Solve for Differential Equations by using regular perturbation method
$begingroup$
carry out regular perturbation calculation for $epsilon$ satisfying
$$x''(t)+x(t)=epsilon x^2(t)$$
correct to the second order in the small parameter $epsilon$.
then use the result to perform a renormalization calculation, straightforwardly
assuming that the lowest order correction to the period of the system
is of order $epsilon^2$, not (the erroneous) $epsilon$.
ordinary-differential-equations perturbation-theory
edited Dec 2 '18 at 10:27
mrtaurho
5,46041237
asked Dec 2 '18 at 10:26
Wilson LinWilson Lin
1
$endgroup$
add a comment |
$begingroup$
carry out regular perturbation calculation for $epsilon$ satisfying
$$x''(t)+x(t)=epsilon x^2(t)$$
correct to the second order in the small parameter $epsilon$.
then use the result to perform a renormalization calculation, straightforwardly
assuming that the lowest order correction to the period of the system
is of order $epsilon^2$, not (the erroneous) $epsilon$.
ordinary-differential-equations perturbation-theory
edited Dec 2 '18 at 10:27
mrtaurho
5,46041237
asked Dec 2 '18 at 10:26
Wilson LinWilson Lin
1
$endgroup$
2
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 10:35
add a comment |
$begingroup$
carry out regular perturbation calculation for $epsilon$ satisfying
$$x''(t)+x(t)=epsilon x^2(t)$$
correct to the second order in the small parameter $epsilon$.
then use the result to perform a renormalization calculation, straightforwardly
assuming that the lowest order correction to the period of the system
is of order $epsilon^2$, not (the erroneous) $epsilon$.
ordinary-differential-equations perturbation-theory
edited Dec 2 '18 at 10:27
mrtaurho
5,46041237
asked Dec 2 '18 at 10:26
Wilson LinWilson Lin
1
$endgroup$
carry out regular perturbation calculation for $epsilon$ satisfying
$$x''(t)+x(t)=epsilon x^2(t)$$
correct to the second order in the small parameter $epsilon$.
then use the result to perform a renormalization calculation, straightforwardly
assuming that the lowest order correction to the period of the system
is of order $epsilon^2$, not (the erroneous) $epsilon$.
ordinary-differential-equations perturbation-theory
ordinary-differential-equations perturbation-theory
edited Dec 2 '18 at 10:27
mrtaurho
5,46041237
asked Dec 2 '18 at 10:26
Wilson LinWilson Lin
1
edited Dec 2 '18 at 10:27
mrtaurho
5,46041237
asked Dec 2 '18 at 10:26
Wilson LinWilson Lin
1
edited Dec 2 '18 at 10:27
mrtaurho
5,46041237
edited Dec 2 '18 at 10:27
mrtaurho
5,46041237
edited Dec 2 '18 at 10:27
mrtaurho
5,46041237
5,46041237
asked Dec 2 '18 at 10:26
Wilson LinWilson Lin
1
asked Dec 2 '18 at 10:26
Wilson LinWilson Lin
1
asked Dec 2 '18 at 10:26
Wilson LinWilson Lin
1
1
2
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 10:35
add a comment |
2
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 10:35
2
2
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 10:35
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 10:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This system is conservative, multiply with $2 x'$ and integrate to get
$$
(x')^2+x^2(1-ϵtfrac23x)=R^2=const.
$$
Write this as $(x')^2+u(x)^2=R^2$. This circle equation can now be parametrized like a circle, with $x'(t)=Rcosvarphi(t)$, and then consequently $Rsinφ(t)=u(x(t))=x(t)sqrt{1-ϵtfrac23x(t)}$.
Now find the inverse function $v$ to $u$ so that $x=v(Rsinφ)$. Then its derivative and the circle equation lead to the identity
$$
Rcosφ=x'=v'(Rsinφ)Rcosφ,φ',
\
1=v'(Rsinφ),φ'.
