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Integrating second order ODE and finding fixed points
Clash Royale CLAN TAG#URR8PPP
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We know that fixed points are such points, where $x'$ = $x''$ = $0$.
Let's say we are given second order ODE $:$ $ x'' + sin(x) = C$
We can rewrite it into system of first order ODEs:
$ begincases x_2 = x_1' \ x_2' = C - sin(x_1) endcases $
Now, according to this equation fixed points occur at $sin(x_1) = C$
But if we take the first integral of the system we will find that $ frac (x_1')^22 - cos(x_1) = C cdot
x_1 + Constant$
Now, points where $sin(x) = C$ don't satisfy the equation above (phase portrait confirms that). Yet the book claims that the fixed points are arrived from first equation.
Could you help wih my confusion? Thanks
differential-equations dynamical-systems
asked 3 hours ago
Deandre Thomson
234
add a comment |Â
up vote
2
down vote
favorite
We know that fixed points are such points, where $x'$ = $x''$ = $0$.
Let's say we are given second order ODE $:$ $ x'' + sin(x) = C$
We can rewrite it into system of first order ODEs:
$ begincases x_2 = x_1' \ x_2' = C - sin(x_1) endcases $
Now, according to this equation fixed points occur at $sin(x_1) = C$
But if we take the first integral of the system we will find that $ frac (x_1')^22 - cos(x_1) = C cdot
x_1 + Constant$
Now, points where $sin(x) = C$ don't satisfy the equation above (phase portrait confirms that). Yet the book claims that the fixed points are arrived from first equation.
Could you help wih my confusion? Thanks
differential-equations dynamical-systems
asked 3 hours ago
Deandre Thomson
234
I meant integral of the original system. Was that your question?
– Deandre Thomson
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We know that fixed points are such points, where $x'$ = $x''$ = $0$.
Let's say we are given second order ODE $:$ $ x'' + sin(x) = C$
We can rewrite it into system of first order ODEs:
$ begincases x_2 = x_1' \ x_2' = C - sin(x_1) endcases $
Now, according to this equation fixed points occur at $sin(x_1) = C$
But if we take the first integral of the system we will find that $ frac (x_1')^22 - cos(x_1) = C cdot
x_1 + Constant$
Now, points where $sin(x) = C$ don't satisfy the equation above (phase portrait confirms that). Yet the book claims that the fixed points are arrived from first equation.
Could you help wih my confusion? Thanks
differential-equations dynamical-systems
asked 3 hours ago
Deandre Thomson
234
We know that fixed points are such points, where $x'$ = $x''$ = $0$.
Let's say we are given second order ODE $:$ $ x'' + sin(x) = C$
We can rewrite it into system of first order ODEs:
$ begincases x_2 = x_1' \ x_2' = C - sin(x_1) endcases $
Now, according to this equation fixed points occur at $sin(x_1) = C$
But if we take the first integral of the system we will find that $ frac (x_1')^22 - cos(x_1) = C cdot
x_1 + Constant$
Now, points where $sin(x) = C$ don't satisfy the equation above (phase portrait confirms that). Yet the book claims that the fixed points are arrived from first equation.
Could you help wih my confusion? Thanks
differential-equations dynamical-systems
differential-equations dynamical-systems
asked 3 hours ago
Deandre Thomson
234
asked 3 hours ago
Deandre Thomson
234
edited 2 hours ago
asked 3 hours ago
Deandre Thomson
234
asked 3 hours ago
Deandre Thomson
234
asked 3 hours ago
Deandre Thomson
234
234
I meant integral of the original system. Was that your question?
– Deandre Thomson
2 hours ago
add a comment |Â
I meant integral of the original system. Was that your question?
– Deandre Thomson
2 hours ago
I meant integral of the original system. Was that your question?
– Deandre Thomson
2 hours ago
I meant integral of the original system. Was that your question?
– Deandre Thomson
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
From my understanding, you have the following system:
$x''(t)+sin(x(t))=C$
for which you would like to find fixed points, where $x''(t)=x'(t)=0$ for all t, such that $x(t)=x_0$ (i.e., x is a constant function which does not change with t).
