Nonlinear equation analysis withe epsilon value [closed]
$begingroup$
Consider the nonlinear equation
$$frac{d^2x}{dt^2}+epsilonsin(x)=0,~~epsilon ll 1\
x(0)=0,~~dot{x}(0)=1$$
and find...
A. The value of $x_0$ as $epsilon$ goes to $0$
B. The first order term $x_1$ and the second order term $x_2$
C. Plot $x(t)$ for $x_0$, $x_1$, and $x_2$
D. Assemble the solution $x=x_0(t)+epsilon x_1(t)+epsilon^2 x_2(t)$. Plot $x(t)=x_0(t)$, $x=x_0(t)+epsilon x_1(t)$ and $x+x_0(t)+epsilon x_1(t)+epsilon^2 x_2(t)$ on the same graph.
So the thought is that you can expand the sine function around $x=x_0$ to obtain the equations you need. Then, using Laplacian transforms, $x_1 and x_2$ are solvable.
calculus linear-algebra ordinary-differential-equations nonlinear-system nonlinear-analysis
asked Dec 7 '18 at 18:03
PascalPascal
15810
$endgroup$
closed as off-topic by MisterRiemann, the_candyman, Davide Giraudo, amWhy, KReiser Dec 8 '18 at 0:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MisterRiemann, Davide Giraudo, amWhy, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the nonlinear equation
$$frac{d^2x}{dt^2}+epsilonsin(x)=0,~~epsilon ll 1\
x(0)=0,~~dot{x}(0)=1$$
and find...
A. The value of $x_0$ as $epsilon$ goes to $0$
B. The first order term $x_1$ and the second order term $x_2$
C. Plot $x(t)$ for $x_0$, $x_1$, and $x_2$
D. Assemble the solution $x=x_0(t)+epsilon x_1(t)+epsilon^2 x_2(t)$. Plot $x(t)=x_0(t)$, $x=x_0(t)+epsilon x_1(t)$ and $x+x_0(t)+epsilon x_1(t)+epsilon^2 x_2(t)$ on the same graph.
So the thought is that you can expand the sine function around $x=x_0$ to obtain the equations you need. Then, using Laplacian transforms, $x_1 and x_2$ are solvable.
calculus linear-algebra ordinary-differential-equations nonlinear-system nonlinear-analysis
asked Dec 7 '18 at 18:03
PascalPascal
15810
$endgroup$
closed as off-topic by MisterRiemann, the_candyman, Davide Giraudo, amWhy, KReiser Dec 8 '18 at 0:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MisterRiemann, Davide Giraudo, amWhy, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What is $x_0, x_1$ and $x_2$? Please, provide more details. Most importantly, add also your efforts. Otherwise, I feel that this question will be closed soon and you won't get any answer.
$endgroup$
– the_candyman
Dec 7 '18 at 18:14
$begingroup$
Where did you find this, or what prompted you to conceive it? It it essential that ε be small? And is there some slick answer? I would say it's not massively different from SHM, as when x gets to the domain in which sine levels-off, it will still be follwing a downward-curving parabola.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 18:38
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Whoops, did not mean to post. I simply wanted to save for later. Regardless, I updated with more information and direction that I am headed.
$endgroup$
– Pascal
Dec 7 '18 at 20:12
2
$begingroup$
@Pascal -- I've done that! It's easy, isn't it, when you're using the fields here as your personal equation-editor ... but who would be so delinquent as to do a thing like that!!? ¶ Whhoops! I've just said ... I've done it!!
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 20:48
add a comment |
$begingroup$
Consider the nonlinear equation
$$frac{d^2x}{dt^2}+epsilonsin(x)=0,~~epsilon ll 1\
x(0)=0,~~dot{x}(0)=1$$
and find...
A. The value of $x_0$ as $epsilon$ goes to $0$
B. The first order term $x_1$ and the second order term $x_2$
C. Plot $x(t)$ for $x_0$, $x_1$, and $x_2$
D. Assemble the solution $x=x_0(t)+epsilon x_1(t)+epsilon^2 x_2(t)$. Plot $x(t)=x_0(t)$, $x=x_0(t)+epsilon x_1(t)$ and $x+x_0(t)+epsilon x_1(t)+epsilon^2 x_2(t)$ on the same graph.
So the thought is that you can expand the sine function around $x=x_0$ to obtain the equations you need. Then, using Laplacian transforms, $x_1 and x_2$ are solvable.
calculus linear-algebra ordinary-differential-equations nonlinear-system nonlinear-analysis
asked Dec 7 '18 at 18:03
PascalPascal
15810
$endgroup$
Consider the nonlinear equation
$$frac{d^2x}{dt^2}+epsilonsin(x)=0,~~epsilon ll 1\
x(0)=0,~~dot{x}(0)=1$$
and find...
