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When to plug in $y = y_0 + epsilon y_1 + epsilon^2 y_2 +cdots$ into the ODE to get the asymptotic series
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Given a second degree ODE with a small parameter $epsilon$,
when can we just plug in $y = y_0 + epsilon y_1 + epsilon^2 y_2 + cdots$ and then solve term by term to get the asymptotic series?
The singular perturbation case, $epsilon y'' + a(x) y' + b(x) y = 0$, we can not plug in this, we have to use the WKB theory and plug in $y = e^y_0/delta + y_1 + delta y_2 + cdots$.
And for the duffing equation $y'' + y + epsilon y^3 = 0$, we have to use multi-scale analysis since the standard perturbation will give us a secular term.
However, for ODEs like $y'' + 2epsilon y'+ y = 0$ and $y'' = epsilon Q(x) y$,we can just plug in the standard series to find the asymptotic.
For the local analysis without $epsilon$ in the ODE, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use. So is there a rule to follow when doing the global analysis with $epsilon$ in the ODE? I guess it will depends on the ODE and also the boundary conditions?
differential-equations asymptotics perturbation-theory
asked Aug 6 at 1:44
Xiao
4,42811333
add a comment |Â
up vote
1
down vote
favorite
Given a second degree ODE with a small parameter $epsilon$,
when can we just plug in $y = y_0 + epsilon y_1 + epsilon^2 y_2 + cdots$ and then solve term by term to get the asymptotic series?
The singular perturbation case, $epsilon y'' + a(x) y' + b(x) y = 0$, we can not plug in this, we have to use the WKB theory and plug in $y = e^y_0/delta + y_1 + delta y_2 + cdots$.
And for the duffing equation $y'' + y + epsilon y^3 = 0$, we have to use multi-scale analysis since the standard perturbation will give us a secular term.
However, for ODEs like $y'' + 2epsilon y'+ y = 0$ and $y'' = epsilon Q(x) y$,we can just plug in the standard series to find the asymptotic.
For the local analysis without $epsilon$ in the ODE, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use. So is there a rule to follow when doing the global analysis with $epsilon$ in the ODE? I guess it will depends on the ODE and also the boundary conditions?
differential-equations asymptotics perturbation-theory
asked Aug 6 at 1:44
Xiao
4,42811333
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a second degree ODE with a small parameter $epsilon$,
when can we just plug in $y = y_0 + epsilon y_1 + epsilon^2 y_2 + cdots$ and then solve term by term to get the asymptotic series?
The singular perturbation case, $epsilon y'' + a(x) y' + b(x) y = 0$, we can not plug in this, we have to use the WKB theory and plug in $y = e^y_0/delta + y_1 + delta y_2 + cdots$.
And for the duffing equation $y'' + y + epsilon y^3 = 0$, we have to use multi-scale analysis since the standard perturbation will give us a secular term.
However, for ODEs like $y'' + 2epsilon y'+ y = 0$ and $y'' = epsilon Q(x) y$,we can just plug in the standard series to find the asymptotic.
For the local analysis without $epsilon$ in the ODE, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use. So is there a rule to follow when doing the global analysis with $epsilon$ in the ODE? I guess it will depends on the ODE and also the boundary conditions?
differential-equations asymptotics perturbation-theory
asked Aug 6 at 1:44
Xiao
4,42811333
Given a second degree ODE with a small parameter $epsilon$,
when can we just plug in $y = y_0 + epsilon y_1 + epsilon^2 y_2 + cdots$ and then solve term by term to get the asymptotic series?
The singular perturbation case, $epsilon y'' + a(x) y' + b(x) y = 0$, we can not plug in this, we have to use the WKB theory and plug in $y = e^y_0/delta + y_1 + delta y_2 + cdots$.
And for the duffing equation $y'' + y + epsilon y^3 = 0$, we have to use multi-scale analysis since the standard perturbation will give us a secular term.
However, for ODEs like $y'' + 2epsilon y'+ y = 0$ and $y'' = epsilon Q(x) y$,we can just plug in the standard series to find the asymptotic.
For the local analysis without $epsilon$ in the ODE, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use. So is there a rule to follow when doing the global analysis with $epsilon$ in the ODE? I guess it will depends on the ODE and also the boundary conditions?
differential-equations asymptotics perturbation-theory
asked Aug 6 at 1:44
Xiao
4,42811333
edited Aug 6 at 15:35
asked Aug 6 at 1:44
Xiao
4,42811333
asked Aug 6 at 1:44
Xiao
4,42811333
asked Aug 6 at 1:44
Xiao
4,42811333
4,42811333
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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up vote
2
down vote
accepted
Side note: assumption of regular perturbation expansion in integer powers of the small parameter is a bit awkwardly restrictive. Assuming merely nonnegative powers of the small parameter allows considerably more flexibility and is still within the realm of regular perturbation theory.
