Finding approximation to stable manifold of saddle point
$begingroup$
I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).
Consider the system $dot{x} = x+e^{-y}, dot{y} = -y$. This system
has a single fixed point, $(-1, 0)$. This is a saddle point. The
unstable manifold is $y=0$, but the stable manifold is some non-linear
curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
> 0)$ and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
expressions for $dy/du$ and equate them.
I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.
I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:
Furthermore, I don't see why they have $frac{dy}{du} = frac{dot{y}}{dot{u}}$ or how they calculated the Taylor approximation for $dot{u}$.
Basically, I'm a bit lost, could someone perhaps give me some hints?
ordinary-differential-equations taylor-expansion dynamical-systems
asked Dec 14 '18 at 21:44
DashermanDasherman
1,152818
$endgroup$
add a comment |
$begingroup$
I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).
Consider the system $dot{x} = x+e^{-y}, dot{y} = -y$. This system
has a single fixed point, $(-1, 0)$. This is a saddle point. The
unstable manifold is $y=0$, but the stable manifold is some non-linear
curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
> 0)$ and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
expressions for $dy/du$ and equate them.
I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.
I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:
Furthermore, I don't see why they have $frac{dy}{du} = frac{dot{y}}{dot{u}}$ or how they calculated the Taylor approximation for $dot{u}$.
Basically, I'm a bit lost, could someone perhaps give me some hints?
ordinary-differential-equations taylor-expansion dynamical-systems
asked Dec 14 '18 at 21:44
DashermanDasherman
1,152818
$endgroup$
add a comment |
$begingroup$
I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).
Consider the system $dot{x} = x+e^{-y}, dot{y} = -y$. This system
has a single fixed point, $(-1, 0)$. This is a saddle point. The
unstable manifold is $y=0$, but the stable manifold is some non-linear
curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
> 0)$ and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
expressions for $dy/du$ and equate them.
I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.
I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:
Furthermore, I don't see why they have $frac{dy}{du} = frac{dot{y}}{dot{u}}$ or how they calculated the Taylor approximation for $dot{u}$.
Basically, I'm a bit lost, could someone perhaps give me some hints?
ordinary-differential-equations taylor-expansion dynamical-systems
asked Dec 14 '18 at 21:44
DashermanDasherman
1,152818
$endgroup$
I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).
Consider the system $dot{x} = x+e^{-y}, dot{y} = -y$. This system
has a single fixed point, $(-1, 0)$. This is a saddle point. The
unstable manifold is $y=0$, but the stable manifold is some non-linear
curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
> 0)$ and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
expressions for $dy/du$ and equate them.
I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.
I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:
Furthermore, I don't see why they have $frac{dy}{du} = frac{dot{y}}{dot{u}}$ or how they calculated the Taylor approximation for $dot{u}$.
Basically, I'm a bit lost, could someone perhaps give me some hints?
ordinary-differential-equations taylor-expansion dynamical-systems
ordinary-differential-equations taylor-expansion dynamical-systems
asked Dec 14 '18 at 21:44
DashermanDasherman
1,152818
asked Dec 14 '18 at 21:44
DashermanDasherman
1,152818
asked Dec 14 '18 at 21:44
DashermanDasherman
1,152818
asked Dec 14 '18 at 21:44
DashermanDasherman
1,152818
asked Dec 14 '18 at 21:44
DashermanDasherman
1,152818
1,152818
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Remember that
$$
frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
$$
If you replace
$$
frac{{rm d}y}{{rm d}t} = -y tag{2}
$$
and
$$
frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
$$
In (1) you will get
$$
frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
$$
Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is
begin{eqnarray}
frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
&=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
end{eqnarray}
You solve for coefficients, and you should get
$$
a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
$$
The black line below is the result
answered Dec 14 '18 at 22:29
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
$endgroup$
– Dasherman
Dec 16 '18 at 14:10
1
$begingroup$
@Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
$endgroup$
– caverac
Dec 16 '18 at 14:17
$begingroup$
Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
$endgroup$
– Dasherman
Dec 16 '18 at 14:27
1
$begingroup$
@Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
$endgroup$
– caverac
Dec 16 '18 at 14:35
1
$begingroup$
@Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
$endgroup$
– caverac
Dec 20 '18 at 13:26
|
show 1 more 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%2f3039914%2ffinding-approximation-to-stable-manifold-of-saddle-point%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Remember that
$$
frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
$$
If you replace
$$
frac{{rm d}y}{{rm d}t} = -y tag{2}
$$
and
$$
frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
$$
In (1) you will get
$$
frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
$$
Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is
begin{eqnarray}
frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
&=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
end{eqnarray}
You solve for coefficients, and you should get
$$
a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
$$
The black line below is the result
answered Dec 14 '18 at 22:29
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
$endgroup$
– Dasherman
Dec 16 '18 at 14:10
1
$begingroup$
@Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
$endgroup$
– caverac
Dec 16 '18 at 14:17