A. The value of $x_0$ as $epsilon$ goes to $0$
B. The first order term $x_1$ and the second order term $x_2$
C. Plot $x(t)$ for $x_0$, $x_1$, and $x_2$
D. Assemble the solution $x=x_0(t)+epsilon x_1(t)+epsilon^2 x_2(t)$. Plot $x(t)=x_0(t)$, $x=x_0(t)+epsilon x_1(t)$ and $x+x_0(t)+epsilon x_1(t)+epsilon^2 x_2(t)$ on the same graph.
So the thought is that you can expand the sine function around $x=x_0$ to obtain the equations you need. Then, using Laplacian transforms, $x_1 and x_2$ are solvable.
calculus linear-algebra ordinary-differential-equations nonlinear-system nonlinear-analysis
calculus linear-algebra ordinary-differential-equations nonlinear-system nonlinear-analysis
asked Dec 7 '18 at 18:03
PascalPascal
15810
asked Dec 7 '18 at 18:03
PascalPascal
15810
edited Dec 7 '18 at 20:11
Pascal
asked Dec 7 '18 at 18:03
PascalPascal
15810
asked Dec 7 '18 at 18:03
PascalPascal
15810
asked Dec 7 '18 at 18:03
PascalPascal
15810
15810
closed as off-topic by MisterRiemann, the_candyman, Davide Giraudo, amWhy, KReiser Dec 8 '18 at 0:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MisterRiemann, Davide Giraudo, amWhy, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by MisterRiemann, the_candyman, Davide Giraudo, amWhy, KReiser Dec 8 '18 at 0:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – MisterRiemann, Davide Giraudo, amWhy, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What is $x_0, x_1$ and $x_2$? Please, provide more details. Most importantly, add also your efforts. Otherwise, I feel that this question will be closed soon and you won't get any answer.
$endgroup$
– the_candyman
Dec 7 '18 at 18:14
$begingroup$
Where did you find this, or what prompted you to conceive it? It it essential that ε be small? And is there some slick answer? I would say it's not massively different from SHM, as when x gets to the domain in which sine levels-off, it will still be follwing a downward-curving parabola.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 18:38
$begingroup$
Whoops, did not mean to post. I simply wanted to save for later. Regardless, I updated with more information and direction that I am headed.
$endgroup$
– Pascal
Dec 7 '18 at 20:12
2
$begingroup$
@Pascal -- I've done that! It's easy, isn't it, when you're using the fields here as your personal equation-editor ... but who would be so delinquent as to do a thing like that!!? ¶ Whhoops! I've just said ... I've done it!!
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 20:48
add a comment |
$begingroup$
What is $x_0, x_1$ and $x_2$? Please, provide more details. Most importantly, add also your efforts. Otherwise, I feel that this question will be closed soon and you won't get any answer.
$endgroup$
– the_candyman
Dec 7 '18 at 18:14
$begingroup$
Where did you find this, or what prompted you to conceive it? It it essential that ε be small? And is there some slick answer? I would say it's not massively different from SHM, as when x gets to the domain in which sine levels-off, it will still be follwing a downward-curving parabola.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 18:38
$begingroup$
Whoops, did not mean to post. I simply wanted to save for later. Regardless, I updated with more information and direction that I am headed.
$endgroup$
– Pascal
Dec 7 '18 at 20:12
2
$begingroup$
@Pascal -- I've done that! It's easy, isn't it, when you're using the fields here as your personal equation-editor ... but who would be so delinquent as to do a thing like that!!? ¶ Whhoops! I've just said ... I've done it!!
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 20:48
$begingroup$
What is $x_0, x_1$ and $x_2$? Please, provide more details. Most importantly, add also your efforts. Otherwise, I feel that this question will be closed soon and you won't get any answer.
$endgroup$
– the_candyman
Dec 7 '18 at 18:14
$begingroup$
What is $x_0, x_1$ and $x_2$? Please, provide more details. Most importantly, add also your efforts. Otherwise, I feel that this question will be closed soon and you won't get any answer.
