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Derive the Green's identities in local and integral form for the bilaplacian. Thank you for any help.

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Could you please provide some background and motivation for you question? What have you done so far, where did you stuck, and what are the definitions of your main objects? –  Yuri Vyatkin Sep 10 '11 at 17:40
    
It looks like a homework question, Green's identities are indeed useful for solving some problems like Dirichlet problem etc. for getting these identities we usually apply Green or divergence (Gauss–Ostrogradsky) theorems. –  Edvin Goey Sep 10 '11 at 18:03

1 Answer 1

Notation:

  • $\Delta$ - Laplacian operator,
  • $\mbox{div}$ is a divergence (of a vector),
  • $\nabla$ is a gradient, $\vartheta$ is a unit (outward) normal vector defined positive away from the region $D$,
  • $\partial D$ is a boudary of $D$.

Let $u$ and $v$ be scalar functions. We use Green's identities for the Laplacian (link). Let us denote $\Delta u = U$, and $\Delta v=V$. Since $\Delta^2$ is a (formal) self-adjoint operator (it should be verified) we have that $L=L^*$ thus $$uLv-vL^*u=uLv-vLu=u \Delta^2 v - v \Delta^2 u= u \Delta V - v \Delta U=$$ $$=\underbrace{(u \Delta V - V \Delta u)}_{{\tiny{\mbox{Green's identity for } \Delta}}} + \underbrace{(U \Delta v - v \Delta U)}_{{\tiny{\mbox{Green's identity for } \Delta}}}+V\Delta u-U\Delta v=$$ $$=\mbox{div} (u \nabla V - V \nabla u)+\mbox{div} (U \nabla v - v \nabla U)+V\Delta u-U\Delta v$$ Since $\mbox{div}$ is a linear operator, $\mbox{div}{(\phi F)}= \nabla(\phi) \cdot F +\phi\mbox{div}(F)$, where $\phi$ a scalar valued function and $F$ is a vector field and $\Delta f= \nabla \cdot \nabla f$ we obtain $$\mbox{div} (u \nabla V) - \mbox{div} (V \nabla u)+\mbox{div} (U \nabla v) - \mbox{div}(v \nabla U)+V\Delta u-U\Delta v= $$ $$=\mbox{div} (u \nabla V)- \mbox{div}(v \nabla U)-\nabla V \cdot \nabla u - V\mbox{div}(\nabla u) + \nabla U \cdot \nabla v + U\mbox{div}(\nabla v)+V\Delta u-U\Delta v=$$ $$\mbox{div} (u \nabla V)- \mbox{div}(v \nabla U)-\nabla V \cdot \nabla u - V\nabla \cdot(\nabla u) + \nabla U \cdot \nabla v + U\nabla \cdot(\nabla v)+V\Delta u-U\Delta v =$$ $$\mbox{div} (u \nabla V)- \mbox{div}(v \nabla U)-\nabla V \cdot \nabla u - V\Delta u + \nabla U \cdot \nabla v + U \Delta v+V\Delta u-U\Delta v =$$ $$\mbox{div} (u \nabla V)- \mbox{div}(v \nabla U)-\nabla V \cdot \nabla u + \nabla U \cdot \nabla v =$$ $$\mbox{div} (u \nabla V)-\nabla u \cdot \nabla V- \mbox{div}(U \nabla v) + \nabla U \cdot \nabla v =$$ $$=\left[\mbox{div}(u \nabla V)-\nabla u\cdot \nabla V\right] -\left[\mbox{div}(v \nabla U)-\nabla v\cdot \nabla U\right]= $$ $$=\mbox{div}\left[u\nabla(\Delta v)-v\nabla(\Delta^2 u) \right]+\left[\nabla v \cdot \nabla(\Delta^2 u)- \nabla u \cdot \nabla(\Delta^2 v)\right]$$ Rewriting the 2nd bracket we get $$\nabla v \cdot \nabla(\Delta^2 u)=\mbox{div}\left[(\nabla v)(\Delta U)\right]-(\Delta v)(\Delta u)$$ $$\nabla u \cdot \nabla(\Delta^2 v)=\mbox{div}\left[(\nabla u)(\Delta v)\right]-(\Delta u)(\Delta v)$$ Hence, $$u \Delta^2 v - v \Delta^2 u=\mbox{div}\left[u\nabla(\Delta^2 v)- v\nabla(\Delta^2 u)+\nabla v(\Delta u)- \nabla u(\Delta v)\right]=\mbox{div}\vec{Q}$$ Finally, by integrating it over the volume $D$ and applying Gauss–Ostrogradsky theorem we obtain $$\int_D \left(u \Delta^2 v - v \Delta^2 u\right) dx=\int_D \mbox{div}\vec{Q}dx=\int_{\partial D}\vartheta\cdot \vec{Q}dS= $$ $$=\int_{\partial D}\left(u \frac{\partial (\Delta v)}{\partial \vartheta}-v\frac{\partial (\Delta u)}{\partial \vartheta}+\Delta u \frac{\partial v}{\partial \vartheta}-\Delta v \frac{\partial u}{\partial \vartheta} \right)dS.$$

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