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
14.7k31130
$endgroup$
$begingroup$
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
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
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oldest
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$begingroup$
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
14.7k31130
$endgroup$
$begingroup$
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
14.7k31130
$endgroup$
$begingroup$
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
14.7k31130
$endgroup$
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
14.7k31130
answered Dec 14 '18 at 22:29
caveraccaverac
14.7k31130
answered Dec 14 '18 at 22:29
caveraccaverac
14.7k31130
answered Dec 14 '18 at 22:29
caveraccaverac
14.7k31130
14.7k31130
$begingroup$
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
|
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