A Multiscale Analysis of Diffusions on Rapidly Varying Surfaces

Abstract

Lateral diffusion of molecules on surfaces plays a very important role in various biological processes, including lipid transport across the cell membrane, synaptic transmission, and other phenomena such as exo- and endocytosis, signal transduction, chemotaxis, and cell growth. In many cases, the surfaces can possess spatial inhomogeneities and/or be rapidly changing shape. Using a generalization of the model for a thermally excited Helfrich elastic membrane, we consider the problem of lateral diffusion on quasi-planar surfaces, possessing both spatial and temporal fluctuations. Using results from homogenization theory, we show that, under the assumption of scale separation between the characteristic length and timescales of the membrane fluctuations and the characteristic scale of the diffusing particle, the lateral diffusion process can be well approximated by a Brownian motion on the plane with constant diffusion tensor $D$ that depends on a highly nonlinear way on the detailed properties of the surface. The effective diffusion tensor will depend on the relative scales of the spatial and temporal fluctuations, and for different scaling regimes, we prove the existence of a macroscopic limit in each case.

Appendix: Formal Multiscale Expansions

In this section, we formally derive the limiting equations in each of the scaling regimes using multiscale expansions.

1.1 Case I

To derive the homogenized equation in this regime, we make the ansatz that the solution $v^\varepsilon$ of (35) is of the form

$$\begin{aligned} v^{\varepsilon } = v_{0}(x,y,t) + \varepsilon v_{1}(x,y,t) + \varepsilon ^{2} v_{2}(x,y,t) + \cdots , \end{aligned}$$

(108)

for smooth $v_{i}:\mathbb {R}^{d}\times {\mathbb {T}}^{d} \times [0,T] \rightarrow \mathbb {R}^d$. Substituting (108) in (35) and identifying equal powers of $\varepsilon$, we obtain the following equations

$$\begin{aligned}&\begin{array}{ll} O\left( \frac{1}{\varepsilon ^2}\right) : &{} \quad {\fancyscript{L}}_{0}v_{0}(x,y,t) = 0,\\ \end{array}\end{aligned}$$

(109)

$$\begin{aligned}&\begin{array}{lr} O\left( \frac{1}{\varepsilon }\right) :&{} \quad {\fancyscript{L}}_{0}v_{1}(x,y,t) = -{\fancyscript{L}}_{1}v_{0}(x,y,t),\\ \end{array}\end{aligned}$$

(110)

$$\begin{aligned}&\begin{array}{lr} O\left( 1\right) : &{}\quad {\fancyscript{L}}_{0}v_{2}(x,y,t) = \frac{\partial v_{0}}{\partial t} -{\fancyscript{L}}_{1}v_{1}(x,y,t) -{\fancyscript{L}}_{2}v_{0}(x,y,t),\\ \end{array} \end{aligned}$$

(111)

for $(x,y,t) \in \mathbb {R}^d \times {\mathbb {T}}^d \times (0,T]$.

Since the nullspace of ${\fancyscript{L}}_{0}$ contains only constants in $y$, Eq. (109) thus implies that $v_{0}$ is a function of $x$ and $t$ only. Equation (110) becomes

$$\begin{aligned} {\fancyscript{L}}_{0}v_{1}(x,y,t) = -F(y)\cdot \nabla _{x}v_{0}(x,t). \end{aligned}$$

(112)

Let $\chi \in C^2({\mathbb {T}}^d; \mathbb {R}^{d})$ be the unique, mean-zero solution of the cell Eq. (39). If we choose $v_1 = \chi \cdot \nabla _x v_0(x,t)$, then it is clear that $v_{1}$ solves (110).

