Multigrid solver

Content

For the sake of simplicity, we will only consider one dimensional boundary problems of the form

\[\begin{cases}-u''(x) = f(x), & x \in (0,1), \\u(x) = g(x), & x \in \{0,1\}. \\\end{cases}\]

This called a one dimensional Poisson equation with Dirichlet boundary conditions. For the discretization we will use the finite difference method (FDM) which leads to solving

\[A_h \widehat{u}_h = \widehat{q}_h,\quad h = \frac{1}{N+1}\]

with

\[A_h =\frac{1}{h^2}\left(\begin{array}{ccccc} 2 & -1 & & & \\ -1 & 2 & -1 & & \\ & \ddots& \ddots& \ddots& \\ & & -1 & 2 & -1 \\ & & & -1 & 2 \\ \end{array}\right)\in \mathbb{R}^{N \times N}\]

and

\[\widehat{q}_h =\left(\begin{array}{l} f_1 + \frac{1}{h^2} g(0) \\ f_2 \\ \vdots \\ f_{N-1} \\ f_N + \frac{1}{h^2} g(1) \\ \end{array}\right)\in \mathbb{R}^{N}\]

The numerical solution \(\widehat{u}_h\) then will approximate the solution of the original problem as follows

\[\widehat{u}_h \approx\left(\begin{array}{c} u(x_1) \\ \vdots \\ u(x_N) \\ \end{array}\right),\quad x_i = i \cdot h \;(1 \leq i \leq N)\]

Numerical representation of the grid functions

The so called grid functions \(\widehat{u}_h\) and \(\widehat{q}_h\) are \(N\) dimensional vectors that give approximations of their continuous counterparts at the interior grid points \(x_1, \dots, x_N\).

However, for our numerical algorithms we will use \(N+2\) dimensional vectors that also take the boundary node into account. We indicate this difference by skipping the hat in the notation. For example, \(\widehat{u}_h\) is represented as

\[u_h \approx\left(\begin{array}{c} u_0 \\ u_1 \\ \vdots \\ u_N \\ u_{N+1} \\ \end{array}\right)\]

Because of the Dirichlet boundary conditions the values \(u_0 = g(0)\) and \(u_{N+1} = g(1)\) are known.

We will also not make actual use of \(\widehat{q}_h\) but instead use only the well-defined values (i.e. values not indicated as arbitrary by \(*\)) of

\[f_h \approx\left(\begin{array}{c} * \\ f_1 \\ \vdots \\ f_N \\ f_{N+1} \\ \end{array}\right)=\left(\begin{array}{c} * \\ f(x_1) \\ \vdots \\ f(x_N) \\ * \\ \end{array}\right)\]

Ingredients needed for a multigrid solver

We will develop step by steps the ingredients needed for a multigrid solver:

We will start with computing the residual and the direct solver.

Computing the residual

For computing the residual we will use \(u_h\) and \(f_h\) and require:

Then we can compute \(\widehat{r}_h = \widehat{q}_h - A_h \widehat{u}_h\) as follows:

In the code given below this algorithm is implemented in function evalResidual (which works on \(r_h\) instead of \(\widehat{r}_h)\).

Direct solver

\[T_N =\left(\begin{array}{ccccc} 2 & -1 & & & \\ -1 & 2 & -1 & & \\ & \ddots& \ddots& \ddots& \\ & & -1 & 2 & -1 \\ & & & -1 & 2 \\ \end{array}\right)\]

the Cholesky factorization is known to be \(T_N = L_N D_N L_N^T\) with

\[L_N =\left(\begin{array}{ccccc} 1 & & & & \\ \ell_1& 1 & & & \\ & \ddots& \ddots & & \\ & & \ell_{N-2}& 1 & \\ & & & \ell_{N-1}& 1 \\ \end{array}\right),\quad \ell_i = -\frac{i}{i+1}\]

and

\[D_N = \text{diag}(d_1, \dots, d_N), \quad d_i = \frac{i+1}{i}\]

Code for exercise

Make yourself familiar with the ingredients:

#include <cmath>

#include <hpc/matvec/axpy.hpp>
#include <hpc/matvec/copy.hpp>
#include <hpc/matvec/densevector.hpp>
#include <hpc/matvec/iterators.hpp>
#include <hpc/matvec/traits.hpp>
#include <hpc/matvec/print.hpp>

//------------------------------------------------------------------------------

using namespace hpc;
using namespace hpc::matvec;


template <typename T>
const T &
assertEqual(const T &a, const T &b)
{
    assert(a==b);
    return a;
}

//------------------------------------------------------------------------------

template <
    typename F, typename G,
    typename T, template<typename> class VectorF,
                template<typename> class VectorU,
                Require< Dense<VectorF<T>>,
                         Dense<VectorU<T>> > = true
>
void
discretizeProblem(const F &f, const G &g, VectorF<T> &f_h, VectorU<T> &u_h)
{
    auto N  = assertEqual(f_h.length(), u_h.length()) - 2;
    auto h  = T(1)/(N+1);

    u_h(0)   = g(0);
    u_h(N+1) = g(1);

    for (std::size_t i=1; i<N; ++i) {
        u_h(i) = 0;
        f_h(i) = f(i*h);
    }
    // actually we don't have to initialize f_h(0), f_h(N+1) so we store some
    // good and evil numbers there:
    f_h(0)   = 42;
    f_h(N+1) = 666;
}

