implicit_incompressible_sph.md

[Implicit Incompressible SPH](@id iisph)

Implicit Incompressible SPH (IISPH) as introduced by [Ihmsen et al. (2013)](@cite Ihmsen2013) is a method that achieves incompressibility by solving the pressure Poisson equation. The resulting linear system is iteratively solved with the relaxed Jacobi method. Unlike the [weakly compressible SPH method](@ref wcsph), incompressible methods determine pressure by enforcing the incompressibility constraint rather than using an equation of state.

Modules = [TrixiParticles]
Pages = [joinpath("schemes", "fluid", "implicit_incompressible_sph", "system.jl")]

Derivation

To derive the linear system of the pressure Poisson equation, we start by discretizing the continuity equation

$$\frac{D\rho}{Dt} = - \rho \nabla \cdot \bm{v}.$$

For a particle i, discretizing the left-hand side of the equation with a forward difference yields

$$\frac{\rho_i(t+ \Delta t) - \rho_i(t)}{\Delta t}.$$

The divergence in the right-hand side is discretized with the SPH discretization for particle i as

$$-\frac{1}{\rho_i} \sum_j m_j \bm{v}_{ij} \nabla W_{ij},$$

where \bm{v}_{ij} = \bm{v}_i - \bm{v}_j.

Together, the following discretized version of the continuity equation for a particle i is achieved:

$$\frac{\rho_i(t + \Delta t) - \rho_i(t)}{\Delta t} = \sum_j m_j \bm{v}_{ij}(t+\Delta t) \nabla W_{ij}.$$

The right-hand side contains the unknown velocities \bm{v}_{ij}(t + \Delta t) at the next time step, making it an implicit formulation.

Using the semi-implicit Euler method, we can obtain the velocity in the next time step as

$$\bm{v}_i(t + \Delta t) = \bm{v}_i(t) + \Delta t \frac{\bm{F}_i^\text{adv}(t) + \bm{F}_i^p(t)}{m_i},$$

where \bm{F}_i^{\text{adv}} denotes all non-pressure forces such as gravity, viscosity, surface tension and more, while \bm{F}_i^pdenotes the unknown pressure forces, which we want to solve for.

Note that the IISPH is an incompressible method, which means that the density of the fluid remain constant over time. By assuming a fixed reference density \rho_0 for all fluid particle over the whole time of the simulation, the density value at the next time step \rho_i(t + \Delta t) also has to be this rest density. So \rho_0 can be plugged in for \rho_i(t + \Delta t) in the formula above.

The goal is to compute the pressure values required to obtain the pressure acceleration that ensures each particle reaches the rest density in the next time step. At the moment these pressure values are unknown in t, but all the non-pressure forces are already known. Therefore a predicted velocity can be calculated, which depends only on the non-pressure forces \bm{F}_i^{\text{adv}}:

$$\bm{v}_i^{\text{adv}}(t+\Delta t)= \bm{v}_i(t) + \Delta t \frac{\bm{F}_i^{\text{adv}}(t)}{m_i},$$

Using this predicted velocity and the continuity equation, a predicted density can be defined in a similar way as

$$\rho_i^{\text{adv}}(t + \Delta t)= \rho_i(t) + \Delta t \sum_j m_j \bm{v}_{ij}^{\text{adv}} \nabla W_{ij}(t).$$

To achieve the rest density, the unknown pressure forces must counteract the compression caused by the non-pressure forces. In other words, they must resolve the deviation between the predicted density and the reference density.

Therefore, the following equation needs to be fulfilled:

$$\Delta t ^2 \sum_j m_j \left( \frac{\bm{F}_i^p(t)}{m_i} - \frac{\bm{F}_j^p(t)}{m_j} \right) \nabla W_{ij}(t) = \rho_0 - \rho_i^{\text{adv}}.$$

This expression is derived by substituting the reference density \rho_0 for \rho_i(t+\Delta t) in the discretized continuity equation and inserting the definitions of the predicted velocity \bm{v}_{ij}(t+\Delta t) and predicted density \rho_i^{\text{adv}}.

The pressure force is defined as:

$$\bm{F}_i^p(t) = -\frac{m_i}{\rho_i} \nabla p_i = -m_i \sum_j m_j \left( \frac{p_i(t)}{\rho_i^2(t)} + \frac{p_j(t)}{\rho_j^2(t)} \right) \nabla W_{ij}.$$

Substituting this definition into the equation yields a linear system \bm{A}(t) \bm{p}(t) = \bm{b}(t) with one equation and one unknown pressure value for each particle. For n particles, this is a linear system with n equations and n unknowns:

$$\sum_j a_{ij} p_j = b_i = \rho_0 - \rho_i^{\text{adv}}.$$

To solve for the pressure values, a relaxed Jacobi scheme is used. This is an iterative numerical method to solve a linear system Ap=b. In each iteration, the new values of p are obtained as computed as

$$p_i^{(k+1)} = (1-\omega) p_i^{(k)} + \omega \left( \frac{1}{a_{ii}} \left( b_i - \sum_{j \neq i} a_{ij} p_j^{(k)} \right)\right).$$

Substituting the right-hand side, we obtain

$$p_i^{l+1} = (1-\omega) p_i^l + \omega \frac{\rho_0 - \rho_i^{\text{adv}} - \sum_{j \neq i} a_{ij}p_j^l}{a_{ii}}.$$

Therefore the diagonal elements a_{ii} and the sum \sum_{j \neq i} a_{ij}p_j^l need to be determined. This can be done efficiently by separating the pressure force acceleration into two components: One that describes the influence of particle i's own pressure on its displacement, and one that describes the displacement due to the pressure values of the neighboring particles j.

