Burgers' equation

Let us consider the Burgers' equation, which describes the dynamics of viscous fluid without the effects of pressure. Despite being unrealistic, it is the simplest description of advective flow with diffusive effects of viscosity. It has the following form: \begin{equation} \frac{\partial \b{u}}{\partial t} + (\b{u}\cdot\nabla)\b{u} = \nu \nabla^2 \b{u} , \end{equation}

where $\b{u}$ is a velocity vector field and $\nu$ is the dynamic viscosity. To solve the equation we will linearize the advection term. There are several tactics of linearization, but we choose the simplest one, which is using the previous velocity as an approximation of the current velociy. Using the implicit Euler method we arrive at the equation

\begin{equation} \b{u}^{n}_i + \Bigl[\b{u}^{n-1}_i\cdot\nabla_i\b{u}^{n} - \nu\nabla^2_i \b{u}^{n}\Bigr]\Delta t = \b{u}^{n-1}_i, \end{equation}

where the subscript $i$ is the index of the domain node, superscripts $n$ and $n-1$ denote the current and the previous time step respectively, $\Delta t$ is the length of the time step. The notation $\nabla_i$ means the gradient operator at node $i$, not the component of the gradient. Same holds for $\nabla^2_i$. Using the Medusa library, this equation can be solved for $\b{u}^{n}_i$.

Solution in 1D

Let's look at an example in 1D with Dirichlet boundary conditions. We are solving the equivalent of equation (2) but with only one spatial variable: \begin{equation} u^{n}_i + \Bigl[u^{n-1}_i\frac{\partial u^{n}}{\partial x} - \nu\frac{\partial^2 u^{n}}{\partial x^2}\Bigr]\Delta t = u^{n-1}_i ,\\ u_i = 0 \text{ on } \partial \Omega . \end{equation}

This will be rewritten into a linear system of equations with Medusa and then solved with Eigen. Let's look at the code (find the full version on our repository). First we construct our 1D domain with length 2l, discretize it with a given step dx and find the closest $n$ nodes of each node which are called stencil nodes or support nodes or simply the stencil.

1 // Build 1D domain
2 BoxShape<Vec1d> domain((-l), (l)); // the interval [-l, l]
3 auto discretization = domain.discretizeWithStep(dx);
4 int domain_size = discretization.size();
5 // Find support nodes
6 FindClosest find_support(n);
7 discretization.findSupport(find_support);

We will use the RBF-FD method with polyharmonic splines of order 3 as Radial Basis Functions (RBFs) augmented with monomials up to 4th order for our approximation engine. We use it in conjunction with stencil nodes to compute the stencil weights (also called shape functions) which are used in approximations of differential operators on our domain.

1 Polyharmonic<double, 3> p;
2 Monomials<Vec1d> mon(4);
3 RBFFD<Polyharmonic<double, 3>, Vec1d, ScaleToClosest> approx(p, mon);
4 auto storage = discretization.computeShapes(approx); 
5 
6 VectorXd u = VectorXd::Zero(domain_size);  // vector for containing the current state

Next we set the initial condition to

\begin{equation} u(x) = \frac{2\nu ak\sin(kx)}{b + a\cos(kx)}, \end{equation} where $a, b, k$ are constants and $\nu$ is viscosity. We chose this function because it has an analytical solution

\begin{equation} u(x, t) = \frac{2\nu ake^{-\nu k^2t}\sin(kx)}{b + ae^{-\nu k^2t}\cos(kx)}, \end{equation}

which we will compare our solution to. It is derived using the Cole-Hopf transformation of a solution of the diffusion equation with the initial profile: $f(x) = b + a\cos(kx)$ for $b>a$.

1     //Set initial condition 
2     for (int i : discretization.all()){ 
3         u(i) = 2*nu*a*k*std::sin(k*discretization.pos(i,0))/(b+a*std::cos(k*discretization.pos(i,0)));

Inside our time-stepping loop, we instantiate the matrix $M$ and the vector $rhs$ representing the linear system. We then construct the differential operators for implicit solving.

1 for (tt = 1; tt <= t_steps; ++tt) {
2     SparseMatrix<double> M(domain_size, domain_size); 
3     VectorXd rhs = VectorXd::Zero(domain_size); 
4     M.reserve(Range<int>(domain_size, n));
5     auto op = storage.implicitOperators(M, rhs);

Now comes the crucial part - setting the equation. In essence, we rewrite equation (3) in Medusa's terms. Terms on the left hand side are written into matrix $M$ and the terms on the right into $rhs$. The terms with the prefix op. are matrices that represent the appropriate differential operators (op.value(i) is a matrix with $A_{ii}$ = 1 and zero everywhere else), vector u[i] represents $\b{u}_i^{n-1}$ - result of the previous step.