Finally, by the Fredholm alternative on ${\fancyscript{L}}_0$, a necessary condition for Eq. (111) to have a solution is that the RHS of (111) has mean zero with respect to the measure $\rho$, that is,

$$\begin{aligned} \frac{\partial v_{0}(x,t)}{\partial t} = \frac{1}{Z}\int _{{\mathbb {T}}^{d}} {\fancyscript{L}}_{1}v_{1}(x,y,t) \rho (y)\,\hbox {d}y + \frac{1}{Z}\int _{{\mathbb {T}}^{d}}{\fancyscript{L}}_{2}v_{0}(x,t) \rho (y)\,\hbox {d}y. \end{aligned}$$

Substituting $v_{0}$ and $v_{1}$, we obtain

$$\begin{aligned} \frac{\partial v_{0}(x,t)}{\partial t}&= \frac{1}{Z}\int _{{\mathbb {T}}^{d}} \nabla _{y}\cdot \left( \sqrt{\left|g \right|(y)}g^{-1}(y)\right) \cdot \nabla _{x}\left( \chi \cdot \nabla _{x}v_{0}(x,t)\right) \, \hbox {d}y \\&\qquad + \frac{2}{Z}\int _{{\mathbb {T}}^{d}}\sqrt{\left|g \right|(y)}g^{-1}(y):\nabla _{x}\nabla _{y}\left( \chi \cdot \nabla _{x}v_{0}(x,t)\right) \,\hbox {d}y \\&\qquad +\frac{1}{Z}\int _{{\mathbb {T}}^{d}}\sqrt{\left|g \right|(y)}g^{-1}(y):\nabla _{x}\nabla _{x}v_{0}(x,t)\,\hbox {d}y. \end{aligned}$$

Integrating the second term by parts with respect to $y$ and simplifying, we obtain

$$\begin{aligned} \frac{\partial v_{0}(x,t)}{\partial t} = \Bigl (\frac{1}{Z}\int _{{\mathbb {T}}^{d}} g^{-1}(y)\bigl (I + \nabla _{y}\chi (y)\bigl ) \sqrt{\left|g \right|(y)}\, \hbox {d}y\Bigl ):\nabla _{x}\nabla _{x}v_{0}(x,t), \end{aligned}$$

where we have used the symmetry of $g^{-1}$. Thus, the homogenized diffusion equation for $v_0$ is

$$\begin{aligned} \frac{\partial v_{0}(x,t)}{\partial t} = D:\nabla _{x}\nabla _{x}v_{0}(x,t), \end{aligned}$$

(113)

where

$$\begin{aligned} D = \frac{1}{Z}\int _{{\mathbb {T}}^{d}} g^{-1}(y)\bigl (I + \nabla _{y}\chi (y)\bigl ) \sqrt{\left|g \right|(y)}\, \hbox {d}y. \end{aligned}$$

Multiplying (39) by $\chi (y)\rho (y)$ and integrating by parts gives

$$\begin{aligned} \int _{{\mathbb {T}}^{d}}\left( I + \nabla _{y}\chi (y)\right) ^{\top } g^{-1}(y)\nabla _{y}\chi (y)\,\sqrt{\left|g \right|(y)}\,\hbox {d}y = 0, \end{aligned}$$

so that the effective diffusion matrix can be written in the following symmetric form

$$\begin{aligned} D = \frac{1}{Z}\int _{{\mathbb {T}}^{d}} \bigl (I + \nabla _{y}\chi (y)\bigl )^\top g^{-1}(y)\bigl (I + \nabla _y \chi (y)\bigl ) \sqrt{\left|g \right|(y)}\, \hbox {d}y. \end{aligned}$$

From the limiting backward Kolmogorov equation (113), we can read off the limiting SDE $\hbox {d}X^0(t) = \sqrt{2D}\,\hbox {d}B(t)$. A rigorous proof of this result can found in Duncan (2013).

1.2 Case II

Analogous to the previous case, we look for solutions $v$ of the form

$$\begin{aligned} v^\varepsilon (x ,\eta , t) = v_{0}(x,\eta ,t) + \varepsilon v_{1}(x,\eta ,t) + \cdots , \end{aligned}$$

for some smooth functions $v_i:\mathbb {R}^d \times \mathbb {R}^K\times [0,T] \rightarrow \mathbb {R}^d$. Substituting this ansatz in (72a) and identifying equal powers of $\varepsilon$, we obtain the following pair of equations

$$\begin{aligned}&\begin{array}{ll} O\left( \frac{1}{\varepsilon }\right) : &{}\quad {\fancyscript{L}}_0 v_0(x, \eta ,t) = 0, \\ \end{array}\end{aligned}$$