The pressure acceleration is given by:

$$\Delta t^2 \frac{\bm{F}_i^p}{m_i} = -\Delta t^2 \sum_j m_j \left( \frac{p_i}{\rho_i^2} + \frac{p_j}{\rho_j^2} \right)\nabla W_{ij} = \underbrace{\left(- \Delta t^2 \sum_j \frac{m_j}{\rho_i^2} \nabla W_{ij} \right) }_{d_{ii}} p_i + \sum_j \underbrace{-\Delta t^2 \frac{m_j}{\rho_j^2} \nabla W_{ij}}_{d_{ij}}p_j.$$

The d_{ii}p_i value describes the displacement of particle i because of the particle i and d_{ij}p_j describes the influence from the neighboring particles j. Using this new values the linear system can be rewritten as

$$\rho_0 - \rho_i^{\text{adv}} = \sum_j m_j \left( d_{ii}p_i + \sum_k d_{ik}p_k - d_{jj}p_j - \sum_k d_{jk}p_k \right) \nabla W_{ij},$$

where the first sum over k loops over all neighbor particles of i and the second sum over k loops over the neighbor particles of j (which is a neighbor of i). So the sum over the neighboring pressure values p_j also includes the pressure values p_i, since i is a neighbor of j. To separate this sum, it can be written as

$$\sum_k d_{jk} p_k = \sum_{k \neq i} d_{jk} p_k + d_{ji} p_i.$$

With this separation, the equation for the linear system can again be rewritten as

$$\rho_0 - \rho_i^{\text{adv}} = p_i \sum_j m_j ( d_{ii} - d_{ji})\nabla W_{ij} + \sum_j m_j \left ( \sum_k d_{ik} p_k - d_{jj} p_j - \sum_{k \neq i} d_{jk}p_k \right) \nabla W_{ij}.$$

In this formulation all coefficients that are getting multiplied with the pressure value p_i are separated from the other. The diagonal elements a_{ii} can therefore be defined as:

$$a_{ii} = \sum_j m_j ( d_{ii} - d_{ji})\nabla W_{ij}.$$

The remaining part of the equation represents the influence of the other pressure values p_j. ​Hence, the final relaxed Jacobi iteration takes the form:

$$p_i^{l+1} = (1 - \omega) p_i^{l} + \omega \frac{1}{a_{ii}} \left( \rho_0 -\rho_i^{\text{adv}} - \sum_j m_j \left( \sum_k d_{ik} p_k^l - d_{jj} p_j^l - \sum_{k \neq i} d_{jk} p_k^l \right) \nabla W_{ij} \right).$$

Because interactions are local, limited to particles within the kernel support defined by the smoothing length, each particle's pressure p_i depends only on its own value and nearby particles. Consequently, the matrix A is sparse (most entries are zero).

The diagonal elements a_{ii} get computed and stored at the beginning of the simulation step and remain unchanged throughout the relaxed Jacobi iterations. The same applies for the d_{ii}. The coefficients d_{ij} are computed, but not stored, as part of the calculation of a_{ii}. During the Jacobi iterations, two loops over all particles are performed: The first updates the values of \sum_j d_{ij} and the second computes the updated pressure values p_i^{l+1}.

The final pressure update follows the equation of the relaxed Jacobi scheme as stated above, with two exceptions:

1. Pressure clamping

To avoid negative pressure values, one can enable pressure clamping. In this case, the pressure update is given by

$$p_i^{l+1} = \max \left(0, (1-\omega) p_i^l + \omega \frac{\rho_0 - \rho_i^{\text{adv}} - \sum_{j \neq i} a_{ij}p_j^l}{a_{ii}}\right).$$

2. Small diagonal elements

If the diagonal element a_{ii} becomes too small or even zero, the simulation may become unstable. This can occur, for example, if a particle is isolated and receives little or no influence from neighboring particles, or due to numerical cancellations. In such cases, the updated pressure value is set to zero.

There are also other options, like setting a_{ii} to the threshold value if it is beneath and then update with the usual equation, or just don't update the pressure value at all, and continue with the old value. The numerical error introduced by this technique remains small, as only isolated or almost isolated particles are affected.

Boundary Handling

The previously introduced formulation only considered interactions between fluid particles and neglected interactions between fluid and boundary particles. To account for boundary interactions, a few modifications to the previous equations are required.