//------------------------------------------------------------------------------

template <
    typename T, template<typename> class Vector,
                Require< Dense<Vector<T>> > = true
>
T
lInfNorm(const Vector<T> &x_h)
{
    T res = 0;

    for (auto [i, xi] : x_h) {
        i=i;
        if (std::abs(xi) > res) {
            res = std::abs(xi);
        }
    }
    return res;
}

//------------------------------------------------------------------------------

template <
    typename U,
    typename T, template<typename> class VectorU,
                template<typename> class VectorE,
                Require< Dense<VectorU<T>>,
                         Dense<VectorE<T>> > = true
>
void
evalError(const U &u, const VectorU<T> &u_h, VectorE<T> &e_h)
{
    auto N  = assertEqual(u_h.length(), e_h.length()) - 2;
    auto h  = T(1)/(N+1);

    for (auto [i, ei] : e_h) {
        ei = u(i*h) - u_h(i);
    }
}

//------------------------------------------------------------------------------

template <
    typename T, template<typename> class VectorU,
                template<typename> class VectorF,
                template<typename> class VectorR,
                Require< Dense<VectorF<T>>,
                         Dense<VectorU<T>>,
                         Dense<VectorR<T>> > = true
>
void
evalResidual(const VectorU<T> &u_h, const VectorF<T> &f_h, VectorR<T> &r_h)
{
    auto N  = assertEqual(u_h.length(), r_h.length()) - 2;
    auto h  = T(1)/(N+1);
    auto h2 = h*h;

    r_h(0) = r_h(N+1) = 0;
    for (std::size_t i=1; i<=N; ++i) {
        r_h(i) = f_h(i) - 1./h2 * (2*u_h(i) - u_h(i-1) - u_h(i+1));
    }
}


//------------------------------------------------------------------------------

template <
    typename T, template<typename> class VectorF,
                template<typename> class VectorU,
                Require< Dense<VectorF<T>>,
                         Dense<VectorU<T>> > = true
>
void
direct_solver(const VectorF<T> &f_h, VectorU<T> &u_h)
{
    auto N  = assertEqual(f_h.length(), u_h.length()) - 2;
    auto h  = T(1)/(N+1);
    auto h2 = h*h;

    // TODO: Your code
}

//------------------------------------------------------------------------------

int
main()
{
    auto f = [](double x) { return -2; };
    auto g = [](double x) { return 1 + 3*x; };
    auto u = [](double x) { return (x+1)*(x+1); };

    std::size_t N = 10;

    DenseVector<double>  f_h(N+2), u_h(N+2), r_h(N+2), e_h(N+2);

    discretizeProblem(f, g, f_h, u_h);

    fmt::printf("f_h =\n"); print(f_h);
    fmt::printf("u_h =\n"); print(u_h);

    fmt::printf("Calling solver: direct_solver(f_h, u_h);\n");
    direct_solver(f_h, u_h);
    evalResidual(u_h, f_h, r_h);

    fmt::printf("u_h =\n"); print(u_h);
    fmt::printf("r_h =\n"); print(r_h);

    fmt::printf("Eval approximation error of u_h:\n");
    evalError(u, u_h, e_h);
    fmt::printf("e_h =\n"); print(e_h);

    fmt::printf("\n\nIn short:\n");
    fmt::printf("lInfNorm(r_h) = %.5e\n", lInfNorm(r_h));
    fmt::printf("lInfNorm(e_h) = %.5e\n", lInfNorm(e_h));
}

Compiling the code

theon$ g++ -Wall -I /home/numerik/pub/hpc/ws18/session25/ -std=c++17 -o direct_solver direct_solver.cpp
direct_solver.cpp: In instantiation of 'void direct_solver(const VectorF&, VectorU&) [with T = double; VectorF = hpc::matvec::DenseVector; VectorU = hpc::matvec::DenseVector; typename std::common_type >::type::is_DenseVector:: value, bool>::type, typename std::enable_if >::type::is_DenseVector:: value, bool>::type>::type  = 1]':
direct_solver.cpp:153:27:   required from here
direct_solver.cpp:129:10: warning: unused variable 'h2' [-Wunused-variable]
     auto h2 = h*h;
          ^~
theon$ ./direct_solver
f_h =
  42.00  -2.00  -2.00  -2.00  -2.00  -2.00  -2.00  -2.00  -2.00  -2.00   0.00 666.00
u_h =
   1.00   0.00   0.00   0.00   0.00   0.00   0.00   0.00   0.00   0.00   0.00   4.00
Calling solver: direct_solver(f_h, u_h);
u_h =
   1.00   0.00   0.00   0.00   0.00   0.00   0.00   0.00   0.00   0.00   0.00   4.00
r_h =
   0.00 119.00  -2.00  -2.00  -2.00  -2.00  -2.00  -2.00  -2.00  -2.00 484.00   0.00
Eval approximation error of u_h:
e_h =
   0.00   1.19   1.40   1.62   1.86   2.12   2.39   2.68   2.98   3.31   3.64   0.00


In short:
lInfNorm(r_h) = 4.84000e+02
lInfNorm(e_h) = 3.64463e+00
theon$ 

Exercise