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?
[Math] How does Calculus of Variational work in Finite Element Method
calculus-of-variationsfinite 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