First, the discretized form of the continuity equation must be adapted for the case in which a neighboring particle is a boundary particle. From now on, we distinguish between neighboring fluid particles (indexed by j) and neighboring boundary particles (indexed by b).

The updated discretized continuity equation becomes:

$$\frac{\rho_i(t + \Delta t) - \rho_i(t)}{\Delta t} = \sum_j m_j \bm{v}_{ij}(t+\Delta t) \nabla W_{ij} + \sum_b m_b \bm{v}_{ib}(t+\Delta t) \nabla W_{ib}.$$

Since boundary particles have zero velocity, the difference between the fluid particle's velocity and the boundary particle's velocity simplifies to just the fluid particle's velocity \bm{v}_{ib}(t+\Delta t) = \bm{v}_{i}(t+\Delta t). Accordingly, the predicted density \rho^{\text{adv}} becomes:

$$\rho_i^{\text{adv}} = \rho_i (t) + \Delta t \sum_j m_j \bm{v}_{ij}^{\text{adv}} \nabla W_{ij}(t) + \Delta t \sum_b m_b \bm{v}_{i}^{\text{adv}} \nabla W_{ib}(t).$$

This leads to the following updated formulation of the linear system:

$$\Delta t^2 \sum_j m_j \left( \frac{\bm{F}_i^p(t)}{m_i} - \frac{\bm{F}_j^p(t)}{m_j} \right) \nabla W_{ij} + \Delta t^2 \sum_b m_b \frac{\bm{F}_i^p(t)}{m_i} \nabla W_{ib} = \rho_0 - \rho_i^{\text{adv}}.$$

Note that, since boundary particles are fixed, the force F_b^p is zero and does not appear in this equation.

The pressure force acting on a fluid particle is computed as:

$$\bm{F}_i^p(t) = -\sum_j m_j \left( \frac{p_i(t)}{\rho_i^2(t)} + \frac{p_j(t)}{\rho_j^2(t)} \right) \nabla W_{ij}(t) - \sum_b m_b \left( \frac{p_i(t)}{\rho_i^2(t)} + \frac{p_b(t)}{\rho_b^2(t)} \right) \nabla W_{ib}(t).$$

This also leads to an updated version of the equation for the diagonal elements:

$$a_{ii} = \sum_j m_j ( d_{ii} - d_{ji})\nabla W_{ij} + \sum_b m_b (-d_{bi}) \nabla W_{ib}.$$

From this point forward, the computation of the coefficients required for the Jacobi scheme (such as d_{ii}, d_{ij} etc.) depends on the specific boundary density evaluation method used in the chosen boundary model.

Pressure Mirroring

When using pressure mirroring, the pressure value p_b of a boundary particle in the equation above is defined to be equal to the pressure of the corresponding fluid particle p_i. In other words, the boundary particle "mirrors" the pressure of the fluid particle interacting with it. As a result, the coefficient that describes the influence of a particle's own pressure value p_i ​must also include contributions from boundary particles. Therefore, the equation for calculating the coefficient d_{ii} must be adjusted as follows:

$$d_{ii} = -\Delta t^2 \sum_j \frac{m_j}{\rho_i^2} \nabla W_{ij} - \Delta t^2 \sum_b \frac{m_b}{\rho_i^2} \nabla W_{ib}.$$

The corresponding relaxed Jacobi iteration for pressure mirroring then becomes:

$$\begin{align*} p_i^{l+1} = (1 - \omega) p_i^l + \omega \frac{1}{a_{ii}} &\left( \rho_0 - \rho_i^{\text{adv}} - \sum_j m_j \left( \sum_k d_{ik} p_k^l - d_{jj}p_j^l - \sum_{k \neq i} d_{jk} p_k^l \right) \nabla W_{ij} \right. \\\ & \quad - \left. \sum_b m_b \sum_j d_{ij} p_j^l \nabla W_{ib} \right). \end{align*}$$

Pressure Zeroing

If pressure zeroing is used instead, the pressure value of a boundary particle p_b ​is assumed to be zero. Consequently, boundary particles do not contribute to the pressure forces acting on fluid particles. In this case, the computation of the coefficient d_{ii} remains unchanged and is given by:

$$d_{ii} = -\Delta t^2 \sum_j \frac{m_j}{\rho_i^2} \nabla W_{ij}.$$

The equation for the relaxed Jacobi iteration remains the same as in the pressure mirroring approach. However, the contribution from boundary particles vanishes due to their zero pressure:

$$\begin{align*} p_i^{l+1} = (1 - \omega) p_i^l + \omega \frac{1}{a_{ii}} &\left( \rho_0 - \rho_i^{\text{adv}} - \sum_j m_j \left( \sum_k d_{ik} p_k^l - d_{jj}p_j^l - \sum_{k \neq i} d_{jk} p_k^l \right) \nabla W_{ij} \right. \\\ & \quad - \left. \sum_b m_b \sum_j d_{ij} p_j^l \nabla W_{ib} \right). \end{align*}$$