Mixing
Solving for moments of $u(x,t)$ given the equation $u_t=xu+fast u$
1
$begingroup$
Given the evolution equation
$$
frac{partial u}{partial t}(x,t)=xu(x,t)+int_{-infty}^infty f(x-y)u(y,t)dy
$$
with initial condition $u(x,t)=delta(x-x_0)$, is there a way to solve for either or both of the quantities
$$
m_0(t)=int_{-infty}^infty u(x,t)dx quadtext{ or }quad m_1(t)=int_{-infty}^infty xu(x,t)dx quad
$$
in closed form, in terms of the positive-valued function $f(cdot)$?
ordinary-differential-equations pde dynamical-systems convolution
asked Jan 11 at 18:00
AlexAlex
868
$endgroup$
add a comment |
1
$begingroup$
Given the evolution equation
$$
frac{partial u}{partial t}(x,t)=xu(x,t)+int_{-infty}^infty f(x-y)u(y,t)dy
$$
with initial condition $u(x,t)=delta(x-x_0)$, is there a way to solve for either or both of the quantities
$$
m_0(t)=int_{-infty}^infty u(x,t)dx quadtext{ or }quad m_1(t)=int_{-infty}^infty xu(x,t)dx quad
$$
in closed form, in terms of the positive-valued function $f(cdot)$?
ordinary-differential-equations pde dynamical-systems convolution
asked Jan 11 at 18:00
AlexAlex
868
$endgroup$
$begingroup$
Have you tried taking the Fourier transform of the PDE to find an explicit solution?
$endgroup$
– Mattos
Jan 12 at 4:43
$begingroup$
@Mattos, thanks for the suggestion. I have tried both the Fourier and Laplace transforms but have not found them to be helpful (yet). The Fourier transform of this equation is $dot{tilde u}=itilde{u}'+tilde ftilde u$, which has the solution $tilde u=expleft{(t-iomega)x_0+iint_{omega+it}^omega tilde f(s)dsright}$. Inverting this transform gives $u=e^{x_0 t}mathcal F^{-1}_{omegato x-x_0}left{iint_{omega+it}^omega tilde f(s)dsright}$. Is there a way to evaluate the inverse Fourier transform of an integral like this?
$endgroup$
– Alex
Jan 12 at 23:14
$begingroup$
Quick correction: Inverting the transform gives $u=e^{x_0t}mathcal F^{-1}_{omegato x-x_0}left{expleft(iint_{omega+it}^omega tilde f(s)dsright)right}$.
$endgroup$
– Alex
Jan 12 at 23:56
$begingroup$
Integrating the original equation over $x in Bbb R$ gives a relationship between $m_0$ and $m_1$
$endgroup$
– Dylan
Jan 14 at 8:33
$begingroup$
@Dylan, thank you, that's a good point. Integrating the original equation gives $m_0'(t)=m_1(t)+F_0 m_0(t)$ where $F_0=int_{-infty}^infty f(x) dx$. Integrating with a kernel of $x$ gives $m_1'=m_2+F_0m_1+F_1m_0$ where $F_1=int_{-infty}^infty xf(x) dx$ and $m_2=int_{-infty}^infty x^2u(x,t) dx$. But in general, the dynamics of $m_i$ always depend on $m_{i+1}$, so it seems like a moment closure will not be possible.
$endgroup$
– Alex
Jan 14 at 19:09
add a comment |
1
1
1
3
$begingroup$
Given the evolution equation
$$
frac{partial u}{partial t}(x,t)=xu(x,t)+int_{-infty}^infty f(x-y)u(y,t)dy
$$
with initial condition $u(x,t)=delta(x-x_0)$, is there a way to solve for either or both of the quantities
$$
m_0(t)=int_{-infty}^infty u(x,t)dx quadtext{ or }quad m_1(t)=int_{-infty}^infty xu(x,t)dx quad
$$
in closed form, in terms of the positive-valued function $f(cdot)$?
ordinary-differential-equations pde dynamical-systems convolution
asked Jan 11 at 18:00
AlexAlex
868
$endgroup$
Given the evolution equation
$$
frac{partial u}{partial t}(x,t)=xu(x,t)+int_{-infty}^infty f(x-y)u(y,t)dy
$$
with initial condition $u(x,t)=delta(x-x_0)$, is there a way to solve for either or both of the quantities
$$
m_0(t)=int_{-infty}^infty u(x,t)dx quadtext{ or }quad m_1(t)=int_{-infty}^infty xu(x,t)dx quad
$$
in closed form, in terms of the positive-valued function $f(cdot)$?
ordinary-differential-equations pde dynamical-systems convolution
ordinary-differential-equations pde dynamical-systems convolution
asked Jan 11 at 18:00
AlexAlex
868
asked Jan 11 at 18:00
AlexAlex
868
asked Jan 11 at 18:00
AlexAlex
868
asked Jan 11 at 18:00
AlexAlex
868
asked Jan 11 at 18:00
AlexAlex
868
868
$begingroup$
Have you tried taking the Fourier transform of the PDE to find an explicit solution?
