DFSPH factor expression with Akinci boundary

[davidAlgis]

Hello,

I would like to know where the DFSPH factor comes from if we use the fluid/solid coupling of Akinci et al [2012] (see function line 1078 of TimeStepDFSPH.cpp) :

$$\alpha_{f_i} = |\sum_{j} m_{f_j} \nabla W_{{f_i}{f_j}}+\sum_{k} \tilde{m}_{b_k} \nabla W_{{f_i}{b_k}}|^2+ \sum_{j} |m_{f_j}\nabla W_{{f_i}{f_j}}|^2$$

where $f_j$ corresponds to the fluid neighbors and $b_k$ to the solid neighbors.

Indeed, I understand well the DFSPH factor, in the case where we have only fluid particles, but if I try to extend the reasoning when we add solid particles (and in particular dynamic solid particles) I have :

$$f^p_{f_i} = - \kappa_{f_i}^v (\sum_j m_{f_j} \nabla W_{{f_i}{f_j}} + \sum_{b_k} \tilde{m}_{b_k} \nabla W_{{f_i}{f_j}})$$

We deduce from the symmetric condition $f^p_{f_i} +\sum_j f^p_{f_j\leftarrow f_i}+\sum_k f^p_{b_k\leftarrow f_i}=0$ and by identification :

$$\begin{cases} f^p_{f_j\leftarrow f_i} = \kappa_{f_i}^v m_{f_j} \nabla W_{{f_i}{f_j}} \\\ f^p_{b_k\leftarrow f_i} = \kappa_{f_i}^v \tilde{m}_{b_k} \nabla W_{{f_i}{f_j}} \end{cases}$$

Then we want to minimize $\frac{D\rho_{f_i}}{Dt}$ :

$$\begin{align} \frac{D\rho_{f_i}}{Dt} &= \sum_j m_{f_j} \mathbf{v_{{f_i}{f_j}}}\cdot \nabla W_{{f_i}{f_j}} + \sum_k \tilde{m}_{b_k} \mathbf{v_{{f_i}{b_k}}}\cdot \nabla W_{{f_i}{b_k}} \\\ &= \Delta t(\sum_j m_{f_j} \left(\frac{\mathbf{f}_{f_i}^p}{\rho_{f_i}}- \frac{\mathbf{f}_{{f_j}\leftarrow {f_i}}^p }{\rho_{f_i}} \right)\cdot \nabla W_{{f_i}{f_j}} + \sum_k \tilde{m}_{b_k} \left(\frac{\mathbf{f}_{f_i}^p}{\rho_{f_i}}- \frac{\mathbf{f}_{{b_k}\leftarrow {f_i}}^p }{\rho_{f_i}} \right)\cdot \nabla W_{{f_i}{f_k}})\\\ &= \frac{\Delta t \kappa_i^v}{\rho_{f_i}}((\sum_j m_{f_j}\sum_j m_{f_j} \nabla W_{{f_i}{f_j}} +\sum_k \tilde{m}_{b_k} \nabla W_{{f_i}{b_k}} +m_j \nabla W_{{f_i}{f_j}} )\cdot \nabla W_{{f_i}{f_j}} +\sum_k \tilde{m}_{b_k}(\sum_j m_{f_j} \nabla W_{{f_i}{f_j}} + \sum_k \tilde{m}_{b_k} \nabla W_{{f_i}{b_k}}+ \tilde{m}_{b_k} \nabla W_{{f_i}{b_k}})\cdot \nabla W_{{f_i}{b_k}}) \end{align}$$

this last expression has a lot of term compared to the result that gives us the expression of $\alpha_i$ above. To retrieve the result we should remove the term where we have $\sum_{f_j}\sum_{b_k}$ and reverse and the term that results from $\frac{\mathbf{f}_{{b_k}\leftarrow {f_i}}^p}{\rho_{f_i}}$. How do you explain this ?

Thanks in advance.

[janbender]

I was not able to follow your last step. However, last year I wrote a paper where I derive DFSPH in a different way with a complete derivation of the boundary terms:

https://animation.rwth-aachen.de/publication/0584/

I hope this helps to understand the boundary handling.

[davidAlgis]

Thanks for your answer, I still don't completely understand, how come the $\alpha_i$ value used in the ($|\sum_{j} m_{f_j} \nabla W_{{f_i}{f_j}}+\sum_{k} \tilde{m}_{b_k} \nabla W_{{f_i}{b_k}}|^2+ \sum_{j} |m_{f_j}\nabla W_{{f_i}{f_j}}|^2$) code seem to not be the same as the one used in the paper ($|\sum_{j} m_{f_j} \nabla W_{{f_i}{f_j}}+\sum_{k} \tilde{m}_{b_k} \nabla W_{{f_i}{b_k}}|^2+ \sum_{j} |m_{f_j}\nabla W_{{f_i}{f_j}}|^2 + m_{f_j} \tilde{m}_{b_k}|\nabla W_{{f_i}{f_j}}|^2$)?

[davidAlgis]

In Consistent SPH Rigid-Fluid Coupling they use $A_{ii} =|\sum_{j} m_{f_j} \nabla W_{{f_i}{f_j}}+\sum_{k} \tilde{m}_{b_k} \nabla W_{{f_i}{b_k}}|^2+ \sum_{j} |m_{f_j}\nabla W_{{f_i}{f_j}}|^2$ (equation 32 in section 3.4.2), but this is based on the approximation of static boundary particles, in case of dynamic boundary particles shouldn't it be $|\sum_{j} m_{f_j} \nabla W_{{f_i}{f_j}}+\sum_{k} \tilde{m}_{b_k} \nabla W_{{f_i}{b_k}}|^2+ \sum_{j} |m_{f_j}\nabla W_{{f_i}{f_j}}|^2 + m_{f_j} \tilde{m}_{b_k}|\nabla W_{{f_i}{f_j}}|^2$ ?