Mixing
How to solve this system differential equations to determine the local flow of a smooth vector field?
$begingroup$
I want to calculate the local flow of the following smooth vector field on $mathbb{R}^2$, $X:mathbb{R}^2to Tmathbb{R}^2$, defined by $X=(x^2-x^1)frac{partial}{partial x^1}-(x^1+x^2)frac{partial}{partial x^2}$.
To do this, I have to resolve the following sistem of differential equations:
begin{cases}
frac{d}{dt}theta^p(t)^1=theta^p(t)^2-theta^p(t)^1 \
frac{d}{dt}theta^p(t)^2=-(theta^p(t)^1+theta^p(t)^2)\
theta^p(0)=p
end{cases}
for all $pinmathbb{R}^2$. (Where $theta^p(t)^i $ is the $i$-component function of $theta^p(t)$ )
I know the solution should be $theta^p(t)=(p^1e^{-t}cos t+p^2e^{-t}sin t,-p^1e^{-t}sin t+p^2e^{-t}cos t)$ defined for all $tin mathbb{R}$ and for all $pinmathbb{R}^2$.
Any hint/ suggestion/ strategy/ observation on how to resolve this?
ordinary-differential-equations multivariable-calculus differential-geometry smooth-manifolds
asked Jan 6 at 14:04
MinatoMinato
465212
$endgroup$
add a comment |
$begingroup$
I want to calculate the local flow of the following smooth vector field on $mathbb{R}^2$, $X:mathbb{R}^2to Tmathbb{R}^2$, defined by $X=(x^2-x^1)frac{partial}{partial x^1}-(x^1+x^2)frac{partial}{partial x^2}$.
To do this, I have to resolve the following sistem of differential equations:
begin{cases}
frac{d}{dt}theta^p(t)^1=theta^p(t)^2-theta^p(t)^1 \
frac{d}{dt}theta^p(t)^2=-(theta^p(t)^1+theta^p(t)^2)\
theta^p(0)=p
end{cases}
for all $pinmathbb{R}^2$. (Where $theta^p(t)^i $ is the $i$-component function of $theta^p(t)$ )
I know the solution should be $theta^p(t)=(p^1e^{-t}cos t+p^2e^{-t}sin t,-p^1e^{-t}sin t+p^2e^{-t}cos t)$ defined for all $tin mathbb{R}$ and for all $pinmathbb{R}^2$.
Any hint/ suggestion/ strategy/ observation on how to resolve this?
ordinary-differential-equations multivariable-calculus differential-geometry smooth-manifolds
asked Jan 6 at 14:04
MinatoMinato
465212
$endgroup$
add a comment |
$begingroup$
I want to calculate the local flow of the following smooth vector field on $mathbb{R}^2$, $X:mathbb{R}^2to Tmathbb{R}^2$, defined by $X=(x^2-x^1)frac{partial}{partial x^1}-(x^1+x^2)frac{partial}{partial x^2}$.
To do this, I have to resolve the following sistem of differential equations:
begin{cases}
frac{d}{dt}theta^p(t)^1=theta^p(t)^2-theta^p(t)^1 \
frac{d}{dt}theta^p(t)^2=-(theta^p(t)^1+theta^p(t)^2)\
theta^p(0)=p
end{cases}
for all $pinmathbb{R}^2$. (Where $theta^p(t)^i $ is the $i$-component function of $theta^p(t)$ )
I know the solution should be $theta^p(t)=(p^1e^{-t}cos t+p^2e^{-t}sin t,-p^1e^{-t}sin t+p^2e^{-t}cos t)$ defined for all $tin mathbb{R}$ and for all $pinmathbb{R}^2$.
Any hint/ suggestion/ strategy/ observation on how to resolve this?
ordinary-differential-equations multivariable-calculus differential-geometry smooth-manifolds
asked Jan 6 at 14:04
MinatoMinato
465212
$endgroup$
I want to calculate the local flow of the following smooth vector field on $mathbb{R}^2$, $X:mathbb{R}^2to Tmathbb{R}^2$, defined by $X=(x^2-x^1)frac{partial}{partial x^1}-(x^1+x^2)frac{partial}{partial x^2}$.