Anyway, the two problematic cases you pointed out are problematic for entirely different reasons. The first case is because the perturbation itself is singular, which means that the $epsilon=0$ problem has no solution, so the $y_0$ you would want to construct is already not defined. This generally occurs when setting the small parameter to zero eliminates the highest order term in the differential equation (because now the general solution should be expected to be "lower dimensional", either literally in the linear case or in a loose sense in the nonlinear case).
The second case is because you want to push the asymptotics to long time scales, but the local approximations you are making have unbounded error on unbounded time scales. On short times there is no problem with regular perturbation expansion, but on long times you need to come to grips with subtle dynamical issues such as periodicity and metastability, which are too complicated for regular perturbation theory to resolve.
answered Aug 6 at 2:16
Ian
65.2k24681
Thank you for the reply! Could you give me some reference for the case when $epsilon$ is expanded at a non integer positive power? I understand for the local analysis without the $epsilon$ term in the ODE, this corresponds to the expansion at regular singular point.
– Xiao
Aug 6 at 15:32
Also for a given ODE with $epsilon$, is there a way to tell which type of expansion we should use? For the local analysis without $epsilon$, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use.
– Xiao
Aug 6 at 15:34
@Xiao I think this is discussed in Holmes Introduction to the Foundations of Applied Mathematics, but it is really all over the place in perturbation theory. In regular perturbation theory the general principle is called dominant balance, and it can be used to select both the $y_k$ and the powers on the $epsilon$'s (i.e. you don't need to assume the powers are integers even when they are, the equation itself can tell you that). Singular perturbations have no general theory.
– Ian
Aug 6 at 15:34
@Xiao As for global analysis, there are theories for that but it is quite sensitive to the precise character of the ODE. In your Duffing example for instance the main point is that $y''=-fracddy (y^2/2+epsilon y^4/4)$ which corresponds to a Hamiltonian system with Hamiltonian $y^2/2+y^4/4$. But you could add a variety of terms to this to break this Hamiltonian structure and then you would need a totally different analysis. Look into dynamical systems theory if you're interested in this kind of thing.
– Ian
Aug 6 at 15:37
Thank you for the answer! And is there a consistent way to check whether an ODE with $epsilon$ is a regular or singular perturbation problem? For example, I think when the highest order derivative is multiplied by $epsilon$, it is always a singular perturbation problem? I know intuitively in the singular perturbation problem, the solution of the ODE for $0<epsilon << 1$ differs drastically from the solution of the ODE when setting $epsilon = 0$.
– Xiao
Aug 6 at 17:50
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Side note: assumption of regular perturbation expansion in integer powers of the small parameter is a bit awkwardly restrictive. Assuming merely nonnegative powers of the small parameter allows considerably more flexibility and is still within the realm of regular perturbation theory.
Anyway, the two problematic cases you pointed out are problematic for entirely different reasons. The first case is because the perturbation itself is singular, which means that the $epsilon=0$ problem has no solution, so the $y_0$ you would want to construct is already not defined. This generally occurs when setting the small parameter to zero eliminates the highest order term in the differential equation (because now the general solution should be expected to be "lower dimensional", either literally in the linear case or in a loose sense in the nonlinear case).
The second case is because you want to push the asymptotics to long time scales, but the local approximations you are making have unbounded error on unbounded time scales. On short times there is no problem with regular perturbation expansion, but on long times you need to come to grips with subtle dynamical issues such as periodicity and metastability, which are too complicated for regular perturbation theory to resolve.
answered Aug 6 at 2:16
Ian
65.2k24681
Thank you for the reply! Could you give me some reference for the case when $epsilon$ is expanded at a non integer positive power? I understand for the local analysis without the $epsilon$ term in the ODE, this corresponds to the expansion at regular singular point.
– Xiao
Aug 6 at 15:32
Also for a given ODE with $epsilon$, is there a way to tell which type of expansion we should use? For the local analysis without $epsilon$, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use.