$endgroup$
– the_candyman
Dec 7 '18 at 18:14
$begingroup$
Where did you find this, or what prompted you to conceive it? It it essential that ε be small? And is there some slick answer? I would say it's not massively different from SHM, as when x gets to the domain in which sine levels-off, it will still be follwing a downward-curving parabola.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 18:38
$begingroup$
Where did you find this, or what prompted you to conceive it? It it essential that ε be small? And is there some slick answer? I would say it's not massively different from SHM, as when x gets to the domain in which sine levels-off, it will still be follwing a downward-curving parabola.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 18:38
$begingroup$
Whoops, did not mean to post. I simply wanted to save for later. Regardless, I updated with more information and direction that I am headed.
$endgroup$
– Pascal
Dec 7 '18 at 20:12
$begingroup$
Whoops, did not mean to post. I simply wanted to save for later. Regardless, I updated with more information and direction that I am headed.
$endgroup$
– Pascal
Dec 7 '18 at 20:12
2
2
$begingroup$
@Pascal -- I've done that! It's easy, isn't it, when you're using the fields here as your personal equation-editor ... but who would be so delinquent as to do a thing like that!!? ¶ Whhoops! I've just said ... I've done it!!
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 20:48
$begingroup$
@Pascal -- I've done that! It's easy, isn't it, when you're using the fields here as your personal equation-editor ... but who would be so delinquent as to do a thing like that!!? ¶ Whhoops! I've just said ... I've done it!!
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 20:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have to solve for the terms in $x=x_0+ϵx_1+ϵ^2x_2+...$, so that comparing the powers of $ϵ$ you get
$$
ddot x_0=0,~~x_0(0)=0,~dot x_0(0)=1\
ddot x_1=-sin x_0,~~x_1(0)=0,~dot x_1(0)=0\
ddot x_2=-cos(x_0),x_1,~~x_2(0)=0,~dot x_2(0)=0\
$$
In another approach, multiply with $2dot x$ and integrate to find
$$
1=dot x^2+2ϵ(1-cos x)=dot x^2+4ϵsin^2frac x2
$$
This is a circle equation that can be parametrized via $dot x=cosphi(t)$, $2sqrtϵsinfrac x2=sinphi(t)$ or $x=2arcsinfrac{sinphi(t)}{2sqrtϵ}$. Compare the expressions for the first derivative
$$
cosϕ(t)=dot x=frac{cosϕ(t),dot ϕ(t)}{sqrt{ϵ-frac14sin^2ϕ(t)}}
$$
Now you can expand $1=dot ϕ(t)left(ϵ-frac14sin^2ϕ(t)right)^{-1/2}$ in powers of $ϵ$ and integrate.
answered Dec 7 '18 at 19:17
LutzLLutzL
57.3k42054
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have to solve for the terms in $x=x_0+ϵx_1+ϵ^2x_2+...$, so that comparing the powers of $ϵ$ you get
$$
ddot x_0=0,~~x_0(0)=0,~dot x_0(0)=1\
ddot x_1=-sin x_0,~~x_1(0)=0,~dot x_1(0)=0\
ddot x_2=-cos(x_0),x_1,~~x_2(0)=0,~dot x_2(0)=0\
$$
In another approach, multiply with $2dot x$ and integrate to find
$$
1=dot x^2+2ϵ(1-cos x)=dot x^2+4ϵsin^2frac x2
$$
This is a circle equation that can be parametrized via $dot x=cosphi(t)$, $2sqrtϵsinfrac x2=sinphi(t)$ or $x=2arcsinfrac{sinphi(t)}{2sqrtϵ}$. Compare the expressions for the first derivative
$$
cosϕ(t)=dot x=frac{cosϕ(t),dot ϕ(t)}{sqrt{ϵ-frac14sin^2ϕ(t)}}
$$
Now you can expand $1=dot ϕ(t)left(ϵ-frac14sin^2ϕ(t)right)^{-1/2}$ in powers of $ϵ$ and integrate.
answered Dec 7 '18 at 19:17
LutzLLutzL
57.3k42054
$endgroup$
add a comment |
$begingroup$
You have to solve for the terms in $x=x_0+ϵx_1+ϵ^2x_2+...$, so that comparing the powers of $ϵ$ you get
$$
ddot x_0=0,~~x_0(0)=0,~dot x_0(0)=1\
ddot x_1=-sin x_0,~~x_1(0)=0,~dot x_1(0)=0\
ddot x_2=-cos(x_0),x_1,~~x_2(0)=0,~dot x_2(0)=0\
$$
In another approach, multiply with $2dot x$ and integrate to find
$$
1=dot x^2+2ϵ(1-cos x)=dot x^2+4ϵsin^2frac x2
$$
This is a circle equation that can be parametrized via $dot x=cosphi(t)$, $2sqrtϵsinfrac x2=sinphi(t)$ or $x=2arcsinfrac{sinphi(t)}{2sqrtϵ}$. Compare the expressions for the first derivative
$$
cosϕ(t)=dot x=frac{cosϕ(t),dot ϕ(t)}{sqrt{ϵ-frac14sin^2ϕ(t)}}
$$
Now you can expand $1=dot ϕ(t)left(ϵ-frac14sin^2ϕ(t)right)^{-1/2}$ in powers of $ϵ$ and integrate.
