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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.
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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
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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.
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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
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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.
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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
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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
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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
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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
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-
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
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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)
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-
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
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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
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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
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-
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
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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
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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)
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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)
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-
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)
∫ ∞−∞
[Tθθ(R2 + dy, θ + β
∗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
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-
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)
∫ ∞−∞
[Tθθ(R2 + dy, θ + β
∗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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