Consider a domain $\Omega$ as shown here:
Boundary:
Displacement: $(u,v)$ (in the $x$ and $y$ coord)
Elasticity tensors: $$ \begin{align*} C_{11} &= \begin{pmatrix} \frac{E (1-\nu)}{(1-2\nu)(1+\nu)} & 0 \\ 0 & \frac{E}{2(1+\nu)} \end{pmatrix}, &\qquad C_{12} &= \begin{pmatrix} 0 & \frac{E \nu}{(1-2\nu)(1+\nu)} \\ \frac{E}{2(1+\nu)} & 0 \end{pmatrix}, \\ C_{21} &= C_{12}^{\rm T}, &\qquad C_{22} &= \begin{pmatrix} \frac{E}{2(1+\nu)} & 0 \\ 0 & \frac{E (1-\nu)}{(1-2\nu)(1+\nu)} \end{pmatrix} \end{align*} $$ Here $E$ is called the modulus of elasticity and $\nu$ is called the Poisson ratio. (In fact, $C$ is a four-dimensional tensor, denoted by $C$, most likely, after Cauchy.)
Equations: $$ \begin{align*} -{\rm div} (C_{11} \nabla u + C_{12} \nabla v) &= 0 \qquad\text{in }\Omega \\ -{\rm div} (C_{21} \nabla u + C_{22} \nabla v) &= 0 \qquad\text{in }\Omega \\ u = v &= 0 \qquad\text{on }\Gamma_1 \\ C_{11} u_n + C_{12} v_n = C_{21} u_n + C_{22} v_n &= 0 \qquad\text{on }\Gamma_2 \\ C_{11} u_n + C_{12} v_n &= 0 \qquad\text{on }\Gamma_3 \\ C_{21} u_n + C_{22} v_n &= \frac{P}{H W} \qquad\text{on }\Gamma_3, \end{align*} $$ where $u_n$ and $v_n$ are normal derivatives of the solution.
Variational formulation:
A good way to go about this problem is to think that your space consists of vector-valued functions $(u,v)\in\mathbb R^2$. Thus, at each node you have two basis functions: such that $(u,v) = (1,0)$ for that node, and $(u,v) = (0,1)$ for that node.
If $u$ is the exact solution and $u_h$ is the approximate solution then one can prove that $$ \|\nabla u_h - \nabla u\|_{L^2} \leq C h \|\nabla^2 u\|_{L^2}, $$ where $h$ is the maximal size of the triangle and $\nabla^2 u$ is the Hessian matrix.
Comes from a truly beautiful argument:
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from IPython.core.display import HTML
def css_styling():
styles = open("./styles/alex.css", "r").read()
return HTML(styles)
css_styling()
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