Mixing
second order SDE vs general diffusion stationary distribution
$begingroup$
A quick follow-on to this question.
Consider the following SDE:
$$
ddot{x} = f(x) - gamma g(x), dot{x} + sigma h(x), xi(t)
tag{1}
$$
Based on [1], we can represent (1) as a system of two first-order equations:
$$
begin{cases}
mathrm{d} X_t = V_t, mathrm{d}t \
mathrm{d} V_t = -gamma g(X_t), mathrm{d}t + f(X_t), mathrm{d}t + sigma h(X_t), mathrm{d}W_t.
end{cases}
tag{2}
$$
Why does the noise contribute to the $V_t$ variables, not the $X_t$ variables? Is it something particular about (1)? Can we rewrite (2) equivalently with noise on the $X_t$ variables?
[1] Burrage, Kevin; Lenane, Ian; Lythe, Grant, Numerical methods for second-order stochastic differential equations, SIAM J. Sci. Comput. 29, No. 1, 245-264 (2007). ZBL1144.65004.
sde
asked Jan 15 at 1:14
jjjjjjjjjjjj
1,157515
$endgroup$
add a comment |
$begingroup$
A quick follow-on to this question.
Consider the following SDE:
$$
ddot{x} = f(x) - gamma g(x), dot{x} + sigma h(x), xi(t)
tag{1}
$$
Based on [1], we can represent (1) as a system of two first-order equations:
$$
begin{cases}
mathrm{d} X_t = V_t, mathrm{d}t \
mathrm{d} V_t = -gamma g(X_t), mathrm{d}t + f(X_t), mathrm{d}t + sigma h(X_t), mathrm{d}W_t.
end{cases}
tag{2}
$$
Why does the noise contribute to the $V_t$ variables, not the $X_t$ variables? Is it something particular about (1)? Can we rewrite (2) equivalently with noise on the $X_t$ variables?
[1] Burrage, Kevin; Lenane, Ian; Lythe, Grant, Numerical methods for second-order stochastic differential equations, SIAM J. Sci. Comput. 29, No. 1, 245-264 (2007). ZBL1144.65004.
sde
asked Jan 15 at 1:14
jjjjjjjjjjjj
1,157515
$endgroup$
add a comment |
$begingroup$
A quick follow-on to this question.
Consider the following SDE:
$$
ddot{x} = f(x) - gamma g(x), dot{x} + sigma h(x), xi(t)
tag{1}
$$
Based on [1], we can represent (1) as a system of two first-order equations:
$$
begin{cases}
mathrm{d} X_t = V_t, mathrm{d}t \
mathrm{d} V_t = -gamma g(X_t), mathrm{d}t + f(X_t), mathrm{d}t + sigma h(X_t), mathrm{d}W_t.
end{cases}
tag{2}
$$
Why does the noise contribute to the $V_t$ variables, not the $X_t$ variables? Is it something particular about (1)? Can we rewrite (2) equivalently with noise on the $X_t$ variables?
[1] Burrage, Kevin; Lenane, Ian; Lythe, Grant, Numerical methods for second-order stochastic differential equations, SIAM J. Sci. Comput. 29, No. 1, 245-264 (2007). ZBL1144.65004.
sde
asked Jan 15 at 1:14
jjjjjjjjjjjj
1,157515
$endgroup$
A quick follow-on to this question.
Consider the following SDE:
$$
ddot{x} = f(x) - gamma g(x), dot{x} + sigma h(x), xi(t)
tag{1}
$$
Based on [1], we can represent (1) as a system of two first-order equations:
$$
begin{cases}
mathrm{d} X_t = V_t, mathrm{d}t \
mathrm{d} V_t = -gamma g(X_t), mathrm{d}t + f(X_t), mathrm{d}t + sigma h(X_t), mathrm{d}W_t.
end{cases}
tag{2}
$$
Why does the noise contribute to the $V_t$ variables, not the $X_t$ variables? Is it something particular about (1)? Can we rewrite (2) equivalently with noise on the $X_t$ variables?
[1] Burrage, Kevin; Lenane, Ian; Lythe, Grant, Numerical methods for second-order stochastic differential equations, SIAM J. Sci. Comput. 29, No. 1, 245-264 (2007). ZBL1144.65004.
sde
sde
asked Jan 15 at 1:14
jjjjjjjjjjjj
1,157515
asked Jan 15 at 1:14
jjjjjjjjjjjj
1,157515
edited Jan 15 at 5:18
jjjjjj
asked Jan 15 at 1:14
jjjjjjjjjjjj
1,157515
asked Jan 15 at 1:14
jjjjjjjjjjjj
1,157515
asked Jan 15 at 1:14
jjjjjjjjjjjj
1,157515
1,157515
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It really depends on the physics hidden behind, namely on where you think the stochastics finds its role. Let me explain with two examples. In both examples, $X_t$ denotes the position vector, $V_t$ the "velocity", and $W_t$ the Wiener process.
