Sound wave equation: Neumann boundary conditions












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$begingroup$


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?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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?










      share|cite|improve this question











      $endgroup$




      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






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      edited Dec 14 '18 at 9:32









      Harry49

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      7,46431340










      asked Dec 13 '18 at 18:45









      Alessandro ZuninoAlessandro Zunino

      17311




      17311






















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          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 $(1)$.






          share|cite|improve this 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 $(1)$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              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 $(1)$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                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 $(1)$.






                share|cite|improve this answer











                $endgroup$



                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 $(1)$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 14 '18 at 18:25

























                answered Dec 14 '18 at 18:09









                Alessandro ZuninoAlessandro Zunino

                17311




                17311






























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