# variational formulation of linear elasticity

First I'm not 100% sure I'm on the good stack for asking my question. I would like to get a bilinear form for linear elasticity that separate a rotational part from a pure divergence part, so starting from the Navier equation $$\mu \nabla^2 \mathbf u +(\mu+\lambda)\nabla(\nabla \cdot \mathbf u) + \mathbf f =0$$

Then I use the vector Laplacian identity $\nabla(\nabla \cdot \mathbf A)= \nabla^2 \mathbf A +\nabla \times \nabla \times\mathbf A$ to write

$$(\lambda +2\mu) \nabla^2 \mathbf u +(\mu+\lambda)\nabla \times (\nabla \times\mathbf u) + \mathbf f =0$$

So multiplying by a test function $\mathbf v$ I get

$$\int_\Omega \bigg((\lambda +2\mu) \nabla^2 \mathbf u \cdot \mathbf v +(\mu+\lambda)\nabla \times (\nabla \times\mathbf u)\cdot \mathbf v + \mathbf f \cdot \mathbf v \bigg) d\Omega=0$$

and using Green formula, $$\int_\Omega \bigg((\lambda +2\mu) (\nabla \mathbf u :\nabla \mathbf v) +(\mu+\lambda)(\nabla \times \mathbf v)\cdot(\nabla \times\mathbf u) + \mathbf f \cdot \mathbf v \bigg) d\Omega=0$$ I'm skeptical about this for, since when I compare the bilinear form $$a(\mathbf u,\mathbf v) := \int_\Omega \bigg((\lambda +2\mu) (\nabla \mathbf u :\nabla \mathbf v) +(\mu+\lambda)(\nabla \times \mathbf v)\cdot(\nabla \times\mathbf u) \bigg) d\Omega$$ With the classical one
$$a(\mathbf u,\mathbf v) := \int_\Omega \sigma(\mathbf u):\varepsilon(\mathbf v) d\Omega$$

I got few different terms. For instance, in 2D the "classical" form (noted $a_1$) gives \begin{eqnarray} a_1(\mathbf u,\mathbf v) &=& \int_\Omega \sigma(\mathbf u):\varepsilon(\mathbf v) d\Omega \\&=& \bigg(2\mu \frac{\partial u_x}{\partial x}+\lambda(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}) \bigg)\frac{\partial v_x}{\partial x} +\mu\bigg(\frac{\partial u_y}{\partial x}+\frac{\partial u_x}{\partial y}\bigg)\bigg(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\bigg) +\bigg(2\mu \frac{\partial u_y}{\partial y}+\lambda(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y})\bigg) \frac{\partial v_y}{\partial y} \\&=&(\lambda+2\mu) \bigg(\frac{\partial u_x}{\partial x} \frac{\partial v_x}{\partial x} +\frac{\partial u_y}{\partial y}\frac{\partial v_y}{\partial y}\bigg) + \mu\bigg(\frac{\partial u_y}{\partial x}+\frac{\partial u_x}{\partial y}\bigg)\bigg(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\bigg) +\lambda\bigg(\frac{\partial u_y}{\partial y}\frac{\partial v_x}{\partial x} +\frac{\partial u_x}{\partial x}\frac{\partial v_y}{\partial y}\bigg)\end{eqnarray}

Whereas, the "rotational" form gives \begin{eqnarray} a_2(\mathbf u,\mathbf v) &=& \int_\Omega \bigg((\lambda +2\mu) (\nabla \mathbf u :\nabla \mathbf v) +(\mu+\lambda)(\nabla \times \mathbf v)\cdot(\nabla \times\mathbf u) \bigg) d\Omega\\ &=& (2\mu+\lambda)\bigg(\frac{\partial u_x}{\partial x}\frac{\partial v_x}{\partial x} + \frac{\partial u_x}{\partial y}\frac{\partial v_x}{\partial y}+\frac{\partial u_y}{\partial x}\frac{\partial v_y}{\partial x} + \frac{\partial u_y}{\partial y}\frac{\partial v_y}{\partial y}\bigg) + (\lambda+\mu)\bigg(\frac{\partial u_y}{\partial x} -\frac{\partial u_x}{\partial y}\bigg) \bigg(\frac{\partial v_x}{\partial y} -\frac{\partial v_y}{\partial x}\bigg) \end{eqnarray} The $(\lambda +\mu)$ terms may be developped and some terms may be factorized with some other terms of $(2\mu+\lambda)$ \begin{eqnarray} a_2(\mathbf u,\mathbf v) &=& (2\mu+\lambda)\bigg(\frac{\partial u_x}{\partial x}\frac{\partial v_x}{\partial x} + \frac{\partial u_y}{\partial y}\frac{\partial v_y}{\partial y}\bigg) +\mu\bigg(\frac{\partial u_x}{\partial y}\frac{\partial v_x}{\partial y} + \frac{\partial u_y}{\partial x}\frac{\partial v_y}{\partial x}\bigg) + (\lambda+\mu)\bigg(\frac{\partial u_y}{\partial x}\frac{\partial v_x}{\partial y} +\frac{\partial u_x}{\partial y}\frac{\partial v_y}{\partial x}\bigg) \end{eqnarray} The $\mu$ terms in two and second expressions may be factorized \begin{eqnarray} a_2(\mathbf u,\mathbf v) &=& (2\mu+\lambda)\bigg(\frac{\partial u_x}{\partial x}\frac{\partial v_x}{\partial x} + \frac{\partial u_y}{\partial y}\frac{\partial v_y}{\partial y}\bigg) +\mu\bigg(\frac{\partial u_y}{\partial x}+\frac{\partial u_x}{\partial y}\bigg)\bigg(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\bigg) + \lambda\bigg(\frac{\partial u_y}{\partial x}\frac{\partial v_x}{\partial y} +\frac{\partial u_x}{\partial y}\frac{\partial v_y}{\partial x}\bigg) \end{eqnarray}

This equation is almost the same as $a_1$ except for the $\lambda$ term. Could someone correct me ?

