Kinetics of self-assembly of inclusions due to lipid membrane ......2020/09/23 · Kinetics of self-assembly of inclusions due to lipid membrane thickness interactions Xinyu Liao1,

Kinetics of self-assembly of inclusions due to lipid membranethickness interactions

Xinyu Liao1, Prashant K. Purohit1,2,

1 Graduate group of Applied Mathematics and Computational Science, University ofPennsylvania, Philadelphia, PA 19104, USA.2 Department of Mechanical Engineering and Applied Mechanics, University ofPennsylvania, Philadelphia, PA 19104, USA.

* [email protected]

Abstract

Self-assembly of proteins on lipid membranes underlies many important processes in cellbiology, such as, exo- and endo-cytosis, assembly of viruses, etc. An attractive force thatcan cause self-assembly is mediated by membrane thickness interactions betweenproteins. The free energy profile associated with this attractive force is a result of theoverlap of thickness deformation fields around the proteins. The thickness deformationfield around proteins of various shapes can be calculated from the solution of aboundary value problem and is relatively well understood. Yet, the time scales overwhich self-assembly occurs has not been explored. In this paper we compute this timescale as a function of the initial distance between two inclusions by viewing theircoalescence as a first passage time problem. The first passage time is computed usingboth Langevin dynamics and a partial differential equation, and both methods arefound to be in excellent agreement. Inclusions of three different shapes are studied andit is found that for two inclusions separated by about hundred nanometers the time tocoalescence is hundreds of milliseconds irrespective of shape. Our Langevin dynamicssimulation of self-assembly required an efficient computation of the interaction energy ofinclusions which was accomplished using a finite difference technique. The interactionenergy profiles obtained using this numerical technique were in excellent agreement withthose from a previously proposed semi-analytical method based on Fourier-Bessel series.The computational strategies described in this paper could potentially lead to efficientmethods to explore the kinetics of self-assembly of proteins on lipid membranes.

Author summary

Self-assembly of proteins on lipid membranes occurs during exo- and endo-cytosis andalso when viruses exit an infected cell. The forces mediating self-assembly of inclusionson membranes have therefore been of long standing interest. However, the kinetics ofself-assembly has received much less attention. As a first step in discerning the kinetics,we examine the time to coalescence of two inclusions on a membrane as a function ofthe distance separating them. We use both Langevin dynamics simulations and apartial differential equation to compute this time scale. We predict that the time tocoalescence is on the scale of hundreds of milliseconds for two inclusions separated byabout hundred nanometers. The deformation moduli of the lipid membrane and themembrane tension can affect this time scale.

September 17, 2020 1/23

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Introduction 1

Self-assembly of proteins on lipid membranes has been a topic of interest for at least the 2last three decades [1–3]. Proteins on membranes self-assemble because they interact 3with each other through forces that have their origins in membrane bending 4deformations [4, 5], membrane thickness deformations [4, 6–11], electrostatics [12] and 5entropic interactions [4, 13]. There is a large literature on this topic that we do not 6attempt to review here [3–6,13–20]. Our interest is in self-assembly caused by 7membrane thickness mediated interactions of proteins. 8

It is well known that lipid bilayers consist of two leaflets with the hydrophobic tails 9of the lipid molecules spanning the membrane thickness. Proteins that are embedded in 10the membrane have hydrophobic peptides placed in such a way as they interact mostly 11with the hydrophobic tails of the lipid molecules. If the thickness of the hydrophobic 12region of a protein is different from that of the lipid membrane then the leaflets deform 13so that the membrane thickness in the vicinity of the protein changes (see Fig. 1(a)). 14The energy cost of the thickness deformation has been estimated analytically by taking 15account of the lipid hydrocarbon chain entropy [9,21]. The result is an energy functional 16written in terms of the deformation field u(x, y) of the half-membrane thickness and its 17gradients [4, 9]. The membrane bending modulus Kb, the membrane thickness modulus 18Kt and the isotropic membrane tension F enter as parameters into this functional. The 19Euler-Lagrange equation obtained by the minimization of this energy functional is a 20fourth order linear partial differential equation (PDE). A series of papers by Phillips, 21Klug, Haselwandter and colleagues [6–8,22] start from this energy functional and utilize 22the linearity of the PDE to computationally analyze allosteric interactions of clusters of 23proteins of various shapes. The key idea is that the thickness deformation fields caused 24by distant proteins can overlap (superimpose) and give rise to interaction forces just as 25defects in elastic solids interact due to the overlap of deformation fields [23]. This idea 26has been in place since at least the mid-1990s [9], but it was computationally extended 27to complex protein shapes and large clusters by the above authors. 28

An important result that emerged from the research on clusters discussed above is 29that the interaction free energy has a maximum when plotted as a function of distance 30between individual proteins (which form a lattice). To the left of the maximum there 31are strong attractive forces between the proteins, while to the right there are weak 32repulsive forces which decay away as the proteins move far apart. The strong attractive 33forces should cause self-assembly if two (or more) proteins happen to come close 34together as they diffuse on the membrane. We are interested in the time scale of the 35self-assembly process. There are few experiments which focus on this time scale, but 36one by Shnyrova et al. [24] found that viral proteins (that did not interact 37electrostatically) on a micron-sized vesicle self-assemble in seconds. 38

Temporal evolution of the self-assembly of viral proteins on a lipid membrane has 39been analyzed in a few recent papers using simulations. Often these simulations can be 40computationally prohibitive, but they do give insight about time scales and 41intermediate states of the cluster of proteins assembling into a virus particle or 42nano-container [1–3,25]. A drawback of these simulations is that they may not be able 43to tackle time scales of seconds over which self-assembly was seen to occur in 44experiments [24]. We will take a different approach in this paper by analyzing 45self-assembly of differently shaped inclusions using Langevin dynamics and the 46corresponding Fokker-Planck equations. In recent work We viewed self-assembly of two 47inclusions as a first-passage time problem which can be quantitatively analyzed using 48the theory of stochastic processes [26]. We implemented this approach in the context of 49interactions based on membrane bending. The analytical calculations (using PDEs) 50in [26] were confined to absorbing boundary conditions on both boundaries. A novelty 51of this work is that we extend the PDE approach to include absorbing and reflecting 52

September 17, 2020 2/23




boundary conditions. 53This paper is arranged as follows. First, we quantify the interaction energy profile of 54

hexagonal, rod- and star-shaped inclusions1. We show that our finite difference 55numerical method for computing energies agrees very well with analytical formulae 56(using Fourier-Bessel series) in most cases. After computing the interaction energies, we 57solve first-passage time problems to find the time scales over which two inclusions 58coalesce due to attractive interactions. We use both Langevin dynamics and the 59Fokker-Planck equation to solve first passage time problems and study both isotropic 60and anisotropic problems with reflecting/absorbing boundary conditions. Finally, we 61summarize our results in the discussion and conclusion sections and point to various 62enrichments that can be implemented following our earlier work [26]. 63