$$
This allows to compute the period of $x$ as
$$
T=int_0^{2pi}v'(Rsinφ),dφ
$$
To find the first terms of the expansion of $v$, bring $u=xsqrt{1-ϵtfrac23x(t)}$ into fixed-point form and applying the binomial series for the power $-1/2$ gives
$$
x=uleft(1-2fracϵ3xright)^{-1/2}=uleft(1+fracϵ3x+frac{ϵ^2}6x^2 + frac{5}{54}(ϵx)^3 + frac{35}{648}(ϵx)^4 +O(ϵ^5)right)
$$
Iterating this relation gives, starting from $x=u+O(ϵ)$,
begin{align}
x&=uleft(1+fracϵ3u+O(ϵ^2)right)
\&vdots\
x=v(u)&=u + frac13ϵu^2 + frac{5}{18}ϵ^2u^3 + frac{8}{27}ϵ^3u^4 + frac{77}{216}ϵ^4u^5 +O(ϵ^5)\
v'(u)&=1 + frac23ϵu + frac{5}{6}ϵ^2u^2 + frac{32}{27}ϵ^3u^3 + frac{385}{216}ϵ^4u^4 +O(ϵ^5)
end{align}
so that, using that odd powers of the sine integrate to zero and the constant part of $sin^{2k}φ$ is $frac{binom{2k}{k}}{2^{2k}}$,
begin{align}
T&=2pi+int_0^{2pi}left[frac23ϵRsinφ + frac{5}{6}(ϵR)^2sin^2φ + frac{32}{27}(ϵR)^3sin^3φ + frac{385}{216}(ϵR)^4sin^4φ +O(ϵ^5)right],dφ
\
&=2pileft(1+frac{5}{6}(ϵR)^2frac{binom{2}{1}}{2^2}+frac{385}{216}(ϵR)^4frac{binom{4}{2}}{2^4}+O(ϵ^6)right)
=2pileft(1+frac{5}{12}(ϵR)^2+frac{385}{576}(ϵR)^4+O(ϵ^6)right).
end{align}
answered Dec 5 '18 at 9:03
LutzLLutzL
58.7k42055
$endgroup$
add a comment |
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$begingroup$
This system is conservative, multiply with $2 x'$ and integrate to get
$$
(x')^2+x^2(1-ϵtfrac23x)=R^2=const.
$$
Write this as $(x')^2+u(x)^2=R^2$. This circle equation can now be parametrized like a circle, with $x'(t)=Rcosvarphi(t)$, and then consequently $Rsinφ(t)=u(x(t))=x(t)sqrt{1-ϵtfrac23x(t)}$.
Now find the inverse function $v$ to $u$ so that $x=v(Rsinφ)$. Then its derivative and the circle equation lead to the identity
$$
Rcosφ=x'=v'(Rsinφ)Rcosφ,φ',
\
1=v'(Rsinφ),φ'.
$$
This allows to compute the period of $x$ as
$$
T=int_0^{2pi}v'(Rsinφ),dφ
$$
To find the first terms of the expansion of $v$, bring $u=xsqrt{1-ϵtfrac23x(t)}$ into fixed-point form and applying the binomial series for the power $-1/2$ gives
$$
x=uleft(1-2fracϵ3xright)^{-1/2}=uleft(1+fracϵ3x+frac{ϵ^2}6x^2 + frac{5}{54}(ϵx)^3 + frac{35}{648}(ϵx)^4 +O(ϵ^5)right)
$$
Iterating this relation gives, starting from $x=u+O(ϵ)$,
begin{align}
x&=uleft(1+fracϵ3u+O(ϵ^2)right)
\&vdots\
x=v(u)&=u + frac13ϵu^2 + frac{5}{18}ϵ^2u^3 + frac{8}{27}ϵ^3u^4 + frac{77}{216}ϵ^4u^5 +O(ϵ^5)\
v'(u)&=1 + frac23ϵu + frac{5}{6}ϵ^2u^2 + frac{32}{27}ϵ^3u^3 + frac{385}{216}ϵ^4u^4 +O(ϵ^5)
end{align}
so that, using that odd powers of the sine integrate to zero and the constant part of $sin^{2k}φ$ is $frac{binom{2k}{k}}{2^{2k}}$,
begin{align}
T&=2pi+int_0^{2pi}left[frac23ϵRsinφ + frac{5}{6}(ϵR)^2sin^2φ + frac{32}{27}(ϵR)^3sin^3φ + frac{385}{216}(ϵR)^4sin^4φ +O(ϵ^5)right],dφ
\
&=2pileft(1+frac{5}{6}(ϵR)^2frac{binom{2}{1}}{2^2}+frac{385}{216}(ϵR)^4frac{binom{4}{2}}{2^4}+O(ϵ^6)right)
=2pileft(1+frac{5}{12}(ϵR)^2+frac{385}{576}(ϵR)^4+O(ϵ^6)right).
end{align}
answered Dec 5 '18 at 9:03
LutzLLutzL
58.7k42055
$endgroup$
add a comment |
$begingroup$
This system is conservative, multiply with $2 x'$ and integrate to get
$$
(x')^2+x^2(1-ϵtfrac23x)=R^2=const.
$$
Write this as $(x')^2+u(x)^2=R^2$. This circle equation can now be parametrized like a circle, with $x'(t)=Rcosvarphi(t)$, and then consequently $Rsinφ(t)=u(x(t))=x(t)sqrt{1-ϵtfrac23x(t)}$.
