Compute the flow $Phi_t$ of $mathbb{X}(x,y)=(y,x)$
$begingroup$
Let $mathbb{X}$ be the vector field on $mathbb{R}^2$ given by $$ mathbb{X}(x,y) = (y,x). $$ Compute the flow $Phi_t$ of
$mathbb{X}$
I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.
The solution he gave was $Phi_t(x,y)=(frac{x+y}{2}e^t+frac{y-x}{2}e^{-t},frac{x+y}{2}e^t +frac{x-y}{2}e^{-t})$
My understanding is that to calculate the flow, you consider: $Phi_t(x,y)=(dot{x},dot{y})=(y,x)$ and try to solve.
Why cant the flow be $Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?
It is probably a problem with definitions, but any help would be greatly appreciated.
calculus ordinary-differential-equations derivatives
asked Dec 27 '14 at 21:02
Sam HoustonSam Houston
1,2141132
$endgroup$
add a comment |
$begingroup$
Let $mathbb{X}$ be the vector field on $mathbb{R}^2$ given by $$ mathbb{X}(x,y) = (y,x). $$ Compute the flow $Phi_t$ of
$mathbb{X}$
I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.
The solution he gave was $Phi_t(x,y)=(frac{x+y}{2}e^t+frac{y-x}{2}e^{-t},frac{x+y}{2}e^t +frac{x-y}{2}e^{-t})$
My understanding is that to calculate the flow, you consider: $Phi_t(x,y)=(dot{x},dot{y})=(y,x)$ and try to solve.
Why cant the flow be $Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?
It is probably a problem with definitions, but any help would be greatly appreciated.
calculus ordinary-differential-equations derivatives
asked Dec 27 '14 at 21:02
Sam HoustonSam Houston
1,2141132
$endgroup$
add a comment |
$begingroup$
Let $mathbb{X}$ be the vector field on $mathbb{R}^2$ given by $$ mathbb{X}(x,y) = (y,x). $$ Compute the flow $Phi_t$ of
$mathbb{X}$
I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.
The solution he gave was $Phi_t(x,y)=(frac{x+y}{2}e^t+frac{y-x}{2}e^{-t},frac{x+y}{2}e^t +frac{x-y}{2}e^{-t})$
My understanding is that to calculate the flow, you consider: $Phi_t(x,y)=(dot{x},dot{y})=(y,x)$ and try to solve.
Why cant the flow be $Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?
It is probably a problem with definitions, but any help would be greatly appreciated.
calculus ordinary-differential-equations derivatives
asked Dec 27 '14 at 21:02
Sam HoustonSam Houston
1,2141132
$endgroup$
Let $mathbb{X}$ be the vector field on $mathbb{R}^2$ given by $$ mathbb{X}(x,y) = (y,x). $$ Compute the flow $Phi_t$ of
$mathbb{X}$
I was reading through an answer on math.stackexchange entitled "Finding the flow of a pushforward of vector field" and the first part of his problem was this question.
The solution he gave was $Phi_t(x,y)=(frac{x+y}{2}e^t+frac{y-x}{2}e^{-t},frac{x+y}{2}e^t +frac{x-y}{2}e^{-t})$
My understanding is that to calculate the flow, you consider: $Phi_t(x,y)=(dot{x},dot{y})=(y,x)$ and try to solve.
Why cant the flow be $Phi(x,y)=((y-x)e^{-t},(x-y)e^{-t})$?
It is probably a problem with definitions, but any help would be greatly appreciated.
calculus ordinary-differential-equations derivatives
calculus ordinary-differential-equations derivatives
asked Dec 27 '14 at 21:02
Sam HoustonSam Houston
1,2141132
asked Dec 27 '14 at 21:02
Sam HoustonSam Houston
1,2141132
asked Dec 27 '14 at 21:02
Sam HoustonSam Houston
1,2141132
asked Dec 27 '14 at 21:02
Sam HoustonSam Houston
1,2141132
asked Dec 27 '14 at 21:02
Sam HoustonSam Houston
1,2141132
1,2141132
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, it is not a problem with definition.
Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
$$
A=begin{bmatrix}
0&1\
1&0
end{bmatrix}
$$
and $z=(x,y)^top$. Hence you need to calculate
$$
e^{tA}
$$
and see the result.
Addition.
There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
$$
e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
$$
Simply by noting that $A^2=I$ I find
$$
e^{tA}=begin{bmatrix}
1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
end{bmatrix}=begin{bmatrix}
cosh t&sinh t\
sinh t&cosh t
end{bmatrix},
$$
where, as usual,
$$
cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
$$
Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.
answered Dec 27 '14 at 21:13
ArtemArtem
11.4k32245
$endgroup$
$begingroup$
yes I remember reading that, thanks for the help, will give it another go
$endgroup$
– Sam Houston
Dec 27 '14 at 21:15
$begingroup$
Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
$endgroup$
– Artem
Dec 27 '14 at 21:16
1
$begingroup$
@Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
$endgroup$
– Artem
Dec 27 '14 at 21:24
1
$begingroup$
Did you hear about the matrix exponent?
$endgroup$
– Artem
Dec 27 '14 at 22:37
1
$begingroup$
See the addition. If you cannot get through it, you should start with the book I advised you.
$endgroup$
– Artem
Dec 27 '14 at 23:14
|
show 4 more comments
$begingroup$
Another way to see this result without needing to use the exponential map is the following:
For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.
answered Nov 26 '18 at 4:43
Brandon Thomas Van OverBrandon Thomas Van Over
3,83721127
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1082792%2fcompute-the-flow-phi-t-of-mathbbxx-y-y-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, it is not a problem with definition.
Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
$$
A=begin{bmatrix}
0&1\
1&0
end{bmatrix}
$$
and $z=(x,y)^top$. Hence you need to calculate
$$
e^{tA}
$$
and see the result.
Addition.
There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
$$
e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
$$
Simply by noting that $A^2=I$ I find
$$
e^{tA}=begin{bmatrix}
1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
end{bmatrix}=begin{bmatrix}
cosh t&sinh t\
sinh t&cosh t
end{bmatrix},
$$
where, as usual,
$$
cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
$$
Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.
answered Dec 27 '14 at 21:13
ArtemArtem
11.4k32245
$endgroup$
$begingroup$
yes I remember reading that, thanks for the help, will give it another go
$endgroup$
– Sam Houston
Dec 27 '14 at 21:15
$begingroup$
Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
$endgroup$
– Artem
Dec 27 '14 at 21:16
1
$begingroup$
@Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
$endgroup$
– Artem
Dec 27 '14 at 21:24
1
$begingroup$
Did you hear about the matrix exponent?
$endgroup$
– Artem
Dec 27 '14 at 22:37
1
$begingroup$
See the addition. If you cannot get through it, you should start with the book I advised you.
$endgroup$
– Artem
Dec 27 '14 at 23:14
|
show 4 more comments
$begingroup$
No, it is not a problem with definition.
Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
$$
A=begin{bmatrix}
0&1\
1&0
end{bmatrix}
$$
and $z=(x,y)^top$. Hence you need to calculate
$$
e^{tA}
$$
and see the result.
Addition.
There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
$$
e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
$$
Simply by noting that $A^2=I$ I find
$$
e^{tA}=begin{bmatrix}
1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
end{bmatrix}=begin{bmatrix}
cosh t&sinh t\
sinh t&cosh t
end{bmatrix},
$$
where, as usual,
$$
cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
$$
Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.
answered Dec 27 '14 at 21:13
ArtemArtem
11.4k32245
$endgroup$
$begingroup$
yes I remember reading that, thanks for the help, will give it another go
$endgroup$
– Sam Houston
Dec 27 '14 at 21:15
$begingroup$
Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
$endgroup$
– Artem
Dec 27 '14 at 21:16
1
$begingroup$
@Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
$endgroup$
– Artem
Dec 27 '14 at 21:24
1
$begingroup$
Did you hear about the matrix exponent?