The first example is the Brownian motion. If we consider the standard Brownian motion, its equation of motion is obviously
$$
{rm d}X_t={rm d}W_t.
$$
Here we believe, in physics, that the stochastics contributes to the position directly. More precisely, we owe the physically-observed stochastic behavior of tiny particles to some mathematically-defined stochastic processes. In this example, there is no properly defined velocity. In other words, if you want to put
$$
{rm d}X_t=V_t,{rm d}t+{rm d}W_t,
$$
it is a must that $V_t=0$, meaning that the velocity is zero. This is of course counterintuitive; after all, all Brownian particles are moving, right? So this really depends on how you define velocity for a stochastic process, or an Ito process for special. Mathematically, there are two ways. The first is to identify the drift term as the velocity, i.e., if
$$
{rm d}X_t=V_t,{rm d}t+sigma_t,{rm d}W_t,
$$
$V_t$ is then the velocity. However, this does not apply to the Brownian motion. The second is to identify the quadratic variation as the velocity, i.e.,
following the above SDE,
$$
{rm d}left<Xright>_t=sigma_t^2,{rm d}t,
$$
and $sigma_t$ is then regarded as the velocity. This applies to the Brownian motion, and $left<Xright>_t$ measures the average kinetic energy.
The second example is your Langevin equation. In this scenario, recall that a Langevin equation models a very large particle whos dynamics is governed by Newton's second law, but with stochastic background forces contributed from tiny Brownian particles. With this understanding, the equation of motion for this very large particle should be
begin{align}
{rm d}X_t&=V_t,{rm d}t,\
{rm d}V_t&=I_{t+{rm d}t}-I_t={rm d}I_t,
end{align}
which is in the form of Newton's second law. Here $I_t$ denotes the impulse of external forces. The only difference is that $I_t$ is no longer deterministic, but yields some stochastic behaviors. Therefore, we put
$$
{rm d}I_t=F_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $F_t$ stands for the conventional forces, and $sigma_t$ indicates the strength of stochastic forces. These stochastic forces arise from the Brownian motion of those tiny particles. Combine the last three equations, and you recover the Langevin equation. In this case, the stochastic factors are all put on the forcing term.
Of course, you may consider
begin{align}
{rm d}X_t&=V_t,{rm d}t+mu_t,{rm d}W_t^1,\
{rm d}V_t&=F_t,{rm d}t+sigma_t,{rm d}W_t^2,
end{align}
where $W_t^1$ and $W_t^2$ are two Wiener processes. This model is no longer Langevin, but for some intermediate particle instead. In other words, the dynamics of this particle is partly governed by Newton's second law with stochastic background forces, but itself also observes some Brownian motion. It really depends on how you interpret the physics hidden behind.
That's it. Hope this could be somewhat helpful for you.
answered Jan 15 at 4:48
hypernovahypernova
4,834414
$endgroup$
$begingroup$
great, thanks for intuition on this. So suppose we're just given a second-order SDE like in (1). From this equation, are we able to think of the dynamics as either (a) a large particle governed by Newton experiencing small random background forces (Langevin) or (b) a large particle governed by Newton with random velocities? That is, can we structurally reformulate (1) as your last display with (a) $mu_t = 0$ and alternatively (b) $sigma_t = 0$?
$endgroup$
– jjjjjj
Jan 15 at 5:12
$begingroup$
Btw, I think my interpretation is off for $(b)$...but I'm just wondering if the rules that govern how we convert a second order SDE to a system of first-order SDEs allow us rewrite (1) equivalently with either noise on the $X$ variables or on the $V$ variables
$endgroup$
– jjjjjj
Jan 15 at 5:17
1
$begingroup$
@jjjjjj: Oops, apologies that I did not tag my equations and have troubled you citing with (a)(b)... Well, mathematically, there is no second order stochastic differential equations (SDEs) (because you can only put ${rm d}X_t$, and ${rm d}^2X_t$ is not defined due to its non-vanishing quadratic variation $left<Xright>_t$). The "second order" term comes from physics, and as far as I see, mostly from Langevin. So, if you encountered a second order equation, given the intuition of Langevin, conventionally you need to put $mu_t=0$, unless the problem context states otherwise.
