How exactly is the Wronskian tied with Cramer's rule in a system of ODEs? The Next CEO of...
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How exactly is the Wronskian tied with Cramer's rule in a system of ODEs?
The Next CEO of Stack OverflowLinear Differential SystemWronskian determinant for $y_2(t)=begin{pmatrix} e^t\ e^t end{pmatrix}$ and $y_1(t)=begin{pmatrix} t^2\ 2t end{pmatrix}$.How can you calculate the derivative of this Wronskian?Prove that the determinant of a given matrix is proportional to the area of the triangle whose corners are the three points.Linear transformations using matricesIs the Wronskian determinant positive or negative?Solving System of Linear Equations with LU Decomposition of $4 times 3$ matrixDetermine independence of functions of r.v.'s and compute joint distributionHow to find the general solution of a system of ODEs?Linear independence of $1, x, x^2$Solving a simple system of equations through matrix operations
$begingroup$
If we have a nonhomogeneous system of linear differential equations:
$$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
where:
$$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
the particular solution involving these three functions could be written:
$$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)
When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
$$c_i(x)=intfrac{W_i}{W}dx$$
Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.
My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?
If I have missed something, please point it out and I'll add it to the post.
Thanks in advance.
linear-algebra ordinary-differential-equations systems-of-equations
$endgroup$
add a comment |
$begingroup$
If we have a nonhomogeneous system of linear differential equations:
$$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
where:
$$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
the particular solution involving these three functions could be written:
$$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)
When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
$$c_i(x)=intfrac{W_i}{W}dx$$
Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.
My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?
If I have missed something, please point it out and I'll add it to the post.
Thanks in advance.
linear-algebra ordinary-differential-equations systems-of-equations
$endgroup$
add a comment |
$begingroup$
If we have a nonhomogeneous system of linear differential equations:
$$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
where:
$$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
the particular solution involving these three functions could be written:
$$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)
When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
$$c_i(x)=intfrac{W_i}{W}dx$$
Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.
My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?
If I have missed something, please point it out and I'll add it to the post.
Thanks in advance.
linear-algebra ordinary-differential-equations systems-of-equations
$endgroup$
If we have a nonhomogeneous system of linear differential equations:
$$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
where:
$$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
the particular solution involving these three functions could be written:
$$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)
When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
$$c_i(x)=intfrac{W_i}{W}dx$$
Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.
My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?
If I have missed something, please point it out and I'll add it to the post.
Thanks in advance.
linear-algebra ordinary-differential-equations systems-of-equations
linear-algebra ordinary-differential-equations systems-of-equations
edited Mar 17 at 17:13
J. W. Tanner
4,1961320
4,1961320
asked Mar 17 at 17:03
Gradient EntropyGradient Entropy
204
204
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write the general solution of the homogeneous linear ODE
$$
tag{1}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
$$
in the matrix form as
$$
Phi(x; x_0) vec{c},
$$
where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).
Now we are looking for a general solution of the nonhomogeneous linear ODE
$$
tag{2}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
$$
in the form
$$
Phi(x, x_0) vec{c}(x).
$$
Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
$$
But
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
$$
and
$$
frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
$$
so after cancellation we obtain
$$
Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
$$
which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.
The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.
EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.
$endgroup$
1
$begingroup$
Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
$endgroup$
– Gradient Entropy
Mar 17 at 21:53
$begingroup$
I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
$endgroup$
– user539887
Mar 17 at 22:00
$begingroup$
One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
$endgroup$
– Gradient Entropy
Mar 17 at 23:01
$begingroup$
It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
$endgroup$
– user539887
Mar 18 at 7:18
add a comment |
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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
$begingroup$
Write the general solution of the homogeneous linear ODE
$$
tag{1}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
$$
in the matrix form as
$$
Phi(x; x_0) vec{c},
$$
where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).
Now we are looking for a general solution of the nonhomogeneous linear ODE
$$
tag{2}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
$$
in the form
$$
Phi(x, x_0) vec{c}(x).
$$
Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
$$
But
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
$$
and
$$
frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
$$
so after cancellation we obtain
$$
Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
$$
which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.
