C++ for Scientists - Technische Universität Dresden

264 CHAPTER 14. NUMERICAL EXERCISES
This is called a boundary value problem.
The goal is to compute the solution u for all x ∈ [0, 1]. Since this is not possible numerically, we
only compute u for a discrete number of x’s, which we call discretization points. We discretize x
as xj = jh for j = 0, . . . , n+1 and h = 1/(n+1). This is called an equidistant distribution. The
smaller h, the closer we are to the continuous problem, i.e. we have more points in [0, 1], but, as
we shall see, the problem becomes more expensive to solve. One method for solving boundary
value problems is to replace the derivative by finite differences. We use finite differences for the
second order derivatives:
Filling this in (14.1), we obtain
d 2 u
dx 2 (xj) ≈ 1
h 2 (−2u(xj) + u(xj−1) + u(xj+1)) .
1
h 2 (−u(xi−1) − u (xi+1) + 2u(xi)) = f(xi) for j = 1, . . . , n . (14.2)
Note that u(x0) = u(xn+1) = 0. Now define the vectors
u = [u(x1), . . . , u(xn)] T
and f = [f(x1), . . . , f(xn)] T .
Putting together (14.2) for j = 1, . . . , n leads to the algebraic system of equations Au = f with
n rows and columns where ⎡
2
⎢ −1
A = ⎢
⎣
−1
2
. ..
−1
. .. . ..
⎤
⎥
⎦
−1 2
.
Note that A is a symmetric tridiagonal matrix. We can show that it is positive definite.
In the algorithms, we need operations on this matrix. We will use two different types of
operations. The first one is the matrix-vector product y = Ax. We write a function for this
with a template argument for the vectors since we do not know beforehand what the type of
the vectors will be.
#ifndef athens poisson 1d hpp
#define athens poisson 1d hpp
#include
#include
namespace athens {
template
void poisson 1d( X const& x, Y& y ) {
assert( glas::size(x)==glas::size(y) ) ;
assert( glas::size(x) > 1 ) ;
y(0) = 2.0∗x(0) − x(1) ;
for ( int i=1; i