[Math] How does Calculus of Variational work in Finite Element Method

calculus-of-variationsfinite element method

I'm learning Finite Element Method. And it is said in a lot of books that Calculus of Variational is the basis of Finite Element Method. But as far as I know, Calculus of Variational is to find a function $f$ which will make the functional $$J=\int_\Omega F(x,y,y')dx$$reach its extremum which means that $$\delta J=0$$
The Finite Element Method is, however, considered as discrete the domain into some tiangles and rewrite the particial difference equation (Take the following equations as example)$$\left\{\begin{aligned} -\Delta u+cu=f\quad ,in \quad\Omega \\ u=g_0\quad ,on\quad\Gamma_D\\\partial_nu=g_1\quad,on\quad\Gamma_N\end{aligned}
\right.$$into its wake/varitional form$$\left\{\begin{aligned} \int_\Omega\nabla u \cdot \nabla v+c\int_\Omega u v=\int_\Omega f v + \int_{\Gamma_N} g_1 v \\ u=g_0\quad ,on\quad\Gamma_D\end{aligned}\right.$$in which u is trial function while v is test function. We donate $u$ in its discrete version $u_h$ with the so called tent function $$u_h=\sum_{j=1}^{N}u_h(p_j)\phi_j$$ in which $p_j$ is the vertex nodes and $\phi_j$ is a "tent function" which means$$\phi_i(p_j)=\delta_{ij}=\left\{\begin{aligned} 1\quad i=j\\0\quad i\not=j\end {aligned}\right.$$Substitute v with $\phi_j$ we get $$\sum_{j \in Ind }(\int_\Omega\nabla \phi_j \cdot \nabla \phi_i+c\int_\Omega \phi_j \phi_i)u_j=\int_\Omega f \phi_i + \int_{\Gamma_N} g_1 \phi_i-\sum_{j \in Dir }(\int_\Omega\nabla \phi_j \cdot \nabla \phi_i+c\int_\Omega \phi_j \phi_i)g_0(p_j)$$
in which Ind denote the independent node set while Dir denote the Dirichlet node set. From the equations above, I cannot find any link between the Calculus of Variational and Finite Element Method. Could anyone please explain how does Calculus of Variational work in Finite Element Method?

Best Answer

The answer to your question comes from your weak/variational form. We can phrase a calculus of variations problem with a specific functional such that $\delta J = 0$ implies satisfying the weak form.

This Wikiversity article explores this further with examples. https://en.wikiversity.org/wiki/Introduction_to_finite_elements/Calculus_of_variations

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