Nonlinear finite elements/Bubnov Galerkin method

(Bubnov)-Galerkin Method for Problem 2

The Bubnov-Galerkin method is the most widely used weighted average method. This method is the basis of most finite element methods.

The finite-dimensional Galerkin form of the problem statement of our second order ODE is :

$\text{(20)}\begin{array}{cc}\text{Find}{u}_h(x)\in {\mathscr{H}}_0^n& \text{such that}\\ & \underset{0}{\overset{1}{\int }}(\frac{d{u}_h}{dx}\frac{d{w}_h}{dx}+u_h{w}_h-x{w}_h)dx=0\text{for all}{w}_h(x)\in {\mathscr{H}}_0^n\\ & u_h(0)=0,u_h(1)=0;w_h(0)=0,w_h(1)=0\end{array}$

Since the basis functions ($N_i$) are known and linearly independent, the approximate solution $u_h$ is completely determined once the constants ($a_i$) are known.

The Galerkin method provides a great way of constructing solutions. But the question is: how do we choose $N_i$ so that these functions are not only linearly independent but arbitrary? Since the solution is expressed as a sum of these functions, the accuracy of our result depends strongly on the choice of $N_i$.

Let the trial solution take the form,

$u_h(x)=\sum _{i=1}^n{a}_i{N}_i(x).$

According to the Bubnov-Galerkin approach, the weighting function also takes a similar form

$w_h(x)=\sum _{j=1}^n{b}_j{N}_j(x).$

Plug these values into the weak form to get

$\underset{0}{\overset{1}{\int }}[\left(\sum _{i=1}^n{a}_i\frac{d{N}_i}{dx}\right)\left(\sum _{j=1}^n{b}_j\frac{d{N}_j}{dx}\right)+\left(\sum _{i=1}^n{a}_i{N}_i\right)\left(\sum _{j=1}^n{b}_j{N}_j\right)-x\left(\sum _{j=1}^n{b}_j{N}_j\right)]dx=0$

or

$\underset{0}{\overset{1}{\int }}\left[\sum _{j=1}^n{b}_j(\frac{d{N}_j}{dx}\sum _{i=1}^n{a}_i\frac{d{N}_i}{dx}+N_j\sum _{i=1}^n{a}_i{N}_i-x{N}_j)\right]dx=0$

or

$\underset{0}{\overset{1}{\int }}\left[\sum _{j=1}^n{b}_j(\sum _{i=1}^n(a_i\frac{d{N}_j}{dx}\frac{d{N}_i}{dx}+a_i{N}_j{N}_i)-x{N}_j)\right]dx=0.$

Taking the sums and constants outside the integrals and rearranging, we get

$\sum _{j=1}^n{b}_j[\sum _{i=1}^n{a}_i\underset{0}{\overset{1}{\int }}(\frac{d{N}_i}{dx}\frac{d{N}_j}{dx}+N_i{N}_j)dx-\underset{0}{\overset{1}{\int }}x{N}_{j}dx]=0.$

Since the $b_j$s are arbitrary, the quantity inside the square brackets must be zero. That is

$\text{(21)}\sum _{i=1}^n{a}_i\underset{0}{\overset{1}{\int }}(\frac{d{N}_i}{dx}\frac{d{N}_j}{dx}+N_i{N}_j)dx-\underset{0}{\overset{1}{\int }}x{N}_{j}dx=0j=1\dots n.$

Let us define

$\text{(22)}{K}_{ji}:=\underset{0}{\overset{1}{\int }}(\frac{d{N}_i}{dx}\frac{d{N}_j}{dx}+N_i{N}_j)dx\text{and}{f}_j:=\underset{0}{\overset{1}{\int }}x{N}_{j}dx.$

Then we get a set of simultaneous linear equations

$\text{(23)}\sum _{i=1}^n{K}_{ji}{a}_i=f_j.$

In matrix form,

$𝐊𝐚=𝐟.$