Table 1. List of parameters

Symbol Description Units Typical values` side length of triangular grid nm 2.5Kb bending modulus pN·nm 82.8 [6]T temperature K 300kB Boltzmann constant N· m· K−1 1.38× 10−23Kt thickness deformation modulus pN· nm−1 248.4 [6]r separations between two inclusions nm 9− 125F applied tension pN· nm−1 0.1− 10a unperturbed bilayer half-thickness nm 1.75 [11]R1(θ1) shape function for the centered inclusion nmR2(θ2) shape function for the moving inclusion nmθ the angle between two inclusions and horizontal line (see Fig 2(b)) radian/degreeu thickness deformation nmui thickness deformation at node i nmAe the area of the triangular element at node i nm

2

R1 radius of the inner boundary w.r.t (r1, θ1) nmR2 radius of the outer boundary w.r.t (r2, θ2) nmub the vector of all nodes determined by Eq (14)-(15) and inside inclusionsua the vector of all nodes that are not in ubu u = [ua

T,ubT]T

φ Energy of the system pN·nmn the number of refined grid elements in one side of the original triangle gridν(νij) translational drag coefficient (tensor) s·pN·nm−1 2.32× 10−5D(Dij) diffusion coefficient (tensor) nm

2· s−1 1.76× 105

1We are limited in the shapes we can explore by the equilateral triangle grid used in our computations.

September 17, 2020 3/23




Energy landscape 64

Analytical solution based on Fourier-Bessel basis 65

(a) (b)

Fig 1. (a) Schematic of bilayer deformations due to a thickness mismatchbetween hydrophobic region of a bilayer leaflet and an embedded protein.(b) The two types of boundary conditions that are used in this work.Dirichlet boundary condition Ui(θi) gives the thickness deformation alongthe boundary of inclusion i, while the slope boundary condition∇u · n̂ = U ′i(θi) determines the derivative along normal directions at eachpoint along the boundary of inclusion i. The top view of the surroundinglipid molecules (green circles) are only shown along the horizontal line, butthey are everywhere on the plane.

(a) (b) (c)

Fig 2. (a) The initial configuration of a system of two inclusions. The fixedinclusion located at the center (blue) has local coordinate: (r1, θ1) and themoving inclusion (purple) has local coordinate: (r2, θ2). (b) The inclusionon the right moves to the green spot and forms an angle θ with thehorizontal line. (c) The energy of the configuration here is the same as theone in (b). Note that the hexagons in (c) are rotated when compared tohexagons in (a).

We consider a circular lipid membrane with radius R2 and two inclusions embedded in 66it. Our first goal is to compute the energy landscape seen by an inclusion interacting 67with another inclusion on a flat membrane. The interactions between the inclusions are 68a result of the overlap of membrane thickness deformation fields in their vicinity. The 69interaction energy will be computed by considering two inclusions, one fixed and the 70other moving as shown in Fig 2(a). The coordinate frame at the fixed inclusion (blue) 71denoted as inclusion 1 (r1, θ1) is set to be the default one. Assume that the moving 72inclusion (purple) denoted as inclusion 2 initially stays in the same orientation as 73inclusion 1 (see Fig 2(a)). To keep the analysis simple, when an inclusion moves we do 74not consider its rotational diffusion. As inclusion 2 moves from the initial configuration 75to the green spot and forms an angle θ with the horizontal line (see Fig 2(b)), the 76

September 17, 2020 4/23




energy of the system can be computed by rotating both inclusions anticlockwise by 77angle θ from the initial configuration (see Fig 2(c)). This interaction energy will enter 78our analysis of the kinetics of the moving inclusion due to Brownian motion. 79

The elastic energy due to thickness deformation is given by [7, 8, 10,27], 80

φ =1

2

∫ {Kb(∇2u)2 +Kt

(ua

)2+ F

[2u

a+ (∇u)2

]}dxdy, (1)

The Euler-Lagrange equation associated with Eq (1) is given by [7], 81

Kb∇4u− F∇2u+Kta2u+

F

a= 0. (2)

Eq (2) can be reduced to the following form using the transformation ū = u+ FaKt , 82

(∇2 − ν+)(∇2 − ν−)ū = 0, ν± =1

2Kb

[F ±

(F 2 − 4KbKt

a2

)1/2]. (3)

First, we consider the case of an infinitely large circular membrane with R2 →∞ 83without applied tension (F = 0). We assume natural boundary condition which means 84that u = ū→ 0 at ∞. Let inclusion 2 be on the right side of inclusion 1. Then, a 85Fourier-Bessel series solution for the thickness deformation field around each inclusion 86i(i = 1, 2) can be obtained, 87

ū±i (ri, θi) = A±i,0K0(

√ν±ri) +

∞∑n=1

A±i,nKn(√ν±ri) cosnθi+B

±i,nKn(

√ν±ri) sinnθi. (4)

For small applied tension F and large membrane size R2, we used Eq (4) as an 88approximation for the solution of ū±i . Since the Euler-Lagrange equation (Eq (2)) is 89linear, the solution for Eq (3) is given by [8], 90

ū = ū+1 (r1, θ1) + ū−1 (r1, θ1) + ū

+2 (r2, θ2) + ū

−2 (r2, θ2), (5)

in which we used the coordinate transformations, 91

r2 =√r2 + r21 − 2rr1 cos θ1 , F1(r1, cos θ1, r), cos θ2 = (−r + r1 cos θ1)/r2,

sin θ2 = r1 sin θ1/r2; r1 =√r2 + r22 + 2rr2 cos θ2 , F2(r2, cos θ2, r),

cos θ1 = (r + r2 cos θ2)/r1, sin θ1 = r2 sin θ2/r1. (6)

In order to efficiently apply the boundary conditions, we rewrite ū2 as a function of 92r1, θ1, r and ū1 as a function of r2, θ2, r, 93

û±1 (r2, θ2, r) = A±1,0K0(

√ν±F2(r2, cos θ2, r)) (7)

+

N∑n=1

A±1,nKn(√ν±F2(r2, cos θ2, r))Tn

(r

r1+r2r1

cos θ2

)+B±1,nKn(

√ν±F2(r2, cos θ2, r))Un−1

(r

r1+r2r1

cos θ2

)r2r1

sin θ2,

û±2 (r1, θ1, r) = A±2,0K0(

√ν±F1(r1, cos θ1, r)) (8)

+N∑n=1

A±2,nKn(√ν±F1(r1, cos θ1, r))Tn

(− rr2

+r1r2

cos θ1

)+B±2,nKn(

√ν±F1(r1, cos θ1, r))Un−1

(− rr2

+r1r2

cos θ1

)r1r2

sin θ1.

