Computational modeling of magnetic particle margination within blood flow through LAMMPS

Abstract

We develop a multiscale and multiphysics computational method to investigate the transport of magnetic particles as drug carriers in blood flow under influence of hydrodynamic interaction and external magnetic field. A hybrid coupling method is proposed to handle red blood cell (RBC)-fluid interface (CFI) and magnetic particle-fluid interface (PFI), respectively. Immersed boundary method (IBM)-based velocity coupling is used to account for CFI, which is validated by tank-treading and tumbling behaviors of a single RBC in simple shear flow. While PFI is captured by IBM-based force coupling, which is verified through movement of a single magnetic particle under non-uniform external magnetic field and breakup of a magnetic chain in rotating magnetic field. These two components are seamlessly integrated within the LAMMPS framework, which is a highly parallelized molecular dynamics solver. In addition, we also implement a parallelized lattice Boltzmann simulator within LAMMPS to handle the fluid flow simulation. Based on the proposed method, we explore the margination behaviors of magnetic particles and magnetic chains within blood flow. We find that the external magnetic field can be used to guide the motion of these magnetic materials and promote their margination to the vascular wall region. Moreover, the scaling performance and speedup test further confirm the high efficiency and robustness of proposed computational method. Therefore, it provides an efficient way to simulate the transport of nanoparticle-based drug carriers within blood flow in a large scale. The simulation results can be applied in the design of efficient drug delivery vehicles that optimally accumulate within diseased tissue, thus providing better imaging sensitivity, therapeutic efficacy and lower toxicity.

Appendix: coarse-grained potential for RBC

First, we use the Eq. (16) to derive the nodal force due to area constraints.

Fig. 13

Sketches of a one and b two adjacent triangular elements of the membrane network

Figure 13 shows the simple triangular element of the membrane network. $A_k$ represents the area of the triangular element. $a_{ij} = p_i - p_j$ and i, j denotes the index from 1 to 3, $p_i$ is the vertex points. $\xi$ is the normal vector of the surface where the element locates, and $\xi = \overrightarrow{a}_{21}\times \overrightarrow{a31}$, where $\times$ is the cross product. Then we have the expression of the area of triangular element

$$\begin{aligned} A_k = \frac{|\overrightarrow{\xi }|}{2} = \frac{\sqrt{\xi _x^2+\xi _y^2+\xi _z^2}}{2}. \end{aligned}$$

(47)

Here, we adopt global area constraint as an example to show how to derive the nodal force. Using Eq. (16), we get

$$\begin{aligned} \nonumber f_{si}= & {} -\frac{\partial [k_a(A_t-A_{t0})^2/(2A_{t0})]}{\partial s_i} = -\frac{k_a(A_t-A_{t0})}{A_{t0}}\frac{\partial A_t}{\partial s_i}\nonumber \\= & {} \beta _a\varSigma _{k\in 1\ldots N_t}\frac{\partial A_k}{\partial s_i} \nonumber \\= & {} \beta _a\varSigma _{k\in 1\ldots N_t}\frac{1}{4A_k}\left( \xi _x^k\frac{\xi _x^k}{\partial s_i}+\xi _y^k\frac{\xi _y^k}{\partial s_i}+\xi _z^k\frac{\xi _z^k}{\partial s_i}\right) , \end{aligned}$$

(48)

where $\beta _a = -\,k_a(A_t-A_{t0}/A_{t0})$, subscript k represents the k-th triangular element. If we set $\alpha = \beta _a/4A_k$, then we have the nodal force expression

$$\begin{aligned} (f_{x1}, f_{y1}, f_{z1})= & {} \alpha (\overrightarrow{\xi }\times \overrightarrow{a}_{32}),\nonumber \\ (f_{x2}, f_{y2}, f_{z2})= & {} \alpha (\overrightarrow{\xi }\times \overrightarrow{a}_{13}),\nonumber \\ (f_{x3}, f_{y3}, f_{z3})= & {} \alpha (\overrightarrow{\xi }\times \overrightarrow{a}_{21}). \end{aligned}$$

(49)

Similar to the global nodal force derivation, the local nodal force can be calculated by simply setting $\alpha = -\,k_d(A_k-A_{k0})/(4A_k-A_{k0})$.

The nodal force due to total volume constraint can be obtained

$$\begin{aligned} f_{si}= & {} -\frac{\partial [k_v(V-V_0)^2]/2V_0}{\partial si} = -\frac{k_v(V-V_0)}{V_0}\frac{\partial V}{\partial si}\nonumber \\= & {} \beta _v\varSigma _{k\in 1...N_t}\frac{\partial V_k}{\partial s_i}, \end{aligned}$$

