Mixing
Wronski-Test for linear ODE
$begingroup$
To test solutions of linear ODEs for lineary independence you can determine the wronski determinant. The theorem says if you have a solution to the linear ODE in the form:
$$
dot{vec{x}}(t) = A cdot vec{x}(t)
$$
you can test for linear independence by showing
$$exists t_0 in Icolon;
operatorname{det}(W(t_0)) neq 0
$$
with $W(x)$ being the Wronski-Matrix:
$$
W(t) = begin{pmatrix}
vec{x_1}(t)& cdots & vec{x_n}(t)
end{pmatrix}
$$ where $x_i$ is the i-th vector of the solution.
If you find just one $t_0$ it follows that:
$$forall t in Icolon;operatorname{det}(W(t)) notequiv 0$$
(With $notequiv 0$ i mean constant not zero).
I don't understand why this implication holds true.
Why can I conclude from the fact that the Wronski determinant is at one point equal zero that it must be everywhere? $ det (W (t)) $ is not constant and it depends on $t$.
I'm happy about any answers!
Many Greetings,
Sebi2020
linear-algebra ordinary-differential-equations fundamental-solution
asked Jan 10 at 23:37
Sebi2020Sebi2020
1225
$endgroup$
add a comment |
$begingroup$
To test solutions of linear ODEs for lineary independence you can determine the wronski determinant. The theorem says if you have a solution to the linear ODE in the form:
$$
dot{vec{x}}(t) = A cdot vec{x}(t)
$$
you can test for linear independence by showing
$$exists t_0 in Icolon;
operatorname{det}(W(t_0)) neq 0
$$
with $W(x)$ being the Wronski-Matrix:
$$
W(t) = begin{pmatrix}
vec{x_1}(t)& cdots & vec{x_n}(t)
end{pmatrix}
$$ where $x_i$ is the i-th vector of the solution.
If you find just one $t_0$ it follows that:
$$forall t in Icolon;operatorname{det}(W(t)) notequiv 0$$
(With $notequiv 0$ i mean constant not zero).
I don't understand why this implication holds true.
Why can I conclude from the fact that the Wronski determinant is at one point equal zero that it must be everywhere? $ det (W (t)) $ is not constant and it depends on $t$.
I'm happy about any answers!
Many Greetings,
Sebi2020
linear-algebra ordinary-differential-equations fundamental-solution
asked Jan 10 at 23:37
Sebi2020Sebi2020
1225
$endgroup$
$begingroup$
I think the issue is that if the Wronskian determinant is nonzero for some $t_0 in I$, we are guaranteed that the solutions are linearly independent. If the solutions were not linearly independent, the Wronskian would always be zero (looking at contrapositive).
$endgroup$
– D.B.
Jan 10 at 23:59
add a comment |
$begingroup$
To test solutions of linear ODEs for lineary independence you can determine the wronski determinant. The theorem says if you have a solution to the linear ODE in the form:
$$
dot{vec{x}}(t) = A cdot vec{x}(t)
$$
you can test for linear independence by showing
$$exists t_0 in Icolon;
operatorname{det}(W(t_0)) neq 0
$$
with $W(x)$ being the Wronski-Matrix:
$$
W(t) = begin{pmatrix}
vec{x_1}(t)& cdots & vec{x_n}(t)
end{pmatrix}
$$ where $x_i$ is the i-th vector of the solution.
If you find just one $t_0$ it follows that:
$$forall t in Icolon;operatorname{det}(W(t)) notequiv 0$$
(With $notequiv 0$ i mean constant not zero).
I don't understand why this implication holds true.
Why can I conclude from the fact that the Wronski determinant is at one point equal zero that it must be everywhere? $ det (W (t)) $ is not constant and it depends on $t$.
I'm happy about any answers!
Many Greetings,
Sebi2020
linear-algebra ordinary-differential-equations fundamental-solution
asked Jan 10 at 23:37
Sebi2020Sebi2020
1225
$endgroup$
To test solutions of linear ODEs for lineary independence you can determine the wronski determinant. The theorem says if you have a solution to the linear ODE in the form:
$$
dot{vec{x}}(t) = A cdot vec{x}(t)
$$
you can test for linear independence by showing
$$exists t_0 in Icolon;
operatorname{det}(W(t_0)) neq 0
$$
with $W(x)$ being the Wronski-Matrix:
$$
W(t) = begin{pmatrix}
vec{x_1}(t)& cdots & vec{x_n}(t)
end{pmatrix}
$$ where $x_i$ is the i-th vector of the solution.