September 17, 2020 5/23




Let h1 = ū+1 + ū

−1 , h2 = ū

+2 + ū

−2 , ĥ1 = û

+1 + û

−1 , ĥ2 = û

+2 + û

−2 . We consider the 94

following type of boundary conditions (see Fig 1(b)), 95

(h1 + ĥ2)(R1(θ1), θ1, r) = U1(θ1)

n̂ ·

(∂(h1 + ĥ2)

∂r1+

1

r1

∂(h1 + ĥ2)

∂θ1

)(R1(θ1), θ1, r) = U ′1(θ1) (9)

(ĥ1 + h2)(R2(θ2), θ2, r) = U2(θ2)

n̂ ·

(∂(ĥ1 + h2)

∂r2+

1

r2

∂(ĥ1 + h2)

∂θ2

)(R2(θ2), θ2, r) = U ′2(θ2). (10)

We can solve for the 4(2N + 1) coefficients 96A±1,0, A

±2,0, A

±1,n, A

±2,n, B

±1,n, B

±2,n, n = 1, 2, · · · , N because Eq (9)-(10) result in a linear 97

system. This determines the full deformation field due to the overlap of the 98deformations caused by both inclusions. The next step is to compute the energy φ(r) 99due to this deformation field. Note that the angular dependence of φ(r) appears in the 100shape functions of two inclusions, R1,R2. 101

Using the divergence theorem, the total energy expression in Eq (1) can be converted 102to the sum of line integrals over the boundary terms, i.e. φ = φ1 + φ2 with φi given by, 103

φi =1

2G0 +

1

2

∫∇ ·[Kb(∇ū)∇2ū−Kbū∇3ū+ Fū∇ū

]dxdy

=1

2G1 −

1

2

∫ 2π0

n̂ ·[Kb(∇ū)∇2ū−Kbū∇3ū+ Fū∇ū

]√R2i (θi) +R′2i (θi)dθi

=1

2G1 −

1

2

∫ 2π0

[U ′i(θ) (Kb(ν+ū+ + ν−ū−) + Fū)−Kbūn̂ · ∇(ν+ū+ + ν−ū−)]√R2i (θi) +R′2i (θi)dθi. (11)

From the first line to the second line we assume the line integral along the outer 104boundary is a constant (which works out to 0 as R2 →∞ and F → 0) w.r.t r and put it 105into the G1 term (both G0 and G1 are constants). The energy φ(r) can be computed 106relatively efficiently using this technique. This is important since φ(r) must be 107computed repeatedly as inclusion 2 moves and r changes due to Brownian motion when 108we solve the first passage time problem. We will also need the forces acting on inclusion 1092 in our analysis later. Eq (11) gives an expression to compute the force analytically, 110which in the special case of an isotropic φ(r) (i.e., no angular dependence) works out to 111

φ′i(r) = −Ri2

∫ 2π0

[U ′i(θ)

(Kb(ν+ū

′+ + ν−ū

′−) + Fū

′)−Kbū′n̂ · ∇(ν+ū+ + ν−ū−)−Kbūn̂ · ∇(ν+ū′+ + ν−ū′−)

]dθi. (12)

When there is only one circular inclusion in the membrane, the thickness deformation 112field in Eq (5) has a closed form solution [7] which can be compared to the simulation 113result of Klingelhoefer et al. [28] who studied radial bilayer thickness profiles for the Gα 114nanopore (among many others). We used the same parameters and boundary conditions 115as they did: a = 34.19Å, U1 = 0.81Å, U

′1 = 0.7, R1 = 10Å and fit their curves by 116

choosing Kt = 120pN· nm−1 and Kb = 2pN· nm. The agreement shown in Fig 3 117justifies the analytical method used here. 118

September 17, 2020 6/23




0 20 40 60 8033.5

34

34.5

35

35.5

36

36.5analytical

simulations

Fig 3. Red squares are data from the simulation by Klingelhoefer et al. [28] and theblack curve is fitted using an analytical solution. The good agreement of the twoprofiles suggests that the energy functional Eq (1) and the associated Euler-Lagrangeequation are a good starting point for estimating interaction energies of inclusions.

Finite difference method based on refined grid 119

The above analysis gives us a semi-analytical technique to compute φ(r). However, not 120all problems can be solved analytically, so a computational method is needed to 121estimate interaction energies. Fortunately, Eq (1) can be minimized using a finite 122difference method. We discretize the membrane using equilateral triangle elements as 123in [29,30]. The thickness deformation at node i is denoted by ui. Since the thickness 124deformation u changes rapidly in the neighborhood of the two inclusions, we used 125refined grids in a region containing the two inclusions and non-refined grids in the 126region far away (see Fig 4(b)) to get accurate solutions with low computational costs. 127Based on the triangular grids used, we study three types of inclusions shown in Fig 4(a). 128

(a) (b)

Fig 4. (a) Three types of inclusions studied in this paper: hexagon(red),star(purple), rod(green). (b) Refined grids are implemented in a regioncontaining two inclusions. Here n = 2.

Using methods similar to those in [29,30] the energy is first written in a discrete 129form and then the thickness deformation field that minimizes this energy is computed. 130

September 17, 2020 7/23




Finally, the minimizer is plugged back into the energy expression, and we have, 131

φ = minui

KbAe∑i

(6∑c=1

uci − 6ui

)24

9`4+∑i

AeKt

(uia

)2+∑i

2FuiAea

+

∑i,j,k

FAe3`2

[(ui − uj)2 + (uj − uk)2 + (uk − ui)2

]= min

uuTMu + λuT1u,(13)

where 1u is a column vector of size len(u) with all entries 1 and λ =2FAea . It can be 132

shown that the boundary condition in [22] can be written in the discrete form, 133

u(ri, θi) = U(θi), ∀i on the boundary (14)uk − 12 (ui + uj)√

3`/2= U ′(θi′), ∀ i, j, k pairs along the boundary (see Fig 4(a)).(15)

Note that ui, uj are given in Eq (14) and thus uk can be sloved from Eq (15) 134immediately. We also assume the inclusions are stiff such that for all nodes i inside the 135inclusions ui are a constant (equal to the value of those at boundary nodes). Hence, 136Eq (13) can be rewritten as, 137