– Xiao
Aug 6 at 15:34
@Xiao I think this is discussed in Holmes Introduction to the Foundations of Applied Mathematics, but it is really all over the place in perturbation theory. In regular perturbation theory the general principle is called dominant balance, and it can be used to select both the $y_k$ and the powers on the $epsilon$'s (i.e. you don't need to assume the powers are integers even when they are, the equation itself can tell you that). Singular perturbations have no general theory.
– Ian
Aug 6 at 15:34
@Xiao As for global analysis, there are theories for that but it is quite sensitive to the precise character of the ODE. In your Duffing example for instance the main point is that $y''=-fracddy (y^2/2+epsilon y^4/4)$ which corresponds to a Hamiltonian system with Hamiltonian $y^2/2+y^4/4$. But you could add a variety of terms to this to break this Hamiltonian structure and then you would need a totally different analysis. Look into dynamical systems theory if you're interested in this kind of thing.
– Ian
Aug 6 at 15:37
Thank you for the answer! And is there a consistent way to check whether an ODE with $epsilon$ is a regular or singular perturbation problem? For example, I think when the highest order derivative is multiplied by $epsilon$, it is always a singular perturbation problem? I know intuitively in the singular perturbation problem, the solution of the ODE for $0<epsilon << 1$ differs drastically from the solution of the ODE when setting $epsilon = 0$.
– Xiao
Aug 6 at 17:50
 |Â
show 1 more comment
up vote
2
down vote
accepted
Side note: assumption of regular perturbation expansion in integer powers of the small parameter is a bit awkwardly restrictive. Assuming merely nonnegative powers of the small parameter allows considerably more flexibility and is still within the realm of regular perturbation theory.
Anyway, the two problematic cases you pointed out are problematic for entirely different reasons. The first case is because the perturbation itself is singular, which means that the $epsilon=0$ problem has no solution, so the $y_0$ you would want to construct is already not defined. This generally occurs when setting the small parameter to zero eliminates the highest order term in the differential equation (because now the general solution should be expected to be "lower dimensional", either literally in the linear case or in a loose sense in the nonlinear case).
The second case is because you want to push the asymptotics to long time scales, but the local approximations you are making have unbounded error on unbounded time scales. On short times there is no problem with regular perturbation expansion, but on long times you need to come to grips with subtle dynamical issues such as periodicity and metastability, which are too complicated for regular perturbation theory to resolve.
answered Aug 6 at 2:16
Ian
65.2k24681
Thank you for the reply! Could you give me some reference for the case when $epsilon$ is expanded at a non integer positive power? I understand for the local analysis without the $epsilon$ term in the ODE, this corresponds to the expansion at regular singular point.
– Xiao
Aug 6 at 15:32
Also for a given ODE with $epsilon$, is there a way to tell which type of expansion we should use? For the local analysis without $epsilon$, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use.
– Xiao
Aug 6 at 15:34
@Xiao I think this is discussed in Holmes Introduction to the Foundations of Applied Mathematics, but it is really all over the place in perturbation theory. In regular perturbation theory the general principle is called dominant balance, and it can be used to select both the $y_k$ and the powers on the $epsilon$'s (i.e. you don't need to assume the powers are integers even when they are, the equation itself can tell you that). Singular perturbations have no general theory.
– Ian
Aug 6 at 15:34
@Xiao As for global analysis, there are theories for that but it is quite sensitive to the precise character of the ODE. In your Duffing example for instance the main point is that $y''=-fracddy (y^2/2+epsilon y^4/4)$ which corresponds to a Hamiltonian system with Hamiltonian $y^2/2+y^4/4$. But you could add a variety of terms to this to break this Hamiltonian structure and then you would need a totally different analysis. Look into dynamical systems theory if you're interested in this kind of thing.
– Ian
Aug 6 at 15:37
Thank you for the answer! And is there a consistent way to check whether an ODE with $epsilon$ is a regular or singular perturbation problem? For example, I think when the highest order derivative is multiplied by $epsilon$, it is always a singular perturbation problem? I know intuitively in the singular perturbation problem, the solution of the ODE for $0<epsilon << 1$ differs drastically from the solution of the ODE when setting $epsilon = 0$.
– Xiao
Aug 6 at 17:50
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Side note: assumption of regular perturbation expansion in integer powers of the small parameter is a bit awkwardly restrictive. Assuming merely nonnegative powers of the small parameter allows considerably more flexibility and is still within the realm of regular perturbation theory.
Anyway, the two problematic cases you pointed out are problematic for entirely different reasons. The first case is because the perturbation itself is singular, which means that the $epsilon=0$ problem has no solution, so the $y_0$ you would want to construct is already not defined. This generally occurs when setting the small parameter to zero eliminates the highest order term in the differential equation (because now the general solution should be expected to be "lower dimensional", either literally in the linear case or in a loose sense in the nonlinear case).