You have already pointed out that any fixed point satisfies $sin(x_0)=C$, and it seems that your confusion stems from attempting to integrate the differential equation with respect to x, done as follows:
$v(t)=x'(t)$
$x''(t)=v'(t)=fracdvdxfracdxdt=vfracdvdx=fracddx(frac12v^2)$
$x''(t)+sin(x(t))=fracddx(frac12v^2)+sin(x)=C implies frac12v^2-cos(x)=Cx+C_2$, where $C_2$ is a constant of integration.
There are two problems with this. First, in our integration, we cannot integrate by x directly, because x is a function of t. For the $sin(x)$ term, for example, the integral becomes $int^t_1_t_0 sin(x(t))x'(t)mathrmdt$, where we then do a u-sub $u=x(t)$ and complete the integral quite simply. However, a fixed point requires $x'(t)=0$, so our entire integral is 0, and in fact the rigorous integrating process yields the equation 0=0, which does not give us any new information about the system. Second, even if the integration were justified, it still would not be useful. We know that $x'(t)=v(t)=0$, which means our integrated expression becomes:
$frac12v^2-cos(x)=Cx+C_2 implies -cos(x_0)=Cx_0+C_2 implies C_2=-cos(x_0)-Cx_0$. We have defined a new constant $C_2$, but we have not learned anything new about $x(t)$ or $x_0$.
In conclusion, your book is right in that $sin(x)=C$ is the only equation you need to fully describe the fixed points of your system.
answered 2 hours ago
Marcus Luebke
675
Thank you very much. But isn't integration the only way to plot the phase portrait?
– Deandre Thomson
1 hour ago
add a comment |Â
up vote
4
down vote
The fixed points occur when $sin(x_1)=C$ and $x_2=0$. When those conditions are met, $x_1'$ and $x_2'$ are both 0. I'm placing a streamline plot with C=0.5 below.
As in Marcus's answer, I get that $U(x(t))= frac x_2^2(t)2 - C, x_1(t) - cos(x_1(t))$ is constant for any solution $x(t)$ of the differential equation (i.e $U(x)= frac x_2^22 - C, x_1 - cos(x_1)$ is a first integral of the system). The fixed points of the ODE occur at the critical points of $U(x)$ which occur exactly when $sin(x_1)=C$ and $x_2=0$.
answered 2 hours ago
irchans
79239
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
From my understanding, you have the following system:
$x''(t)+sin(x(t))=C$
for which you would like to find fixed points, where $x''(t)=x'(t)=0$ for all t, such that $x(t)=x_0$ (i.e., x is a constant function which does not change with t).
You have already pointed out that any fixed point satisfies $sin(x_0)=C$, and it seems that your confusion stems from attempting to integrate the differential equation with respect to x, done as follows:
$v(t)=x'(t)$
$x''(t)=v'(t)=fracdvdxfracdxdt=vfracdvdx=fracddx(frac12v^2)$
$x''(t)+sin(x(t))=fracddx(frac12v^2)+sin(x)=C implies frac12v^2-cos(x)=Cx+C_2$, where $C_2$ is a constant of integration.
There are two problems with this. First, in our integration, we cannot integrate by x directly, because x is a function of t. For the $sin(x)$ term, for example, the integral becomes $int^t_1_t_0 sin(x(t))x'(t)mathrmdt$, where we then do a u-sub $u=x(t)$ and complete the integral quite simply. However, a fixed point requires $x'(t)=0$, so our entire integral is 0, and in fact the rigorous integrating process yields the equation 0=0, which does not give us any new information about the system. Second, even if the integration were justified, it still would not be useful. We know that $x'(t)=v(t)=0$, which means our integrated expression becomes:
$frac12v^2-cos(x)=Cx+C_2 implies -cos(x_0)=Cx_0+C_2 implies C_2=-cos(x_0)-Cx_0$. We have defined a new constant $C_2$, but we have not learned anything new about $x(t)$ or $x_0$.
In conclusion, your book is right in that $sin(x)=C$ is the only equation you need to fully describe the fixed points of your system.
answered 2 hours ago
Marcus Luebke
675
Thank you very much. But isn't integration the only way to plot the phase portrait?
– Deandre Thomson
1 hour ago
add a comment |Â
up vote
2
down vote
accepted
From my understanding, you have the following system:
$x''(t)+sin(x(t))=C$
for which you would like to find fixed points, where $x''(t)=x'(t)=0$ for all t, such that $x(t)=x_0$ (i.e., x is a constant function which does not change with t).
