$nabla cdot (b nabla c) = 0$ where $b$ and $c$ are unknown












1














Description:



I want to solve the equation
$$ nabla cdot left( b nabla c right) = 0 $$
for $b(x,y)$ and $c(x,y)$ in the domain $x,y geq 0$. This partial differential equation is supplemented by
$$ b nabla c = vec{g} $$
where $vec{g}(x,y)$ is a known vector that has non-zero curl. Furthermore, the values of $b$ and $c$ are known on the boundary $(x = 0, y = 0)$. As a consequence, the normal derivatives of $c(x,y)$ on the boundary are also known:
$$ nabla times vec{g} = R(x,y) ne 0 \
c = 0 hspace{4mm} text{at} hspace{4mm} x=0 \
c = 0 hspace{4mm} text{at} hspace{4mm} y=0 \ $$



What solution methods to this problem exist? Can you point me to any resources? If I try to solve this numerically, which software packages can you recommend? Thanks.



It is possible to derive the first-order pde's
$$ vec{g} times nabla c = 0 \ nabla ln(b) times vec{g} = R ; . $$
Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$. So the interior of the domain is not connected to the boundary. I was hoping there are other approaches.






pde vectors vector-analysis boundary-value-problem







share|cite|improve this question





























    1














    Description:



    I want to solve the equation
    $$ nabla cdot left( b nabla c right) = 0 $$
    for $b(x,y)$ and $c(x,y)$ in the domain $x,y geq 0$. This partial differential equation is supplemented by
    $$ b nabla c = vec{g} $$
    where $vec{g}(x,y)$ is a known vector that has non-zero curl. Furthermore, the values of $b$ and $c$ are known on the boundary $(x = 0, y = 0)$. As a consequence, the normal derivatives of $c(x,y)$ on the boundary are also known:
    $$ nabla times vec{g} = R(x,y) ne 0 \
    c = 0 hspace{4mm} text{at} hspace{4mm} x=0 \
    c = 0 hspace{4mm} text{at} hspace{4mm} y=0 \ $$



    What solution methods to this problem exist? Can you point me to any resources? If I try to solve this numerically, which software packages can you recommend? Thanks.



    It is possible to derive the first-order pde's
    $$ vec{g} times nabla c = 0 \ nabla ln(b) times vec{g} = R ; . $$
    Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$. So the interior of the domain is not connected to the boundary. I was hoping there are other approaches.






    pde vectors vector-analysis boundary-value-problem







    share|cite|improve this question



























      1












      1








      1


      1





      Description:



      I want to solve the equation
      $$ nabla cdot left( b nabla c right) = 0 $$
      for $b(x,y)$ and $c(x,y)$ in the domain $x,y geq 0$. This partial differential equation is supplemented by
      $$ b nabla c = vec{g} $$
      where $vec{g}(x,y)$ is a known vector that has non-zero curl. Furthermore, the values of $b$ and $c$ are known on the boundary $(x = 0, y = 0)$. As a consequence, the normal derivatives of $c(x,y)$ on the boundary are also known:
      $$ nabla times vec{g} = R(x,y) ne 0 \
      c = 0 hspace{4mm} text{at} hspace{4mm} x=0 \
      c = 0 hspace{4mm} text{at} hspace{4mm} y=0 \ $$



      What solution methods to this problem exist? Can you point me to any resources? If I try to solve this numerically, which software packages can you recommend? Thanks.



      It is possible to derive the first-order pde's
      $$ vec{g} times nabla c = 0 \ nabla ln(b) times vec{g} = R ; . $$
      Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$. So the interior of the domain is not connected to the boundary. I was hoping there are other approaches.






      pde vectors vector-analysis boundary-value-problem







      share|cite|improve this question















      Description:



      I want to solve the equation
      $$ nabla cdot left( b nabla c right) = 0 $$
      for $b(x,y)$ and $c(x,y)$ in the domain $x,y geq 0$. This partial differential equation is supplemented by
      $$ b nabla c = vec{g} $$
      where $vec{g}(x,y)$ is a known vector that has non-zero curl. Furthermore, the values of $b$ and $c$ are known on the boundary $(x = 0, y = 0)$. As a consequence, the normal derivatives of $c(x,y)$ on the boundary are also known:
      $$ nabla times vec{g} = R(x,y) ne 0 \
      c = 0 hspace{4mm} text{at} hspace{4mm} x=0 \
      c = 0 hspace{4mm} text{at} hspace{4mm} y=0 \ $$



      What solution methods to this problem exist? Can you point me to any resources? If I try to solve this numerically, which software packages can you recommend? Thanks.