φ = minua

[uaub

]T [Maa MabMTab Mbb

] [uaub

]+ λ

[uaub

]T[1a1b

]= uTa Maaua + 2u

Ta Mabub + u

Tb Mbbub + λu

Ta 1a + λu

Tb 1b. (16)

Taking∂(uTMu+λuT1u)

∂ua= 0, we get ūa = −Maa−1(Mabub + λ21a) at which Eq (16) is 138

minimized where 1a is a column vector of size len(ua) with all entries 1. Then, we can 139write the minimized total energy as, 140

φ =

(λ

21Ta + u

Tb M

Tab

)M−1aa

(Mabub +

λ

21a

)− 2

(λ

21Ta + u

Tb M

Tab

)M−1aa Mabub

+uTb Mbbub − λ(λ

21Ta + u

Tb M

Tab

)M−1aa 1a + λu

Tb 1b. (17)

Applications to hexagon, rod and star shaped inclusions 141

We now focus on the interaction of two hexagon shaped inclusions on a lipid membrane 142which has a rotational periodicity of π/3. We use Eq (17) derived from our numerical 143method and Eq (11) derived from the analytical method, to compute the interaction 144energy of two inclusions with distance r and then make comparisons. Agreement of the 145results from the two methods is expected. As shown in Fig 5(a), the energy computed 146using the analytical method for two inclusions separated by distance r in two different 147orientations (shown in the inset differing by a rotation of π/6) are almost the same. 148Hence, we can simplify our model and consider the energy landscape generated by two 149hexagon inclusions as being almost isotropic (insensitive to rotation). In Fig 5(b), we fix 150the shape of two hexagons (see inset of the figure), and show that as the number of grid 151elements per side n increases, the match between the energy computed from the 152numerical method and analytical method gets better, justifying our numerical approach 153of using refined grids near the inclusions. From Fig 5(c) we learn that as applied tension 154increases, the attraction at small separations (around r = R1 = 7 nm) becomes weaker 155but the repulsive force at around 9− 10nm also becomes weaker. Please check those to 156be sure. 157

September 17, 2020 8/23




6 8 10 12 14 16-15

-10

-5

0

5

10

(a)

6 8 10 12 14 16

-20

-15

-10

-5

0

5

10numerical (n=10)

numerical (n=15)

numerical (n=20)

analytical

(b)

6 8 10 12 14 16 18-20

-15

-10

-5

0

5

100.1pN

1pN

10pN

(c)

Fig 5. (a) The energy computed by analytical method using Eq (11)between two configurations of hexagon inclusions up to a rotation by π/6under F = 1pN·nm−1. (b) The energy of the configuration underF = 1pN·nm−1 computed numerically using Eq (17) converged to theenergy computed by Eq (11) as the number of elements increases. (c) Acomparison of the energy profiles for three different applied tension:0.1pN·nm−1, 1pN·nm−1, 10pN·nm−1.

8 10 12 14 16 18 20-80

-60

-40

-20

0

20

0°

16.1°

30°

49.1°

60°

79.1°

90°

(a)

8 10 12 14 16 18-60

-40

-20

0

20

0° numerical

0° analytical

90° numerical

90° analytical

(b)

Fig 6. (a) The energy of two shaped rod inclusions with different θ underapplied tension 1pN·nm−1. (b) The energy computed by analytical methodusing Eq (11) fits the one computed by numerical method using Eq (17)both with θ = 0◦ and θ = 90◦ under applied tension 1pN·nm−1.

Fig 6(a) shows that the energy landscape of two rod shaped inclusions is anisotropic 158- at small separations the force is repulsive at θ = 0◦ and becomes attractive at some 159angle around 40◦ < θ < 50◦. The attraction increases as θ goes up to 90◦. This 160behavior of the energy of two rod shaped inclusions is reminiscent of the energy from 161out-of-plane deflection for two rods [26] on a lipid membrane. Fig 6(b) shows that the 162energy computed by the numerical method and analytical method again agree very well 163which gives us confidence in the numerical method. 164

September 17, 2020 9/23




8 10 12 14 16 18 20-80

-60

-40

-20

0

20

0°

16.1°

30°

49.1°

60°

79.1°

90°

(a)

8 10 12 14 16 18 20-80

-60

-40

-20

0

20

0°

16.1°

30°

49.1°

60°

79.1°

90°

(b)

Fig 7. (a) The energy of two rod shape inclusions under different θ withapplied tension 0.1pN·nm−1. (b) The energy of two rod shape inclusionsunder different θ with applied tension 10pN·nm−1.

Next we compute the interaction energy of rod shaped inclusions for tensions 165F = 0.1pN·nm−1 and F = 10pN·nm−1. The comparison between Fig 7 and Fig 6(a) 166shows that as applied tension increases, the force becomes weaker at short separations, 167which implies that elastic interactions could be weakened by strong applied tension. 168Physically, this is reasonable, since high tension will tend to make the membrane flatter 169so that the thickness is more uniform everywhere. 170

10 12 14 16 18 20-150

-100

-50

0

50

(a)

8 10 12 14 16 18-60

-40

-20

0

20

0.1pN

1pN

10pN

(b)

Fig 8. (a) Solid lines are the energy computed by numerical method usingEq (17) and dashed lines are the energy computed by analytical methodusing Eq (11). The applied tension is 1pN·nm−1. (b) The energy computedby numerical method under three applied tensions: 0.1pN·nm−1,1pN·nm−1, 10pN·nm−1.

Next, we apply both methods to compute the interaction energy of two star shaped 171inclusions in Fig 8. Just as in the case of hexagons, we consider various orientations of 172the star shaped inclusions as shown in the inset of Fig 8(a). The match between the 173analytical method and numerical method is not as good in this case because the star 174shaped inclusion has 12 vertices at which the derivative along normal directions are 175discontinuous. Since in the analytical method we used Fourier-Bessel series to 176approximate the contour (R1,R2) and the derivative along normal directions to the 177boundaries, it requires a large number of terms N to obtain a good approximation. This 178is computationally not feasible for symbolic operations in MATLAB. Thus, we have 179greater confidence in our finite difference numerical method to compute interaction 180energies in complex geometries. In Fig 8(b) we use our numerical method to compute 181

September 17, 2020 10/23




the interaction energies for star shaped inclusions for various values of F . The trends 182are similar to those seen for hexagon shaped inclusions. 183

First passage time for isotropic inclusions under 184mixed boundary condition 185

Our main goal in this paper is to study the kinetics of an inclusion diffusing in an 186energy landscape resulting from elastic interactions with another inclusion. Efficient 187methods to compute the energy landscape developed above are a pre-requisite to this 188exercise. We will now use these methods to solve first passage time problems. 189