Now find the inverse function $v$ to $u$ so that $x=v(Rsinφ)$. Then its derivative and the circle equation lead to the identity
$$
Rcosφ=x'=v'(Rsinφ)Rcosφ,φ',
\
1=v'(Rsinφ),φ'.
$$
This allows to compute the period of $x$ as
$$
T=int_0^{2pi}v'(Rsinφ),dφ
$$
To find the first terms of the expansion of $v$, bring $u=xsqrt{1-ϵtfrac23x(t)}$ into fixed-point form and applying the binomial series for the power $-1/2$ gives
$$
x=uleft(1-2fracϵ3xright)^{-1/2}=uleft(1+fracϵ3x+frac{ϵ^2}6x^2 + frac{5}{54}(ϵx)^3 + frac{35}{648}(ϵx)^4 +O(ϵ^5)right)
$$
Iterating this relation gives, starting from $x=u+O(ϵ)$,
begin{align}
x&=uleft(1+fracϵ3u+O(ϵ^2)right)
\&vdots\
x=v(u)&=u + frac13ϵu^2 + frac{5}{18}ϵ^2u^3 + frac{8}{27}ϵ^3u^4 + frac{77}{216}ϵ^4u^5 +O(ϵ^5)\
v'(u)&=1 + frac23ϵu + frac{5}{6}ϵ^2u^2 + frac{32}{27}ϵ^3u^3 + frac{385}{216}ϵ^4u^4 +O(ϵ^5)
end{align}
so that, using that odd powers of the sine integrate to zero and the constant part of $sin^{2k}φ$ is $frac{binom{2k}{k}}{2^{2k}}$,
begin{align}
T&=2pi+int_0^{2pi}left[frac23ϵRsinφ + frac{5}{6}(ϵR)^2sin^2φ + frac{32}{27}(ϵR)^3sin^3φ + frac{385}{216}(ϵR)^4sin^4φ +O(ϵ^5)right],dφ
\
&=2pileft(1+frac{5}{6}(ϵR)^2frac{binom{2}{1}}{2^2}+frac{385}{216}(ϵR)^4frac{binom{4}{2}}{2^4}+O(ϵ^6)right)
=2pileft(1+frac{5}{12}(ϵR)^2+frac{385}{576}(ϵR)^4+O(ϵ^6)right).
end{align}
answered Dec 5 '18 at 9:03
LutzLLutzL
58.7k42055
$endgroup$
add a comment |
$begingroup$
This system is conservative, multiply with $2 x'$ and integrate to get
$$
(x')^2+x^2(1-ϵtfrac23x)=R^2=const.
$$
Write this as $(x')^2+u(x)^2=R^2$. This circle equation can now be parametrized like a circle, with $x'(t)=Rcosvarphi(t)$, and then consequently $Rsinφ(t)=u(x(t))=x(t)sqrt{1-ϵtfrac23x(t)}$.
Now find the inverse function $v$ to $u$ so that $x=v(Rsinφ)$. Then its derivative and the circle equation lead to the identity
$$
Rcosφ=x'=v'(Rsinφ)Rcosφ,φ',
\
1=v'(Rsinφ),φ'.
