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Green-St Venant strain tensor is defined by $E(u)={1\over 2}[\nabla u+(\nabla u)^T+(\nabla u)^T\nabla u]$, where $\nabla u$ is the displacement gradient.

Show that

$u\in H^1(\Omega), E(u)\in L^r(\Omega), r\ge1\Rightarrow u\in W^{1,2r}(\Omega)$.

Here $H^1, W^{1,k}$ are standard Sobolev spaces, $\Omega$ is bounded domain in $R^3$.

It's an exercise(1.16) in P.G.Ciarlet's book "Mathematical Elasticity, Vol. I : Three-Dimensional Elasticity", cited as 'due to Luc Tartar'.

I don't have a clue on it. Any idea and/or comment are very much appreciated.

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    $\begingroup$ Consider the trace of E. $\endgroup$
    – Michael Renardy
    Commented Apr 24, 2012 at 14:30
  • $\begingroup$ ok, let me try: consider the diagonal element, i.e., for each i=1,2,3, $E_{ii}={\partial u_i\over\partial x_i}+\sum_k |{\partial u_k\over\partial x_i }|^2$. 1. $r\le 2$ $u\in H^1\Rightarrow u_i\over\partial x_i \in L^2\Rightarrow u_i\over\partial x_i\in L^r$. $E_{ii}\in L^r\Rightarrow \sum_k |{\partial u_k\over\partial x_i }|^2\in L^r\Rightarrow \nabla u\in L^{2r}$. By the classical Poincare inequality here en.wikipedia.org/wiki/Poincar%C3%A9_inequality and note $u\in L^1$, we have $u\in L^{2r}$, done. $\endgroup$
    – pde_bk
    Commented Apr 24, 2012 at 19:07
  • $\begingroup$ 2. for r>2 $E_{ii}\in L^r\Rightarrow E_{ii}\in L^2\Rightarrow |\nabla u|^2\in L^2$ since ${\partial u_i\over\partial x_i}\in L^2$. Hence $\nabla u\in L^4$ and by Poincare inequality again, we have $u\in L^4$. Hence if $r<\leq 4$, the proof is completed. Otherwise start with $u\in L^4$ and continue. $\endgroup$
    – pde_bk
    Commented Apr 24, 2012 at 19:15

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Answer given by pde_bk in comments:

Consider the diagonal entries. For each $i=1,2,3$, $$ E_{ii}(u) = \partial_i u_i + \frac{1}{2}\sum_{k=1}^3 \left|\partial_i u_k\right|^{2} \quad\quad (\star). $$ Suppose first $1\leq r \leq 2$. As $u\in W^{1,2}$, $\partial_i u_i \in L^r$.

Thus if $E_{ii} \in L^{r}$ for each $i=1,2,3$ then $$\sum_{i=1}^3\sum_{k=1}^3 \left|\partial_i u_k\right|^2 = \left|\nabla u\right|^2 \in L^r, \quad\quad(\star\star)$$ and therefore $u\in W^{1,2r}$.

By induction suppose now $E\in L^r$ with $2^{n}<r\leq2^{n+1}$ (we just did $n=0$), and that the result holds for $r=2^n$. Since $E\in L^{r}\subset L^{2^{n}}$, we have from the induction hypothesis $\nabla u \in L^{2^{n+1}}$, and therefore $\nabla u \in L^{r}$. Using ($\star$) and ($\star\star$), the conclusion follows.

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