PS: the rotational being defined by \begin{eqnarray} \nabla \times \mathbf u = \left| \begin{array}{l} \partial_x\\ \partial_y \\ \partial_z \end{array} \right. \times \left| \begin{array}{l} u_x\\ u_y \\ u_z \end{array} \right. = \left| \begin{array}{l} \partial_y u_z -\partial_z u_y\\ \partial_z u_x -\partial_x u_z \\ \partial_x u_y -\partial_y u_x \end{array} \right. \end{eqnarray}

We have \begin{eqnarray} (\nabla \times \nabla \times \mathbf u ) \cdot \mathbf v = \left| \begin{array}{l} \partial_y (\partial_x u_y -\partial_y u_x) - \partial_z(\partial_z u_x -\partial_x u_z)\\ \partial_z (\partial_y u_z -\partial_z u_y) - \partial_x(\partial_x u_y -\partial_y u_x)\\ \partial_x (\partial_z u_x -\partial_x u_z) - \partial_y (\partial_y u_z -\partial_z u_y) \end{array} \right. \cdot \left| \begin{array}{l} v_x \\ v_y\\ v_z \end{array} \right. \end{eqnarray}

And using integration by part (but I suspect that the flaw stands here), \begin{eqnarray} \int_\Omega (\nabla \times \nabla \times \mathbf u ) \cdot \mathbf v &=& -\int_\Omega \partial_y v_x(\partial_x u_y -\partial_y u_x) - \partial_z v_x (\partial_z u_x -\partial_x u_z)\\ &&+\partial_z v_y(\partial_y u_z -\partial_z u_y) - \partial_x v_y (\partial_x u_y -\partial_y u_x)\\ && +\partial_x v_z(\partial_z u_x -\partial_x u_z) - \partial_y v_z(\partial_y u_z -\partial_z u_y)\\ &=& - \int_\Omega (\partial_y v_x - \partial_x v_y )(\partial_x u_y -\partial_y u_x) + (\partial_z v_y-\partial_y v_z) (\partial_y u_z -\partial_z u_y) +(\partial_x v_z- \partial_z v_x )(\partial_z u_x -\partial_x u_z)\\ &=& \int_\Omega (\nabla \times \mathbf v )\cdot (\nabla \times \mathbf u) \end{eqnarray}

Thank you very much.

• It would be useful to explain what differences you get. What did you try? Dec 15, 2012 at 19:23
• Wolfgang, I have edited the question.
– Tom
Dec 17, 2012 at 10:17
• Although divergence and gradient are defined independent of dimension, this is not the case for the curl. The vector calculus you apply is defined for 3D vector fields. (There is a cross product on 2D vectors, but that is not used here.) Dec 17, 2012 at 13:42
• Also, the last term of your first $a_2$ formula is not symmetric with respect to $u$ and $v$, although the bilinear form obviously is. Dec 17, 2012 at 13:46
• Christian, I did the maths in 3D but with the addtional hypothesis that vector field is zero on the z axis. I know that the form I obtain is not symmetric with respect to derivatives, and this is why is is obviously wrong. But I don't know why.
– Tom
Dec 17, 2012 at 14:26

Your derivation is correct up to minor imprecisions: \begin{multline} \int_\Omega \bigg((\lambda +2\mu) \nabla^2 \mathbf u \cdot \mathbf v +(\mu+\lambda)\nabla \times (\nabla \times\mathbf u)\cdot \mathbf v + \mathbf f \cdot \mathbf v \bigg) d\Omega = \\ = -\int_\Omega \bigg((\lambda +2\mu) \nabla \mathbf u :\nabla \mathbf v +(\mu+\lambda)(\nabla \times \mathbf v)\cdot(\nabla \times\mathbf u) \bigg)d\Omega + \\ + \int_{\partial \Omega} \bigg((\lambda +2\mu) (\nabla \mathbf u \,\mathbf n) \cdot \mathbf v + (\mu+\lambda) \: \mathbf n \times (\nabla \times \mathbf u) \cdot \mathbf v \bigg) d \partial \Omega + \int_\Omega \big( \mathbf f \cdot \mathbf v \big) d\Omega \end{multline} where $\partial \Omega$ is the boundary and $\mathbf n$ is the unit outward normal.
By inspecting the boundary term it is clear that this is not linear elasticity, unless you have Dirichlet b.c. on the whole boundary $\partial \Omega$. The point here is that if you want to recover the proper Neumann b.c. you are not free to integrate by parts at your will: you have to transfer to the test function the correct derivatives.