$$
This allows to compute the period of $x$ as
$$
T=int_0^{2pi}v'(Rsinφ),dφ
$$
To find the first terms of the expansion of $v$, bring $u=xsqrt{1-ϵtfrac23x(t)}$ into fixed-point form and applying the binomial series for the power $-1/2$ gives
$$
x=uleft(1-2fracϵ3xright)^{-1/2}=uleft(1+fracϵ3x+frac{ϵ^2}6x^2 + frac{5}{54}(ϵx)^3 + frac{35}{648}(ϵx)^4 +O(ϵ^5)right)
$$
Iterating this relation gives, starting from $x=u+O(ϵ)$,
begin{align}
x&=uleft(1+fracϵ3u+O(ϵ^2)right)
\&vdots\
x=v(u)&=u + frac13ϵu^2 + frac{5}{18}ϵ^2u^3 + frac{8}{27}ϵ^3u^4 + frac{77}{216}ϵ^4u^5 +O(ϵ^5)\
v'(u)&=1 + frac23ϵu + frac{5}{6}ϵ^2u^2 + frac{32}{27}ϵ^3u^3 + frac{385}{216}ϵ^4u^4 +O(ϵ^5)
end{align}
so that, using that odd powers of the sine integrate to zero and the constant part of $sin^{2k}φ$ is $frac{binom{2k}{k}}{2^{2k}}$,
begin{align}
T&=2pi+int_0^{2pi}left[frac23ϵRsinφ + frac{5}{6}(ϵR)^2sin^2φ + frac{32}{27}(ϵR)^3sin^3φ + frac{385}{216}(ϵR)^4sin^4φ +O(ϵ^5)right],dφ
\
&=2pileft(1+frac{5}{6}(ϵR)^2frac{binom{2}{1}}{2^2}+frac{385}{216}(ϵR)^4frac{binom{4}{2}}{2^4}+O(ϵ^6)right)
=2pileft(1+frac{5}{12}(ϵR)^2+frac{385}{576}(ϵR)^4+O(ϵ^6)right).
end{align}
answered Dec 5 '18 at 9:03
LutzLLutzL
58.7k42055
$endgroup$
This system is conservative, multiply with $2 x'$ and integrate to get
$$
(x')^2+x^2(1-ϵtfrac23x)=R^2=const.
$$
Write this as $(x')^2+u(x)^2=R^2$. This circle equation can now be parametrized like a circle, with $x'(t)=Rcosvarphi(t)$, and then consequently $Rsinφ(t)=u(x(t))=x(t)sqrt{1-ϵtfrac23x(t)}$.
Now find the inverse function $v$ to $u$ so that $x=v(Rsinφ)$. Then its derivative and the circle equation lead to the identity
$$
Rcosφ=x'=v'(Rsinφ)Rcosφ,φ',
\
1=v'(Rsinφ),φ'.
$$
This allows to compute the period of $x$ as
$$
T=int_0^{2pi}v'(Rsinφ),dφ
$$
To find the first terms of the expansion of $v$, bring $u=xsqrt{1-ϵtfrac23x(t)}$ into fixed-point form and applying the binomial series for the power $-1/2$ gives
$$
x=uleft(1-2fracϵ3xright)^{-1/2}=uleft(1+fracϵ3x+frac{ϵ^2}6x^2 + frac{5}{54}(ϵx)^3 + frac{35}{648}(ϵx)^4 +O(ϵ^5)right)
$$
Iterating this relation gives, starting from $x=u+O(ϵ)$,
begin{align}
x&=uleft(1+fracϵ3u+O(ϵ^2)right)
\&vdots\
x=v(u)&=u + frac13ϵu^2 + frac{5}{18}ϵ^2u^3 + frac{8}{27}ϵ^3u^4 + frac{77}{216}ϵ^4u^5 +O(ϵ^5)\
v'(u)&=1 + frac23ϵu + frac{5}{6}ϵ^2u^2 + frac{32}{27}ϵ^3u^3 + frac{385}{216}ϵ^4u^4 +O(ϵ^5)
end{align}
so that, using that odd powers of the sine integrate to zero and the constant part of $sin^{2k}φ$ is $frac{binom{2k}{k}}{2^{2k}}$,
begin{align}
T&=2pi+int_0^{2pi}left[frac23ϵRsinφ + frac{5}{6}(ϵR)^2sin^2φ + frac{32}{27}(ϵR)^3sin^3φ + frac{385}{216}(ϵR)^4sin^4φ +O(ϵ^5)right],dφ
\
&=2pileft(1+frac{5}{6}(ϵR)^2frac{binom{2}{1}}{2^2}+frac{385}{216}(ϵR)^4frac{binom{4}{2}}{2^4}+O(ϵ^6)right)
=2pileft(1+frac{5}{12}(ϵR)^2+frac{385}{576}(ϵR)^4+O(ϵ^6)right).
end{align}
answered Dec 5 '18 at 9:03
LutzLLutzL
58.7k42055
answered Dec 5 '18 at 9:03
LutzLLutzL
58.7k42055
answered Dec 5 '18 at 9:03
LutzLLutzL
58.7k42055
answered Dec 5 '18 at 9:03
LutzLLutzL
58.7k42055
58.7k42055
add a comment |
add a comment |
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$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 10:35