We consider a circular membrane of size R2 = 125 nm with a circular inclusion of 190size 2.5 nm fixed at the center. Another circular inclusion of the same size is diffusing 191around driven by stochastic forces. Recall from the energy landscape that there are 192attractive interactions between inclusions when the separations are small. Hence, if the 193moving inclusion comes close enough to the static one at the center then it will be 194strongly attracted. Therefore, we assume that at R1 = 7 nm there is an absorbing wall 195at which the moving inclusion will disappear by being attracted towards the center. We 196assume that at R2 = 125 nm (far away) there is a reflecting wall where the moving 197inclusion will be bounced back. Note that problems in which both boundaries are 198absorbing were solved elsewhere [26]. The exercise we will perform now is as follows. 199We place the second inclusion randomly on a circle of radius r = y at time t = 0 and let 200it diffuse around. At some time t = Tin when the inclusion hits the inner boundary for 201the first time we stop it from diffusing and record Tin. We repeat this experiment a 202large number of times and record Tin for each repetition. The mean value of Tin is the 203mean first passage time T1. Our goal is to find T1(y) as a function of the initial 204condition r = y. This can be done analytically or through a Langevin dynamics 205simulation. We will use both methods in the following. 206

To estimate T1(y) analytically we first need to compute survival probabilities. Let p 207be the probability density (for finding the inclusion) at position r and angle θ given 208

initial condition r = y, θ = α and P (r, t|y) =∫ 2π0p(r, t, θ|y, α)dθ. The probability 209

density p is independent of θ since neither the energy landscape nor the diffusion (or 210drag) coefficient of the inclusion depends on it. As a result, the Fokker-Planck equation 211for the evolution of this probability is in the following isotropic form [26], 212

∂P

∂t=

∂

∂r

[1

ν

∂φ

∂rP +D

∂P

∂r

]+

1

r

[1

ν

∂φ

∂rP +D

∂P

∂r

], (18)

with Dirichlet boundary condition at the inner boundary and Robin boundary condition 213at the outer boundary [31], 214

P (R1, t) = 0,

(kBT

∂P

∂r+∂φ

∂rP

)∣∣∣∣(R2,t)

= 0, ∀t ≥ 0. (19)

In the above D is a diffusion coefficient of the inclusion in the lipid membrane and ν is 215a drag coefficient which are related by the Nernst-Einstein relation νD = kBT [26]. Let 216S(y, t) be the survival probability, 217

S(y, t) =

∫ R2R1

P (r, t|y)rdr. (20)

Then, we can get the first passage time density, 218

f(y, t) = −∂S(y, t)∂t

= −∫ R2R1

∂P (r, t|y)∂t

rdr. (21)

September 17, 2020 11/23




The existence of the first moment of P (r, t|y) with respect to time t can be shown from 219the fundamental solution constructed by Itô in [32]. Then, tP (r, t|y)→ 0 as t→∞. 220Accordingly, the first passage time T1(y) can be derived from Eq (21), 221

T1(y) =

∫ ∞0

f(y, t)tdt = −∫ ∞0

∫ R2R1

∂P (r, t|y)∂t

rdrtdt

=

∫ R2R1

∫ ∞0

P (r, t|y)dtrdr =∫ R2R1

g1(r, y)rdr, (22)

where g1 is defined by, 222

g1(r, y) =

∫ ∞0

P (r, t|y)dt. (23)

Theorem 1: The ODE for T1(y) with a reflecting wall at the outer boundary and an 223absorbing wall at the inner boundary is 224

∂2T1(y)

∂y2+

(− 1kBT

∂φ

∂y+

1

y

)∂T1(y)

∂y+

1

D= 0, (24)

with boundary conditions, 225

T1(R1) = 0, T′1(R2) = 0. (25)

Proof: See Proof of Theorem 1. 226Next, we describe how to estimate T1(y) using Langevin dynamics simulations. The 227

overdamped version of the Langevin equation in an isotropic setting is given by [26], 228

dri = −1

ν

∂φ

∂ridt+

√2kBTdt

νξi, (26)

where i represents two perpendicular directions of the motion. ν is the translational 229drag coefficient of a circular inclusion given by the Saffman-Delbrck model [33]. 230ξi ∼ N (0, 1), a normally distributed random variable with mean 0 and variance 1, 231represents the stochastic force along direction i. We initially put the moving particle 232somewhere at r = y, and choose a time step dt that ensures convergence of the Lagenvin 233dynamics simulation. Then, for each time step dt, we perform the calculation in 234Eq (26), updating the position of the moving inclusion. We record the time at which 235the moving particle hits the absorbing wall at R1. We run 8000 simulations and then 236take an average to estimate the first passage time. For more details the readers are 237referred to [26]. Fig 5(a) and Fig 8(a) show that the φ(r) for hexagon and star 238inclusions can be regarded as nearly isotropic. We use Eq (24) to numerically solve for 239the first passage time and compare the results obtained from the two methods. 240

September 17, 2020 12/23




0 50 100 1500

0.05

0.1

0.15

0.2

0.1pN

1pN

10pN

(a)

0 50 100 1500

0.05

0.1

0.15

0.2

0.1pN

1pN

10pN

(b)

Fig 9. The first passage time for hexagon shaped inclusions is computedusing (a) Langevin dynamics simulations in Eq (26), (b) ODE in Eq (24)under three applied tensions 0.1pN·nm−1, 1pN·nm−1, 10pN·nm−1.

Fig 9 shows that the first passage time for hexagonal inclusion derived from the two 241methods are in good agreement. As the applied tension increases, the first passage time 242is reduced at most r that are not close to R1. At first glance this might seem 243counter-intuitive because from Fig 5(c) we know that at small separations (close to R1) 244the attraction force becomes weaker as applied tension increases. However, there is a 245stronger repulsive force at around r = 9− 10nm under large applied tension which slows 246the motion of the moving particle from a large starting separation. 247

0 50 100 1500

0.05

0.1

0.15

0.2

0.1pN

1pN

10pN

(a)

0 50 100 1500

0.05

0.1

0.15

0.2

0.1pN

1pN

10pN

(b)

Fig 10. The first passage time for star-shaped inclusions are computed by(a) Langevin dynamics simulations in Eq (26), (b) ODE in Eq (24) underthree applied tensions 0.1pN·nm−1, 1pN·nm−1, 10pN·nm−1.