$endgroup$
– hypernova
Jan 15 at 5:38
1
$begingroup$
@jjjjjj: By the way, perhaps you have heard about the Fokker-Plank equation, which gives the governing equation of the probability density function (PDF) of $left(X_t,V_tright)$. You may figure out that your (a) and (b) are two distinct cases, because they lead to different PDEs to govern their respective PDFs. Therefore, when you encounter a second order stochastic equation, you had better be careful and always ask for its physical nature. Otherwise, the resulting SDE might not be giving what you expect.
$endgroup$
– hypernova
Jan 15 at 5:46
1
$begingroup$
@jjjjjj: Er... I meant, (1) conventionally means (a). By "convention" I meant (1) is treated as Langevin, and Langevin has a clear physical context. There is no need to check FPEs every time. You can try (b) but (1) does not corresponds to (b). This is because you have $ddot{x}$ in (1). Once you put (b), $x$ is no longer twice differentiable with respect to $t$, making $ddot{x}$, and hence the whole (1), ill-defined.
$endgroup$
– hypernova
Jan 15 at 8:13
|
show 3 more comments
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$begingroup$
It really depends on the physics hidden behind, namely on where you think the stochastics finds its role. Let me explain with two examples. In both examples, $X_t$ denotes the position vector, $V_t$ the "velocity", and $W_t$ the Wiener process.
The first example is the Brownian motion. If we consider the standard Brownian motion, its equation of motion is obviously
$$
{rm d}X_t={rm d}W_t.
$$
Here we believe, in physics, that the stochastics contributes to the position directly. More precisely, we owe the physically-observed stochastic behavior of tiny particles to some mathematically-defined stochastic processes. In this example, there is no properly defined velocity. In other words, if you want to put
$$
{rm d}X_t=V_t,{rm d}t+{rm d}W_t,
$$
it is a must that $V_t=0$, meaning that the velocity is zero. This is of course counterintuitive; after all, all Brownian particles are moving, right? So this really depends on how you define velocity for a stochastic process, or an Ito process for special. Mathematically, there are two ways. The first is to identify the drift term as the velocity, i.e., if
$$
{rm d}X_t=V_t,{rm d}t+sigma_t,{rm d}W_t,
$$
$V_t$ is then the velocity. However, this does not apply to the Brownian motion. The second is to identify the quadratic variation as the velocity, i.e.,
following the above SDE,
$$
{rm d}left<Xright>_t=sigma_t^2,{rm d}t,
$$
and $sigma_t$ is then regarded as the velocity. This applies to the Brownian motion, and $left<Xright>_t$ measures the average kinetic energy.
The second example is your Langevin equation. In this scenario, recall that a Langevin equation models a very large particle whos dynamics is governed by Newton's second law, but with stochastic background forces contributed from tiny Brownian particles. With this understanding, the equation of motion for this very large particle should be
begin{align}
{rm d}X_t&=V_t,{rm d}t,\
{rm d}V_t&=I_{t+{rm d}t}-I_t={rm d}I_t,
end{align}
which is in the form of Newton's second law. Here $I_t$ denotes the impulse of external forces. The only difference is that $I_t$ is no longer deterministic, but yields some stochastic behaviors. Therefore, we put
$$
{rm d}I_t=F_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $F_t$ stands for the conventional forces, and $sigma_t$ indicates the strength of stochastic forces. These stochastic forces arise from the Brownian motion of those tiny particles. Combine the last three equations, and you recover the Langevin equation. In this case, the stochastic factors are all put on the forcing term.
Of course, you may consider
begin{align}
{rm d}X_t&=V_t,{rm d}t+mu_t,{rm d}W_t^1,\
{rm d}V_t&=F_t,{rm d}t+sigma_t,{rm d}W_t^2,
end{align}
where $W_t^1$ and $W_t^2$ are two Wiener processes. This model is no longer Langevin, but for some intermediate particle instead. In other words, the dynamics of this particle is partly governed by Newton's second law with stochastic background forces, but itself also observes some Brownian motion. It really depends on how you interpret the physics hidden behind.
That's it. Hope this could be somewhat helpful for you.
answered Jan 15 at 4:48
hypernovahypernova
4,834414
$endgroup$
$begingroup$
great, thanks for intuition on this. So suppose we're just given a second-order SDE like in (1). From this equation, are we able to think of the dynamics as either (a) a large particle governed by Newton experiencing small random background forces (Langevin) or (b) a large particle governed by Newton with random velocities? That is, can we structurally reformulate (1) as your last display with (a) $mu_t = 0$ and alternatively (b) $sigma_t = 0$?