$begingroup$
Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
$endgroup$
– Dasherman
Dec 16 '18 at 14:27
1
$begingroup$
@Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
$endgroup$
– caverac
Dec 16 '18 at 14:35
1
$begingroup$
@Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
$endgroup$
– caverac
Dec 20 '18 at 13:26
|
show 1 more comment
$begingroup$
Remember that
$$
frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
$$
If you replace
$$
frac{{rm d}y}{{rm d}t} = -y tag{2}
$$
and
$$
frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
$$
In (1) you will get
$$
frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
$$
Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is
begin{eqnarray}
frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
&=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
end{eqnarray}
You solve for coefficients, and you should get
$$
a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
$$
The black line below is the result
answered Dec 14 '18 at 22:29
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
$endgroup$
– Dasherman
Dec 16 '18 at 14:10
1
$begingroup$
@Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
$endgroup$
– caverac
Dec 16 '18 at 14:17
$begingroup$
Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
$endgroup$
– Dasherman
Dec 16 '18 at 14:27
1
$begingroup$
@Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
$endgroup$
– caverac
Dec 16 '18 at 14:35
1
$begingroup$
@Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
$endgroup$
– caverac
Dec 20 '18 at 13:26
|
show 1 more comment
$begingroup$
Remember that
$$
frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
$$
If you replace
$$
frac{{rm d}y}{{rm d}t} = -y tag{2}
$$
and
$$
frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
$$
In (1) you will get
$$
frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
$$
Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is
begin{eqnarray}
frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
&=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
end{eqnarray}
You solve for coefficients, and you should get
$$
a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
$$
The black line below is the result
answered Dec 14 '18 at 22:29
caveraccaverac
14.8k31130
$endgroup$
Remember that
$$
frac{{rm d}y}{{rm d}u} = frac{{rm d}y / {rm d}t}{{rm d}u / {rm d}t} tag{1}
$$
If you replace
$$
frac{{rm d}y}{{rm d}t} = -y tag{2}
$$
and
$$
frac{{rm d}u}{{rm d}t} = frac{{rm d}x}{{rm d}t} = u - 1 + left(1 - y + frac{y^2}{2} - cdots right) tag{3}
$$
In (1) you will get
$$
frac{{rm d}y}{{rm d}u} = -frac{y}{u - y + y^2/2 - y^3/6 + cdots} tag{4}
$$
Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is
begin{eqnarray}
frac{{rm d}y}{{rm d}u} &=& frac{a_1}{a_1 - 1} + frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + cdots \
&=& a_1 + 2a_2 u + 3a_3 u^3 + cdots
end{eqnarray}
You solve for coefficients, and you should get
$$
a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
$$
The black line below is the result
answered Dec 14 '18 at 22:29
caveraccaverac
14.8k31130
answered Dec 14 '18 at 22:29
caveraccaverac
14.8k31130
answered Dec 14 '18 at 22:29
caveraccaverac
14.8k31130
answered Dec 14 '18 at 22:29
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
$endgroup$
– Dasherman
Dec 16 '18 at 14:10
1
$begingroup$
@Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
$endgroup$
– caverac
Dec 16 '18 at 14:17
$begingroup$
Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
$endgroup$
– Dasherman
Dec 16 '18 at 14:27
1
$begingroup$
@Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
$endgroup$
– caverac
Dec 16 '18 at 14:35
1
$begingroup$
@Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
$endgroup$
– caverac
Dec 20 '18 at 13:26
|
show 1 more comment
$begingroup$
Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
$endgroup$
– Dasherman
Dec 16 '18 at 14:10
1
$begingroup$
@Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
$endgroup$
– caverac
Dec 16 '18 at 14:17
$begingroup$
Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
$endgroup$
– Dasherman
Dec 16 '18 at 14:27
1
$begingroup$
@Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
$endgroup$
– caverac
Dec 16 '18 at 14:35
1
$begingroup$
@Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
$endgroup$
– caverac
Dec 20 '18 at 13:26
$begingroup$
Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
$endgroup$
– Dasherman
Dec 16 '18 at 14:10
$begingroup$
Thank you for your answer! How did you find the Taylor expansion around $u=0$? If I substitute the equation for $y(u)$ into $(4)$, I get a function of $u$ that is $0$ for $u=0$, rather than the constant term $frac{a_1}{a_1-1}$ you find.
$endgroup$
– Dasherman
Dec 16 '18 at 14:10
1
1
$begingroup$
@Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
$endgroup$
– caverac
Dec 16 '18 at 14:17
$begingroup$
@Dasherman It will be zero in the numerator and in the denominator, take the limit when $uto 0$
$endgroup$
– caverac
Dec 16 '18 at 14:17
$begingroup$
Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
$endgroup$
– Dasherman
Dec 16 '18 at 14:27
$begingroup$
Oh, I see. Thank you! I only checked the numerator. Do you also happen to know what the 'correct' way is to treat the $O(u^3)$ term when we do an approximation like this? Should I treat it as $0$ or should I treat it as a term $a_3u^3$?
$endgroup$
– Dasherman
Dec 16 '18 at 14:27
1
1
$begingroup$
@Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
$endgroup$
– caverac
Dec 16 '18 at 14:35
$begingroup$
@Dasherman It means just to truncate the series at $u^2$, ignore the higher order terms. In my solution above I actually went up to $u^3$ and ignored $u^4$, $u^5$, $cdots$
$endgroup$
– caverac
Dec 16 '18 at 14:35
1
1
$begingroup$
@Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
$endgroup$
– caverac
Dec 20 '18 at 13:26
$begingroup$
@Dasherman Because if you select the root $a_1 = 0$ you will see that $a_2 = 0$, $a_3 = 0$
$endgroup$
– caverac
Dec 20 '18 at 13:26
|
show 1 more 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%2f3039914%2ffinding-approximation-to-stable-manifold-of-saddle-point%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