The second case is because you want to push the asymptotics to long time scales, but the local approximations you are making have unbounded error on unbounded time scales. On short times there is no problem with regular perturbation expansion, but on long times you need to come to grips with subtle dynamical issues such as periodicity and metastability, which are too complicated for regular perturbation theory to resolve.
answered Aug 6 at 2:16
Ian
65.2k24681
Side note: assumption of regular perturbation expansion in integer powers of the small parameter is a bit awkwardly restrictive. Assuming merely nonnegative powers of the small parameter allows considerably more flexibility and is still within the realm of regular perturbation theory.
Anyway, the two problematic cases you pointed out are problematic for entirely different reasons. The first case is because the perturbation itself is singular, which means that the $epsilon=0$ problem has no solution, so the $y_0$ you would want to construct is already not defined. This generally occurs when setting the small parameter to zero eliminates the highest order term in the differential equation (because now the general solution should be expected to be "lower dimensional", either literally in the linear case or in a loose sense in the nonlinear case).
The second case is because you want to push the asymptotics to long time scales, but the local approximations you are making have unbounded error on unbounded time scales. On short times there is no problem with regular perturbation expansion, but on long times you need to come to grips with subtle dynamical issues such as periodicity and metastability, which are too complicated for regular perturbation theory to resolve.
answered Aug 6 at 2:16
Ian
65.2k24681
answered Aug 6 at 2:16
Ian
65.2k24681
answered Aug 6 at 2:16
Ian
65.2k24681
answered Aug 6 at 2:16
Ian
65.2k24681
65.2k24681
Thank you for the reply! Could you give me some reference for the case when $epsilon$ is expanded at a non integer positive power? I understand for the local analysis without the $epsilon$ term in the ODE, this corresponds to the expansion at regular singular point.
– Xiao
Aug 6 at 15:32
Also for a given ODE with $epsilon$, is there a way to tell which type of expansion we should use? For the local analysis without $epsilon$, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use.
– Xiao
Aug 6 at 15:34
@Xiao I think this is discussed in Holmes Introduction to the Foundations of Applied Mathematics, but it is really all over the place in perturbation theory. In regular perturbation theory the general principle is called dominant balance, and it can be used to select both the $y_k$ and the powers on the $epsilon$'s (i.e. you don't need to assume the powers are integers even when they are, the equation itself can tell you that). Singular perturbations have no general theory.
– Ian
Aug 6 at 15:34
@Xiao As for global analysis, there are theories for that but it is quite sensitive to the precise character of the ODE. In your Duffing example for instance the main point is that $y''=-fracddy (y^2/2+epsilon y^4/4)$ which corresponds to a Hamiltonian system with Hamiltonian $y^2/2+y^4/4$. But you could add a variety of terms to this to break this Hamiltonian structure and then you would need a totally different analysis. Look into dynamical systems theory if you're interested in this kind of thing.
– Ian
Aug 6 at 15:37
Thank you for the answer! And is there a consistent way to check whether an ODE with $epsilon$ is a regular or singular perturbation problem? For example, I think when the highest order derivative is multiplied by $epsilon$, it is always a singular perturbation problem? I know intuitively in the singular perturbation problem, the solution of the ODE for $0<epsilon << 1$ differs drastically from the solution of the ODE when setting $epsilon = 0$.
– Xiao
Aug 6 at 17:50
 |Â
show 1 more comment
Thank you for the reply! Could you give me some reference for the case when $epsilon$ is expanded at a non integer positive power? I understand for the local analysis without the $epsilon$ term in the ODE, this corresponds to the expansion at regular singular point.
– Xiao
Aug 6 at 15:32
Also for a given ODE with $epsilon$, is there a way to tell which type of expansion we should use? For the local analysis without $epsilon$, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use.
– Xiao
Aug 6 at 15:34
@Xiao I think this is discussed in Holmes Introduction to the Foundations of Applied Mathematics, but it is really all over the place in perturbation theory. In regular perturbation theory the general principle is called dominant balance, and it can be used to select both the $y_k$ and the powers on the $epsilon$'s (i.e. you don't need to assume the powers are integers even when they are, the equation itself can tell you that). Singular perturbations have no general theory.