(114)

$$\begin{aligned}&\begin{array}{ll} O(1): &{}\quad \frac{\partial v(x, \eta , t)}{\partial t} = {\fancyscript{L}}_{0}v_1(x,\eta , t) + {\fancyscript{L}}_{1}v_0(x, \eta , t),\\ \end{array} \end{aligned}$$

(115)

where $(x, \eta , t) \in \mathbb {R}^d \times \mathbb {R}^K \times (0,T].$

The $O(\frac{1}{\varepsilon })$ equation immediately implies that $v_0$ is independent of the fast-scale fluctuations. The second equation then becomes

$$\begin{aligned} {\fancyscript{L}}_0 v_1(x, \eta , t) = \frac{\partial v(x, \eta , t)}{\partial t} - {\fancyscript{L}}_1 v_0(x, \eta , t). \end{aligned}$$

Applying the Fredholm alternative, a necessary condition for the existence of a solution $v_1$ is that the RHS is orthogonal to the invariant measure $\rho _\eta$, that is,

$$\begin{aligned} \frac{\partial v_0}{\partial t}(x,t) = \left[ \int _{\mathbb {R}^K}F(x,\eta )\rho _\eta (\eta )\right] \cdot \nabla v_0(x,t) + \left[ \int _{\mathbb {R}^K}\varSigma (x,\eta )\rho _\eta (\eta )\right] :\nabla \nabla v_0(x,t), \end{aligned}$$

which is the backward equation for SDE (74).

1.3 Case III

We make the ansatz that

$$\begin{aligned} v^\varepsilon = v_{0} + \varepsilon v_{1} + \varepsilon ^{2} v_{2} + \cdots , \end{aligned}$$

for some smooth functions $v_i:\mathbb {R}^d \times {\mathbb {T}}^d \times \mathbb {R}^K \times [0,T] \rightarrow \mathbb {R}$. Substituting $v^\varepsilon$ in (82a) and identifying equal powers of $\varepsilon$, we obtain the following equations

$$\begin{aligned}&\begin{array}{ll} O\left( \frac{1}{\varepsilon ^{2}}\right) : &{}\qquad {\fancyscript{L}}_{0}v_{0} = 0, \\ \end{array}\end{aligned}$$

(116)

$$\begin{aligned}&\begin{array}{ll} O\left( \frac{1}{\varepsilon }\right) : &{}\qquad {\fancyscript{L}}_{0}v_{1} = -{\fancyscript{L}}_{\eta }v_{0} -{\fancyscript{L}}_{1}v_{0} ,\\ \end{array}\end{aligned}$$

(117)

$$\begin{aligned}&\begin{array}{ll} O(1): \qquad \qquad &{}{\fancyscript{L}}_{0}v_{2} = -\left( \frac{\partial v_{0} }{\partial t} - {\fancyscript{L}}_{\eta }v_{1} - {\fancyscript{L}}_{1}v_{1} - {\fancyscript{L}}_{2}v_{0}\right) .\\ \end{array} \end{aligned}$$

(118)

The first equation implies that $v_{0} \in {\fancyscript{N}}[{\fancyscript{L}}_{0}]$ so that $v_{0}$ is a constant in $y$. The second equation thus becomes

$$\begin{aligned} {\fancyscript{L}}_{0}v_1(x,y, \eta , t) = \left( {\fancyscript{L}}_{\eta }v_{0}(x, \eta , t) + F(y,\eta )\cdot \nabla _{x}v_{0}(x, \eta , t)\right) . \end{aligned}$$

By the Fredholm alternative applied to ${\fancyscript{L}}_{0}$, we require that the RHS is centered with respect to $\sqrt{\left|g \right|}$, for each fixed $x$ and $\eta$ that is,

$$\begin{aligned} \int _{{\mathbb {T}}^{d}}\left( F(y,\eta )\cdot \nabla _{x}v_{0}(x,\eta , t) + {\fancyscript{L}}_{\eta }v_{0}(x, \eta , t) \right) \sqrt{\left|g \right|(y,\eta )}\,\hbox {d}y = 0. \end{aligned}$$

