$L^2$ norm of a PDE











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Iam working on the following problem:



Let $Omega subseteq mathbb{R}^n$ and $v$ be the solution of $(-Delta+q-lambda)v=F$ on $Omega$ such that $q$ is bounded and $v=exp(-ax)u$ where $uin H_{0}^{1}(Omega)cap H^{2}(Omega)$



For a large $Xin mathbb{R}$, we define a cut-off function $chi_{X}$ on $Omega_{X}$:
$chi_{X}=begin{cases} 1 & text{on }[0,X-1]\ 1leqchi_{X}leq0 & text{on} (X-1,X)\
0 & text{on} [X,infty)end{cases}$



Iam trying to find $L^2(Omega_{X})$-norm of $(chi_{X}-1)(-Delta v)+(chi_{X}-1)(q-lambda)v$ ..



I startred by the following:
$|(chi_{X}-1)(-Delta v)+(chi_{X}-1)(q-lambda)v|^2_{L^2(Omega_{X})}=\ |(chi_{X}-1)(-Delta+q-lambda)v|^2_{L^2(Omega_{X})}=\ int_{Omega_{X}}(chi_{X}-1)^2((-Delta+q-lambda)v)^2$



Now Iam not sure how I can finish the calculation, I know $(chi_{X}-1)$ is bounded on $Omega_{X}$ hence
$ int_{Omega_{X}}(chi_{X}-1)^2((-Delta+q-lambda)v)^2= int_{Omega_{X}}|chi_{X}-1|^2((-Delta+q-lambda)v)^2leq C int_{Omega_{X}}((-Delta+q-lambda)v)^2$
Are these steps correct?
and how I can find the $L^2$ norm on $Omega_{X}$ for the laplacian term?






functional-analysis operator-theory norm







share|cite|improve this question


























    up vote
    1
    down vote

    favorite












    Iam working on the following problem:



    Let $Omega subseteq mathbb{R}^n$ and $v$ be the solution of $(-Delta+q-lambda)v=F$ on $Omega$ such that $q$ is bounded and $v=exp(-ax)u$ where $uin H_{0}^{1}(Omega)cap H^{2}(Omega)$



    For a large $Xin mathbb{R}$, we define a cut-off function $chi_{X}$ on $Omega_{X}$:
    $chi_{X}=begin{cases} 1 & text{on }[0,X-1]\ 1leqchi_{X}leq0 & text{on} (X-1,X)\
    0 & text{on} [X,infty)end{cases}$



    Iam trying to find $L^2(Omega_{X})$-norm of $(chi_{X}-1)(-Delta v)+(chi_{X}-1)(q-lambda)v$ ..



    I startred by the following:
    $|(chi_{X}-1)(-Delta v)+(chi_{X}-1)(q-lambda)v|^2_{L^2(Omega_{X})}=\ |(chi_{X}-1)(-Delta+q-lambda)v|^2_{L^2(Omega_{X})}=\ int_{Omega_{X}}(chi_{X}-1)^2((-Delta+q-lambda)v)^2$



    Now Iam not sure how I can finish the calculation, I know $(chi_{X}-1)$ is bounded on $Omega_{X}$ hence
    $ int_{Omega_{X}}(chi_{X}-1)^2((-Delta+q-lambda)v)^2= int_{Omega_{X}}|chi_{X}-1|^2((-Delta+q-lambda)v)^2leq C int_{Omega_{X}}((-Delta+q-lambda)v)^2$
    Are these steps correct?
    and how I can find the $L^2$ norm on $Omega_{X}$ for the laplacian term?






    functional-analysis operator-theory norm







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Iam working on the following problem:



      Let $Omega subseteq mathbb{R}^n$ and $v$ be the solution of $(-Delta+q-lambda)v=F$ on $Omega$ such that $q$ is bounded and $v=exp(-ax)u$ where $uin H_{0}^{1}(Omega)cap H^{2}(Omega)$



      For a large $Xin mathbb{R}$, we define a cut-off function $chi_{X}$ on $Omega_{X}$:
      $chi_{X}=begin{cases} 1 & text{on }[0,X-1]\ 1leqchi_{X}leq0 & text{on} (X-1,X)\
      0 & text{on} [X,infty)end{cases}$



      Iam trying to find $L^2(Omega_{X})$-norm of $(chi_{X}-1)(-Delta v)+(chi_{X}-1)(q-lambda)v$ ..



      I startred by the following:
      $|(chi_{X}-1)(-Delta v)+(chi_{X}-1)(q-lambda)v|^2_{L^2(Omega_{X})}=\ |(chi_{X}-1)(-Delta+q-lambda)v|^2_{L^2(Omega_{X})}=\ int_{Omega_{X}}(chi_{X}-1)^2((-Delta+q-lambda)v)^2$



      Now Iam not sure how I can finish the calculation, I know $(chi_{X}-1)$ is bounded on $Omega_{X}$ hence
      $ int_{Omega_{X}}(chi_{X}-1)^2((-Delta+q-lambda)v)^2= int_{Omega_{X}}|chi_{X}-1|^2((-Delta+q-lambda)v)^2leq C int_{Omega_{X}}((-Delta+q-lambda)v)^2$
      Are these steps correct?
      and how I can find the $L^2$ norm on $Omega_{X}$ for the laplacian term?






      functional-analysis operator-theory norm







      share|cite|improve this question













      Iam working on the following problem:



      Let $Omega subseteq mathbb{R}^n$ and $v$ be the solution of $(-Delta+q-lambda)v=F$ on $Omega$ such that $q$ is bounded and $v=exp(-ax)u$ where $uin H_{0}^{1}(Omega)cap H^{2}(Omega)$



      For a large $Xin mathbb{R}$, we define a cut-off function $chi_{X}$ on $Omega_{X}$:
      $chi_{X}=begin{cases} 1 & text{on }[0,X-1]\ 1leqchi_{X}leq0 & text{on} (X-1,X)\
      0 & text{on} [X,infty)end{cases}$



      Iam trying to find $L^2(Omega_{X})$-norm of $(chi_{X}-1)(-Delta v)+(chi_{X}-1)(q-lambda)v$ ..



      I startred by the following:
      $|(chi_{X}-1)(-Delta v)+(chi_{X}-1)(q-lambda)v|^2_{L^2(Omega_{X})}=\ |(chi_{X}-1)(-Delta+q-lambda)v|^2_{L^2(Omega_{X})}=\ int_{Omega_{X}}(chi_{X}-1)^2((-Delta+q-lambda)v)^2$



      Now Iam not sure how I can finish the calculation, I know $(chi_{X}-1)$ is bounded on $Omega_{X}$ hence
      $ int_{Omega_{X}}(chi_{X}-1)^2((-Delta+q-lambda)v)^2= int_{Omega_{X}}|chi_{X}-1|^2((-Delta+q-lambda)v)^2leq C int_{Omega_{X}}((-Delta+q-lambda)v)^2$
      Are these steps correct?
      and how I can find the $L^2$ norm on $Omega_{X}$ for the laplacian term?






      functional-analysis operator-theory norm




      functional-analysis operator-theory norm






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 22 at 16:28









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