– Ian
Aug 6 at 15:34
@Xiao As for global analysis, there are theories for that but it is quite sensitive to the precise character of the ODE. In your Duffing example for instance the main point is that $y''=-fracddy (y^2/2+epsilon y^4/4)$ which corresponds to a Hamiltonian system with Hamiltonian $y^2/2+y^4/4$. But you could add a variety of terms to this to break this Hamiltonian structure and then you would need a totally different analysis. Look into dynamical systems theory if you're interested in this kind of thing.
– Ian
Aug 6 at 15:37
Thank you for the answer! And is there a consistent way to check whether an ODE with $epsilon$ is a regular or singular perturbation problem? For example, I think when the highest order derivative is multiplied by $epsilon$, it is always a singular perturbation problem? I know intuitively in the singular perturbation problem, the solution of the ODE for $0<epsilon << 1$ differs drastically from the solution of the ODE when setting $epsilon = 0$.
– Xiao
Aug 6 at 17:50
Thank you for the reply! Could you give me some reference for the case when $epsilon$ is expanded at a non integer positive power? I understand for the local analysis without the $epsilon$ term in the ODE, this corresponds to the expansion at regular singular point.
– Xiao
Aug 6 at 15:32
Thank you for the reply! Could you give me some reference for the case when $epsilon$ is expanded at a non integer positive power? I understand for the local analysis without the $epsilon$ term in the ODE, this corresponds to the expansion at regular singular point.
– Xiao
Aug 6 at 15:32
Also for a given ODE with $epsilon$, is there a way to tell which type of expansion we should use? For the local analysis without $epsilon$, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use.
– Xiao
Aug 6 at 15:34
Also for a given ODE with $epsilon$, is there a way to tell which type of expansion we should use? For the local analysis without $epsilon$, there are ways to classify regular points, regular singular points, and irregular singular points, and for each case, we know exactly what type of expansions we should use.
– Xiao
Aug 6 at 15:34
@Xiao I think this is discussed in Holmes Introduction to the Foundations of Applied Mathematics, but it is really all over the place in perturbation theory. In regular perturbation theory the general principle is called dominant balance, and it can be used to select both the $y_k$ and the powers on the $epsilon$'s (i.e. you don't need to assume the powers are integers even when they are, the equation itself can tell you that). Singular perturbations have no general theory.
– Ian
Aug 6 at 15:34
@Xiao I think this is discussed in Holmes Introduction to the Foundations of Applied Mathematics, but it is really all over the place in perturbation theory. In regular perturbation theory the general principle is called dominant balance, and it can be used to select both the $y_k$ and the powers on the $epsilon$'s (i.e. you don't need to assume the powers are integers even when they are, the equation itself can tell you that). Singular perturbations have no general theory.
– Ian
Aug 6 at 15:34
@Xiao As for global analysis, there are theories for that but it is quite sensitive to the precise character of the ODE. In your Duffing example for instance the main point is that $y''=-fracddy (y^2/2+epsilon y^4/4)$ which corresponds to a Hamiltonian system with Hamiltonian $y^2/2+y^4/4$. But you could add a variety of terms to this to break this Hamiltonian structure and then you would need a totally different analysis. Look into dynamical systems theory if you're interested in this kind of thing.
– Ian
Aug 6 at 15:37
@Xiao As for global analysis, there are theories for that but it is quite sensitive to the precise character of the ODE. In your Duffing example for instance the main point is that $y''=-fracddy (y^2/2+epsilon y^4/4)$ which corresponds to a Hamiltonian system with Hamiltonian $y^2/2+y^4/4$. But you could add a variety of terms to this to break this Hamiltonian structure and then you would need a totally different analysis. Look into dynamical systems theory if you're interested in this kind of thing.
– Ian
Aug 6 at 15:37
Thank you for the answer! And is there a consistent way to check whether an ODE with $epsilon$ is a regular or singular perturbation problem? For example, I think when the highest order derivative is multiplied by $epsilon$, it is always a singular perturbation problem? I know intuitively in the singular perturbation problem, the solution of the ODE for $0<epsilon << 1$ differs drastically from the solution of the ODE when setting $epsilon = 0$.
– Xiao
Aug 6 at 17:50
Thank you for the answer! And is there a consistent way to check whether an ODE with $epsilon$ is a regular or singular perturbation problem? For example, I think when the highest order derivative is multiplied by $epsilon$, it is always a singular perturbation problem? I know intuitively in the singular perturbation problem, the solution of the ODE for $0<epsilon << 1$ differs drastically from the solution of the ODE when setting $epsilon = 0$.
– Xiao
Aug 6 at 17:50
 |Â
show 1 more comment
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