The first passage time computed by the two methods is also in good agreement when 248the inclusions are star shaped. The order of the first passage time is the same as the 249hexagonal inclusions and similar arguments for the shape of the curves can be made 250here. 251

September 17, 2020 13/23




First passage time for anisotropic inclusions under 252mixed boundary condition 253

For two non-circular inclusions, the corresponding Fokker-Planck equation is a partial 254differential equation of parabolic type [26], 255

∂p

∂t=

∂

∂xi

[ν−1ij

∂φ

∂xjp

]+

∂2

∂xi∂xj

[Dijp

]=

1

νa

(∂2φ

∂x21p+

∂φ

∂x1

∂p

∂x1

)+

1

νb

(∂2φ

∂x22p+

∂φ

∂x2

∂p

∂x2

)+Da

∂2p

∂x21+Db

∂2p

∂x22.(27)

Accordingly, we need to redefine the first passage time in Eq (22) which is now given by, 256

T1(y, α) =

∫ ∞0

f(y, α, t)tdt = −∫ ∞0

∫ R2R1

∫ 2π0

∂p(r, θ, t|y, α)∂t

rdrdθtdt (28)

=

∫ R2R1

∫ 2π0

∫ ∞0

p(r, θ, t|y, α)dtdθrdr =∫ R2R1

∫ 2π0

q1(r, θ|y, α)dθrdr,(29)

where tp(r, θ, t|y, α)→ 0 as t→∞ is implemented in the first equation of the second 257line and q1 is defined by, 258

q1(r, θ|y, α) =∫ ∞0

p(r, θ, t|y, α)dt. (30)

Theorem 2: The PDE for T1(y, α) with a reflecting wall at the outer boundary and an 259absorbing wall at the inner boundary is given below, 260

(Da cos

2 α+Db sin2 α) ∂2T1∂y2

+

(Da

sin2 α

y2+Db

cos2 α

y2

)∂2T1∂α2

+(−Da

sin 2α

y+Db

sin 2α

y

)∂2T1∂y∂α

+

[Da

sin2 α

y+Db

cos2 α

y+

1

νa

(sin 2α ∂φ∂α

2y

− cos2 α∂φ∂y

)− 1νb

(sin 2α ∂φ∂α

2y+ sin2 α

∂φ

∂y

)]∂T1∂y

+

[Da

sin 2α

y2−Db

sin 2α

y2

+1

νa

(sin 2α∂φ∂y

2y−

sin2 α ∂φ∂αy2

)− 1νb

(sin 2α∂φ∂y

2y+

cos2 α ∂φ∂αy2

)]∂T1∂α

+ 1 = 0,(31)

with boundary conditions 261

T1(R1, α) = 0,∂T1∂y

(R2, α) = 0, T1(y, 0) = T1(y, 2π). (32)

Proof: See Proof of Theorem 2. 262The overdamped Langevin equation in an anisotropic setting is given by [26], 263

dri = −ν−1ij∂φ

∂rjdt+

√2kBTdt

νiiξi (no sum in the second term), (33)

where i represents two perpendicular directions of the motion and νij and ξi are the 264diffusion tensor and random force tensor (for more details see [26]). Fig 6(a) shows that 265the interaction energy φ(r) for rod shaped inclusions depends on θ (it is anisotropic). In 266the Langevin dynamics calculations, for each initial position y we use Eq (33) to run 267

September 17, 2020 14/23




8000 simulations with a reflecting wall at R2 and an absorbing wall at R1 for four 268θ = 0◦, 30◦, 60◦, 90◦ and then take an average (for each θ separately) to estimate the 269first passage time. We also use Eq (31) to numerically solve the first passage time and 270compare the results derived from the two methods for F = 0.1, 1, 10 pN/nm. 271

0 50 100 1500

0.05

0.1

0.15

0/6 pi

1/6 pi

2/6 pi

3/6 pi

(a)

0 50 100 1500

0.05

0.1

0.15

0/6 pi

1/6 pi

2/6 pi

3/6 pi

(b)

Fig 11. The first passage time for rod-shaped inclusions computed from (a)Langevin dynamics using Eq (33), (b) PDE using Eq (31).

The good agreement between the first passage time solved from the PDE in Eq (31) 272and estimated by Langevin equation once again shows that our methods are work well. 273As shown in Fig 11, as the initial angle increases from 0◦ to 90◦, the first passage time 274decreases at small separations but increases at large separations. This can be similarly 275explained by the fact that stronger attractive force near R1 pulls the moving particle to 276be absorbed faster from smaller initial separations while stronger repulsive force around 27712− 16 nm leads to a larger first passage time when the particle is initially located at a 278large distance. 279

0 50 100 1500

0.05

0.1

0.15

0/6 pi

1/6 pi

2/6 pi

3/6 pi

(a)

0 50 100 1500

0.05

0.1

0.15

0/6 pi

1/6 pi

2/6 pi

3/6 pi

(b)

Fig 12. The first passage time for rod-shaped inclusions computed from (a)Langevin dynamics using Eq (33), (b) PDE using Eq (31).

The result of the first passage time under 1pN/nm applied tension in Fig 12 is 280similar to the one under 0.1pN/nm applied tension. 281

September 17, 2020 15/23




0 50 100 1500

0.05

0.1

0.15

0/6 pi

1/6 pi

2/6 pi

3/6 pi

(a)

0 50 100 1500

0.05

0.1

0.15

0/6 pi

1/6 pi

2/6 pi

3/6 pi

(b)

Fig 13. The first passage time computed from (a) Langevin dynamics usingEq (33), (b) PDE using Eq (31).

Compared to the results under 0.1,1pN·nm−1 applied tension, the first passage time 282is reduced under 10pN·nm−1 applied tension. The order of magnitude of the first 283passage time under all three tensions is similar. 284

Discussion 285

This paper has two major parts. In the first part we use a finite difference method to 286compute the interaction energy of two inclusions due to membrane thickness 287deformations. In the second part we use the computed energy landscape to solve first 288passage time problems. Our method to compute energies is different from the analytical 289method in [7,8] which uses perturbation theory to study thickness mediated interactions 290between two anisotropic inclusions; we implement an approach to compute the energy 291using the divergence theorem which is more general and can deal with strongly 292anisotropic inclusions. The advantage of analytical methods in both [7, 8] and this work 293is that they can compute the energy accurately at small applied tension F if enough 294terms in the Fourier-Bessel series are used. However, it is time consuming to compute 295the coefficients in the Fourier-Bessel series and this becomes computationally infeasible 296when the inclusions are strongly anisotropic. On the other hand, our numerical method 297is able to handle arbitrary values of F and can efficiently compute the interaction 298energy of two inclusions for different separations r given a fixed set of parameters 299(Kb,Kt, a etc.) which are stored in a pre-calculated stiffness matrix. 300