You have already pointed out that any fixed point satisfies $sin(x_0)=C$, and it seems that your confusion stems from attempting to integrate the differential equation with respect to x, done as follows:
$v(t)=x'(t)$
$x''(t)=v'(t)=fracdvdxfracdxdt=vfracdvdx=fracddx(frac12v^2)$
$x''(t)+sin(x(t))=fracddx(frac12v^2)+sin(x)=C implies frac12v^2-cos(x)=Cx+C_2$, where $C_2$ is a constant of integration.
There are two problems with this. First, in our integration, we cannot integrate by x directly, because x is a function of t. For the $sin(x)$ term, for example, the integral becomes $int^t_1_t_0 sin(x(t))x'(t)mathrmdt$, where we then do a u-sub $u=x(t)$ and complete the integral quite simply. However, a fixed point requires $x'(t)=0$, so our entire integral is 0, and in fact the rigorous integrating process yields the equation 0=0, which does not give us any new information about the system. Second, even if the integration were justified, it still would not be useful. We know that $x'(t)=v(t)=0$, which means our integrated expression becomes:
$frac12v^2-cos(x)=Cx+C_2 implies -cos(x_0)=Cx_0+C_2 implies C_2=-cos(x_0)-Cx_0$. We have defined a new constant $C_2$, but we have not learned anything new about $x(t)$ or $x_0$.
In conclusion, your book is right in that $sin(x)=C$ is the only equation you need to fully describe the fixed points of your system.
answered 2 hours ago
Marcus Luebke
675
Thank you very much. But isn't integration the only way to plot the phase portrait?
– Deandre Thomson
1 hour ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
From my understanding, you have the following system:
$x''(t)+sin(x(t))=C$
for which you would like to find fixed points, where $x''(t)=x'(t)=0$ for all t, such that $x(t)=x_0$ (i.e., x is a constant function which does not change with t).
You have already pointed out that any fixed point satisfies $sin(x_0)=C$, and it seems that your confusion stems from attempting to integrate the differential equation with respect to x, done as follows:
$v(t)=x'(t)$
$x''(t)=v'(t)=fracdvdxfracdxdt=vfracdvdx=fracddx(frac12v^2)$
$x''(t)+sin(x(t))=fracddx(frac12v^2)+sin(x)=C implies frac12v^2-cos(x)=Cx+C_2$, where $C_2$ is a constant of integration.
There are two problems with this. First, in our integration, we cannot integrate by x directly, because x is a function of t. For the $sin(x)$ term, for example, the integral becomes $int^t_1_t_0 sin(x(t))x'(t)mathrmdt$, where we then do a u-sub $u=x(t)$ and complete the integral quite simply. However, a fixed point requires $x'(t)=0$, so our entire integral is 0, and in fact the rigorous integrating process yields the equation 0=0, which does not give us any new information about the system. Second, even if the integration were justified, it still would not be useful. We know that $x'(t)=v(t)=0$, which means our integrated expression becomes:
$frac12v^2-cos(x)=Cx+C_2 implies -cos(x_0)=Cx_0+C_2 implies C_2=-cos(x_0)-Cx_0$. We have defined a new constant $C_2$, but we have not learned anything new about $x(t)$ or $x_0$.
In conclusion, your book is right in that $sin(x)=C$ is the only equation you need to fully describe the fixed points of your system.
answered 2 hours ago
Marcus Luebke
675
From my understanding, you have the following system:
$x''(t)+sin(x(t))=C$
for which you would like to find fixed points, where $x''(t)=x'(t)=0$ for all t, such that $x(t)=x_0$ (i.e., x is a constant function which does not change with t).
You have already pointed out that any fixed point satisfies $sin(x_0)=C$, and it seems that your confusion stems from attempting to integrate the differential equation with respect to x, done as follows:
$v(t)=x'(t)$
$x''(t)=v'(t)=fracdvdxfracdxdt=vfracdvdx=fracddx(frac12v^2)$
$x''(t)+sin(x(t))=fracddx(frac12v^2)+sin(x)=C implies frac12v^2-cos(x)=Cx+C_2$, where $C_2$ is a constant of integration.