The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.
EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.
$endgroup$
1
$begingroup$
Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
$endgroup$
– Gradient Entropy
Mar 17 at 21:53
$begingroup$
I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
$endgroup$
– user539887
Mar 17 at 22:00
$begingroup$
One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
$endgroup$
– Gradient Entropy
Mar 17 at 23:01
$begingroup$
It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
$endgroup$
– user539887
Mar 18 at 7:18
add a comment |
$begingroup$
Write the general solution of the homogeneous linear ODE
$$
tag{1}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
$$
in the matrix form as
$$
Phi(x; x_0) vec{c},
$$
where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).
Now we are looking for a general solution of the nonhomogeneous linear ODE
$$
tag{2}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
$$
in the form
$$
Phi(x, x_0) vec{c}(x).
$$
Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
$$
But
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
$$
and
$$
frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
$$
so after cancellation we obtain
$$
Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
$$
which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.
The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.
EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.
$endgroup$
1
$begingroup$
Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
$endgroup$
– Gradient Entropy
Mar 17 at 21:53
$begingroup$
I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
$endgroup$
– user539887
Mar 17 at 22:00
$begingroup$
One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
$endgroup$
– Gradient Entropy
Mar 17 at 23:01
$begingroup$
It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
$endgroup$
– user539887
Mar 18 at 7:18
add a comment |
$begingroup$
Write the general solution of the homogeneous linear ODE
$$
tag{1}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
$$
in the matrix form as
$$
Phi(x; x_0) vec{c},
$$
where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).
Now we are looking for a general solution of the nonhomogeneous linear ODE
$$
tag{2}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
$$
in the form
$$
Phi(x, x_0) vec{c}(x).
$$
Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
$$
But
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
$$
and
$$
frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
$$
so after cancellation we obtain
$$
Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
$$
which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.
The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.
EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.
$endgroup$
Write the general solution of the homogeneous linear ODE
$$
tag{1}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
$$
in the matrix form as
$$
Phi(x; x_0) vec{c},
$$
where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).
Now we are looking for a general solution of the nonhomogeneous linear ODE
$$
tag{2}
frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
$$
in the form
$$
Phi(x, x_0) vec{c}(x).
$$
Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
$$
But
$$
frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
$$
and
$$
frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
$$
so after cancellation we obtain
$$
Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
$$
which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.
The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.
EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.
edited Mar 17 at 21:59
answered Mar 17 at 21:17
user539887user539887
1,7811414
1,7811414
1
$begingroup$
Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
$endgroup$
– Gradient Entropy
Mar 17 at 21:53
$begingroup$
I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
$endgroup$
– user539887
Mar 17 at 22:00
$begingroup$
One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
$endgroup$
– Gradient Entropy
Mar 17 at 23:01
$begingroup$
It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
$endgroup$
– user539887
Mar 18 at 7:18
add a comment |
1
$begingroup$
Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
$endgroup$
– Gradient Entropy
Mar 17 at 21:53
$begingroup$
I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
$endgroup$
– user539887
Mar 17 at 22:00
$begingroup$
One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
$endgroup$
– Gradient Entropy
Mar 17 at 23:01
$begingroup$
It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
$endgroup$
– user539887
Mar 18 at 7:18
1
1
$begingroup$
Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
$endgroup$
– Gradient Entropy
Mar 17 at 21:53
$begingroup$
Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
$endgroup$
– Gradient Entropy
Mar 17 at 21:53
$begingroup$
I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
$endgroup$
– user539887
Mar 17 at 22:00
$begingroup$
I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
$endgroup$
– user539887
Mar 17 at 22:00
$begingroup$
One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
$endgroup$
– Gradient Entropy
Mar 17 at 23:01
$begingroup$
One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
$endgroup$
– Gradient Entropy
Mar 17 at 23:01
$begingroup$
It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
$endgroup$
– user539887
Mar 18 at 7:18
$begingroup$
It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
$endgroup$
– user539887
Mar 18 at 7:18
add a comment |
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