In the second part of the paper we compute the time to coalescence of two inclusions 301of various shapes as a function of the distance separating them. We use both Langevin 302dynamics and a PDE to arrive at our estimates. For two inclusions separated by about 303125 nm we predict that the time to coalescence is hundreds of milliseconds irrespective 304of the shape of the inclusion. The time to coalescence with only membrane bending 305interactions was of similar magnitude as shown in [26]. The order of magnitude of the 306time to coalescence is the same even though the attractive force due to membrane 307thickness interactions is stronger than that due to membrane bending interactions in [26] 308at small separations. The reason is that even with membrane thickness interactions the 309attractive force decays to zero quickly and Brownian motion dominates the kinetics of 310the moving particle in most regions, just as in [26]. Therefore, at small separations the 311first passage time with thickness mediated interactions is smaller than that with 312out-of-plane bending interactions, but is not very different at large separations. 313

September 17, 2020 16/23




Conclusion 314

In this paper we have analyzed the temporal self-assembly of inclusions due to 315interactions mediated by membrane thickness variations. A major accomplishment of 316the paper is to show that the results from Langevin dynamics simulations agree well 317with those obtained from a PDE for the first passage time. The approach based on the 318PDE is much faster than the Langevin dynamics simulation and could open new ways 319to study the process of self-assembly. This is a step beyond earlier studies which focused 320on the energy landscape of clusters of proteins, but did not look into kinetics. Some 321papers based on molecular simulation did consider the temporal process, but to the best 322of our knowledge most did not reach the time scales calculated in this paper. We close 323this paper by mentioning some effects that we did not consider. First, hydrodynamic 324interactions between inclusions (based on the Oseen tensor) were shown to speed up 325self-assembly in [26] and they are expected to have a similar effect here. Second, the 326temporal behavior of a cluster of inclusions are not studied in this paper due to 327limitations of computational power, but we expect the overall behavior to be similar to 328the clusters studied in our earlier work [26]. Third, only a limited set of inclusion shapes 329are considered in this paper, but it is found that the time to coalescence does not 330depend strongly on shape. We leave it to future work to add these refinements and 331extend this type of analysis to important functional proteins such as ion-channels [22]. 332

Supporting information 333

Proof of Theorem 1 Following techniques in [26, 31, 34], we integrate Eq (18) for P 334over all t ≥ 0, 335∫ ∞

0

∂P

∂tdt =

∂

∂r

[1

ν

∂φ

∂rg1 +D

∂g1∂r

]+

1

r

[1

ν

∂φ

∂rg1 +D

∂g1∂r

](34)

−1rδ(r − y) = Lrg1(r, y), (35)

where 1r δ(r − y) is the initial condition and the second order linear differential operator 336Lr : D(Lr) ⊂ C2([R1, R2])→ C2([R1, R2]) is defined as, 337

Lr =∂

∂r

[1

ν

∂φ

∂r+D

∂

∂r

]+

1

r

[1

ν

∂φ

∂r+D

∂

∂r

], (36)

with domain 338

D(Lr) ={v1 ∈ C2([R1, R2]) |v1(R1) = 0, kBTv′1(R2) + φ′(R2)v1(R2) = 0

}. (37)

Using the method in [26], we can get the adjoint operator L∗r which satisfies 339〈v2,Lrv1〉 = 〈L∗rv2, v1〉 , ∀v1 ∈ D(Lr), v2 ∈ D(L∗r), 340

L∗r = −1

ν

∂φ

∂r

∂

∂r+D

∂2

∂r2+

1

ν

∂φ

∂r

1

r−D

∂ 1r∂r. (38)

with domain 341

D(L∗r) ={v2 ∈ C2([R1, R2])

∣∣∣∣v2(R1) = 0, v2(R2)R2 − v′2(R2) = 0}, (39)

and the inner product is defined as, 342

〈v1, v2〉 =∫ R2R1

v1v2dr, ∀v1 ∈ D(Lr), v2 ∈ D(L∗r). (40)

September 17, 2020 17/23




Proofs for the existence of the solutions of second order inhomogeneous linear ordinary 343differential equation are well known. Hence, we can find a u0 ∈ C2([R1, R2]) s.t. 344L∗ru0(r) = r. Then, it follows from Eq (22) that, 345

T1(y) =

∫ R2R1

(L∗ru0(r)) g1(r, y)dr =∫ R2R1

u0(r) (Lrg1(r, y)) dr

= −∫ R2R1

u0(r)1

rδ(r − y)dr = −u0(y)

1

y, (41)

=⇒ Ly∗yT1(y) = −y. (42)

Using Eq (38), we can derive Eq (24), a second order ODE for T1(y). The boundary 346condition of T1(y) at the absorbing wall is straightforward [31,34]: T1(R1) = 0. For the 347boundary condition at the reflecting wall, we appeal to the Langevin equation in 348Eq (26). If the particle sits at position R2, decomposing the overdamped Langevin 349equation [26] into radial direction and angular direction, we have, 350

dy = −1ν

∂φ

∂ydt+

√2kBTdt

νξy, (43)

dθ = −1ν

1

y

∂φ

∂θdt+

1

y

√2kBTdt

νξθ. (44)

After time dt, the particle can only move to R2 + dy(dy < 0) along the radial direction 351because of the reflecting wall at R2. The motion along the angular direction can be 352neglected because T1(y) does not have dependence on angular direction. Note that dy is 353a random variable depending on ξy and dt with constraint R1 ≤ R2 + dy ≤ R2. Then, 354we can write 355

T1(R2) = dt+ C1(dt)

∫ 0C2(dt)

T1(R2 + dy)G(ξy)dξy

= dt+ C1(dt)

∫ 0C2(dt)

(T1(R2) + T′1(R2 + ηdy)dy)G(ξy)dξy

= dt+ T1(R2) + C1(dt)

∫ 0C2(dt)

(T ′1(R2) + ηdyT′′1 (R2 + βdy ))dyG(ξy)dξy

= dt+ T1(R2)−1

ν

∂φ

∂yT ′1(R2)dt+ C1(dt)

∫ 0C2(dt)

T ′1(R2)

√2kBTdt

νξyG(ξy)dξy

+C1(dt)

∫ 0C2(dt)

ηdydy

T ′′1 (R2 + βdy )

(2kBTdt

νξ2y + o(dt)

)G(ξy)dξy (45)

where we used mean value theorem twice to reach to Eq (45) with 356R2 + dy < R2 + ηdy < R2 + βdy < R2. Note that βdy depends on ηdy and thus depends 357on dy. C2(dt) is the value to satisfy R2 + dy = R1 for given dt and ξy. C1(dt) is the 358scaling factor such that the integral of probability density equals 1: 359

C1(dt)∫ 0C2(dt)