There are two problems with this. First, in our integration, we cannot integrate by x directly, because x is a function of t. For the $sin(x)$ term, for example, the integral becomes $int^t_1_t_0 sin(x(t))x'(t)mathrmdt$, where we then do a u-sub $u=x(t)$ and complete the integral quite simply. However, a fixed point requires $x'(t)=0$, so our entire integral is 0, and in fact the rigorous integrating process yields the equation 0=0, which does not give us any new information about the system. Second, even if the integration were justified, it still would not be useful. We know that $x'(t)=v(t)=0$, which means our integrated expression becomes:
$frac12v^2-cos(x)=Cx+C_2 implies -cos(x_0)=Cx_0+C_2 implies C_2=-cos(x_0)-Cx_0$. We have defined a new constant $C_2$, but we have not learned anything new about $x(t)$ or $x_0$.
In conclusion, your book is right in that $sin(x)=C$ is the only equation you need to fully describe the fixed points of your system.
answered 2 hours ago
Marcus Luebke
675
answered 2 hours ago
Marcus Luebke
675
answered 2 hours ago
Marcus Luebke
675
answered 2 hours ago
Marcus Luebke
675
675
Thank you very much. But isn't integration the only way to plot the phase portrait?
– Deandre Thomson
1 hour ago
add a comment |Â
Thank you very much. But isn't integration the only way to plot the phase portrait?
– Deandre Thomson
1 hour ago
Thank you very much. But isn't integration the only way to plot the phase portrait?
– Deandre Thomson
1 hour ago
Thank you very much. But isn't integration the only way to plot the phase portrait?
– Deandre Thomson
1 hour ago
add a comment |Â
up vote
4
down vote
The fixed points occur when $sin(x_1)=C$ and $x_2=0$. When those conditions are met, $x_1'$ and $x_2'$ are both 0. I'm placing a streamline plot with C=0.5 below.
As in Marcus's answer, I get that $U(x(t))= frac x_2^2(t)2 - C, x_1(t) - cos(x_1(t))$ is constant for any solution $x(t)$ of the differential equation (i.e $U(x)= frac x_2^22 - C, x_1 - cos(x_1)$ is a first integral of the system). The fixed points of the ODE occur at the critical points of $U(x)$ which occur exactly when $sin(x_1)=C$ and $x_2=0$.
answered 2 hours ago
irchans
79239
add a comment |Â
up vote
4
down vote
The fixed points occur when $sin(x_1)=C$ and $x_2=0$. When those conditions are met, $x_1'$ and $x_2'$ are both 0. I'm placing a streamline plot with C=0.5 below.
As in Marcus's answer, I get that $U(x(t))= frac x_2^2(t)2 - C, x_1(t) - cos(x_1(t))$ is constant for any solution $x(t)$ of the differential equation (i.e $U(x)= frac x_2^22 - C, x_1 - cos(x_1)$ is a first integral of the system). The fixed points of the ODE occur at the critical points of $U(x)$ which occur exactly when $sin(x_1)=C$ and $x_2=0$.
answered 2 hours ago
irchans
79239
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The fixed points occur when $sin(x_1)=C$ and $x_2=0$. When those conditions are met, $x_1'$ and $x_2'$ are both 0. I'm placing a streamline plot with C=0.5 below.
As in Marcus's answer, I get that $U(x(t))= frac x_2^2(t)2 - C, x_1(t) - cos(x_1(t))$ is constant for any solution $x(t)$ of the differential equation (i.e $U(x)= frac x_2^22 - C, x_1 - cos(x_1)$ is a first integral of the system). The fixed points of the ODE occur at the critical points of $U(x)$ which occur exactly when $sin(x_1)=C$ and $x_2=0$.
answered 2 hours ago
irchans
79239
The fixed points occur when $sin(x_1)=C$ and $x_2=0$. When those conditions are met, $x_1'$ and $x_2'$ are both 0. I'm placing a streamline plot with C=0.5 below.
As in Marcus's answer, I get that $U(x(t))= frac x_2^2(t)2 - C, x_1(t) - cos(x_1(t))$ is constant for any solution $x(t)$ of the differential equation (i.e $U(x)= frac x_2^22 - C, x_1 - cos(x_1)$ is a first integral of the system). The fixed points of the ODE occur at the critical points of $U(x)$ which occur exactly when $sin(x_1)=C$ and $x_2=0$.
answered 2 hours ago
irchans
79239
edited 1 hour ago
answered 2 hours ago
irchans
79239
answered 2 hours ago
irchans
79239
answered 2 hours ago
irchans
79239
79239
add a comment |Â
add a comment |Â
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I meant integral of the original system. Was that your question?
– Deandre Thomson
2 hours ago