$endgroup$
– jjjjjj
Jan 15 at 5:12
$begingroup$
Btw, I think my interpretation is off for $(b)$...but I'm just wondering if the rules that govern how we convert a second order SDE to a system of first-order SDEs allow us rewrite (1) equivalently with either noise on the $X$ variables or on the $V$ variables
$endgroup$
– jjjjjj
Jan 15 at 5:17
1
$begingroup$
@jjjjjj: Oops, apologies that I did not tag my equations and have troubled you citing with (a)(b)... Well, mathematically, there is no second order stochastic differential equations (SDEs) (because you can only put ${rm d}X_t$, and ${rm d}^2X_t$ is not defined due to its non-vanishing quadratic variation $left<Xright>_t$). The "second order" term comes from physics, and as far as I see, mostly from Langevin. So, if you encountered a second order equation, given the intuition of Langevin, conventionally you need to put $mu_t=0$, unless the problem context states otherwise.
$endgroup$
– hypernova
Jan 15 at 5:38
1
$begingroup$
@jjjjjj: By the way, perhaps you have heard about the Fokker-Plank equation, which gives the governing equation of the probability density function (PDF) of $left(X_t,V_tright)$. You may figure out that your (a) and (b) are two distinct cases, because they lead to different PDEs to govern their respective PDFs. Therefore, when you encounter a second order stochastic equation, you had better be careful and always ask for its physical nature. Otherwise, the resulting SDE might not be giving what you expect.
$endgroup$
– hypernova
Jan 15 at 5:46
1
$begingroup$
@jjjjjj: Er... I meant, (1) conventionally means (a). By "convention" I meant (1) is treated as Langevin, and Langevin has a clear physical context. There is no need to check FPEs every time. You can try (b) but (1) does not corresponds to (b). This is because you have $ddot{x}$ in (1). Once you put (b), $x$ is no longer twice differentiable with respect to $t$, making $ddot{x}$, and hence the whole (1), ill-defined.
$endgroup$
– hypernova
Jan 15 at 8:13
|
show 3 more comments
$begingroup$
It really depends on the physics hidden behind, namely on where you think the stochastics finds its role. Let me explain with two examples. In both examples, $X_t$ denotes the position vector, $V_t$ the "velocity", and $W_t$ the Wiener process.
The first example is the Brownian motion. If we consider the standard Brownian motion, its equation of motion is obviously
$$
{rm d}X_t={rm d}W_t.
$$
Here we believe, in physics, that the stochastics contributes to the position directly. More precisely, we owe the physically-observed stochastic behavior of tiny particles to some mathematically-defined stochastic processes. In this example, there is no properly defined velocity. In other words, if you want to put
$$
{rm d}X_t=V_t,{rm d}t+{rm d}W_t,
$$
it is a must that $V_t=0$, meaning that the velocity is zero. This is of course counterintuitive; after all, all Brownian particles are moving, right? So this really depends on how you define velocity for a stochastic process, or an Ito process for special. Mathematically, there are two ways. The first is to identify the drift term as the velocity, i.e., if
$$
{rm d}X_t=V_t,{rm d}t+sigma_t,{rm d}W_t,
$$
$V_t$ is then the velocity. However, this does not apply to the Brownian motion. The second is to identify the quadratic variation as the velocity, i.e.,
following the above SDE,
$$
{rm d}left<Xright>_t=sigma_t^2,{rm d}t,
$$
and $sigma_t$ is then regarded as the velocity. This applies to the Brownian motion, and $left<Xright>_t$ measures the average kinetic energy.
The second example is your Langevin equation. In this scenario, recall that a Langevin equation models a very large particle whos dynamics is governed by Newton's second law, but with stochastic background forces contributed from tiny Brownian particles. With this understanding, the equation of motion for this very large particle should be
begin{align}
{rm d}X_t&=V_t,{rm d}t,\
{rm d}V_t&=I_{t+{rm d}t}-I_t={rm d}I_t,
end{align}
which is in the form of Newton's second law. Here $I_t$ denotes the impulse of external forces. The only difference is that $I_t$ is no longer deterministic, but yields some stochastic behaviors. Therefore, we put
$$
{rm d}I_t=F_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $F_t$ stands for the conventional forces, and $sigma_t$ indicates the strength of stochastic forces. These stochastic forces arise from the Brownian motion of those tiny particles. Combine the last three equations, and you recover the Langevin equation. In this case, the stochastic factors are all put on the forcing term.
Of course, you may consider
begin{align}
{rm d}X_t&=V_t,{rm d}t+mu_t,{rm d}W_t^1,\
{rm d}V_t&=F_t,{rm d}t+sigma_t,{rm d}W_t^2,
end{align}
where $W_t^1$ and $W_t^2$ are two Wiener processes. This model is no longer Langevin, but for some intermediate particle instead. In other words, the dynamics of this particle is partly governed by Newton's second law with stochastic background forces, but itself also observes some Brownian motion. It really depends on how you interpret the physics hidden behind.