G(ξy)dξy = 1 where G(ξy) =e−ξ2y/2√

2π. After some re-arrangements and 360

dividing by dt on both sides, 361

−1 = −1ν

∂φ

∂yT ′1(R2) + T

′1(R2)

√2kBT

νdtC1(dt)

∫ 0C2(dt)

ξyG(ξy)dξy

+C1(dt)

∫ 0C2(dt)

ηdydy

T ′′1 (R2 + βdy )

(2kBT

νξ2y + o(1)

)G(ξy)dξy. (46)

September 17, 2020 18/23




As dt→ 0, C1 → 2, C2 → −∞. Note thatηdydy < 1 and we have |T

′′1 (R2 + βdy )| < M 362

for some M because T1 is C2. Then if we take t→∞ the third term in RHS of Eq (46) 363

can be bounded as, 364

limt→∞

∣∣∣∣∣C1(dt)∫ 0C2(dt)

ηdydy

T ′′1 (R2 + βdy )

(2kBT

νξ2y + o(1)

)G(ξy)dξy

∣∣∣∣∣≤ lim

t→∞C1(dt)

∫ 0−∞

∣∣∣∣ηdydy T ′′1 (R2 + βdy )(

2kBT

νξ2y + o(1)

)G(ξy)

∣∣∣∣ dξy≤ 2M lim

t→∞C1(dt)

∫ 0−∞

∣∣∣∣(2kBTν ξ2y)G(ξy)

∣∣∣∣ dξy≤ 4M

∫ 0−∞

∣∣∣∣(2kBTν ξ2y)G(ξy)

∣∣∣∣ dξy< ∞. (47)

The first term in the RHS of Eq (46) is independent of dt and thus is finite as t→∞. 365For the second term in RHS of Eq (46), limt→∞

∣∣∣C1(dt) ∫ 0C2(dt) ξyG(ξy)dξy∣∣∣

Then, we can derive a second order PDE for T1(y) (Eq (31)). For boundary conditions, 380we just need to worry about the reflecting wall. For anisotropic case, dy < 0. dθ could 381be either positive or negative. Similarly we can write, 382

T (R2, θ) = dt+ C1(dt)

∫ 0C2(dt)

∫ ∞−∞

T (R2 + dy, θ + dθ)G(ξθ)dξθG(ξy)dξy

= dt+ C1(dt)

∫ 0C2(dt)

∫ ∞−∞

[T (R2 + dy, θ) + Tθ(R2 + dy, θ + η

∗dy,dθ

)dθ]·(54)

G(ξθ)dξθG(ξy)dξy

= dt+ C1(dt)

∫ 0C2(dt)

[T (R2, θ) + Ty(R2 + ηdy , θ)dy

]G(ξy)dξy

+C1(dt)

∫ 0C2(dt)

∫ ∞−∞

[Tθ(R2 + dy, θ + η

∗dy,dθ

)dθ]G(ξθ)dξθG(ξy)dξy

= dt+ C1(dt)

∫ 0C2(dt)

(Ty(R2, θ) + ηdyTyy(R2 + βdy , θ))dyG(ξy)dξy

+C1(dt)

∫ 0C2(dt)

∫ ∞−∞

[Tθ(R2 + dy, θ)dθ]G(ξθ)dξθG(ξy)dξy + C1(dt) ·∫ 0C2(dt)

∫ ∞−∞

[Tθθ(R2 + dy, θ + β

∗dy,dθ

) (dθ)2η∗dy,dθdθ

]G(ξθ)dξθG(ξy)dξy

+T (R2, θ)

= dt+ T (R2, θ) + C1(dt)

∫ 0C2(dt)

ηdydy

Tyy(R2 + βdy , θ)(dy)2G(ξy)dξy

−1ν

∂φ

∂yTy(R2, θ)dt+ C1(dt)

∫ 0C2(dt)

Ty(R2, θ)

√2kBTdt

νξyG(ξy)dξy

−C1(dt)∫ 0C2(dt)

∫ ∞−∞

[Tθ(R2 + dy, θ)

1

ν

1

R2

∂φ

∂θdt

]G(ξθ)dξθG(ξy)dξy

+C1(dt)

∫ 0C2(dt)

∫ ∞−∞


∗dy,dθ

)

(2kBTdt

R22νξ2θ + o(dt)

)η∗dy,dθdθ

]G(ξθ)dξθG(ξy)dξy (55)

+C1(dt)

∫ 0C2(dt)

Tθ(R2 + dy, θ)1

R2

√2kBTdt

ν

∫ ∞−∞

ξθG(ξθ)dξθG(ξy)dξy,

where in the process to Eq (55) we used mean value theorem three times with 383θ < θ+ β∗dy,dθ < θ+ η

∗dy,dθ

< θ+ dθ if dθ > 0 and θ+ dθ < θ+ η∗dy,dθ < θ+ β∗dy,dθ

< θ if 384

September 17, 2020 20/23




dθ < 0. After some re-arrangements and dividing by dt on both sides, we get 385

−1 = C1(dt)∫ 0C2(dt)

ηdydy

Tyy(R2 + βdy , θ)

(2kBT

νξ2y + o(1)

)G(ξy)dξy

+C1(dt)

∫ 0C2(dt)

Ty(R2, θ)

√2kBT

νdtξyG(ξy)dξy

−C1(dt)∫ 0C2(dt)

[Tθ(R2 + dy, θ)

1

ν

1

R2

∂φ

∂θ

]G(ξy)dξy −

1

ν

∂φ

∂yTy(R2, θ)

+C1(dt)

∫ 0C2(dt)

∫ ∞−∞


∗dy,dθ

)

(2kBT

νR22ξ2θ + o(1)

)η∗dy,dθdθ

]·

G(ξθ)dξθG(ξy)dξy

+C1(dt)

∫ 0C2(dt)

Tθ(R2 + dy, θ)1

R2

√2kBT

νdt

∫ ∞−∞

ξθG(ξθ)dξθG(ξy)dξy. (56)

Using the continuity of Tθ, Tyy and Tθθ and the fact thatηdydy ,

η∗dy,dθdθ < 1, it is clear that 386

the terms in the first, third and fourth line of Eq (56) are finite as dt→ 0. The term in 387the last line is vanishing due to

∫∞−∞ ξθG(ξθ)dξθ = 0. Since the LHS of Eq (56) is finite 388

also, the term in the second line of Eq (56) must also be finite as dt→ 0. Accordingly, 389Ty(R2, θ) = 0 follows from limdt→∞

√2kBTνdt →∞. 390

Acknowledgments 391

We acknowledge support for this work through an NSF grant CMMI 1662101. 392

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