2D finite difference boundary conditions for radial direction












1












$begingroup$


I am trying to solve Poisson's equation in an axisymmetric cylindrical domain using finite difference. So I start with my differential equation and boundary conditions and discretize them. However, I'm having trouble thinking of how to discretize the radial portion



begin{array}{lll}
frac{1}{r} frac{partial}{partial r} left(r frac{partial u}{partial r} right) + frac{partial^2 u}{partial z^2} = -frac{rho}{varepsilon_0} & longrightarrow & frac{r_{i+1/2} u_{i+1, j} - 2 r_{i} u_{i,j} + r_{i-1/2} u_{i-1, j}}{r_i Delta r^2} + frac{u_{i, j+1} - 2 u_{i, j} + u_{i, j - 1}}{Delta z^2} = -frac{rho_{i,j}}{varepsilon_0} \
left. frac{partial u}{partial r} right|_{r = 0} = 0 & longrightarrow & u_{0,j} = frac{1}{3} left(4 u_{1,j} - u_{2,j} right) ~ \
lim_{rtoinfty} u(r, z) = 0 & longrightarrow & ? ~ \
u(r, l(r)) = f_3(r) & longrightarrow & u_{i, l(r_i)/Delta z} = f_3(r_i) \
u(r, h) = f_4(r) & longrightarrow & u_{i, N} = f_4(r_i) \
end{array}



where



$$
l(r) = left{begin{aligned}
&h - frac{H}{R} r &&: r le R\
&0 &&: r > R
end{aligned}
right.$$
Looking at my old class notes, the professor mentioned a method called irregular singular points as a method to better approximate boundary conditions in (semi-)infinite domains but, I don't understand how I would apply this to 2D systems or to radial systems.



I would appreciate any suggestions.










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$endgroup$

















    1












    $begingroup$


    I am trying to solve Poisson's equation in an axisymmetric cylindrical domain using finite difference. So I start with my differential equation and boundary conditions and discretize them. However, I'm having trouble thinking of how to discretize the radial portion



    begin{array}{lll}
    frac{1}{r} frac{partial}{partial r} left(r frac{partial u}{partial r} right) + frac{partial^2 u}{partial z^2} = -frac{rho}{varepsilon_0} & longrightarrow & frac{r_{i+1/2} u_{i+1, j} - 2 r_{i} u_{i,j} + r_{i-1/2} u_{i-1, j}}{r_i Delta r^2} + frac{u_{i, j+1} - 2 u_{i, j} + u_{i, j - 1}}{Delta z^2} = -frac{rho_{i,j}}{varepsilon_0} \
    left. frac{partial u}{partial r} right|_{r = 0} = 0 & longrightarrow & u_{0,j} = frac{1}{3} left(4 u_{1,j} - u_{2,j} right) ~ \
    lim_{rtoinfty} u(r, z) = 0 & longrightarrow & ? ~ \
    u(r, l(r)) = f_3(r) & longrightarrow & u_{i, l(r_i)/Delta z} = f_3(r_i) \
    u(r, h) = f_4(r) & longrightarrow & u_{i, N} = f_4(r_i) \
    end{array}



    where



    $$
    l(r) = left{begin{aligned}
    &h - frac{H}{R} r &&: r le R\
    &0 &&: r > R
    end{aligned}
    right.$$
    Looking at my old class notes, the professor mentioned a method called irregular singular points as a method to better approximate boundary conditions in (semi-)infinite domains but, I don't understand how I would apply this to 2D systems or to radial systems.



    I would appreciate any suggestions.










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I am trying to solve Poisson's equation in an axisymmetric cylindrical domain using finite difference. So I start with my differential equation and boundary conditions and discretize them. However, I'm having trouble thinking of how to discretize the radial portion



      begin{array}{lll}
      frac{1}{r} frac{partial}{partial r} left(r frac{partial u}{partial r} right) + frac{partial^2 u}{partial z^2} = -frac{rho}{varepsilon_0} & longrightarrow & frac{r_{i+1/2} u_{i+1, j} - 2 r_{i} u_{i,j} + r_{i-1/2} u_{i-1, j}}{r_i Delta r^2} + frac{u_{i, j+1} - 2 u_{i, j} + u_{i, j - 1}}{Delta z^2} = -frac{rho_{i,j}}{varepsilon_0} \
      left. frac{partial u}{partial r} right|_{r = 0} = 0 & longrightarrow & u_{0,j} = frac{1}{3} left(4 u_{1,j} - u_{2,j} right) ~ \
      lim_{rtoinfty} u(r, z) = 0 & longrightarrow & ? ~ \
      u(r, l(r)) = f_3(r) & longrightarrow & u_{i, l(r_i)/Delta z} = f_3(r_i) \
      u(r, h) = f_4(r) & longrightarrow & u_{i, N} = f_4(r_i) \
      end{array}



      where



      $$
      l(r) = left{begin{aligned}
      &h - frac{H}{R} r &&: r le R\
      &0 &&: r > R
      end{aligned}
      right.$$
      Looking at my old class notes, the professor mentioned a method called irregular singular points as a method to better approximate boundary conditions in (semi-)infinite domains but, I don't understand how I would apply this to 2D systems or to radial systems.



