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$L^2$-norm of the solution of a 2-order elliptic nonlinear equation



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Unique weak solution to the biharmonic equationBounding the solution of a wave equation in 3 dimensionsShow an equation only has harmonic solutionSource Solution to heat equation Sense of distributionIs the parabolic heat equation with pure neumann conditions well posed?Asymmetric solution of heat equation with heat sourceQuestions on Nonlinear Elliptic Theory by SchauderCan superposition be used in the PDE $ nabla cdot [sigma(x,y,z)nablaphi]=-fracpartial rhopartial t delta(x_s)delta(y_s)delta(z_s) $Reference about eigenvalue of Kirchhoff type operatorUniqueness of PDE Solution using Energy Method










0












$begingroup$


When $1<p<2^*$, $E>0$ is a constant, assuming the positive solution of equation below is $U_E$
$$
-Delta u + Eu - |u|^p-1u =0, ~~~~xin mathbb R^n (nge1)
$$

as I read in Evans' book, the term $Eu$ can be treat as heat source, and the solution is the heat distribution, so I think $E$ is bigger, the $||U_E||_L^2$ is bigger. But I don't know how to prove it, any way to start it. Obviously,
$$
fracddE||U_E||_L^2^2 =fracddEleft[-frac1Eint |nabla U_E|^2+frac1Eint |U_E|^p+1right]
$$

but the right part is still hard to know whether it is positive, how should I do ?










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    When $1<p<2^*$, $E>0$ is a constant, assuming the positive solution of equation below is $U_E$
    $$
    -Delta u + Eu - |u|^p-1u =0, ~~~~xin mathbb R^n (nge1)
    $$

    as I read in Evans' book, the term $Eu$ can be treat as heat source, and the solution is the heat distribution, so I think $E$ is bigger, the $||U_E||_L^2$ is bigger. But I don't know how to prove it, any way to start it. Obviously,
    $$
    fracddE||U_E||_L^2^2 =fracddEleft[-frac1Eint |nabla U_E|^2+frac1Eint |U_E|^p+1right]
    $$

    but the right part is still hard to know whether it is positive, how should I do ?










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      When $1<p<2^*$, $E>0$ is a constant, assuming the positive solution of equation below is $U_E$
      $$
      -Delta u + Eu - |u|^p-1u =0, ~~~~xin mathbb R^n (nge1)
      $$

      as I read in Evans' book, the term $Eu$ can be treat as heat source, and the solution is the heat distribution, so I think $E$ is bigger, the $||U_E||_L^2$ is bigger. But I don't know how to prove it, any way to start it. Obviously,
      $$
      fracddE||U_E||_L^2^2 =fracddEleft[-frac1Eint |nabla U_E|^2+frac1Eint |U_E|^p+1right]
      $$

      but the right part is still hard to know whether it is positive, how should I do ?










      share|cite|improve this question









      $endgroup$




      When $1<p<2^*$, $E>0$ is a constant, assuming the positive solution of equation below is $U_E$
      $$
      -Delta u + Eu - |u|^p-1u =0, ~~~~xin mathbb R^n (nge1)
      $$

      as I read in Evans' book, the term $Eu$ can be treat as heat source, and the solution is the heat distribution, so I think $E$ is bigger, the $||U_E||_L^2$ is bigger. But I don't know how to prove it, any way to start it. Obviously,
      $$
      fracddE||U_E||_L^2^2 =fracddEleft[-frac1Eint |nabla U_E|^2+frac1Eint |U_E|^p+1right]
      $$

      but the right part is still hard to know whether it is positive, how should I do ?







      pde elliptic-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 2 at 8:16









      lanse7ptylanse7pty

      1,8711823




      1,8711823




















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