If you find just one $t_0$ it follows that:
$$forall t in Icolon;operatorname{det}(W(t)) notequiv 0$$
(With $notequiv 0$ i mean constant not zero).
I don't understand why this implication holds true.
Why can I conclude from the fact that the Wronski determinant is at one point equal zero that it must be everywhere? $ det (W (t)) $ is not constant and it depends on $t$.
I'm happy about any answers!
Many Greetings,
Sebi2020
linear-algebra ordinary-differential-equations fundamental-solution
linear-algebra ordinary-differential-equations fundamental-solution
asked Jan 10 at 23:37
Sebi2020Sebi2020
1225
asked Jan 10 at 23:37
Sebi2020Sebi2020
1225
edited Jan 10 at 23:43
Sebi2020
asked Jan 10 at 23:37
Sebi2020Sebi2020
1225
asked Jan 10 at 23:37
Sebi2020Sebi2020
1225
asked Jan 10 at 23:37
Sebi2020Sebi2020
1225
1225
$begingroup$
I think the issue is that if the Wronskian determinant is nonzero for some $t_0 in I$, we are guaranteed that the solutions are linearly independent. If the solutions were not linearly independent, the Wronskian would always be zero (looking at contrapositive).
$endgroup$
– D.B.
Jan 10 at 23:59
add a comment |
$begingroup$
I think the issue is that if the Wronskian determinant is nonzero for some $t_0 in I$, we are guaranteed that the solutions are linearly independent. If the solutions were not linearly independent, the Wronskian would always be zero (looking at contrapositive).
$endgroup$
– D.B.
Jan 10 at 23:59
$begingroup$
I think the issue is that if the Wronskian determinant is nonzero for some $t_0 in I$, we are guaranteed that the solutions are linearly independent. If the solutions were not linearly independent, the Wronskian would always be zero (looking at contrapositive).
$endgroup$
– D.B.
Jan 10 at 23:59
$begingroup$
I think the issue is that if the Wronskian determinant is nonzero for some $t_0 in I$, we are guaranteed that the solutions are linearly independent. If the solutions were not linearly independent, the Wronskian would always be zero (looking at contrapositive).
$endgroup$
– D.B.
Jan 10 at 23:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This matter may be resolved via Liouville's formula, which affirms that if $X(t)$ an $n times n$ matrix solution of the differential equation
$dot X(t) = A(t)X(t) tag 1$
the determinant of $det(X(t))$ of $X(t)$ is given by
$det(X(t)) = det(X(t_0)) exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ); tag 2$
since
$exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ) ne 0, ; forall t_0, t in I, tag 3$
we see that
$det(X(t)) ne 0 Longleftrightarrow det(X(t_0)) ne 0. tag 4$
Now we may conclude with the simple observation that we may take
$W(t) = det(X(t)) = det(vec x_1(t), vec x_2(t), ldots, vec x_n(t)). tag 5$
answered Jan 10 at 23:59
Robert LewisRobert Lewis
46k23066
$endgroup$
1
$begingroup$
But is it only a implication from the liouville formula or is there another explanation? Our lecturer said that the reason for this is that the solutions form a vector room. But I cannot explain myself why this should be a proof for $det(X(t)) neq 0 Leftrightarrow det(X(t_0))neq 0$ whereas the liouville formula shows that. I think the vector room property explains only the need to look for solutions which form a fundamental system.
$endgroup$
– Sebi2020
Jan 11 at 14:06
$begingroup$
@Sebi2020: is a "vector room" the same as a vector space?
$endgroup$
– Robert Lewis
Jan 11 at 16:14
1
$begingroup$
Yes. English is not my native language, so I used the wrong translation for "vector space". But I wanted to say vector space.
$endgroup$
– Sebi2020
Jan 11 at 20:57
1
$begingroup$
I think I don't quite understand it. I struggle with this: " if the $x_i$ are linearly dependent at some point, so that $sum a_i x_i(t_0) = 0$, then we must have the unique solution $x(t) = sum a_ix_i(t) = 0$ everywhere". This is what I dont understand. Which property of linear equations or vector spaces proves that? I would thought that $a_ix_i(t_0)$ maybe resides to $operatorname{ker} x(t_0)$ but that this dosn't have to be true for $x(t), tneq0$. Is it just the existence and uniqueness theorem saying that there can only be one solution with $x(t) = 0$?
$endgroup$
– Sebi2020
Jan 11 at 22:07
1
$begingroup$
Thank you! Your're explanations were very helpful.
