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Green's identities for Laplacian-squared

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data
    a) Derive Green's identities in local and integral form for the partial differential operator ##\triangle^2##.
    b) Compute the adjoint operator ##(\triangle^2)^*##.

    2. Relevant information
    ##U\subset\mathbb{R}^n##, ##u:U\to\mathbb{R}##, differential operator ##Lu##.
    In the lecture notes it is given that:
    First Green's identity: ##\int_U\left[(Lu)v-u(L^*v)\right]dx=0## for any U and ##u,v\in C_0^\infty(U)##.
    Second Green's identity: ##\int_U\left[(Lu)v-u(L^*v)\right]dx=\int_{\partial U}QndS## for some field Q and ##u,v\in C^\infty(U)##.

    3. The attempt at a solution
    I haven't done PDEs in a long while, do I get it right that the order of actions to solve the exercise would be:
    1) Assume ##u,v\in C_0^\infty(U)##, by partial integration obtain ##\int_U(\triangle^2u)vdx= \int_Uu(\triangle^2v)dx## (am not fully sure if this holds, will need to verify) meaning via 1st Green's identity that ##(\triangle^2)^*=\triangle^2##.
    2) Plug this into 2nd Green's identity, assume ##u,v\in C^\infty(U)## and get the expression for Q.
     
  2. jcsd
  3. Feb 14, 2013 #2
    Using the following identity$$\int_U\frac{\partial^2 u}{\partial x_i^2}vdx=-\int_{\partial U}\frac{\partial u}{\partial x_i}vn_i dS+\int_{\partial U}u\frac{\partial v}{\partial x_i}n_idS-\int_Uu\frac{\partial^2v}{\partial x_i^2}dx$$
    (where ##n_i## is i-th coordinate of the orthogonal vector) I did some writing out
    $$\int_U(\bigtriangleup^2u)vdx=\int_U \bigtriangleup\left(\sum_{j=1}^n\frac{\partial^2u}{\partial x_j^2}\right)vdx=\int_U\left(\sum_{i=1}^n\sum_{j=1}^n\frac{\partial^4u}{\partial x_i^2\partial x_j^2}\right)vdx=\sum_{i=1}^n\sum_{j=1}^n\int_U \frac{\partial^4u}{\partial x_i^2\partial x_j^2}vdx=$$
    $$=\sum_{i=1}^n\sum_{j=1}^n\left(-\int_{\partial U}\frac{\partial^3u}{\partial x_i\partial x_j^2}vn_idS+\int_{\partial U}\frac{\partial^2u}{\partial x_j^2}\frac{\partial v}{\partial x_i}n_idS-\int_U\frac{\partial^2u}{\partial x_j^2}\frac{\partial^2v}{\partial x_i^2}dx\right)=$$
    $$=\sum_{i=1}^n\sum_{j=1}^n\left(\int_{\partial U}\left(\frac{\partial^2u}{\partial x_j^2}\frac{\partial v}{\partial x_i}-\frac{\partial^3u}{\partial x_i\partial x_j^2}v\right)n_idS+\int_{\partial U}\frac{\partial u}{\partial x_j}\frac{\partial^2v}{\partial x_i^2}n_jdS-\int_{\partial U}u\frac{\partial^3v}{\partial x_i^2\partial x_j}n_jdS+\int_Uu\frac{\partial^4v}{\partial x_i^2\partial x_j^2}dx\right)=$$
    $$=\sum_{i=1}^n\sum_{j=1}^n\int_{\partial U}\left(\left(\frac{\partial^2u}{\partial x_j^2}\frac{\partial v}{\partial x_i}-\frac{\partial^3u}{\partial x_i\partial x_j^2}v\right)n_i+\left(\frac{\partial u}{\partial x_j}\frac{\partial^2v}{\partial x_i^2}-u\frac{\partial^3v}{\partial x_i^2\partial x_j}\right)n_j\right)dS+\sum_{i=1}^n\sum_{j=1}^n \int_Uu\frac{\partial^4v}{\partial x_i^2\partial x_j^2}dx=$$
    $$=\sum_{i=1}^n\sum_{j=1}^n\int_{\partial U}\left(\left(\frac{\partial^2u}{\partial x_j^2}\frac{\partial v}{\partial x_i}-\frac{\partial^3u}{\partial x_i\partial x_j^2}v\right)n_i+\left(\frac{\partial u}{\partial x_j}\frac{\partial^2v}{\partial x_i^2}-u\frac{\partial^3v}{\partial x_i^2\partial x_j}\right)n_j\right)dS+\int_Uu(\bigtriangleup^2v)dx$$

    Could anyone suggest a way to simplify the double sum in the last expression?

    EDIT: nevermind, seem to have worked it out myself
     
    Last edited: Feb 14, 2013
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