That's it. Hope this could be somewhat helpful for you.
answered Jan 15 at 4:48
hypernovahypernova
4,834414
$endgroup$
$begingroup$
great, thanks for intuition on this. So suppose we're just given a second-order SDE like in (1). From this equation, are we able to think of the dynamics as either (a) a large particle governed by Newton experiencing small random background forces (Langevin) or (b) a large particle governed by Newton with random velocities? That is, can we structurally reformulate (1) as your last display with (a) $mu_t = 0$ and alternatively (b) $sigma_t = 0$?
$endgroup$
– jjjjjj
Jan 15 at 5:12
$begingroup$
Btw, I think my interpretation is off for $(b)$...but I'm just wondering if the rules that govern how we convert a second order SDE to a system of first-order SDEs allow us rewrite (1) equivalently with either noise on the $X$ variables or on the $V$ variables
$endgroup$
– jjjjjj
Jan 15 at 5:17
1
$begingroup$
@jjjjjj: Oops, apologies that I did not tag my equations and have troubled you citing with (a)(b)... Well, mathematically, there is no second order stochastic differential equations (SDEs) (because you can only put ${rm d}X_t$, and ${rm d}^2X_t$ is not defined due to its non-vanishing quadratic variation $left<Xright>_t$). The "second order" term comes from physics, and as far as I see, mostly from Langevin. So, if you encountered a second order equation, given the intuition of Langevin, conventionally you need to put $mu_t=0$, unless the problem context states otherwise.
$endgroup$
– hypernova
Jan 15 at 5:38
1
$begingroup$
@jjjjjj: By the way, perhaps you have heard about the Fokker-Plank equation, which gives the governing equation of the probability density function (PDF) of $left(X_t,V_tright)$. You may figure out that your (a) and (b) are two distinct cases, because they lead to different PDEs to govern their respective PDFs. Therefore, when you encounter a second order stochastic equation, you had better be careful and always ask for its physical nature. Otherwise, the resulting SDE might not be giving what you expect.
$endgroup$
– hypernova
Jan 15 at 5:46
1
$begingroup$
@jjjjjj: Er... I meant, (1) conventionally means (a). By "convention" I meant (1) is treated as Langevin, and Langevin has a clear physical context. There is no need to check FPEs every time. You can try (b) but (1) does not corresponds to (b). This is because you have $ddot{x}$ in (1). Once you put (b), $x$ is no longer twice differentiable with respect to $t$, making $ddot{x}$, and hence the whole (1), ill-defined.
$endgroup$
– hypernova
Jan 15 at 8:13
|
show 3 more comments
$begingroup$
It really depends on the physics hidden behind, namely on where you think the stochastics finds its role. Let me explain with two examples. In both examples, $X_t$ denotes the position vector, $V_t$ the "velocity", and $W_t$ the Wiener process.
The first example is the Brownian motion. If we consider the standard Brownian motion, its equation of motion is obviously
$$
{rm d}X_t={rm d}W_t.
$$
Here we believe, in physics, that the stochastics contributes to the position directly. More precisely, we owe the physically-observed stochastic behavior of tiny particles to some mathematically-defined stochastic processes. In this example, there is no properly defined velocity. In other words, if you want to put
$$
{rm d}X_t=V_t,{rm d}t+{rm d}W_t,
$$
it is a must that $V_t=0$, meaning that the velocity is zero. This is of course counterintuitive; after all, all Brownian particles are moving, right? So this really depends on how you define velocity for a stochastic process, or an Ito process for special. Mathematically, there are two ways. The first is to identify the drift term as the velocity, i.e., if
$$
{rm d}X_t=V_t,{rm d}t+sigma_t,{rm d}W_t,
$$
$V_t$ is then the velocity. However, this does not apply to the Brownian motion. The second is to identify the quadratic variation as the velocity, i.e.,
following the above SDE,
$$
{rm d}left<Xright>_t=sigma_t^2,{rm d}t,
$$
and $sigma_t$ is then regarded as the velocity. This applies to the Brownian motion, and $left<Xright>_t$ measures the average kinetic energy.
The second example is your Langevin equation. In this scenario, recall that a Langevin equation models a very large particle whos dynamics is governed by Newton's second law, but with stochastic background forces contributed from tiny Brownian particles. With this understanding, the equation of motion for this very large particle should be
begin{align}
{rm d}X_t&=V_t,{rm d}t,\
{rm d}V_t&=I_{t+{rm d}t}-I_t={rm d}I_t,
end{align}
which is in the form of Newton's second law. Here $I_t$ denotes the impulse of external forces. The only difference is that $I_t$ is no longer deterministic, but yields some stochastic behaviors. Therefore, we put
$$
{rm d}I_t=F_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $F_t$ stands for the conventional forces, and $sigma_t$ indicates the strength of stochastic forces. These stochastic forces arise from the Brownian motion of those tiny particles. Combine the last three equations, and you recover the Langevin equation. In this case, the stochastic factors are all put on the forcing term.