The first term in the above integral is clearly $0$. Since ${\fancyscript{L}}_{\eta }v_{0}$ is independent of $y$, the centering condition becomes

$$\begin{aligned} Z(\eta ){\fancyscript{L}}_{\eta }v_{0}(x, \eta , t) = 0. \end{aligned}$$

Since $Z > 1$, it follows that $v_{0} \in {\fancyscript{N}}[{\fancyscript{L}}_{\eta }]$ is a sufficient condition for the centering condition to hold, which we therefore will assume. By ergodicity of the Ornstein–Uhlenbeck process $\eta (t)$ over $\mathbb {R}^{K}$, it follows that $v_{0}$ is also independent of $\eta$ so that $v_{0}$ is a function of $x$ only. The second equation thus becomes

$$\begin{aligned} {\fancyscript{L}}_{0}v_1 = F(y,\eta )\cdot \nabla _{x}v_{0}. \end{aligned}$$

Let $\chi (\cdot , \eta )$ be the unique, mean-zero solution of the cell equation solution by the Fredholm alternative, since the centering condition holds. Choosing $v_{1} = \chi \cdot \nabla _{x}v_{0}$, it is clear that $v_1$ solves the $O(\frac{1}{\varepsilon })$ equation.

We now consider the $O(1)$ equation. By the Fredholm alternative, a necessary condition for the existence of a unique solution $v_{2}$ is that the RHS is centered with respect to the invariant measure of ${\fancyscript{L}}_{0}$, that is,

$$\begin{aligned} \frac{\partial v_{0}}{\partial t} =\int _{{\mathbb {T}}^{d}}\left( {\fancyscript{L}}_{\eta }v_{1} + {\fancyscript{L}}_{1}v_{1} + {\fancyscript{L}}_{2}v_{0}\right) \rho _{y}\, \hbox {d}y, \end{aligned}$$

which, substituting the definitions of the ${\fancyscript{L}}_{i}$’s and $v_{j}$’s, can be written as follows:

$$\begin{aligned} \frac{\partial v_{0}}{\partial t}&= \int _{{\mathbb {T}}^{d}} F\otimes \chi \, \rho _{y}\,\hbox {d}y: \nabla _{x}\nabla _{x}v_{0} \end{aligned}$$

(119a)

$$\begin{aligned}&\quad + \int _{{\mathbb {T}}^{d}} g^{-1}\nabla _{y}\chi \rho _{y} + \nabla _{y}\chi \, g^{-1}\rho _{y}\,\hbox {d}y: \nabla _{x}\nabla _{x}v_{0} \end{aligned}$$

(119b)

$$\begin{aligned}&\quad + \int _{{\mathbb {T}}^{d}} g^{-1}\rho _{y}\,\hbox {d}y:\nabla _{x}\nabla _{x}v_{0} \end{aligned}$$

(119c)

$$\begin{aligned}&\quad + \int _{{\mathbb {T}}^{d}} {\fancyscript{L}}_{\eta }\chi \rho _{y}\,\hbox {d}y\, \nabla _{x}v_{0}. \end{aligned}$$

(119d)

First, we note that

$$\begin{aligned}&\int _{{\mathbb {T}}^{d}} F(y,\eta )\otimes \chi (y, \eta ) \sqrt{\left|g \right|(y, \eta )}\,\hbox {d}y = -\int _{{\mathbb {T}}^{d}} {\fancyscript{L}}_{0}\chi (y, \eta )\otimes \chi (y, \eta ) \sqrt{\left|g \right|(y, \eta )}\,\hbox {d}y \\&\quad = \int _{{\mathbb {T}}^{d}}\nabla _{y}\chi (y, \eta ) g^{-1}(y, \eta ) \nabla _{y}\chi (y, \eta )^\top \sqrt{\left|g \right|(y,\eta )}\, \hbox {d}y, \end{aligned}$$

so that we can write (119) as

$$\begin{aligned} \frac{\partial v_{0}}{\partial t}&= \int _{{\mathbb {T}}^{d}}\left( I + \nabla _{y}\chi \right) g^{-1}\left( I + \nabla _{y}\chi \right) ^\top \rho _{y}\, \hbox {d}y: \nabla _{x}\nabla _{x}v_{0}\\&+ \int _{{\mathbb {T}}^{d}} {\fancyscript{L}}_{\eta }\chi \rho _{y}\,\hbox {d}y \cdot \nabla _{x}v_{0}. \end{aligned}$$

