Beyond Finite Differences 2

Let’s consider the Galerkin method, which provides another approach to determining the coefficients $c_1, ..., c_n$ in the approximation $\displaystyle \sum\limits_{j=1}^{n}c_{j}\phi_{j}(x),$ where $\phi_{i}(0)=\phi_{i}(1)=0$ for $i= 1, ..., n.$

We define the operator

$\displaystyle R(\square) =\square'' + q(x)\square - f(x).$

Given that $y(x)$ is the exact solution of the differential equation

$\displaystyle {\square}''(x) + q(x){\square}(x)=f(x),$

we have

$\displaystyle R(y(x)) = 0.$

This implies that

$\displaystyle \int\limits_{0}^{1}R(y(x))\phi_i(x)\;dx = 0, \quad i = 1, ..., n.$

However, since

$\displaystyle R\left(\sum\limits_{j=1}^{n}c_j \phi_j(x)\right) \ne 0$

in general, approximating $y(x)$ by $\displaystyle \sum\limits_{j=1}^{n}c_j \phi_j(x)$ requires finding $c_1, ..., c_n$ such that

$\displaystyle \int\limits_{0}^{1}R\left(\sum\limits_{j=1}^{n}c_j \phi_j(x)\right)\phi_i(x)\;dx = 0, \quad i=1,..., n.$

This condition is equivalent to

$\displaystyle \int\limits_{0}^{1} \left(\sum\limits_{j=1}^{n}c_j\phi_{j}''(x) + q(x)\sum\limits_{j=1}^{n}c_j\phi_{j}(x)-f(x)\right)\phi_{i}(x)\;dx=0.$

Expanding the expression gives

$\displaystyle \sum\limits_{j=1}^{n}c_j\int\limits_{0}^{1}(\phi_{j}''(x) + q(x)\phi_{j}(x))\phi_{i}(x)\;dx - \int\limits_{0}^{1}f(x)\phi_i(x)\;dx = 0$

$\displaystyle \implies \sum\limits_{j=1}^{n}c_j\int\limits_{0}^{1}\phi_{j}''(x)\phi_{i}(x) + q(x)\phi_{j}(x)\phi_{i}(x)\;dx = \int\limits_{0}^{1}f(x)\phi_i(x)\;dx$

$\displaystyle \implies \sum\limits_{j=1}^{n}c_j\left(\underline{\int\limits_{0}^{1}\phi_{j}''(x)\phi_{i}(x)\;dx} + \int\limits_{0}^{1}q(x)\phi_{j}(x)\phi_{i}(x)\;dx\right) = \int\limits_{0}^{1}f(x)\phi_i(x)\;dx$

$\displaystyle \implies \sum\limits_{j=1}^{n}c_j\left(\underline{\phi_{j}'(x)\phi_i(x)\bigg|^{1}_{0} - \int\limits_{0}^{1}\phi_i'(x)\phi_{j}'(x)\;dx} + \int\limits_{0}^{1}q(x)\phi_{i}(x)\phi_{j}(x)\;dx\right) = \int\limits_{0}^{1}f(x)\phi_i(x)\;dx.$

Since $\phi_i(0) = \phi_i(1) = 0$ , we are left with:

$\displaystyle \sum\limits_{j=1}^{n}c_j\left(\displaystyle -\int\limits_{0}^{1} \phi_{i}'(x)\phi_{j}'(x)\;dx + \int\limits_{0}^{1}q(x)\phi_i(x)\phi_{j}(x)\; dx \right) = \int\limits_{0}^{1}f(x)\phi_i(x)\; dx.\quad(*)$

For $i, j=1, ..., n$ , we define the coefficients

$\displaystyle a_{ij} = -\int\limits_{0}^{1} \phi_{i}'(x)\phi_{j}'(x)\;dx + \int\limits_{0}^{1}q(x)\phi_i(x)\phi_{j}(x)\;dx$

and

$\displaystyle f_i = \int\limits_{0}^{1}f(x)\phi_i(x)\; dx.$

With these definitions, (*) can be expressed as a system of linear equations:

$\displaystyle \bold{Ac} = \bold{f}\quad\quad\quad(**)$

where $\bold{A}$ is a $n \times n$ matrix with entries $a_{ij}, \bold{c} = (c_1, ..., c_n)^T,$ and $\bold{f} = (f_1, ..., f_n)^T$ . Solving (**) yields the coefficients $\bold{c}$ .

Fig. 1 $y'' + y = x^3, y(0)=y(1)=0$ solved by the Galerkin method with $\phi_j(x) = x^j(1-x)$

Fig. 2 $y'' + y = x^3, y(0)=y(1)=0$ solved by the Galerkin method with $\phi_j(x) = \sin(j\pi x)$

Exercise-1 Verify that $\displaystyle y''(x)+q(x)y(x)=f(x) \Longleftrightarrow R(y(x)) = 0 \Rightarrow \int\limits_{0}^{1}R(y(x))\phi_{i}(x)\;dx = 0.$

Exercise-2 Solve $y''+x y = x^3, y(0)=y(1)=0$ by the Galerkin method. (Hint: Fig. 1)

Beyond Finite Differences 2

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