      I would appreciate any suggestions.










      share|cite|improve this question











      $endgroup$




      I am trying to solve Poisson's equation in an axisymmetric cylindrical domain using finite difference. So I start with my differential equation and boundary conditions and discretize them. However, I'm having trouble thinking of how to discretize the radial portion



      begin{array}{lll}
      frac{1}{r} frac{partial}{partial r} left(r frac{partial u}{partial r} right) + frac{partial^2 u}{partial z^2} = -frac{rho}{varepsilon_0} & longrightarrow & frac{r_{i+1/2} u_{i+1, j} - 2 r_{i} u_{i,j} + r_{i-1/2} u_{i-1, j}}{r_i Delta r^2} + frac{u_{i, j+1} - 2 u_{i, j} + u_{i, j - 1}}{Delta z^2} = -frac{rho_{i,j}}{varepsilon_0} \
      left. frac{partial u}{partial r} right|_{r = 0} = 0 & longrightarrow & u_{0,j} = frac{1}{3} left(4 u_{1,j} - u_{2,j} right) ~ \
      lim_{rtoinfty} u(r, z) = 0 & longrightarrow & ? ~ \
      u(r, l(r)) = f_3(r) & longrightarrow & u_{i, l(r_i)/Delta z} = f_3(r_i) \
      u(r, h) = f_4(r) & longrightarrow & u_{i, N} = f_4(r_i) \
      end{array}



      where



      $$
      l(r) = left{begin{aligned}
      &h - frac{H}{R} r &&: r le R\
      &0 &&: r > R
      end{aligned}
      right.$$
      Looking at my old class notes, the professor mentioned a method called irregular singular points as a method to better approximate boundary conditions in (semi-)infinite domains but, I don't understand how I would apply this to 2D systems or to radial systems.



      I would appreciate any suggestions.







      polar-coordinates boundary-value-problem finite-differences






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      edited Aug 14 '15 at 17:49







      user1543042

















      asked Aug 14 '15 at 14:28









      user1543042user1543042

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          The actual boundary condition should be



          $$lim_{r rightarrow infty} frac{partial u}{partial r} = 0$$



          which I approximated at some finite point using the backward second order finite difference.



          The actual boundary condition can be seen to be correct by a simple thought experiment by setting $H = 0$ (a rectangular grid) with $rho = 0$ and we must recover the solution 1D cartesian solution to Laplace's equation.






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            $begingroup$

            The actual boundary condition should be



            $$lim_{r rightarrow infty} frac{partial u}{partial r} = 0$$



            which I approximated at some finite point using the backward second order finite difference.



            The actual boundary condition can be seen to be correct by a simple thought experiment by setting $H = 0$ (a rectangular grid) with $rho = 0$ and we must recover the solution 1D cartesian solution to Laplace's equation.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The actual boundary condition should be



              $$lim_{r rightarrow infty} frac{partial u}{partial r} = 0$$



              which I approximated at some finite point using the backward second order finite difference.



              The actual boundary condition can be seen to be correct by a simple thought experiment by setting $H = 0$ (a rectangular grid) with $rho = 0$ and we must recover the solution 1D cartesian solution to Laplace's equation.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The actual boundary condition should be



                $$lim_{r rightarrow infty} frac{partial u}{partial r} = 0$$



                which I approximated at some finite point using the backward second order finite difference.



                The actual boundary condition can be seen to be correct by a simple thought experiment by setting $H = 0$ (a rectangular grid) with $rho = 0$ and we must recover the solution 1D cartesian solution to Laplace's equation.






                share|cite|improve this answer









                $endgroup$



                The actual boundary condition should be



                $$lim_{r rightarrow infty} frac{partial u}{partial r} = 0$$



                which I approximated at some finite point using the backward second order finite difference.



                The actual boundary condition can be seen to be correct by a simple thought experiment by setting $H = 0$ (a rectangular grid) with $rho = 0$ and we must recover the solution 1D cartesian solution to Laplace's equation.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 28 '15 at 1:50









                user1543042user1543042

                330111




                330111






























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