Averaging with respect to the invariant measure $\rho _{\eta }$ of ${\fancyscript{L}}_{1}$, we derive the effective diffusion equation

$$\begin{aligned} \frac{\partial v_{0}}{\partial t}&= \int \int _{{\mathbb {T}}^{d}}\left( I + \nabla _{y}\chi \right) g^{-1}\left( I + \nabla _{y}\chi \right) ^\top \rho _{y}\, \rho _{\eta }\, \hbox {d}y\,\hbox {d}\eta : \nabla _{x}\nabla _{x}v_{0}\\&+ \int \int _{{\mathbb {T}}^{d}} {\fancyscript{L}}_{\eta }\chi \rho _{y}\rho _{\eta }\,\hbox {d}y\, \hbox {d}\eta \cdot \nabla _{x}v_{0}, \end{aligned}$$

or more compactly

$$\begin{aligned} \frac{\partial v_{0}}{\partial t} = D: \nabla _{x}\nabla _{x}v_{0} + V\cdot \nabla _{x}v_{0}, \end{aligned}$$

where $D$ and $V$ are given by (86) and (87), respectively. From the limiting backward Kolmogorov equation, we can read off the limiting SDE (85) for the process $X^{\varepsilon }(t)$.

1.4 Case IV

We look for solutions $v$ of the form

$$\begin{aligned} v^\varepsilon = v_0 + \varepsilon v_1 + \varepsilon ^2 v_2 + \cdots \end{aligned}$$

of (103a) for some smooth functions $v_i:\mathbb {R}^d \times {\mathbb {T}}^d \times \mathbb {R}^K \times [0,T]\rightarrow \mathbb {R}$. Substituting this ansatz in (103a) and equating equal powers of $\varepsilon$, we obtain the following three equations

$$\begin{aligned}&\begin{array}{ll} O\left( \frac{1}{\varepsilon ^{2}}\right) : &{}\qquad \fancyscript{G}v_{0} = 0, \\ \end{array}\end{aligned}$$

(120)

$$\begin{aligned}&\begin{array}{ll} O\left( \frac{1}{\varepsilon }\right) :&{}\qquad \fancyscript{G}v_{1} = -{\fancyscript{L}}_{1}v_{0},\\ \end{array}\end{aligned}$$

(121)

$$\begin{aligned}&\begin{array}{ll} O(1):&\qquad \fancyscript{G}v_{2} = -\left( \frac{\partial v_{0}}{\partial t} - {\fancyscript{L}}_{1}v_{1} - {\fancyscript{L}}_{2}v_{0}\right) . \end{array} \end{aligned}$$

(122)

As the fast process is ergodic, the first equation implies that $v_0$ is independent of $y$ and $\eta$. The second equation thus becomes

$$\begin{aligned} \fancyscript{G}v_1 = -F(y, \eta )\cdot \nabla _x v_0. \end{aligned}$$

Since we are assuming Assumption (101), there exists a unique solution of the Poisson problem (100), by Proposition 8. By choosing $v_1 = \chi \cdot \nabla _x v_0$, we see that the second equation is satisfied.

By Proposition 8, a sufficient condition for the final equation to have a solution is that the RHS is orthogonal to the measure $\rho (\hbox {d}y, \hbox {d}\eta )$, (assuming that the RHS grows at most polynomially), that is,

$$\begin{aligned} \frac{\partial v_0}{\partial t}(y, \eta )&= \int F(y, \eta )\cdot \nabla _x v_1 \, \rho (\hbox {d}y, \hbox {d}\eta )+ \int 2\varSigma (y, \eta ):\nabla _x \nabla _y v_1 \, \rho (\hbox {d}y, \hbox {d}\eta ) \\&\quad + \int \varSigma (y,\eta ):\nabla _x\nabla _x v_0 \, \rho (\hbox {d}y, \hbox {d}\eta ), \end{aligned}$$