      It is possible to derive the first-order pde's
      $$ vec{g} times nabla c = 0 \ nabla ln(b) times vec{g} = R ; . $$
      Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$. So the interior of the domain is not connected to the boundary. I was hoping there are other approaches.






      pde vectors vector-analysis boundary-value-problem




      pde vectors vector-analysis boundary-value-problem






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 22 at 20:49

























      asked Oct 30 at 8:57









      Crenguta

      406




      406






















          1 Answer
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          You can start by noticing that we need
          $$
          begin{split}
          0 &= nablatimesnabla c = nablatimesfrac gb
          = nablafrac1b times g + frac1b nablatimes g
          = -frac{nabla b}{b^2} times g + frac Rb .
          end{split}
          $$
          Multiplying by $b$ you get
          $$
          nabla(log b) times g = R.
          $$
          This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
          You can find more pieces of information here, here, and here.



          Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.






          share|cite|improve this answer





















          • Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
            – Crenguta
            Nov 6 at 7:47











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          1 Answer
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          active

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          You can start by noticing that we need
          $$
          begin{split}
          0 &= nablatimesnabla c = nablatimesfrac gb
          = nablafrac1b times g + frac1b nablatimes g
          = -frac{nabla b}{b^2} times g + frac Rb .
          end{split}
          $$
          Multiplying by $b$ you get
          $$
          nabla(log b) times g = R.
          $$
          This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
          You can find more pieces of information here, here, and here.



          Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.






          share|cite|improve this answer





















          • Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
            – Crenguta
            Nov 6 at 7:47
















          1














          You can start by noticing that we need
          $$
          begin{split}
          0 &= nablatimesnabla c = nablatimesfrac gb
          = nablafrac1b times g + frac1b nablatimes g
          = -frac{nabla b}{b^2} times g + frac Rb .
          end{split}
          $$
          Multiplying by $b$ you get
          $$
          nabla(log b) times g = R.
          $$
          This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
          You can find more pieces of information here, here, and here.



          Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.






          share|cite|improve this answer





















          • Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
            – Crenguta
            Nov 6 at 7:47














          1












          1








          1






          You can start by noticing that we need
          $$
          begin{split}
          0 &= nablatimesnabla c = nablatimesfrac gb
          = nablafrac1b times g + frac1b nablatimes g
          = -frac{nabla b}{b^2} times g + frac Rb .
          end{split}
          $$
          Multiplying by $b$ you get
          $$
          nabla(log b) times g = R.
          $$
          This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
          You can find more pieces of information here, here, and here.



          Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.






          share|cite|improve this answer












          You can start by noticing that we need
          $$
          begin{split}
          0 &= nablatimesnabla c = nablatimesfrac gb
          = nablafrac1b times g + frac1b nablatimes g
          = -frac{nabla b}{b^2} times g + frac Rb .
          end{split}
          $$
          Multiplying by $b$ you get
          $$
          nabla(log b) times g = R.
          $$
          This is a linear first order partial differential equation, whose solution is of the form $b=exp(tilde f+f_0)$, where $tilde f$ is a particular solution to the inhomogeneous PDE $nabla f times g = R$ and $f_0$ is a solution to the homogeneous one. The PDE is solvable as long as $g$ is not orthogonal to $partialOmega$ (where $Omega$ is your domain, the positive quadrant) and boundary value for $b$ is compatible with the characteristic lines.
          You can find more pieces of information here, here, and here.



          Once you have found $b$, $g/b$ is irrotational, therefore you find $c$ such that $nabla c=g/b$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 5 at 14:17









          Federico

          4,429512




          4,429512












          • Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
            – Crenguta
            Nov 6 at 7:47


















          • Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
            – Crenguta
            Nov 6 at 7:47
















          Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
          – Crenguta
          Nov 6 at 7:47




          Unfortunately, the characteristic lines of the pde's are tangential to the boundary on $partial Omega$.
          – Crenguta
          Nov 6 at 7:47


















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