answered Dec 7 '18 at 19:17
LutzLLutzL
57.3k42054
$endgroup$
add a comment |
$begingroup$
You have to solve for the terms in $x=x_0+ϵx_1+ϵ^2x_2+...$, so that comparing the powers of $ϵ$ you get
$$
ddot x_0=0,~~x_0(0)=0,~dot x_0(0)=1\
ddot x_1=-sin x_0,~~x_1(0)=0,~dot x_1(0)=0\
ddot x_2=-cos(x_0),x_1,~~x_2(0)=0,~dot x_2(0)=0\
$$
In another approach, multiply with $2dot x$ and integrate to find
$$
1=dot x^2+2ϵ(1-cos x)=dot x^2+4ϵsin^2frac x2
$$
This is a circle equation that can be parametrized via $dot x=cosphi(t)$, $2sqrtϵsinfrac x2=sinphi(t)$ or $x=2arcsinfrac{sinphi(t)}{2sqrtϵ}$. Compare the expressions for the first derivative
$$
cosϕ(t)=dot x=frac{cosϕ(t),dot ϕ(t)}{sqrt{ϵ-frac14sin^2ϕ(t)}}
$$
Now you can expand $1=dot ϕ(t)left(ϵ-frac14sin^2ϕ(t)right)^{-1/2}$ in powers of $ϵ$ and integrate.
answered Dec 7 '18 at 19:17
LutzLLutzL
57.3k42054
$endgroup$
You have to solve for the terms in $x=x_0+ϵx_1+ϵ^2x_2+...$, so that comparing the powers of $ϵ$ you get
$$
ddot x_0=0,~~x_0(0)=0,~dot x_0(0)=1\
ddot x_1=-sin x_0,~~x_1(0)=0,~dot x_1(0)=0\
ddot x_2=-cos(x_0),x_1,~~x_2(0)=0,~dot x_2(0)=0\
$$
In another approach, multiply with $2dot x$ and integrate to find
$$
1=dot x^2+2ϵ(1-cos x)=dot x^2+4ϵsin^2frac x2
$$
This is a circle equation that can be parametrized via $dot x=cosphi(t)$, $2sqrtϵsinfrac x2=sinphi(t)$ or $x=2arcsinfrac{sinphi(t)}{2sqrtϵ}$. Compare the expressions for the first derivative
$$
cosϕ(t)=dot x=frac{cosϕ(t),dot ϕ(t)}{sqrt{ϵ-frac14sin^2ϕ(t)}}
$$
Now you can expand $1=dot ϕ(t)left(ϵ-frac14sin^2ϕ(t)right)^{-1/2}$ in powers of $ϵ$ and integrate.
answered Dec 7 '18 at 19:17
LutzLLutzL
57.3k42054
edited Dec 7 '18 at 19:25
answered Dec 7 '18 at 19:17
LutzLLutzL
57.3k42054
answered Dec 7 '18 at 19:17
LutzLLutzL
57.3k42054
answered Dec 7 '18 at 19:17
LutzLLutzL
57.3k42054
57.3k42054
add a comment |
add a comment |
$begingroup$
What is $x_0, x_1$ and $x_2$? Please, provide more details. Most importantly, add also your efforts. Otherwise, I feel that this question will be closed soon and you won't get any answer.
$endgroup$
– the_candyman
Dec 7 '18 at 18:14
$begingroup$
Where did you find this, or what prompted you to conceive it? It it essential that ε be small? And is there some slick answer? I would say it's not massively different from SHM, as when x gets to the domain in which sine levels-off, it will still be follwing a downward-curving parabola.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 18:38
$begingroup$
Whoops, did not mean to post. I simply wanted to save for later. Regardless, I updated with more information and direction that I am headed.
$endgroup$
– Pascal
Dec 7 '18 at 20:12
2
$begingroup$
@Pascal -- I've done that! It's easy, isn't it, when you're using the fields here as your personal equation-editor ... but who would be so delinquent as to do a thing like that!!? ¶ Whhoops! I've just said ... I've done it!!
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 20:48