Sound wave equation: Neumann boundary conditions
$begingroup$
In this paper it's described the solution of the damped wave equation in cylindrical coordinates
$$ nabla^2left(c^2rho_1+nufrac{partialrho_1}{partial t}right)-frac{partial^2rho_1}{partial t^2}=0$$
where $rho_1$ is the difference of the density relative to the unperturbed state $rho_0$.
The applied boundary condition is
$$ mathbf{v}big|_{r=r_0}=v_Acos(omega t)mathbf{hat{r}}$$
where $v$ is the velocity of the fluid.
They claim that this boundary condition can we rewritten as
begin{equation}
frac{partialrho_1}{partial r} bigg|_{r=r_0}=frac{rho_0v_Aomega c^2}{nu^2omega^2+c^4}sin(omega t)-frac{rho_0v_Aomega^2 nu}{nu^2omega^2+c^4}cos(omega t)hspace{1cm} (1)
end{equation}
just imposing $nablatimes mathbf{v}=mathbf{0}$ and using the equations for the conservation of mass and momentum
$$ frac{partialrho_1}{partial t} + nablacdot(rho_0 mathbf{v}) =0$$
$$ frac{partial}{partial t}(rho_0 mathbf{v})+c^2nablarho_1+nabla cdot mathbf{D}_1=mathbf{0}$$
where $mathbf{D}_1$ is the viscous stress tensor.
It is possible to prove that, if $nablatimes mathbf{v}=0$, then $nabla cdot mathbf{D}_1 = -nunabla^2mathbf{v} $.
I tried hard but I've not been able to prove equation $(1)$. Do you know how to proceed?
multivariable-calculus pde wave-equation fluid-dynamics
$endgroup$
add a comment |
$begingroup$
In this paper it's described the solution of the damped wave equation in cylindrical coordinates
$$ nabla^2left(c^2rho_1+nufrac{partialrho_1}{partial t}right)-frac{partial^2rho_1}{partial t^2}=0$$
where $rho_1$ is the difference of the density relative to the unperturbed state $rho_0$.
The applied boundary condition is
$$ mathbf{v}big|_{r=r_0}=v_Acos(omega t)mathbf{hat{r}}$$
where $v$ is the velocity of the fluid.
They claim that this boundary condition can we rewritten as
begin{equation}
frac{partialrho_1}{partial r} bigg|_{r=r_0}=frac{rho_0v_Aomega c^2}{nu^2omega^2+c^4}sin(omega t)-frac{rho_0v_Aomega^2 nu}{nu^2omega^2+c^4}cos(omega t)hspace{1cm} (1)
end{equation}
just imposing $nablatimes mathbf{v}=mathbf{0}$ and using the equations for the conservation of mass and momentum
$$ frac{partialrho_1}{partial t} + nablacdot(rho_0 mathbf{v}) =0$$
$$ frac{partial}{partial t}(rho_0 mathbf{v})+c^2nablarho_1+nabla cdot mathbf{D}_1=mathbf{0}$$
where $mathbf{D}_1$ is the viscous stress tensor.
It is possible to prove that, if $nablatimes mathbf{v}=0$, then $nabla cdot mathbf{D}_1 = -nunabla^2mathbf{v} $.
I tried hard but I've not been able to prove equation $(1)$. Do you know how to proceed?
multivariable-calculus pde wave-equation fluid-dynamics
$endgroup$
add a comment |
$begingroup$
In this paper it's described the solution of the damped wave equation in cylindrical coordinates
$$ nabla^2left(c^2rho_1+nufrac{partialrho_1}{partial t}right)-frac{partial^2rho_1}{partial t^2}=0$$
where $rho_1$ is the difference of the density relative to the unperturbed state $rho_0$.
The applied boundary condition is
$$ mathbf{v}big|_{r=r_0}=v_Acos(omega t)mathbf{hat{r}}$$
where $v$ is the velocity of the fluid.
They claim that this boundary condition can we rewritten as
begin{equation}
frac{partialrho_1}{partial r} bigg|_{r=r_0}=frac{rho_0v_Aomega c^2}{nu^2omega^2+c^4}sin(omega t)-frac{rho_0v_Aomega^2 nu}{nu^2omega^2+c^4}cos(omega t)hspace{1cm} (1)
end{equation}
just imposing $nablatimes mathbf{v}=mathbf{0}$ and using the equations for the conservation of mass and momentum
$$ frac{partialrho_1}{partial t} + nablacdot(rho_0 mathbf{v}) =0$$
$$ frac{partial}{partial t}(rho_0 mathbf{v})+c^2nablarho_1+nabla cdot mathbf{D}_1=mathbf{0}$$
where $mathbf{D}_1$ is the viscous stress tensor.