$endgroup$
– Mattos
Jan 12 at 4:43
$begingroup$
@Mattos, thanks for the suggestion. I have tried both the Fourier and Laplace transforms but have not found them to be helpful (yet). The Fourier transform of this equation is $dot{tilde u}=itilde{u}'+tilde ftilde u$, which has the solution $tilde u=expleft{(t-iomega)x_0+iint_{omega+it}^omega tilde f(s)dsright}$. Inverting this transform gives $u=e^{x_0 t}mathcal F^{-1}_{omegato x-x_0}left{iint_{omega+it}^omega tilde f(s)dsright}$. Is there a way to evaluate the inverse Fourier transform of an integral like this?
$endgroup$
– Alex
Jan 12 at 23:14
$begingroup$
Quick correction: Inverting the transform gives $u=e^{x_0t}mathcal F^{-1}_{omegato x-x_0}left{expleft(iint_{omega+it}^omega tilde f(s)dsright)right}$.
$endgroup$
– Alex
Jan 12 at 23:56
$begingroup$
Integrating the original equation over $x in Bbb R$ gives a relationship between $m_0$ and $m_1$
$endgroup$
– Dylan
Jan 14 at 8:33
$begingroup$
@Dylan, thank you, that's a good point. Integrating the original equation gives $m_0'(t)=m_1(t)+F_0 m_0(t)$ where $F_0=int_{-infty}^infty f(x) dx$. Integrating with a kernel of $x$ gives $m_1'=m_2+F_0m_1+F_1m_0$ where $F_1=int_{-infty}^infty xf(x) dx$ and $m_2=int_{-infty}^infty x^2u(x,t) dx$. But in general, the dynamics of $m_i$ always depend on $m_{i+1}$, so it seems like a moment closure will not be possible.
$endgroup$
– Alex
Jan 14 at 19:09
add a comment |
$begingroup$
Have you tried taking the Fourier transform of the PDE to find an explicit solution?
$endgroup$
– Mattos
Jan 12 at 4:43
$begingroup$
@Mattos, thanks for the suggestion. I have tried both the Fourier and Laplace transforms but have not found them to be helpful (yet). The Fourier transform of this equation is $dot{tilde u}=itilde{u}'+tilde ftilde u$, which has the solution $tilde u=expleft{(t-iomega)x_0+iint_{omega+it}^omega tilde f(s)dsright}$. Inverting this transform gives $u=e^{x_0 t}mathcal F^{-1}_{omegato x-x_0}left{iint_{omega+it}^omega tilde f(s)dsright}$. Is there a way to evaluate the inverse Fourier transform of an integral like this?
$endgroup$
– Alex
Jan 12 at 23:14
$begingroup$
Quick correction: Inverting the transform gives $u=e^{x_0t}mathcal F^{-1}_{omegato x-x_0}left{expleft(iint_{omega+it}^omega tilde f(s)dsright)right}$.
$endgroup$
– Alex
Jan 12 at 23:56
$begingroup$
Integrating the original equation over $x in Bbb R$ gives a relationship between $m_0$ and $m_1$
$endgroup$
– Dylan
Jan 14 at 8:33
$begingroup$
@Dylan, thank you, that's a good point. Integrating the original equation gives $m_0'(t)=m_1(t)+F_0 m_0(t)$ where $F_0=int_{-infty}^infty f(x) dx$. Integrating with a kernel of $x$ gives $m_1'=m_2+F_0m_1+F_1m_0$ where $F_1=int_{-infty}^infty xf(x) dx$ and $m_2=int_{-infty}^infty x^2u(x,t) dx$. But in general, the dynamics of $m_i$ always depend on $m_{i+1}$, so it seems like a moment closure will not be possible.
$endgroup$
– Alex
Jan 14 at 19:09
$begingroup$
Have you tried taking the Fourier transform of the PDE to find an explicit solution?
$endgroup$
– Mattos
Jan 12 at 4:43
$begingroup$
Have you tried taking the Fourier transform of the PDE to find an explicit solution?
$endgroup$
– Mattos
Jan 12 at 4:43
$begingroup$
@Mattos, thanks for the suggestion. I have tried both the Fourier and Laplace transforms but have not found them to be helpful (yet). The Fourier transform of this equation is $dot{tilde u}=itilde{u}'+tilde ftilde u$, which has the solution $tilde u=expleft{(t-iomega)x_0+iint_{omega+it}^omega tilde f(s)dsright}$. Inverting this transform gives $u=e^{x_0 t}mathcal F^{-1}_{omegato x-x_0}left{iint_{omega+it}^omega tilde f(s)dsright}$. Is there a way to evaluate the inverse Fourier transform of an integral like this?