To do this, I have to resolve the following sistem of differential equations:
begin{cases}
frac{d}{dt}theta^p(t)^1=theta^p(t)^2-theta^p(t)^1 \
frac{d}{dt}theta^p(t)^2=-(theta^p(t)^1+theta^p(t)^2)\
theta^p(0)=p
end{cases}
for all $pinmathbb{R}^2$. (Where $theta^p(t)^i $ is the $i$-component function of $theta^p(t)$ )
I know the solution should be $theta^p(t)=(p^1e^{-t}cos t+p^2e^{-t}sin t,-p^1e^{-t}sin t+p^2e^{-t}cos t)$ defined for all $tin mathbb{R}$ and for all $pinmathbb{R}^2$.
Any hint/ suggestion/ strategy/ observation on how to resolve this?
ordinary-differential-equations multivariable-calculus differential-geometry smooth-manifolds
ordinary-differential-equations multivariable-calculus differential-geometry smooth-manifolds
asked Jan 6 at 14:04
MinatoMinato
465212
asked Jan 6 at 14:04
MinatoMinato
465212
asked Jan 6 at 14:04
MinatoMinato
465212
asked Jan 6 at 14:04
MinatoMinato
465212
asked Jan 6 at 14:04
MinatoMinato
465212
465212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint. Your system can be written as:
$$begin{pmatrix}dot{theta_p^1}\dot{theta_p^2}end{pmatrix}=begin{pmatrix}-1& 1\-1&-1end{pmatrix}begin{pmatrix}theta_p^1\theta_p^2end{pmatrix},$$
which is a first order linear equation in $mathbb{R}^2$ and notice that one has:
$$expleft(tbegin{pmatrix}-1& 1\-1&-1end{pmatrix}right)=begin{pmatrix}e^{-t}cos(t)&e^{-t}sin(t)\-e^{-t}sin(t)&e^{-t}cos(t)end{pmatrix},$$
whence the result.
Remark. I must explain how I compute this matrix exponential, I diagonalize it over $mathbb{C}$, finding its eigenvalues are $(-1-i)t$ and $(-1+i)t$ and the change of basis matrix is given by $begin{pmatrix}i&-i\1&1end{pmatrix}$.
Reminder. Let $A$ be a square matrix, then one has:
$$dot{x}=Axifffrac{mathrm{d}}{mathrm{d}t}e^{tA}x=0iff x=x(0)e^{-tA},$$
the point being that the derivative of $tmapsto e^{tA}$ is $tmapsto Ae^{tA}$ and that $e^{-tA}e^{tA}=I$.
answered Jan 6 at 14:21
C. FalconC. Falcon
15.1k41950
$endgroup$
$begingroup$
Thank-you very much! I don't know the theory behind your answer, and Here on Wikipedia it ist not mentioned the case in which the eigenvalues are complex. Can you also provide me a reference for the theory behind your answer?
$endgroup$
– Minato
Jan 7 at 11:11
$begingroup$
I have added a little addendum on the resolution of a linear system of differential equations of order $1$. Regarding the complex eigenvalues case, there is no more to it than the real eigenvalues case, the point is that $exp(PAP^{-1})=Pexp(A)P^{-1}$.
$endgroup$
– C. Falcon
Jan 7 at 11:40
add a comment |
$begingroup$
The method in C. Falcon's solution applies in a much broader setting (all systems of linear, constant-coefficient, homogeneous o.d.e.s of $n$ functions in $n$ variables). But if you don't have the machinery at hand, here's a naive method:
Hint Differentiating, say, the first equation in the system gives
$$ddottheta{}^1 = -dottheta{}^1 + dottheta{}^2 = -(-theta{}^1 + theta^2) - (theta^1 + theta^2) = -2theta^2 = -2 (dottheta{}^1 + theta^1) ,$$
and rearranging gives the o.d.e.
$$boxed{ddottheta{}^1 + 2 dottheta{}^1 + 2 theta^1 = 0}$$
in $theta^1$ alone. (For legibility we've denoted $theta^i := theta^p(t)^i$.) NB this is a second-order equation, so to find a unique solution, we need two conditions, say, on $theta^1(0)$ and $dottheta{}^1(0)$---can you find these?
answered Jan 7 at 12:38
TravisTravis
60k767147
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint. Your system can be written as:
$$begin{pmatrix}dot{theta_p^1}\dot{theta_p^2}end{pmatrix}=begin{pmatrix}-1& 1\-1&-1end{pmatrix}begin{pmatrix}theta_p^1\theta_p^2end{pmatrix},$$
which is a first order linear equation in $mathbb{R}^2$ and notice that one has:
$$expleft(tbegin{pmatrix}-1& 1\-1&-1end{pmatrix}right)=begin{pmatrix}e^{-t}cos(t)&e^{-t}sin(t)\-e^{-t}sin(t)&e^{-t}cos(t)end{pmatrix},$$
whence the result.
