Variational problem of Laplace equation











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We know that:
$Delta u=-f, inOmega$
$u=0$ ,on $partialOmega$



Is equivalent to this variational problem:



find the minimum of $J(u)$ in $C^2(Omega)cap C^1(barOmega)$, where $J(u)=int_Omega[frac{1}{2}(u_x^2+u_y^2)-uf]dxdy$.



My question is:



why we need u belongs to $C^2(Omega)cap C^1(barOmega)$? specifically, why we need $u,Du$ continuos up to the boundary of $Omega$? Why can't we just use $uin C^2(Omega)cap C^1(Omega)$?






functional-analysis pde sobolev-spaces







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    up vote
    1
    down vote

    favorite
    1












    We know that:
    $Delta u=-f, inOmega$
    $u=0$ ,on $partialOmega$



    Is equivalent to this variational problem:



    find the minimum of $J(u)$ in $C^2(Omega)cap C^1(barOmega)$, where $J(u)=int_Omega[frac{1}{2}(u_x^2+u_y^2)-uf]dxdy$.



    My question is:



    why we need u belongs to $C^2(Omega)cap C^1(barOmega)$? specifically, why we need $u,Du$ continuos up to the boundary of $Omega$? Why can't we just use $uin C^2(Omega)cap C^1(Omega)$?






    functional-analysis pde sobolev-spaces







    share|cite|improve this question









    New contributor




    chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      1
      down vote

      favorite
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      up vote
      1
      down vote

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      We know that:
      $Delta u=-f, inOmega$
      $u=0$ ,on $partialOmega$



      Is equivalent to this variational problem:



      find the minimum of $J(u)$ in $C^2(Omega)cap C^1(barOmega)$, where $J(u)=int_Omega[frac{1}{2}(u_x^2+u_y^2)-uf]dxdy$.



      My question is:



      why we need u belongs to $C^2(Omega)cap C^1(barOmega)$? specifically, why we need $u,Du$ continuos up to the boundary of $Omega$? Why can't we just use $uin C^2(Omega)cap C^1(Omega)$?






      functional-analysis pde sobolev-spaces







      share|cite|improve this question









      New contributor




      chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      We know that:
      $Delta u=-f, inOmega$
      $u=0$ ,on $partialOmega$



      Is equivalent to this variational problem:



      find the minimum of $J(u)$ in $C^2(Omega)cap C^1(barOmega)$, where $J(u)=int_Omega[frac{1}{2}(u_x^2+u_y^2)-uf]dxdy$.



      My question is:



      why we need u belongs to $C^2(Omega)cap C^1(barOmega)$? specifically, why we need $u,Du$ continuos up to the boundary of $Omega$? Why can't we just use $uin C^2(Omega)cap C^1(Omega)$?






      functional-analysis pde sobolev-spaces




      functional-analysis pde sobolev-spaces






      share|cite|improve this question









      New contributor




      chloe hj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      edited yesterday









      Mefitico

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          Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.






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            Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.






            share|cite|improve this answer

























              up vote
              2
              down vote













              Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.






              share|cite|improve this answer























                up vote
                2
                down vote










                up vote
                2
                down vote









                Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.






                share|cite|improve this answer












                Typically the requirement is $uin C^2(Omega)cap C^0(bar{Omega})$ if you want a classical solution; you need this because you have boundary conditions and you want u to be continuous on the boundary (otherwise you could just find a solution without BC in the interior and define it as $0$ on the boundary). The (stronger) requirement $uin C^2(Omega)cap C^1(bar{Omega})$ is a bit weird actually, I've never heard about it, but it guarantees continuity at $partialOmega$ . However it may be useless.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 22 hours ago









                Marco

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