-system linearization
asked Jul 21 at 21:27
Enlightened One
4,60231244
Consider the following coupled dynamical system,
$$dot x = f(x,y)$$
$$dot y = h(x,y)$$
For example, $dot x$ could be the dynamics of the lung, $dot y$ could be the dynamics of the heart.
Suppose we assign state $z_1 = x$, $z_2 = y$, then we have a new system given by $dot z = [dot z_1, dot z_2]^T = [dot x, dot y]^T = mathbff$.
Linearizing this system, yields the Jacobian matrix:
$$J[mathbff] = beginbmatrix dfracpartial fpartial x & dfracpartial fpartial y \ dfracpartial hpartial x & dfracpartial hpartial y endbmatrix$$
Normally, in order to analyze the behaviour system around an equilibrium point, we solve the Jacobian around an equilibrium point $(bar x, bar y)$, then we find the eigenvalues associated with the Jacobian, and deduce the behaviour of the nonlinear system in a local neighborhood of that equilibrium point.
However, let's say, I am only interested in the dynamics of the lung, i.e., $dot x$ and I know the point $(bar x, bar y)$ which I want to linearize around.
It is possible to understand the convergence behaviour of the solution of $dot x = f(x,y)$ only by looking at $dfracpartial fpartial x|_(bar x, bar y)$ alone? Suppose that $dfracpartial fpartial x|_(bar x, bar y) = -1$, then can we say that $x(t)$ will converge towards the equilibrium point $bar x$ in a small neighborhood around $bar x$?
In other words: can we analyze a subsystem of a coupled system by only looking at the sub-matrix of the Jacobian, and not the entire Jacobian?
Note: this is trivially true if the system is decoupled, e.g.,
$$dot x= - x$$ $$dot y = y$$
I guess a refinement of the question would be, what types of coupled system can we recover this result for decoupled system? When does the coupling terms not matter in the overall analysis? Or, can the coupling be "weak" enough that do not affect the result provided using linearization?
differential-equations analysis dynamical-systems nonlinear-system linearization
asked Jul 21 at 21:27
Enlightened One
4,60231244
edited Jul 21 at 21:56
asked Jul 21 at 21:27
Enlightened One
4,60231244
asked Jul 21 at 21:27
Enlightened One
4,60231244
asked Jul 21 at 21:27
Enlightened One
4,60231244
4,60231244
add a comment |Â
add a comment |Â
1 Answer
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What you are really asking is this. Is there a relationship between a diagonal entry of a matrix, and the eigenvalues of the matrix. Unless the matrix is upper or lower triangular, the answer is generally "no."
answered Jul 22 at 0:04
Stephen Montgomery-Smith
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
up vote
0
down vote
What you are really asking is this. Is there a relationship between a diagonal entry of a matrix, and the eigenvalues of the matrix. Unless the matrix is upper or lower triangular, the answer is generally "no."
answered Jul 22 at 0:04
Stephen Montgomery-Smith
17.4k12047
add a comment |Â
up vote
0
down vote
What you are really asking is this. Is there a relationship between a diagonal entry of a matrix, and the eigenvalues of the matrix. Unless the matrix is upper or lower triangular, the answer is generally "no."
answered Jul 22 at 0:04
Stephen Montgomery-Smith
17.4k12047
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What you are really asking is this. Is there a relationship between a diagonal entry of a matrix, and the eigenvalues of the matrix. Unless the matrix is upper or lower triangular, the answer is generally "no."
answered Jul 22 at 0:04
Stephen Montgomery-Smith
17.4k12047
What you are really asking is this. Is there a relationship between a diagonal entry of a matrix, and the eigenvalues of the matrix. Unless the matrix is upper or lower triangular, the answer is generally "no."
answered Jul 22 at 0:04
Stephen Montgomery-Smith
17.4k12047
answered Jul 22 at 0:04
Stephen Montgomery-Smith
17.4k12047
answered Jul 22 at 0:04
Stephen Montgomery-Smith
17.4k12047
answered Jul 22 at 0:04
Stephen Montgomery-Smith
17.4k12047
17.4k12047
add a comment |Â
add a comment |Â
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