$endgroup$
– Sebi2020
Jan 11 at 23:23
|
show 5 more comments
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$begingroup$
This matter may be resolved via Liouville's formula, which affirms that if $X(t)$ an $n times n$ matrix solution of the differential equation
$dot X(t) = A(t)X(t) tag 1$
the determinant of $det(X(t))$ of $X(t)$ is given by
$det(X(t)) = det(X(t_0)) exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ); tag 2$
since
$exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ) ne 0, ; forall t_0, t in I, tag 3$
we see that
$det(X(t)) ne 0 Longleftrightarrow det(X(t_0)) ne 0. tag 4$
Now we may conclude with the simple observation that we may take
$W(t) = det(X(t)) = det(vec x_1(t), vec x_2(t), ldots, vec x_n(t)). tag 5$
answered Jan 10 at 23:59
Robert LewisRobert Lewis
46k23066
$endgroup$
1
$begingroup$
But is it only a implication from the liouville formula or is there another explanation? Our lecturer said that the reason for this is that the solutions form a vector room. But I cannot explain myself why this should be a proof for $det(X(t)) neq 0 Leftrightarrow det(X(t_0))neq 0$ whereas the liouville formula shows that. I think the vector room property explains only the need to look for solutions which form a fundamental system.
$endgroup$
– Sebi2020
Jan 11 at 14:06
$begingroup$
@Sebi2020: is a "vector room" the same as a vector space?
$endgroup$
– Robert Lewis
Jan 11 at 16:14
1
$begingroup$
Yes. English is not my native language, so I used the wrong translation for "vector space". But I wanted to say vector space.
$endgroup$
– Sebi2020
Jan 11 at 20:57
1
$begingroup$
I think I don't quite understand it. I struggle with this: " if the $x_i$ are linearly dependent at some point, so that $sum a_i x_i(t_0) = 0$, then we must have the unique solution $x(t) = sum a_ix_i(t) = 0$ everywhere". This is what I dont understand. Which property of linear equations or vector spaces proves that? I would thought that $a_ix_i(t_0)$ maybe resides to $operatorname{ker} x(t_0)$ but that this dosn't have to be true for $x(t), tneq0$. Is it just the existence and uniqueness theorem saying that there can only be one solution with $x(t) = 0$?
$endgroup$
– Sebi2020
Jan 11 at 22:07
1
$begingroup$
Thank you! Your're explanations were very helpful.
$endgroup$
– Sebi2020
Jan 11 at 23:23
|
show 5 more comments
$begingroup$
This matter may be resolved via Liouville's formula, which affirms that if $X(t)$ an $n times n$ matrix solution of the differential equation
$dot X(t) = A(t)X(t) tag 1$
the determinant of $det(X(t))$ of $X(t)$ is given by
$det(X(t)) = det(X(t_0)) exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ); tag 2$
since
$exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ) ne 0, ; forall t_0, t in I, tag 3$
we see that
$det(X(t)) ne 0 Longleftrightarrow det(X(t_0)) ne 0. tag 4$
Now we may conclude with the simple observation that we may take
$W(t) = det(X(t)) = det(vec x_1(t), vec x_2(t), ldots, vec x_n(t)). tag 5$
answered Jan 10 at 23:59
Robert LewisRobert Lewis
46k23066
$endgroup$
1
$begingroup$
But is it only a implication from the liouville formula or is there another explanation? Our lecturer said that the reason for this is that the solutions form a vector room. But I cannot explain myself why this should be a proof for $det(X(t)) neq 0 Leftrightarrow det(X(t_0))neq 0$ whereas the liouville formula shows that. I think the vector room property explains only the need to look for solutions which form a fundamental system.
$endgroup$
– Sebi2020
Jan 11 at 14:06
$begingroup$
@Sebi2020: is a "vector room" the same as a vector space?
$endgroup$
– Robert Lewis
Jan 11 at 16:14
1
$begingroup$
Yes. English is not my native language, so I used the wrong translation for "vector space". But I wanted to say vector space.
$endgroup$
– Sebi2020
Jan 11 at 20:57
1
$begingroup$
I think I don't quite understand it. I struggle with this: " if the $x_i$ are linearly dependent at some point, so that $sum a_i x_i(t_0) = 0$, then we must have the unique solution $x(t) = sum a_ix_i(t) = 0$ everywhere". This is what I dont understand. Which property of linear equations or vector spaces proves that? I would thought that $a_ix_i(t_0)$ maybe resides to $operatorname{ker} x(t_0)$ but that this dosn't have to be true for $x(t), tneq0$. Is it just the existence and uniqueness theorem saying that there can only be one solution with $x(t) = 0$?