Remark. I must explain how I compute this matrix exponential, I diagonalize it over $mathbb{C}$, finding its eigenvalues are $(-1-i)t$ and $(-1+i)t$ and the change of basis matrix is given by $begin{pmatrix}i&-i\1&1end{pmatrix}$.
Reminder. Let $A$ be a square matrix, then one has:
$$dot{x}=Axifffrac{mathrm{d}}{mathrm{d}t}e^{tA}x=0iff x=x(0)e^{-tA},$$
the point being that the derivative of $tmapsto e^{tA}$ is $tmapsto Ae^{tA}$ and that $e^{-tA}e^{tA}=I$.
answered Jan 6 at 14:21
C. FalconC. Falcon
15.1k41950
$endgroup$
$begingroup$
Thank-you very much! I don't know the theory behind your answer, and Here on Wikipedia it ist not mentioned the case in which the eigenvalues are complex. Can you also provide me a reference for the theory behind your answer?
$endgroup$
– Minato
Jan 7 at 11:11
$begingroup$
I have added a little addendum on the resolution of a linear system of differential equations of order $1$. Regarding the complex eigenvalues case, there is no more to it than the real eigenvalues case, the point is that $exp(PAP^{-1})=Pexp(A)P^{-1}$.
$endgroup$
– C. Falcon
Jan 7 at 11:40
add a comment |
$begingroup$
Hint. Your system can be written as:
$$begin{pmatrix}dot{theta_p^1}\dot{theta_p^2}end{pmatrix}=begin{pmatrix}-1& 1\-1&-1end{pmatrix}begin{pmatrix}theta_p^1\theta_p^2end{pmatrix},$$
which is a first order linear equation in $mathbb{R}^2$ and notice that one has:
$$expleft(tbegin{pmatrix}-1& 1\-1&-1end{pmatrix}right)=begin{pmatrix}e^{-t}cos(t)&e^{-t}sin(t)\-e^{-t}sin(t)&e^{-t}cos(t)end{pmatrix},$$
whence the result.
Remark. I must explain how I compute this matrix exponential, I diagonalize it over $mathbb{C}$, finding its eigenvalues are $(-1-i)t$ and $(-1+i)t$ and the change of basis matrix is given by $begin{pmatrix}i&-i\1&1end{pmatrix}$.
Reminder. Let $A$ be a square matrix, then one has:
$$dot{x}=Axifffrac{mathrm{d}}{mathrm{d}t}e^{tA}x=0iff x=x(0)e^{-tA},$$
the point being that the derivative of $tmapsto e^{tA}$ is $tmapsto Ae^{tA}$ and that $e^{-tA}e^{tA}=I$.
answered Jan 6 at 14:21
C. FalconC. Falcon
15.1k41950
$endgroup$
$begingroup$
Thank-you very much! I don't know the theory behind your answer, and Here on Wikipedia it ist not mentioned the case in which the eigenvalues are complex. Can you also provide me a reference for the theory behind your answer?
$endgroup$
– Minato
Jan 7 at 11:11
$begingroup$
I have added a little addendum on the resolution of a linear system of differential equations of order $1$. Regarding the complex eigenvalues case, there is no more to it than the real eigenvalues case, the point is that $exp(PAP^{-1})=Pexp(A)P^{-1}$.
$endgroup$
– C. Falcon
Jan 7 at 11:40
add a comment |
$begingroup$
Hint. Your system can be written as:
$$begin{pmatrix}dot{theta_p^1}\dot{theta_p^2}end{pmatrix}=begin{pmatrix}-1& 1\-1&-1end{pmatrix}begin{pmatrix}theta_p^1\theta_p^2end{pmatrix},$$
which is a first order linear equation in $mathbb{R}^2$ and notice that one has:
$$expleft(tbegin{pmatrix}-1& 1\-1&-1end{pmatrix}right)=begin{pmatrix}e^{-t}cos(t)&e^{-t}sin(t)\-e^{-t}sin(t)&e^{-t}cos(t)end{pmatrix},$$
whence the result.
Remark. I must explain how I compute this matrix exponential, I diagonalize it over $mathbb{C}$, finding its eigenvalues are $(-1-i)t$ and $(-1+i)t$ and the change of basis matrix is given by $begin{pmatrix}i&-i\1&1end{pmatrix}$.
