# Deriving the weak form for linear elasticity equation

Consider the BVP from linear elasticity: \begin{align*} -\mu \Delta\textbf{u} - (\lambda + \mu) \nabla(\nabla \cdot \textbf{u})) &= \textbf{f}, \text{ in } \Omega \subset\mathbb{R}^2, \\ \textbf{u} &= \textbf{g}_D, \text{ on } \partial \Omega. \end{align*} First, we multiply by a test function and integrate both sides, \begin{align*} \int_{\Omega}( -\mu \textbf{v} \Delta \textbf{u} - (\lambda + \mu) \textbf{v} \nabla (\nabla\cdot \textbf{u})))dX = \int_{\Omega} \textbf{v}\textbf{f} dX. \end{align*} Now I am not sure what the best way to proceed is. I think I should break everything up into components, but I am not confident. I think I will end up using the divergence theorem (twice?). Would anyone be able to help me work out the details? I think the $\Delta \textbf{u}$ term should be simple, but I have no clue what identities to use for $\nabla (\nabla \cdot \textbf{u})$. I appreciate any help.

## 1 Answer

Edit: First equation it should be $$\nabla b$$ and not $$\nabla v$$. And it not possible to edit just one letter. So the edit.

Let us start with some background in tensor calculus.

For arbitrary domain $$D$$, tensor $$A_{ij}$$ and vector $$b_i$$ the following Green's theorem holds: $$\int_D \mathrm{div} A \cdot b \,\mathrm{d}x = \int_{\partial D} (An)\cdot b \,\mathrm{d}s - \int_D A : \nabla b\,\mathrm{d}x,$$ where $$n$$ is the normal vector to the boundary $$\partial D$$ and $$:$$ denotes the double dot product, i.e. for tensors $$C_{ij}$$ and $$D_{ij}$$ we have $$C : D = \sum_{i,j=1}^3 C_{ij} D_{ij}.$$ The divergence of a tensor field $$A_{ij}$$ is defined as the vector with components $$(\mathrm{div} A)_i = \sum_{j=1}^3 \frac{\partial A_{ij}}{\partial x_j}$$ The gradient of a vector field $$b_i$$ is the tensor with components $$(\nabla b)_{ij} = \frac{\partial b_i}{\partial x_j}.$$

Back to linear elasticity. People usually write the governing equations in the following way: $$-\mathrm{div}\,\sigma(u) = f,$$ where the stress tensor is $$\sigma(u) = 2 \mu \,\epsilon(u) + \lambda \,\mathrm{tr}\,\epsilon(u) I,$$ and the strain tensor is $$\epsilon(u) = \frac12(\nabla u + \nabla u^T).$$ Then you multiply the first equation with a vectorial test function and integrate over the domain $$\Omega$$ to get $$-\int_\Omega \mathrm{div}\, \sigma(u) \cdot v \,\mathrm{d}x = \int_\Omega f \cdot v \,\mathrm{d}x.$$ Next we use the Green's theorem to get $$-\int_{\partial \Omega} \sigma(u)n \cdot v \,\mathrm{d}s + \int_\Omega \sigma(u) : \nabla v \,\mathrm{d}x = \int_\Omega f \cdot v \,\mathrm{d}x.$$ Since $$\sigma$$ happens to be symmetric we can equivalently write $$-\int_{\partial \Omega} \sigma(u)n \cdot v \,\mathrm{d}s + \int_\Omega \sigma(u) : (\frac12(\nabla v + \nabla v^T)) \,\mathrm{d}x = \int_\Omega f \cdot v \,\mathrm{d}x,$$ or $$-\int_{\partial \Omega} \sigma(u)n \cdot v \,\mathrm{d}s + \int_\Omega \sigma(u) : \epsilon(v) \,\mathrm{d}x = \int_\Omega f \cdot v \,\mathrm{d}x,$$ which is the weak formulation of linear elasticity.

• It is defined in the second equation.
– knl
Dec 17, 2017 at 15:27