$endgroup$
– Sebi2020
Jan 11 at 22:07
1
$begingroup$
Thank you! Your're explanations were very helpful.
$endgroup$
– Sebi2020
Jan 11 at 23:23
|
show 5 more comments
$begingroup$
This matter may be resolved via Liouville's formula, which affirms that if $X(t)$ an $n times n$ matrix solution of the differential equation
$dot X(t) = A(t)X(t) tag 1$
the determinant of $det(X(t))$ of $X(t)$ is given by
$det(X(t)) = det(X(t_0)) exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ); tag 2$
since
$exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ) ne 0, ; forall t_0, t in I, tag 3$
we see that
$det(X(t)) ne 0 Longleftrightarrow det(X(t_0)) ne 0. tag 4$
Now we may conclude with the simple observation that we may take
$W(t) = det(X(t)) = det(vec x_1(t), vec x_2(t), ldots, vec x_n(t)). tag 5$
answered Jan 10 at 23:59
Robert LewisRobert Lewis
46k23066
$endgroup$
This matter may be resolved via Liouville's formula, which affirms that if $X(t)$ an $n times n$ matrix solution of the differential equation
$dot X(t) = A(t)X(t) tag 1$
the determinant of $det(X(t))$ of $X(t)$ is given by
$det(X(t)) = det(X(t_0)) exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ); tag 2$
since
$exp left ( displaystyle int_{t_0}^t text{trace}(A(s)) ; ds right ) ne 0, ; forall t_0, t in I, tag 3$
we see that
$det(X(t)) ne 0 Longleftrightarrow det(X(t_0)) ne 0. tag 4$
Now we may conclude with the simple observation that we may take
$W(t) = det(X(t)) = det(vec x_1(t), vec x_2(t), ldots, vec x_n(t)). tag 5$
answered Jan 10 at 23:59
Robert LewisRobert Lewis
46k23066
answered Jan 10 at 23:59
Robert LewisRobert Lewis
46k23066
answered Jan 10 at 23:59
Robert LewisRobert Lewis
46k23066
answered Jan 10 at 23:59
Robert LewisRobert Lewis
46k23066
46k23066
1
$begingroup$
But is it only a implication from the liouville formula or is there another explanation? Our lecturer said that the reason for this is that the solutions form a vector room. But I cannot explain myself why this should be a proof for $det(X(t)) neq 0 Leftrightarrow det(X(t_0))neq 0$ whereas the liouville formula shows that. I think the vector room property explains only the need to look for solutions which form a fundamental system.
$endgroup$
– Sebi2020
Jan 11 at 14:06
$begingroup$
@Sebi2020: is a "vector room" the same as a vector space?
$endgroup$
– Robert Lewis
Jan 11 at 16:14
1
$begingroup$
Yes. English is not my native language, so I used the wrong translation for "vector space". But I wanted to say vector space.
$endgroup$
– Sebi2020
Jan 11 at 20:57
1
$begingroup$
I think I don't quite understand it. I struggle with this: " if the $x_i$ are linearly dependent at some point, so that $sum a_i x_i(t_0) = 0$, then we must have the unique solution $x(t) = sum a_ix_i(t) = 0$ everywhere". This is what I dont understand. Which property of linear equations or vector spaces proves that? I would thought that $a_ix_i(t_0)$ maybe resides to $operatorname{ker} x(t_0)$ but that this dosn't have to be true for $x(t), tneq0$. Is it just the existence and uniqueness theorem saying that there can only be one solution with $x(t) = 0$?
$endgroup$
– Sebi2020
Jan 11 at 22:07
1
$begingroup$
Thank you! Your're explanations were very helpful.
$endgroup$
– Sebi2020
Jan 11 at 23:23
|
show 5 more comments
1
$begingroup$
But is it only a implication from the liouville formula or is there another explanation? Our lecturer said that the reason for this is that the solutions form a vector room. But I cannot explain myself why this should be a proof for $det(X(t)) neq 0 Leftrightarrow det(X(t_0))neq 0$ whereas the liouville formula shows that. I think the vector room property explains only the need to look for solutions which form a fundamental system.
$endgroup$
– Sebi2020
Jan 11 at 14:06
$begingroup$
@Sebi2020: is a "vector room" the same as a vector space?