Reminder. Let $A$ be a square matrix, then one has:
$$dot{x}=Axifffrac{mathrm{d}}{mathrm{d}t}e^{tA}x=0iff x=x(0)e^{-tA},$$
the point being that the derivative of $tmapsto e^{tA}$ is $tmapsto Ae^{tA}$ and that $e^{-tA}e^{tA}=I$.
answered Jan 6 at 14:21
C. FalconC. Falcon
15.1k41950
$endgroup$
Hint. Your system can be written as:
$$begin{pmatrix}dot{theta_p^1}\dot{theta_p^2}end{pmatrix}=begin{pmatrix}-1& 1\-1&-1end{pmatrix}begin{pmatrix}theta_p^1\theta_p^2end{pmatrix},$$
which is a first order linear equation in $mathbb{R}^2$ and notice that one has:
$$expleft(tbegin{pmatrix}-1& 1\-1&-1end{pmatrix}right)=begin{pmatrix}e^{-t}cos(t)&e^{-t}sin(t)\-e^{-t}sin(t)&e^{-t}cos(t)end{pmatrix},$$
whence the result.
Remark. I must explain how I compute this matrix exponential, I diagonalize it over $mathbb{C}$, finding its eigenvalues are $(-1-i)t$ and $(-1+i)t$ and the change of basis matrix is given by $begin{pmatrix}i&-i\1&1end{pmatrix}$.
Reminder. Let $A$ be a square matrix, then one has:
$$dot{x}=Axifffrac{mathrm{d}}{mathrm{d}t}e^{tA}x=0iff x=x(0)e^{-tA},$$
the point being that the derivative of $tmapsto e^{tA}$ is $tmapsto Ae^{tA}$ and that $e^{-tA}e^{tA}=I$.
answered Jan 6 at 14:21
C. FalconC. Falcon
15.1k41950
edited Jan 7 at 12:06
answered Jan 6 at 14:21
C. FalconC. Falcon
15.1k41950
answered Jan 6 at 14:21
C. FalconC. Falcon
15.1k41950
answered Jan 6 at 14:21
C. FalconC. Falcon
15.1k41950
15.1k41950
$begingroup$
Thank-you very much! I don't know the theory behind your answer, and Here on Wikipedia it ist not mentioned the case in which the eigenvalues are complex. Can you also provide me a reference for the theory behind your answer?
$endgroup$
– Minato
Jan 7 at 11:11
$begingroup$
I have added a little addendum on the resolution of a linear system of differential equations of order $1$. Regarding the complex eigenvalues case, there is no more to it than the real eigenvalues case, the point is that $exp(PAP^{-1})=Pexp(A)P^{-1}$.
$endgroup$
– C. Falcon
Jan 7 at 11:40
add a comment |
$begingroup$
Thank-you very much! I don't know the theory behind your answer, and Here on Wikipedia it ist not mentioned the case in which the eigenvalues are complex. Can you also provide me a reference for the theory behind your answer?
$endgroup$
– Minato
Jan 7 at 11:11
$begingroup$
I have added a little addendum on the resolution of a linear system of differential equations of order $1$. Regarding the complex eigenvalues case, there is no more to it than the real eigenvalues case, the point is that $exp(PAP^{-1})=Pexp(A)P^{-1}$.
$endgroup$
– C. Falcon
Jan 7 at 11:40
$begingroup$
Thank-you very much! I don't know the theory behind your answer, and Here on Wikipedia it ist not mentioned the case in which the eigenvalues are complex. Can you also provide me a reference for the theory behind your answer?
$endgroup$
– Minato
Jan 7 at 11:11
$begingroup$
Thank-you very much! I don't know the theory behind your answer, and Here on Wikipedia it ist not mentioned the case in which the eigenvalues are complex. Can you also provide me a reference for the theory behind your answer?
$endgroup$
– Minato
Jan 7 at 11:11
$begingroup$
I have added a little addendum on the resolution of a linear system of differential equations of order $1$. Regarding the complex eigenvalues case, there is no more to it than the real eigenvalues case, the point is that $exp(PAP^{-1})=Pexp(A)P^{-1}$.
$endgroup$
– C. Falcon
Jan 7 at 11:40
$begingroup$
I have added a little addendum on the resolution of a linear system of differential equations of order $1$. Regarding the complex eigenvalues case, there is no more to it than the real eigenvalues case, the point is that $exp(PAP^{-1})=Pexp(A)P^{-1}$.