1     // Setting the equation interior
2     for (int i : interior) {
3         op.value(i) + (u[i] * dt) * op.der1(i, 0) + (-dt*nu) * op.der2(i, 0, 0) = u[i];  
4 }

Boundary conditions are handled in the same way. The analytical solution is defined on an infinite domain, but we see from equation (5) that $u = 0$ for $kx = z\pi, z \in \mathbb{Z}$. So we will be able to enforce Dirichlet boundary conditions on a finite domain, but this will limit our choice of $k$ to $\frac{z\pi}{l}, z \in \mathbb{Z}$, where $l$ is the length of the domain.

1     // Setting Dirichlet boundary conditions
2     for (int i : boundary) {
3         op.value(i) = 0;  
4 }

We then solve the equation and update u for use in the next time step. The solution is also saved into an output file with HDF5.

 1     M.makeCompressed();
 2     BiCGSTAB<SparseMatrix<double, RowMajor>> solver;
 3     solver.compute(M); 
 4 
 5     // Solve the linear system and update the state
 6     VectorXd u2 = solver.solve(rhs);
 7     u = u2;
 8 
 9     if (tt%(t_factor) == 0) {
10         // Saving state in output file
11         hdf_out.openGroup("/step" + std::to_string(t_save));
12         hdf_out.writeSparseMatrix("M", M);
13         hdf_out.writeDoubleAttribute("TimeStep", t_save);
14         std::cout<< "Time step " << t_save << " of " << t_steps/t_factor << "." << std::endl;
15         hdf_out.writeDoubleAttribute("time",(t_0 + tt * dt));
16         ++t_save;
17         hdf_out.writeDoubleArray("u", u);
18     }

Results in 1D

For the simulation bellow we used the following parameters: $\nu = 0.05, l = 1, a = 4, b = 4.1, k = 2\pi$, $n = 13$, $m = 4$, $dx = 2\cdot 10^{-3}$, $dt = 10^{-6}$.

Convergence

Spatial discretization

We fix the time step to $dt = 10^{-6}$, vary the monomial augmentation order $m \in \{2, 4, 6, 8\}$ and analyze the convergence with increasing number of discretization nodes $N$.

After some initial oscillation the maximum error decreases approximately as $\left(\frac{1}{N}\right)^k$, which is proportional to $dx^k$, where $k$ is the fit parameter for the function: $\log_{10}e_\infty = -k\log_{10}N + c$ for every $m$. But eventually the error plateaus because for finer spatial discretizations we are bounded by the time step.

Temporal discretization

Now we fix $N$ and $m$ to 1024 and 4 respectively and vary the size of the time step from $10^{-3}$ to $10^{-6}$. The error decreases linearly with the time step.

Solution in 2D

In 2D the Burgers' equation (1) gives a system of two coupled nonlinear equations: \begin{gather} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = \nu\nabla^2u,\\ \frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = \nu\nabla^2v, \end{gather}

where $u$ and $v$ are velocities in the $x$ and $y$ direction respectively. We discretize equations (4) and (5) in the same way as in the 1D case: \begin{gather} u^{n}_i + \Bigl[u^{n-1}_i\frac{\partial u^{n}}{\partial x} + v^{n-1}_i\frac{\partial u^{n}}{\partial y} - \nu\nabla^2u^{n}\Bigr]\Delta t = u^{n-1}_i ,\\ v^{n}_i + \Bigl[u^{n-1}_i\frac{\partial v^{n}}{\partial x} + v^{n-1}_i\frac{\partial v^{n}}{\partial y} - \nu\nabla^2v^{n}\Bigr]\Delta t = v^{n-1}_i . \end{gather}

Find the full version of the code in our repository. We will solve each equation separately every time step, this will enable us to partialy parallelize the algorithm. First we contruct a 2D square domain with side length of 1, fill it with nodes and find 45 closest nodes of each node.

