C++ for Scientists - Technische Universität Dresden

272 CHAPTER 14. NUMERICAL EXERCISES
14.2 The 2D Poisson equation
The 2D Poisson equation is
− ∂2u ∂x2 − ∂2u = f
∂y2 where u(x, y) is the solution and f the excitation and (x, y) ∈ [0, 1] × [0, 1]. We impose the
boundary conditions
u(0, y) = u(1, y) = y(x, 0) = y(x, 1) = 0 .
We discretize the x as xj = jh for j = 1, . . . , n and h = 1/n. Similarly, yj = jh. We use finite
differences for the second order derivatives. This produces the equation
1
h 2 (−u(xi−1, yj)−u(xi, yj−1)−u (xi+1, yj)−u(xi, yj+1)+4u(xi, yj)) = f(xi, yj) for i, j = 1, . . . , n .
This leads to the algebraic system of equations Au = f with n 2 rows and columns.
Recall the example exercise of §14.1. We do exactly the same exercise. Since the matrix is not
tridiagonal, we cannot use the LAPACK routine pttrf any longer. We use the LAPACK routine
sytrf for a full matrix instead. See the documentation on
boost-sandbox/libs/numeric/bindings/lapack/doc/index.html.
For a 2D problem the solution vector u can be represented as a matrix. The row index corresponds
to the variable x and the column index to a variable y.
In particular, you develop the functions inverse iteration, poisson 2d for the matrix vector product,
scaled poisson for the scaled matrix vector product, and richardson. Give for each templated
argument the conceptual conditions. make a plot of the eigenvector using Gnuplot’s splot (for
plotting surfaces).
14.3 The solution of a system of differential equations
In this exercise, we write a function for the computation of a time step of a system of differential
equations using Runge-Kutta methods.
14.3.1 Explicit time integration
Methods for the solution of the differential equation
˙u = f(u) u(0) = u0
operate time step by time step, i.e. the time is discretized and given the solution at time step
tj, we compute the solution at time step tj+1 = tj + h where h is small.