A Diffusion Approximation Based on Renewal Processes with Applications to Strongly Biased Run–Tumble Motion

Abstract

We consider organisms which use a renewal strategy such as run–tumble when moving in space, for example to perform chemotaxis in chemical gradients. We derive a diffusion approximation for the motion, applying a central limit theorem due to Anscombe for renewal-reward processes; this theorem has not previously been applied in this context. Our results extend previous work, which has established the mean drift but not the diffusivity. For a classical model of tumble rates applied to chemotaxis, we find that the resulting chemotactic drift saturates to the swimming velocity of the organism when the chemical gradients grow increasingly steep. The dispersal becomes anisotropic in steep gradients, with larger dispersal across the gradient than along the gradient. In contrast to one-dimensional settings, strong bias increases dispersal. We next include Brownian rotation in the model and find that, in limit of high chemotactic sensitivity, the chemotactic drift is 64 % of the swimming velocity, independent of the magnitude of the Brownian rotation. We finally derive characteristic timescales of the motion that can be used to assess whether the diffusion limit is justified in a given situation. The proposed technique for obtaining diffusion approximations is conceptually and computationally simple, and applicable also when statistics of the motion is obtained empirically or through Monte Carlo simulation of the motion.

Appendices

Appendix 1: Proof of Proposition 1

Define the scalar stochastic process

$$\begin{aligned} z_i = c' x_i - \frac{c'\mu }{\tau } t_i \end{aligned}$$

where c is an arbitrary column vector and $c'$ denotes transpose. $z_i$ is a scalar auxiliary process which has been constructed so as to have mean $\mathbf {E}z_i = 0$, thus enabling us to apply Anscombe’s theorem. Its variance is

$$\begin{aligned} \mathbf {V}z_i = \sigma ^2 = c ' \left( {\varSigma }_x - \frac{1}{\tau } \left( \mu \rho ' + \rho \mu '\right) + \frac{\mu \mu ' }{ \tau ^2} \sigma ^2_t \right) c = c'{\varSigma }c~~. \end{aligned}$$

Define

$$\begin{aligned} S_n = \sum _{i=1}^n z_i \end{aligned}$$

and note that by the law of large numbers, $t^{-1} N(t) \rightarrow \tau ^{-1}$ almost surely. We may thus apply Anscombe’s theorem to conclude that $S_{N(t)}$ is asymptotically normally distributed. Now, $S_n$ is closely related to the cumulated reward $\bar{X}_T$ which is our object of investigation:

$$\begin{aligned} S_{N(t)} = c' \bar{X}_t - \frac{c' \mu }{\tau } T_{N(t)} = c' \bar{X}_t - \frac{c' \mu }{\tau } t + \frac{c'\mu }{\tau } A_t \end{aligned}$$

where

$$\begin{aligned} A_t = t - T_{N(t)} \end{aligned}$$

is the age, i.e., the time that has passed since last tumble. The process $A_t$ is bounded in probability, and hence

$$\begin{aligned} \frac{A_t}{\sqrt{t}} \rightarrow 0 \hbox { in probability as } t\rightarrow \infty . \end{aligned}$$

Thus, the asymptotic distribution of the sum $S_{N(t)}$ implies a similar property of $\bar{X}_t$:

$$\begin{aligned} c ' \frac{\bar{X}_t - \frac{\mu }{\tau } t}{\sqrt{t/\tau }} = \frac{S_{N(t)}}{\sqrt{t/\tau }} - \frac{\mu }{\sqrt{\tau }} \frac{A_t}{\sqrt{t}} ~~ \rightarrow N(0,\sigma ) \hbox { in distribution. } \end{aligned}$$

Since this holds for any c, $\bar{X}_t$ itself must be asymptotically normally distributed:

$$\begin{aligned} \frac{1}{\sqrt{t}} \left( \bar{X}_t - \frac{\mu }{\tau } t \right) \rightarrow N(0,{\varSigma }/\tau ) \text{ in } \text{ distribution. } \end{aligned}$$