Of course, you may consider
begin{align}
{rm d}X_t&=V_t,{rm d}t+mu_t,{rm d}W_t^1,\
{rm d}V_t&=F_t,{rm d}t+sigma_t,{rm d}W_t^2,
end{align}
where $W_t^1$ and $W_t^2$ are two Wiener processes. This model is no longer Langevin, but for some intermediate particle instead. In other words, the dynamics of this particle is partly governed by Newton's second law with stochastic background forces, but itself also observes some Brownian motion. It really depends on how you interpret the physics hidden behind.
That's it. Hope this could be somewhat helpful for you.
answered Jan 15 at 4:48
hypernovahypernova
4,834414
$endgroup$
It really depends on the physics hidden behind, namely on where you think the stochastics finds its role. Let me explain with two examples. In both examples, $X_t$ denotes the position vector, $V_t$ the "velocity", and $W_t$ the Wiener process.
The first example is the Brownian motion. If we consider the standard Brownian motion, its equation of motion is obviously
$$
{rm d}X_t={rm d}W_t.
$$
Here we believe, in physics, that the stochastics contributes to the position directly. More precisely, we owe the physically-observed stochastic behavior of tiny particles to some mathematically-defined stochastic processes. In this example, there is no properly defined velocity. In other words, if you want to put
$$
{rm d}X_t=V_t,{rm d}t+{rm d}W_t,
$$
it is a must that $V_t=0$, meaning that the velocity is zero. This is of course counterintuitive; after all, all Brownian particles are moving, right? So this really depends on how you define velocity for a stochastic process, or an Ito process for special. Mathematically, there are two ways. The first is to identify the drift term as the velocity, i.e., if
$$
{rm d}X_t=V_t,{rm d}t+sigma_t,{rm d}W_t,
$$
$V_t$ is then the velocity. However, this does not apply to the Brownian motion. The second is to identify the quadratic variation as the velocity, i.e.,
following the above SDE,
$$
{rm d}left<Xright>_t=sigma_t^2,{rm d}t,
$$
and $sigma_t$ is then regarded as the velocity. This applies to the Brownian motion, and $left<Xright>_t$ measures the average kinetic energy.
The second example is your Langevin equation. In this scenario, recall that a Langevin equation models a very large particle whos dynamics is governed by Newton's second law, but with stochastic background forces contributed from tiny Brownian particles. With this understanding, the equation of motion for this very large particle should be
begin{align}
{rm d}X_t&=V_t,{rm d}t,\
{rm d}V_t&=I_{t+{rm d}t}-I_t={rm d}I_t,
end{align}
which is in the form of Newton's second law. Here $I_t$ denotes the impulse of external forces. The only difference is that $I_t$ is no longer deterministic, but yields some stochastic behaviors. Therefore, we put
$$
{rm d}I_t=F_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $F_t$ stands for the conventional forces, and $sigma_t$ indicates the strength of stochastic forces. These stochastic forces arise from the Brownian motion of those tiny particles. Combine the last three equations, and you recover the Langevin equation. In this case, the stochastic factors are all put on the forcing term.
Of course, you may consider
begin{align}
{rm d}X_t&=V_t,{rm d}t+mu_t,{rm d}W_t^1,\
{rm d}V_t&=F_t,{rm d}t+sigma_t,{rm d}W_t^2,
end{align}
where $W_t^1$ and $W_t^2$ are two Wiener processes. This model is no longer Langevin, but for some intermediate particle instead. In other words, the dynamics of this particle is partly governed by Newton's second law with stochastic background forces, but itself also observes some Brownian motion. It really depends on how you interpret the physics hidden behind.
That's it. Hope this could be somewhat helpful for you.
answered Jan 15 at 4:48
hypernovahypernova
4,834414
answered Jan 15 at 4:48
hypernovahypernova
4,834414
answered Jan 15 at 4:48
hypernovahypernova
4,834414
answered Jan 15 at 4:48
hypernovahypernova
4,834414
4,834414
$begingroup$
great, thanks for intuition on this. So suppose we're just given a second-order SDE like in (1). From this equation, are we able to think of the dynamics as either (a) a large particle governed by Newton experiencing small random background forces (Langevin) or (b) a large particle governed by Newton with random velocities? That is, can we structurally reformulate (1) as your last display with (a) $mu_t = 0$ and alternatively (b) $sigma_t = 0$?