$endgroup$
– Robert Lewis
Jan 11 at 16:14
1
$begingroup$
Yes. English is not my native language, so I used the wrong translation for "vector space". But I wanted to say vector space.
$endgroup$
– Sebi2020
Jan 11 at 20:57
1
$begingroup$
I think I don't quite understand it. I struggle with this: " if the $x_i$ are linearly dependent at some point, so that $sum a_i x_i(t_0) = 0$, then we must have the unique solution $x(t) = sum a_ix_i(t) = 0$ everywhere". This is what I dont understand. Which property of linear equations or vector spaces proves that? I would thought that $a_ix_i(t_0)$ maybe resides to $operatorname{ker} x(t_0)$ but that this dosn't have to be true for $x(t), tneq0$. Is it just the existence and uniqueness theorem saying that there can only be one solution with $x(t) = 0$?
$endgroup$
– Sebi2020
Jan 11 at 22:07
1
$begingroup$
Thank you! Your're explanations were very helpful.
$endgroup$
– Sebi2020
Jan 11 at 23:23
1
1
$begingroup$
But is it only a implication from the liouville formula or is there another explanation? Our lecturer said that the reason for this is that the solutions form a vector room. But I cannot explain myself why this should be a proof for $det(X(t)) neq 0 Leftrightarrow det(X(t_0))neq 0$ whereas the liouville formula shows that. I think the vector room property explains only the need to look for solutions which form a fundamental system.
$endgroup$
– Sebi2020
Jan 11 at 14:06
$begingroup$
But is it only a implication from the liouville formula or is there another explanation? Our lecturer said that the reason for this is that the solutions form a vector room. But I cannot explain myself why this should be a proof for $det(X(t)) neq 0 Leftrightarrow det(X(t_0))neq 0$ whereas the liouville formula shows that. I think the vector room property explains only the need to look for solutions which form a fundamental system.
$endgroup$
– Sebi2020
Jan 11 at 14:06
$begingroup$
@Sebi2020: is a "vector room" the same as a vector space?
$endgroup$
– Robert Lewis
Jan 11 at 16:14
$begingroup$
@Sebi2020: is a "vector room" the same as a vector space?
$endgroup$
– Robert Lewis
Jan 11 at 16:14
1
1
$begingroup$
Yes. English is not my native language, so I used the wrong translation for "vector space". But I wanted to say vector space.
$endgroup$
– Sebi2020
Jan 11 at 20:57
$begingroup$
Yes. English is not my native language, so I used the wrong translation for "vector space". But I wanted to say vector space.
$endgroup$
– Sebi2020
Jan 11 at 20:57
1
1
$begingroup$
I think I don't quite understand it. I struggle with this: " if the $x_i$ are linearly dependent at some point, so that $sum a_i x_i(t_0) = 0$, then we must have the unique solution $x(t) = sum a_ix_i(t) = 0$ everywhere". This is what I dont understand. Which property of linear equations or vector spaces proves that? I would thought that $a_ix_i(t_0)$ maybe resides to $operatorname{ker} x(t_0)$ but that this dosn't have to be true for $x(t), tneq0$. Is it just the existence and uniqueness theorem saying that there can only be one solution with $x(t) = 0$?
$endgroup$
– Sebi2020
Jan 11 at 22:07
$begingroup$
I think I don't quite understand it. I struggle with this: " if the $x_i$ are linearly dependent at some point, so that $sum a_i x_i(t_0) = 0$, then we must have the unique solution $x(t) = sum a_ix_i(t) = 0$ everywhere". This is what I dont understand. Which property of linear equations or vector spaces proves that? I would thought that $a_ix_i(t_0)$ maybe resides to $operatorname{ker} x(t_0)$ but that this dosn't have to be true for $x(t), tneq0$. Is it just the existence and uniqueness theorem saying that there can only be one solution with $x(t) = 0$?
$endgroup$
– Sebi2020
Jan 11 at 22:07
1
1
$begingroup$
Thank you! Your're explanations were very helpful.
$endgroup$
– Sebi2020
Jan 11 at 23:23
$begingroup$
Thank you! Your're explanations were very helpful.
$endgroup$
– Sebi2020
Jan 11 at 23:23
|
show 5 more comments
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$begingroup$
I think the issue is that if the Wronskian determinant is nonzero for some $t_0 in I$, we are guaranteed that the solutions are linearly independent. If the solutions were not linearly independent, the Wronskian would always be zero (looking at contrapositive).
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– D.B.
Jan 10 at 23:59