PDE- Galerkin method and example

 
The Galerkin method is a numerical technique that transforms boundary value problems, particularly partial differential equations (PDEs), into a system of linear algebraic equations by projecting the problem onto a finite-dimensional subspace.
Steps of the Galerkin Method


 Problem Setup: 

Start with the weak form of the differential equation integrated against a test function.

Choice of Basis Functions: 

Select basis functions $\{\phi_i\}_{i=1}^n$ that satisfy the boundary conditions.
 Approximation of the Solution: 

Assume $u(x) \approx u_n(x) = \sum_{i=1}^n c_i \phi_i(x)$.
 Galerkin Projection: 

Ensure the residual is orthogonal to the span of the basis functions:
  $$\int \text{Residual} \cdot \phi_j \, dx = 0 \quad \text{for all } j.$$

Example: One-Dimensional Poisson Equation

Consider the problem
$$-u''(x) = f(x), \quad u(0) = u(1) = 0.$$
on the interval $[0,1]$.
Step 1: Weak Form

Multiply by a test function $v(x)$ and integrate by parts to get:
$$\int_0^1 u'(x) v'(x) \, dx = \int_0^1 f(x) v(x) \, dx.$$
Step 2: Discretization

Choose linear basis functions and assume:
$$u(x) \approx u_n(x) = \sum_{i=1}^n c_i \phi_i(x).$$
Step 3: Galerkin Projection

Insert the approximation into the weak form using basis functions as test functions:
$$\sum_{i=1}^n c_i \int_0^1 \phi_i'(x) \phi_j'(x) \, dx = \int_0^1 f(x) \phi_j(x) \, dx \quad \text{for all } j.$$
Step 4: Matrix Formulation

Define the stiffness matrix $\mathbf{A}$ and load vector $\mathbf{b}$ as follows:
$$A_{ij} = \int_0^1 \phi_i'(x) \phi_j'(x) \, dx, \quad b_j = \int_0^1 f(x) \phi_j(x) \, dx.$$
This leads to the linear system:
$$\mathbf{Ac} = \mathbf{b},$$
where $\mathbf{c}$ is the vector of coefficients.
\subsection*{Example: Very simple and concrete}
Take a differential equation
$$\frac{d^2y}{dx^2} + x+ y = 0 , 0 \le x \le 1$$
with boundary condition $y(0) = y(1) = 0$

Example:

Step 1
Take a trail solution $y(x) = a_0+a_1x+a_2 x^2$. Always take the number of constants one greater than the highest differential power. Here we have second order differential. So, we take $2+1=3$ constants $a_0,a_1,a_2$.
Apply boundary conditions $x=0,y=0$. This leads to $a_0 =0$. Then apply boundary condition $x = 1, y = 0$ leads to $a_1 = -a_2$. Then the trail function becomes $y(x) = a_2(x^2-x)$. This is one parameter solution - namely $a_2$.

Step 2
Compute Weighting function
$$W(x) = \frac{\partial{y}}{a_2} = x^2 - x$$

Step 3
Compute domain residuals. Simply substitute the trail function into original differential equation.
$$\begin{align*}   R_d &= \frac{d^2y}{dx^2} + x + y \\   R_d(x) &= \frac{d a_2(x^2-x)}{d x^2} + x + a_2(x^2-x) \end{align*}$$
 Step 4
Minimization of domain residual:
$$\int_0^1 W(x) R_d(x) dx = 0$$
Compute minimization. You will get $a_2 = -\frac{5}{98}$.
Then the solution is $y(x) =  -\frac{5}{98}(x - x^2)$.
Other formulations
For the differential equation $-u{''} = f$, we can write this as first order by setting
$$\sigma = -u' , \sigma' = f$$
The pair $\sigma,f$ can be characterized variationally as unique critical point of the functional
$$I(\sigma,u) = \int_{-1}^1 (\frac{1}{2}\sigma^2 - u \sigma') dx +  \int_{-1}^1 fu dx$$
over $H^1(-1,1) \times L^2(-1,1)$
Note if we take the original $J(x)$ the Energy functional and substitute, we get above.
$$\begin{align*}   J(u) &= \frac{1}{2}\int_{-1}^1 |u'(x)|^2 dx - \int_{-1}^1f(x)u(x)dx \\        &=  \frac{1}{2}\int_{-1}^1\sigma^2 -  \int_{-1}^1 f(x) u(x) dx \\        &= \frac{1}{2}\int_{-1}^1\sigma^2 + f(x)\int_{-1}^1u(x)dx - \int_{-1}^1\int_{-1}^1u(x) dx f'(x) dx \\        &= \frac{1}{2}\int_{-1}^1\sigma^2 + 0 \\        &= \frac{1}{2}\int_{-1}^1\sigma^2 + \int_{-1}^1f(x)u(x)dx  - \int_{-1}^1f(x)u(x)dx \\        &= \int_{-1}^1\frac{1}{2}\sigma^2 - f \sigma' + \int_{-1}^1f(x)u(x)dx \end{align*}$$