$endgroup$
– jjjjjj
Jan 15 at 5:12
$begingroup$
Btw, I think my interpretation is off for $(b)$...but I'm just wondering if the rules that govern how we convert a second order SDE to a system of first-order SDEs allow us rewrite (1) equivalently with either noise on the $X$ variables or on the $V$ variables
$endgroup$
– jjjjjj
Jan 15 at 5:17
1
$begingroup$
@jjjjjj: Oops, apologies that I did not tag my equations and have troubled you citing with (a)(b)... Well, mathematically, there is no second order stochastic differential equations (SDEs) (because you can only put ${rm d}X_t$, and ${rm d}^2X_t$ is not defined due to its non-vanishing quadratic variation $left<Xright>_t$). The "second order" term comes from physics, and as far as I see, mostly from Langevin. So, if you encountered a second order equation, given the intuition of Langevin, conventionally you need to put $mu_t=0$, unless the problem context states otherwise.
$endgroup$
– hypernova
Jan 15 at 5:38
1
$begingroup$
@jjjjjj: By the way, perhaps you have heard about the Fokker-Plank equation, which gives the governing equation of the probability density function (PDF) of $left(X_t,V_tright)$. You may figure out that your (a) and (b) are two distinct cases, because they lead to different PDEs to govern their respective PDFs. Therefore, when you encounter a second order stochastic equation, you had better be careful and always ask for its physical nature. Otherwise, the resulting SDE might not be giving what you expect.
$endgroup$
– hypernova
Jan 15 at 5:46
1
$begingroup$
@jjjjjj: Er... I meant, (1) conventionally means (a). By "convention" I meant (1) is treated as Langevin, and Langevin has a clear physical context. There is no need to check FPEs every time. You can try (b) but (1) does not corresponds to (b). This is because you have $ddot{x}$ in (1). Once you put (b), $x$ is no longer twice differentiable with respect to $t$, making $ddot{x}$, and hence the whole (1), ill-defined.
$endgroup$
– hypernova
Jan 15 at 8:13
|
show 3 more comments
$begingroup$
great, thanks for intuition on this. So suppose we're just given a second-order SDE like in (1). From this equation, are we able to think of the dynamics as either (a) a large particle governed by Newton experiencing small random background forces (Langevin) or (b) a large particle governed by Newton with random velocities? That is, can we structurally reformulate (1) as your last display with (a) $mu_t = 0$ and alternatively (b) $sigma_t = 0$?
$endgroup$
– jjjjjj
Jan 15 at 5:12
$begingroup$
Btw, I think my interpretation is off for $(b)$...but I'm just wondering if the rules that govern how we convert a second order SDE to a system of first-order SDEs allow us rewrite (1) equivalently with either noise on the $X$ variables or on the $V$ variables
$endgroup$
– jjjjjj
Jan 15 at 5:17
1
$begingroup$
@jjjjjj: Oops, apologies that I did not tag my equations and have troubled you citing with (a)(b)... Well, mathematically, there is no second order stochastic differential equations (SDEs) (because you can only put ${rm d}X_t$, and ${rm d}^2X_t$ is not defined due to its non-vanishing quadratic variation $left<Xright>_t$). The "second order" term comes from physics, and as far as I see, mostly from Langevin. So, if you encountered a second order equation, given the intuition of Langevin, conventionally you need to put $mu_t=0$, unless the problem context states otherwise.
$endgroup$
– hypernova
Jan 15 at 5:38
1
$begingroup$
@jjjjjj: By the way, perhaps you have heard about the Fokker-Plank equation, which gives the governing equation of the probability density function (PDF) of $left(X_t,V_tright)$. You may figure out that your (a) and (b) are two distinct cases, because they lead to different PDEs to govern their respective PDFs. Therefore, when you encounter a second order stochastic equation, you had better be careful and always ask for its physical nature. Otherwise, the resulting SDE might not be giving what you expect.
$endgroup$
– hypernova
Jan 15 at 5:46
1
$begingroup$
@jjjjjj: Er... I meant, (1) conventionally means (a). By "convention" I meant (1) is treated as Langevin, and Langevin has a clear physical context. There is no need to check FPEs every time. You can try (b) but (1) does not corresponds to (b). This is because you have $ddot{x}$ in (1). Once you put (b), $x$ is no longer twice differentiable with respect to $t$, making $ddot{x}$, and hence the whole (1), ill-defined.
$endgroup$
– hypernova
Jan 15 at 8:13
$begingroup$
great, thanks for intuition on this. So suppose we're just given a second-order SDE like in (1). From this equation, are we able to think of the dynamics as either (a) a large particle governed by Newton experiencing small random background forces (Langevin) or (b) a large particle governed by Newton with random velocities? That is, can we structurally reformulate (1) as your last display with (a) $mu_t = 0$ and alternatively (b) $sigma_t = 0$?