which we can rewrite as

$$\begin{aligned} \frac{\partial v_0}{\partial t} = D:\nabla _x\nabla _x v_0, \end{aligned}$$

where the effective diffusion tensor $D$ is given by

$$\begin{aligned} D = \int \left[ \frac{1}{\sqrt{\left|g \right|}}\nabla _y\cdot \left( \sqrt{\left|g \right|}g^{-1}\right) \otimes \chi + g^{-1}\nabla _y \chi ^\top + \nabla _y \chi g^{-1} + g^{-1}\, \right] \rho (\hbox {d}y, \hbox {d}\eta ) \end{aligned}$$

Note that the first term on the RHS

$$\begin{aligned} \displaystyle \int \frac{1}{\sqrt{\left|g \right|}}\nabla _y\cdot \left( \sqrt{\left|g \right|}g^{-1}\right) \otimes \chi \, \rho (\hbox {d}y, \hbox {d}\eta ):\nabla _x \nabla _x v_0, \end{aligned}$$

can be rewritten as $\fancyscript{K}:\nabla _x \nabla _x v_0,$ where

$$\begin{aligned} \fancyscript{K} = \text{ Sym }\left[ \displaystyle \int \frac{1}{\sqrt{\left|g \right|}}\nabla _y\cdot \left( \sqrt{\left|g \right|}g^{-1}\right) \otimes \chi \, \rho (\hbox {d}y, \hbox {d}\eta )\right] , \end{aligned}$$

where $\text{ Sym }\left[ \cdot \right]$ denotes the symmetric part of the matrix. Let $e \in \mathbb {R}^d$ be a unit vector and consider

$$\begin{aligned} \fancyscript{K}^e := e \cdot \fancyscript{K} e = \displaystyle \int \frac{1}{\sqrt{\left|g \right|}}\nabla _y\cdot \left( \sqrt{\left|g \right|}g^{-1}e\right) \chi ^e\, \rho (\hbox {d}y, \hbox {d}\eta ), \end{aligned}$$

where $\chi ^e = \chi \cdot e$. Noting that

$$\begin{aligned} -\fancyscript{G}\chi ^e = \frac{1}{\sqrt{\left|g \right|}}\nabla _y\cdot \left( \sqrt{\left|g \right|}g^{-1}e\right) , \end{aligned}$$

it follows that

$$\begin{aligned} \fancyscript{K}^e = \int \chi ^e\left( -\fancyscript{G}\chi ^e\right) \, \rho (\hbox {d}y, \hbox {d}\eta ), \end{aligned}$$

which, by (102), can be written as

$$\begin{aligned} \fancyscript{K}^e = \int \nabla _y \chi ^e\cdot g^{-1} \nabla _y \chi ^e \,\rho (\hbox {d}y, \hbox {d}\eta ) + \int \nabla _\eta \chi ^e \cdot \varGamma \Pi \nabla _\eta \chi ^e \, \rho (\hbox {d}y, \hbox {d}\eta ), \end{aligned}$$

so that

$$\begin{aligned} e\cdot D e \!=\! \int \left( e + \nabla _y\chi ^e\right) \cdot g^{-1}\left( e \!+\! \nabla _y\chi ^e\right) \, \rho (\hbox {d}y, \hbox {d}\eta ) \!+\! \int \nabla _\eta \chi ^e \cdot \varGamma \Pi \nabla _\eta \chi ^e \,\rho (\hbox {d}y, \hbox {d}\eta ), \end{aligned}$$

or in matrix notation

$$\begin{aligned} D = \int \left( I + \nabla _y\chi \right) g^{-1}\left( I + \nabla _y\chi \right) ^\top \, \rho (\hbox {d}y, \hbox {d}\eta ) + \int \nabla _\eta \chi \, \varGamma \Pi \nabla _\eta \chi ^\top \,\rho (\hbox {d}y, \hbox {d}\eta ), \end{aligned}$$

as required. A rigorous proof of this result can be found in Duncan (2013).