1 BoxShape<Vec2d> b((0, 0), (1, 1));
2 double step = 0.05;
3 auto discretization = b.discretizeBoundaryWithStep(dr);
4 GeneralFill<Vec2d> fill; fill.seed(0);
5 discretization.fill(fill, step);
6 // Find support nodes, excluding boundary nodes from supports of boundary nodes for better stability of neumann BC
7     discretization.findSupport(FindClosest(n).forNodes(discretization.interior()));
8     discretization.findSupport(FindClosest(n).forNodes(discretization.boundary()).searchAmong(discretization.interior()).forceSelf(true));
9 int domain_size = discretization.size();

The approximation engine is created and shape functions are computed in the same way as in the 1D example, except the template parameter is Vec2d instead of Vec1d. We will be simulating the following initial and boundary conditions:

\begin{gather*} u(x,y,0) = sin(\pi x)cos(\pi y),\\ v(x,y,0) = cos(\pi x)sin(\pi y),\\ u(0,y,t)=u(1,y,t)=v(x,0,t)=v(x,1,t)=0,\\ \frac{\partial u}{\partial y}\bigg|_{y = 0} = \frac{\partial u}{\partial y}\bigg|_{y = 1} = \frac{\partial v}{\partial x}\bigg|_{x = 0} = \frac{\partial v}{\partial x}\bigg|_{x = 1} = 0. \end{gather*}

The analytical solution is given here . In the time stepping loop we use #pragma omp parallel sections to enable concurrent solving for $u$ and $v$. In each section we define separate matrices Mu or Mv, rhs vectors rhu and rhv, and implicit operators opu and opv which will write into their respective matrices. The code bellow includes only the section for $u$. The types -1 ,-2 , -3 ,-4 correspond to the left, right, bottom and top boundaries. At the end of each section the velocities are updated.

 1 for (tt = 1; tt <= t_steps; ++tt) {
 2 
 3 double t = tt * dt;
 4 
 5 #pragma omp parallel sections shared(domain_size, discretization, storage, dt, u, v) 
 6 {
 7     ////Solving for u////
 8     #pragma omp section
 9     {
10         SparseMatrix<double, RowMajor> Mu(domain_size, domain_size); 
11         VectorXd rhu = VectorXd::Zero(domain_size);  
12         auto opu = storage.implicitOperators(Mu, rhu);        
13         Mu.reserve(storage.supportSizes());
14 
15         // Setting the equation 
16         #pragma omp parallel for
17         for (int i : discretization.interior()) {
18             opu.value(i) + (u[i]*dt) * opu.der1(i, 0) + (v[i]*dt) * opu.der1(i, 1) + (-1/rndls*dt) * opu.lap(i) = u[i];  
19         }
20         // Boundary conditions
21         for (int i : discretization.types() == -3){
22             opu.value(i) = analytical_u(discretization.pos(i,0), discretization.pos(i,1), t, 1.0/rndls);
23         }
24         for (int i : discretization.types() == -4){
25             opu.value(i) = analytical_u(discretization.pos(i,0), discretization.pos(i,1), t, 1.0/rndls);
26         }
27         for (int i : discretization.types() == -1){
28             opu.value(i) = 0.0;
29         }
30         for (int i : discretization.types() == -2){
31             opu.value(i) = 0.0;
32         }
33 
34         Mu.makeCompressed(); 
35         BiCGSTAB<SparseMatrix<double, RowMajor>> solveru; // Creating solver object
36         solveru.compute(Mu); // Initialize the iterative solver
37         VectorXd u2 = solveru.solve(rhu);
38         u = u2; // update velocity vector
39     }

Results in 2D

We use the RBF-FD method with PHS radial basis function $\phi(r) = r^3$ augmented with monomials. For the simulation bellow we used parameters $\nu = 1$, $n$ = 45, $m = 4$, $dr = 0.02$, $dt = 10^{-5}$. The animations below show the velocities in both directions and the magnitude of the velocity.

Convergence

Spatial discretization

We fix the time step to $dt = 10^{-6}$ and change the augmentation monomial order. The rate at which the maximum error decreases is close to $\left(\frac{1}{\sqrt{N}}\right)^{k}$ as shown on the graph bellow. The distance between neighbouring nodes in 2D is approximately proportional to $\frac{1}{\sqrt{N}}$ if the node distribution is uniform. For finer discretizations and bigger $m$ the error becomes approximately constant as we are again bounded by the time step.

Temporal discretization

We fix $m = 4$, $dr = 0.02$ giving us $N = 753$ and vary the time step size. The error decreases linearily as in the 1D case.