$endgroup$
– Alex
Jan 12 at 23:14
$begingroup$
@Mattos, thanks for the suggestion. I have tried both the Fourier and Laplace transforms but have not found them to be helpful (yet). The Fourier transform of this equation is $dot{tilde u}=itilde{u}'+tilde ftilde u$, which has the solution $tilde u=expleft{(t-iomega)x_0+iint_{omega+it}^omega tilde f(s)dsright}$. Inverting this transform gives $u=e^{x_0 t}mathcal F^{-1}_{omegato x-x_0}left{iint_{omega+it}^omega tilde f(s)dsright}$. Is there a way to evaluate the inverse Fourier transform of an integral like this?
$endgroup$
– Alex
Jan 12 at 23:14
$begingroup$
Quick correction: Inverting the transform gives $u=e^{x_0t}mathcal F^{-1}_{omegato x-x_0}left{expleft(iint_{omega+it}^omega tilde f(s)dsright)right}$.
$endgroup$
– Alex
Jan 12 at 23:56
$begingroup$
Quick correction: Inverting the transform gives $u=e^{x_0t}mathcal F^{-1}_{omegato x-x_0}left{expleft(iint_{omega+it}^omega tilde f(s)dsright)right}$.
$endgroup$
– Alex
Jan 12 at 23:56
$begingroup$
Integrating the original equation over $x in Bbb R$ gives a relationship between $m_0$ and $m_1$
$endgroup$
– Dylan
Jan 14 at 8:33
$begingroup$
Integrating the original equation over $x in Bbb R$ gives a relationship between $m_0$ and $m_1$
$endgroup$
– Dylan
Jan 14 at 8:33
$begingroup$
@Dylan, thank you, that's a good point. Integrating the original equation gives $m_0'(t)=m_1(t)+F_0 m_0(t)$ where $F_0=int_{-infty}^infty f(x) dx$. Integrating with a kernel of $x$ gives $m_1'=m_2+F_0m_1+F_1m_0$ where $F_1=int_{-infty}^infty xf(x) dx$ and $m_2=int_{-infty}^infty x^2u(x,t) dx$. But in general, the dynamics of $m_i$ always depend on $m_{i+1}$, so it seems like a moment closure will not be possible.
$endgroup$
– Alex
Jan 14 at 19:09
$begingroup$
@Dylan, thank you, that's a good point. Integrating the original equation gives $m_0'(t)=m_1(t)+F_0 m_0(t)$ where $F_0=int_{-infty}^infty f(x) dx$. Integrating with a kernel of $x$ gives $m_1'=m_2+F_0m_1+F_1m_0$ where $F_1=int_{-infty}^infty xf(x) dx$ and $m_2=int_{-infty}^infty x^2u(x,t) dx$. But in general, the dynamics of $m_i$ always depend on $m_{i+1}$, so it seems like a moment closure will not be possible.
$endgroup$
– Alex
Jan 14 at 19:09
add a comment |
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$begingroup$
Have you tried taking the Fourier transform of the PDE to find an explicit solution?
$endgroup$
– Mattos
Jan 12 at 4:43
$begingroup$
@Mattos, thanks for the suggestion. I have tried both the Fourier and Laplace transforms but have not found them to be helpful (yet). The Fourier transform of this equation is $dot{tilde u}=itilde{u}'+tilde ftilde u$, which has the solution $tilde u=expleft{(t-iomega)x_0+iint_{omega+it}^omega tilde f(s)dsright}$. Inverting this transform gives $u=e^{x_0 t}mathcal F^{-1}_{omegato x-x_0}left{iint_{omega+it}^omega tilde f(s)dsright}$. Is there a way to evaluate the inverse Fourier transform of an integral like this?
$endgroup$
– Alex
Jan 12 at 23:14
$begingroup$
Quick correction: Inverting the transform gives $u=e^{x_0t}mathcal F^{-1}_{omegato x-x_0}left{expleft(iint_{omega+it}^omega tilde f(s)dsright)right}$.
$endgroup$
– Alex
Jan 12 at 23:56
$begingroup$
Integrating the original equation over $x in Bbb R$ gives a relationship between $m_0$ and $m_1$
$endgroup$
– Dylan
Jan 14 at 8:33
$begingroup$
@Dylan, thank you, that's a good point. Integrating the original equation gives $m_0'(t)=m_1(t)+F_0 m_0(t)$ where $F_0=int_{-infty}^infty f(x) dx$. Integrating with a kernel of $x$ gives $m_1'=m_2+F_0m_1+F_1m_0$ where $F_1=int_{-infty}^infty xf(x) dx$ and $m_2=int_{-infty}^infty x^2u(x,t) dx$. But in general, the dynamics of $m_i$ always depend on $m_{i+1}$, so it seems like a moment closure will not be possible.
$endgroup$
– Alex
Jan 14 at 19:09