$endgroup$
– Artem
Dec 27 '14 at 22:37
1
$begingroup$
See the addition. If you cannot get through it, you should start with the book I advised you.
$endgroup$
– Artem
Dec 27 '14 at 23:14
|
show 4 more comments
$begingroup$
No, it is not a problem with definition.
Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
$$
A=begin{bmatrix}
0&1\
1&0
end{bmatrix}
$$
and $z=(x,y)^top$. Hence you need to calculate
$$
e^{tA}
$$
and see the result.
Addition.
There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
$$
e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
$$
Simply by noting that $A^2=I$ I find
$$
e^{tA}=begin{bmatrix}
1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
end{bmatrix}=begin{bmatrix}
cosh t&sinh t\
sinh t&cosh t
end{bmatrix},
$$
where, as usual,
$$
cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
$$
Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.
answered Dec 27 '14 at 21:13
ArtemArtem
11.4k32245
$endgroup$
No, it is not a problem with definition.
Hint. The flow of the linear vector field $Az$ is given by its matrix exponent $e^{tA}z$. In your case
$$
A=begin{bmatrix}
0&1\
1&0
end{bmatrix}
$$
and $z=(x,y)^top$. Hence you need to calculate
$$
e^{tA}
$$
and see the result.
Addition.
There are different methods to calculate $e^{tA}$, but here I can use simply the defintion
$$
e^{tA}=I+tA+frac{t^2}{2!}A^2+ldots
$$
Simply by noting that $A^2=I$ I find
$$
e^{tA}=begin{bmatrix}
1+frac{t^2}{2!}+frac{t^4}{4!}+ldots& t+frac{t^3}{3!}+ldots\
t+frac{t^3}{3!}+ldots&1+frac{t^2}{2!}+frac{t^4}{4!}+ldots
end{bmatrix}=begin{bmatrix}
cosh t&sinh t\
sinh t&cosh t
end{bmatrix},
$$
where, as usual,
$$
cosh t=frac{e^t+e^{-t}}{2},quad sinh t=frac{e^t-e^{-t}}{2}.
$$
Now you can see you flow by multiplying $e^{tA}$ by the vector of initial conditions.
answered Dec 27 '14 at 21:13
ArtemArtem
11.4k32245
edited Dec 27 '14 at 23:14
answered Dec 27 '14 at 21:13
ArtemArtem
11.4k32245
answered Dec 27 '14 at 21:13
ArtemArtem
11.4k32245
answered Dec 27 '14 at 21:13
ArtemArtem
11.4k32245
11.4k32245
$begingroup$
yes I remember reading that, thanks for the help, will give it another go
$endgroup$
– Sam Houston
Dec 27 '14 at 21:15
$begingroup$
Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
$endgroup$
– Artem
Dec 27 '14 at 21:16
1
$begingroup$
@Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
$endgroup$
– Artem
Dec 27 '14 at 21:24
1
$begingroup$
Did you hear about the matrix exponent?
$endgroup$
– Artem
Dec 27 '14 at 22:37
1
$begingroup$
See the addition. If you cannot get through it, you should start with the book I advised you.
$endgroup$
– Artem
Dec 27 '14 at 23:14
|
show 4 more comments
$begingroup$
yes I remember reading that, thanks for the help, will give it another go
$endgroup$
– Sam Houston
Dec 27 '14 at 21:15
$begingroup$
Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
$endgroup$
– Artem
Dec 27 '14 at 21:16
1
$begingroup$
@Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
$endgroup$
– Artem
Dec 27 '14 at 21:24
1
$begingroup$
Did you hear about the matrix exponent?
$endgroup$
– Artem
Dec 27 '14 at 22:37
1
$begingroup$
See the addition. If you cannot get through it, you should start with the book I advised you.