$endgroup$
– jjjjjj
Jan 15 at 5:12
$begingroup$
great, thanks for intuition on this. So suppose we're just given a second-order SDE like in (1). From this equation, are we able to think of the dynamics as either (a) a large particle governed by Newton experiencing small random background forces (Langevin) or (b) a large particle governed by Newton with random velocities? That is, can we structurally reformulate (1) as your last display with (a) $mu_t = 0$ and alternatively (b) $sigma_t = 0$?
$endgroup$
– jjjjjj
Jan 15 at 5:12
$begingroup$
Btw, I think my interpretation is off for $(b)$...but I'm just wondering if the rules that govern how we convert a second order SDE to a system of first-order SDEs allow us rewrite (1) equivalently with either noise on the $X$ variables or on the $V$ variables
$endgroup$
– jjjjjj
Jan 15 at 5:17
$begingroup$
Btw, I think my interpretation is off for $(b)$...but I'm just wondering if the rules that govern how we convert a second order SDE to a system of first-order SDEs allow us rewrite (1) equivalently with either noise on the $X$ variables or on the $V$ variables
$endgroup$
– jjjjjj
Jan 15 at 5:17
1
1
$begingroup$
@jjjjjj: Oops, apologies that I did not tag my equations and have troubled you citing with (a)(b)... Well, mathematically, there is no second order stochastic differential equations (SDEs) (because you can only put ${rm d}X_t$, and ${rm d}^2X_t$ is not defined due to its non-vanishing quadratic variation $left<Xright>_t$). The "second order" term comes from physics, and as far as I see, mostly from Langevin. So, if you encountered a second order equation, given the intuition of Langevin, conventionally you need to put $mu_t=0$, unless the problem context states otherwise.
$endgroup$
– hypernova
Jan 15 at 5:38
$begingroup$
@jjjjjj: Oops, apologies that I did not tag my equations and have troubled you citing with (a)(b)... Well, mathematically, there is no second order stochastic differential equations (SDEs) (because you can only put ${rm d}X_t$, and ${rm d}^2X_t$ is not defined due to its non-vanishing quadratic variation $left<Xright>_t$). The "second order" term comes from physics, and as far as I see, mostly from Langevin. So, if you encountered a second order equation, given the intuition of Langevin, conventionally you need to put $mu_t=0$, unless the problem context states otherwise.
$endgroup$
– hypernova
Jan 15 at 5:38
1
1
$begingroup$
@jjjjjj: By the way, perhaps you have heard about the Fokker-Plank equation, which gives the governing equation of the probability density function (PDF) of $left(X_t,V_tright)$. You may figure out that your (a) and (b) are two distinct cases, because they lead to different PDEs to govern their respective PDFs. Therefore, when you encounter a second order stochastic equation, you had better be careful and always ask for its physical nature. Otherwise, the resulting SDE might not be giving what you expect.
$endgroup$
– hypernova
Jan 15 at 5:46
$begingroup$
@jjjjjj: By the way, perhaps you have heard about the Fokker-Plank equation, which gives the governing equation of the probability density function (PDF) of $left(X_t,V_tright)$. You may figure out that your (a) and (b) are two distinct cases, because they lead to different PDEs to govern their respective PDFs. Therefore, when you encounter a second order stochastic equation, you had better be careful and always ask for its physical nature. Otherwise, the resulting SDE might not be giving what you expect.
$endgroup$
– hypernova
Jan 15 at 5:46
1
1
$begingroup$
@jjjjjj: Er... I meant, (1) conventionally means (a). By "convention" I meant (1) is treated as Langevin, and Langevin has a clear physical context. There is no need to check FPEs every time. You can try (b) but (1) does not corresponds to (b). This is because you have $ddot{x}$ in (1). Once you put (b), $x$ is no longer twice differentiable with respect to $t$, making $ddot{x}$, and hence the whole (1), ill-defined.
$endgroup$
– hypernova
Jan 15 at 8:13
$begingroup$
@jjjjjj: Er... I meant, (1) conventionally means (a). By "convention" I meant (1) is treated as Langevin, and Langevin has a clear physical context. There is no need to check FPEs every time. You can try (b) but (1) does not corresponds to (b). This is because you have $ddot{x}$ in (1). Once you put (b), $x$ is no longer twice differentiable with respect to $t$, making $ddot{x}$, and hence the whole (1), ill-defined.
$endgroup$
– hypernova
Jan 15 at 8:13
|
show 3 more comments
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