It is possible to prove that, if $nablatimes mathbf{v}=0$, then $nabla cdot mathbf{D}_1 = -nunabla^2mathbf{v} $.
I tried hard but I've not been able to prove equation $(1)$. Do you know how to proceed?
multivariable-calculus pde wave-equation fluid-dynamics
$endgroup$
In this paper it's described the solution of the damped wave equation in cylindrical coordinates
$$ nabla^2left(c^2rho_1+nufrac{partialrho_1}{partial t}right)-frac{partial^2rho_1}{partial t^2}=0$$
where $rho_1$ is the difference of the density relative to the unperturbed state $rho_0$.
The applied boundary condition is
$$ mathbf{v}big|_{r=r_0}=v_Acos(omega t)mathbf{hat{r}}$$
where $v$ is the velocity of the fluid.
They claim that this boundary condition can we rewritten as
begin{equation}
frac{partialrho_1}{partial r} bigg|_{r=r_0}=frac{rho_0v_Aomega c^2}{nu^2omega^2+c^4}sin(omega t)-frac{rho_0v_Aomega^2 nu}{nu^2omega^2+c^4}cos(omega t)hspace{1cm} (1)
end{equation}
just imposing $nablatimes mathbf{v}=mathbf{0}$ and using the equations for the conservation of mass and momentum
$$ frac{partialrho_1}{partial t} + nablacdot(rho_0 mathbf{v}) =0$$
$$ frac{partial}{partial t}(rho_0 mathbf{v})+c^2nablarho_1+nabla cdot mathbf{D}_1=mathbf{0}$$
where $mathbf{D}_1$ is the viscous stress tensor.
It is possible to prove that, if $nablatimes mathbf{v}=0$, then $nabla cdot mathbf{D}_1 = -nunabla^2mathbf{v} $.
I tried hard but I've not been able to prove equation $(1)$. Do you know how to proceed?
multivariable-calculus pde wave-equation fluid-dynamics
multivariable-calculus pde wave-equation fluid-dynamics
edited Dec 14 '18 at 9:32
Harry49
7,46431340
7,46431340
asked Dec 13 '18 at 18:45
Alessandro ZuninoAlessandro Zunino
17311
17311
add a comment |
add a comment |
1 Answer
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$begingroup$
You can combine the two conservation equations to obtain
$$ rho_0frac{partial v}{partial t}+c^2frac{partial rho_1}{partial r}-nufrac{partial}{partial t}frac{partial rho_1}{partial r}bigg|_{r=r_0} = 0$$
using the Fourier transform method is possible to solve this differential equation for the variable $frac{partial rho_1}{partial r}bigg|_{r=r_0}$ obtaining then equation $(1)$.
$endgroup$
add a comment |
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$begingroup$
You can combine the two conservation equations to obtain
$$ rho_0frac{partial v}{partial t}+c^2frac{partial rho_1}{partial r}-nufrac{partial}{partial t}frac{partial rho_1}{partial r}bigg|_{r=r_0} = 0$$
using the Fourier transform method is possible to solve this differential equation for the variable $frac{partial rho_1}{partial r}bigg|_{r=r_0}$ obtaining then equation $(1)$.
$endgroup$
add a comment |
$begingroup$
You can combine the two conservation equations to obtain
$$ rho_0frac{partial v}{partial t}+c^2frac{partial rho_1}{partial r}-nufrac{partial}{partial t}frac{partial rho_1}{partial r}bigg|_{r=r_0} = 0$$
using the Fourier transform method is possible to solve this differential equation for the variable $frac{partial rho_1}{partial r}bigg|_{r=r_0}$ obtaining then equation $(1)$.
$endgroup$
add a comment |
$begingroup$
You can combine the two conservation equations to obtain
$$ rho_0frac{partial v}{partial t}+c^2frac{partial rho_1}{partial r}-nufrac{partial}{partial t}frac{partial rho_1}{partial r}bigg|_{r=r_0} = 0$$
using the Fourier transform method is possible to solve this differential equation for the variable $frac{partial rho_1}{partial r}bigg|_{r=r_0}$ obtaining then equation $(1)$.
$endgroup$
You can combine the two conservation equations to obtain
$$ rho_0frac{partial v}{partial t}+c^2frac{partial rho_1}{partial r}-nufrac{partial}{partial t}frac{partial rho_1}{partial r}bigg|_{r=r_0} = 0$$
using the Fourier transform method is possible to solve this differential equation for the variable $frac{partial rho_1}{partial r}bigg|_{r=r_0}$ obtaining then equation $(1)$.
edited Dec 14 '18 at 18:25
answered Dec 14 '18 at 18:09
Alessandro ZuninoAlessandro Zunino
17311
17311
add a comment |
add a comment |
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