(50)

where $V_k = \frac{\overrightarrow{\xi }^{k} \cdot \overrightarrow{b}^k}{6}$, and $\overrightarrow{b}^k = (p_1^k+p_2^k+p_3^k)/3$ is the center of the mass of k-th triangular element. Then the nodal force could be written as

$$\begin{aligned} \nonumber (f_{x1}, f_{y1}, f_{z1})= & {} \frac{\beta _v}{6}(\overrightarrow{\xi }/3 + \overrightarrow{b}\times \overrightarrow{a}_{32}),\\ \nonumber (f_{x2}, f_{y2}, f_{z2})= & {} \frac{\beta _v}{6}(\overrightarrow{\xi }/3 + \overrightarrow{b}\times \overrightarrow{a}_{13}),\\ (f_{x3}, f_{y3}, f_{z3})= & {} \frac{\beta _v}{6}(\overrightarrow{\xi }/3 + \overrightarrow{b}\times \overrightarrow{a}_{21}). \end{aligned}$$

(51)

In Fig. 13b, the normal vectors of the two triangular element are $\overrightarrow{\xi } = \overrightarrow{a}_{21}\times \overrightarrow{a}_{21}$ and $\overrightarrow{\zeta } = \overrightarrow{a}_{34}\times \overrightarrow{a}_{24}$. $A_1$ and $A_2$ are the area of the two triangles, respectively. Then we can calculate the nodal force contributed by the bending potential as

$$\begin{aligned} f_{si} = -\frac{\partial [k_b(1-\cos (\theta -\theta _0))]}{\partial si} = -\,k_b\sin (\theta -\theta _0)\frac{\partial \theta }{\partial si}.\nonumber \\ \end{aligned}$$

(52)

As $\theta$ is the dihedral angle, it can be expressed as $\cos \theta = \frac{\overrightarrow{\xi }\cdot \overrightarrow{\zeta }}{|\overrightarrow{\xi }||\overrightarrow{\zeta }|}$. Then the derivation of $\theta$ with respect to $s_i$ is

$$\begin{aligned} \frac{\partial \theta }{\partial s_i} = \frac{\partial \left[ \arccos \left( \frac{\overrightarrow{\xi }\cdot \overrightarrow{\zeta }}{|\overrightarrow{\xi }||\overrightarrow{\zeta }|}\right) \right] }{\partial s_i} = -\frac{1}{\sqrt{1-\cos ^2\theta }}\frac{\frac{\overrightarrow{\xi }\cdot \overrightarrow{\zeta }}{|\overrightarrow{\xi }||\overrightarrow{\zeta }|}}{\partial s_i}.\nonumber \\ \end{aligned}$$

(53)

According to this analytical expression, we can obtain the results of nodal force exerted on the four vertex points shown in Fig. 13b as following

$$\begin{aligned} \nonumber (f_{x1}, f_{y1}, f_{z1})= & {} k_{11}(\overrightarrow{\xi }\times \overrightarrow{a}_{32})+k_{12}(\overrightarrow{\zeta }\times \overrightarrow{a}_{32}),\nonumber \\ \nonumber (f_{x2}, f_{y2}, f_{z2})= & {} k_{11}(\overrightarrow{\xi }\times \overrightarrow{a}_{13})\nonumber \\&+\,k_{12}(\overrightarrow{\xi }\times \overrightarrow{a}_{34}+\overrightarrow{\zeta }\times \overrightarrow{a}_{13})\nonumber \\&+\,k_{22}(\overrightarrow{\zeta }\times \overrightarrow{a}_{34}),\nonumber \\ \nonumber (f_{x3}, f_{y3}, f_{z3})= & {} k_{11}(\overrightarrow{\xi }\times \overrightarrow{a}_{21})\nonumber \\&+\,k_{12}(\overrightarrow{\xi }\times \overrightarrow{a}_{42}+\overrightarrow{\zeta }\times \overrightarrow{a}_{21})\nonumber \\&+\,k_{22}(\overrightarrow{\zeta }\times \overrightarrow{a}_{42}),\nonumber \\ (f_{x4}, f_{y4}, f_{z4})= & {} k_{12}(\overrightarrow{\xi }\times \overrightarrow{a}_{23})+k_{22}(\overrightarrow{\zeta }\times \overrightarrow{a}_{23}), \end{aligned}$$

(54)

where $k_{11} = -\,\beta _b\frac{\cos \theta }{|\overrightarrow{\xi }|^2},k_{12} = \beta _b\frac{1}{|\overrightarrow{\xi }||\overrightarrow{\zeta }|},k_{22} = -\,\beta _b\frac{\cos \theta }{|\overrightarrow{\zeta }|^2}$, and $\beta _b = \frac{ k_b(\sin \theta \cos \theta _0-\cos \theta \sin \theta _0)}{\sqrt{1-\cos ^2\theta }}$. Here, because $\theta \in (0,\pi ]$, the sign of $\sin \theta$ can be either positive or negative. It is defined by the sign of $\mathbb {S} = (\overrightarrow{\xi }-\overrightarrow{\zeta })\cdot (\overrightarrow{b}^1-\overrightarrow{b}^2)$, where $\overrightarrow{b}^1$ and $\overrightarrow{b}^2$ are the vectors of center of mass of triangles 1 and 2, respectively. If $\mathbb {S}\ge 0$, $\sin \theta = \sqrt{1-\cos ^2\theta }$ and $\sin \theta = -\,\sqrt{1-\cos ^2\theta }$ for $\mathbb {S} \le 0$.