$endgroup$
– Artem
Dec 27 '14 at 23:14
$begingroup$
yes I remember reading that, thanks for the help, will give it another go
$endgroup$
– Sam Houston
Dec 27 '14 at 21:15
$begingroup$
yes I remember reading that, thanks for the help, will give it another go
$endgroup$
– Sam Houston
Dec 27 '14 at 21:15
$begingroup$
Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
$endgroup$
– Artem
Dec 27 '14 at 21:16
$begingroup$
Another hint to see where you are a little confused: not every solution to the system of ODE will give you the flow. Only those, where you can simply replace two arbitrary constants with two initial conditions.
$endgroup$
– Artem
Dec 27 '14 at 21:16
1
1
$begingroup$
@Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
$endgroup$
– Artem
Dec 27 '14 at 21:24
$begingroup$
@Dansmith I am not sure that I can give you an ultimate reference. The flows of linear constant coefficients vector fields are linear transformation groups defined simply by $e^{tA}$. There is not much to learn here. Try Arnold's ODE, if you can get through his deceivably easy-to-understand writings, you will learn the general theory of flows, transformation groups, and vector fieleds.
$endgroup$
– Artem
Dec 27 '14 at 21:24
1
1
$begingroup$
Did you hear about the matrix exponent?
$endgroup$
– Artem
Dec 27 '14 at 22:37
$begingroup$
Did you hear about the matrix exponent?
$endgroup$
– Artem
Dec 27 '14 at 22:37
1
1
$begingroup$
See the addition. If you cannot get through it, you should start with the book I advised you.
$endgroup$
– Artem
Dec 27 '14 at 23:14
$begingroup$
See the addition. If you cannot get through it, you should start with the book I advised you.
$endgroup$
– Artem
Dec 27 '14 at 23:14
|
show 4 more comments
$begingroup$
Another way to see this result without needing to use the exponential map is the following:
For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.
answered Nov 26 '18 at 4:43
Brandon Thomas Van OverBrandon Thomas Van Over
3,83721127
$endgroup$
add a comment |
$begingroup$
Another way to see this result without needing to use the exponential map is the following:
For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.
answered Nov 26 '18 at 4:43
Brandon Thomas Van OverBrandon Thomas Van Over
3,83721127
$endgroup$
add a comment |
$begingroup$
Another way to see this result without needing to use the exponential map is the following:
For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.
answered Nov 26 '18 at 4:43
Brandon Thomas Van OverBrandon Thomas Van Over
3,83721127
$endgroup$
Another way to see this result without needing to use the exponential map is the following:
For $Y = xfrac{partial}{partial y} + yfrac{partial}{ partial x}$ we obtain the differential equations $x'(t) = y$ and $y'(t) = x$. We note that differentiating either of the equations gives us that $y''(t)=y(t)$ and similarly for $x(t)$. We know that the general solution to each of these equations is $y(t)= d_1e^t +d_2e^{-t}$ and $x(t) = c_1e^t +c_2e^{-t}$. Now our global flow must satisfy $x(0)=y$ and $y(0)=x$, and thus we impose the conditions that $c_1 +c_2 =y$ and $d_1+d_2 =x$. Now given that $x'(t) = y(t)$ we have the following relationship between coefficients: $c_1=d_1$ and $c_2 =-d_2$. Plugging these in to the system of equations above and solving for $c_1$ and $c_2$ we obtain that $c_1 = frac{x+y}{2}$, and $c_2 = frac{y-x}{2}$. Thus our flow becomes $theta_t(x, y) = (frac{x+y}{2}e^t + frac{y-x}{2}e^{-t}, frac{x+y}{2}e^t-frac{y-x}{2}e^{-t})$.
answered Nov 26 '18 at 4:43
Brandon Thomas Van OverBrandon Thomas Van Over
3,83721127
answered Nov 26 '18 at 4:43
Brandon Thomas Van OverBrandon Thomas Van Over
3,83721127
answered Nov 26 '18 at 4:43
Brandon Thomas Van OverBrandon Thomas Van Over
3,83721127
answered Nov 26 '18 at 4:43
Brandon Thomas Van OverBrandon Thomas Van Over
3,83721127
3,83721127
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1082792%2fcompute-the-flow-phi-t-of-mathbbxx-y-y-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