$endgroup$
– C. Falcon
Jan 7 at 11:40
add a comment |
$begingroup$
The method in C. Falcon's solution applies in a much broader setting (all systems of linear, constant-coefficient, homogeneous o.d.e.s of $n$ functions in $n$ variables). But if you don't have the machinery at hand, here's a naive method:
Hint Differentiating, say, the first equation in the system gives
$$ddottheta{}^1 = -dottheta{}^1 + dottheta{}^2 = -(-theta{}^1 + theta^2) - (theta^1 + theta^2) = -2theta^2 = -2 (dottheta{}^1 + theta^1) ,$$
and rearranging gives the o.d.e.
$$boxed{ddottheta{}^1 + 2 dottheta{}^1 + 2 theta^1 = 0}$$
in $theta^1$ alone. (For legibility we've denoted $theta^i := theta^p(t)^i$.) NB this is a second-order equation, so to find a unique solution, we need two conditions, say, on $theta^1(0)$ and $dottheta{}^1(0)$---can you find these?
answered Jan 7 at 12:38
TravisTravis
60k767147
$endgroup$
add a comment |
$begingroup$
The method in C. Falcon's solution applies in a much broader setting (all systems of linear, constant-coefficient, homogeneous o.d.e.s of $n$ functions in $n$ variables). But if you don't have the machinery at hand, here's a naive method:
Hint Differentiating, say, the first equation in the system gives
$$ddottheta{}^1 = -dottheta{}^1 + dottheta{}^2 = -(-theta{}^1 + theta^2) - (theta^1 + theta^2) = -2theta^2 = -2 (dottheta{}^1 + theta^1) ,$$
and rearranging gives the o.d.e.
$$boxed{ddottheta{}^1 + 2 dottheta{}^1 + 2 theta^1 = 0}$$
in $theta^1$ alone. (For legibility we've denoted $theta^i := theta^p(t)^i$.) NB this is a second-order equation, so to find a unique solution, we need two conditions, say, on $theta^1(0)$ and $dottheta{}^1(0)$---can you find these?
answered Jan 7 at 12:38
TravisTravis
60k767147
$endgroup$
add a comment |
$begingroup$
The method in C. Falcon's solution applies in a much broader setting (all systems of linear, constant-coefficient, homogeneous o.d.e.s of $n$ functions in $n$ variables). But if you don't have the machinery at hand, here's a naive method:
Hint Differentiating, say, the first equation in the system gives
$$ddottheta{}^1 = -dottheta{}^1 + dottheta{}^2 = -(-theta{}^1 + theta^2) - (theta^1 + theta^2) = -2theta^2 = -2 (dottheta{}^1 + theta^1) ,$$
and rearranging gives the o.d.e.
$$boxed{ddottheta{}^1 + 2 dottheta{}^1 + 2 theta^1 = 0}$$
in $theta^1$ alone. (For legibility we've denoted $theta^i := theta^p(t)^i$.) NB this is a second-order equation, so to find a unique solution, we need two conditions, say, on $theta^1(0)$ and $dottheta{}^1(0)$---can you find these?
answered Jan 7 at 12:38
TravisTravis
60k767147
$endgroup$
The method in C. Falcon's solution applies in a much broader setting (all systems of linear, constant-coefficient, homogeneous o.d.e.s of $n$ functions in $n$ variables). But if you don't have the machinery at hand, here's a naive method:
Hint Differentiating, say, the first equation in the system gives
$$ddottheta{}^1 = -dottheta{}^1 + dottheta{}^2 = -(-theta{}^1 + theta^2) - (theta^1 + theta^2) = -2theta^2 = -2 (dottheta{}^1 + theta^1) ,$$
and rearranging gives the o.d.e.
$$boxed{ddottheta{}^1 + 2 dottheta{}^1 + 2 theta^1 = 0}$$
in $theta^1$ alone. (For legibility we've denoted $theta^i := theta^p(t)^i$.) NB this is a second-order equation, so to find a unique solution, we need two conditions, say, on $theta^1(0)$ and $dottheta{}^1(0)$---can you find these?
answered Jan 7 at 12:38
TravisTravis
60k767147
answered Jan 7 at 12:38
TravisTravis
60k767147
answered Jan 7 at 12:38
TravisTravis
60k767147
answered Jan 7 at 12:38
TravisTravis
60k767147
60k767147
add a comment |
add a comment |
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