$\square$

Appendix 2: Proof of Theorem 2

We aim to apply Proposition 1 and therefore pursue the statistics of a single run. For simplicity, we rescale time and length so that the swimming speed is 1 and the base tumble rate is 1, and express the tumble rate as a function of the angle $\theta$ between v and $e_z$. Then, the tumble rate is

$$\begin{aligned} \lambda (\theta ) = e^{-\beta \cos \theta } \end{aligned}$$

(4)

At each tumble, a new run direction is chosen, uniformally with respect to solid angle, i.e., the angle $\theta _i$ between the z-axis and the velocity during run i is distributed between 0 and $\pi$ with density:

$$\begin{aligned} \phi (\theta ) = \frac{1}{2} \sin \theta ~~\hbox { for } 0\le \theta \le \pi . \end{aligned}$$

The duration $t_i$ of a run is, when conditioning on the angle $\theta _i$, exponentially distributed with mean $1/\lambda (\theta _i)$. We then obtain the unconditional mean:

$$\begin{aligned} \tau = \mathbf {E}t_i = \mathbf {E}\left( \mathbf {E}\{ t_i | \theta _i \} \right) = \mathbf {E}\frac{1}{\lambda (\theta _i)} = \int _0^\pi \frac{\phi (\theta )}{\lambda (\theta )} ~d\theta = \frac{1}{2\beta } \left( e^\beta - e^{-\beta } \right) = \frac{\sinh \beta }{\beta }~~. \end{aligned}$$

This mean duration is a timescale which characterizes the process. It is examined in Sect. 4 and compared to other timescales, and graphed in Fig. 2.

Likewise, conditional on the angle $\theta _i$, the second moment of the run duration is $\mathbf {E}\{t_i^2 | \theta _i\} = 2/\lambda ^2(\theta _i)$. We thus find the unconditional second moment:

$$\begin{aligned} \mathbf {E}t_i^2 = \mathbf {E}\left( \mathbf {E}\left\{ t_i^2 | \theta _i \right\} \right) = \mathbf {E}\frac{2}{\lambda ^2(\theta _i)} = \int _0^\pi \frac{2\phi (\theta )}{\lambda ^2(\theta )} ~d\theta = \frac{1}{2\beta } \left( e^{2\beta }- e^{-2\beta } \right) = \frac{\sinh 2 \beta }{\beta } \end{aligned}$$

from which we obtain the variance of the run duration:

$$\begin{aligned} \sigma ^2_t = \mathbf {V}t_i = \mathbf {E}t_i^2 - (\mathbf {E}t_i)^2 = \frac{\sinh 2\beta }{\beta } - \frac{\sinh ^2 \beta }{\beta ^2} \end{aligned}$$

We now turn to the statistics of the displacement during a run, which we—with a slight abuse of notation—represent with coordinates $(x_i,y_i,z_i)$. We have the geometric relationship $z_i =t_i \cdot \cos \theta _i$ which yields the mean displacement in the z-direction:

$$\begin{aligned} \mu _z = \mathbf {E}z_i = \mathbf {E}( \mathbf {E}\{ z_i | \theta _i \} ) =\mathbf {E}\frac{\cos \theta _i}{\lambda (\theta _i)} = \int _0^\pi \frac{\phi (\theta )}{\lambda (\theta )} \cos \theta ~d\theta = \frac{ \cosh \beta }{\beta } - \frac{\sinh \beta }{\beta ^2} \end{aligned}$$

Rotational symmetry yields $\mu _x=\mu _y=0$. Proposition 1 now gives the drift in the diffusion approximation:

$$\begin{aligned} v_x = v_y = 0, \quad v_z = \frac{\mu _z}{\tau } = \coth \beta - \frac{1}{\beta } \end{aligned}$$

as shown in Fig. 1.

Next, we need the second-order properties of the displacement during a run. Rotational symmetry and Pythagoras’ theorem yield

$$\begin{aligned} \mathbf {E}x_i^2 = \mathbf {E}y_i^2 \hbox { and } x_i^2 + y_i^2 = t_i^2\sin ^2 \theta _i. \end{aligned}$$

Combining these two, we obtain the variance of displacement in the direction of the x-axis (or the y-axis):

$$\begin{aligned} \mathbf {E}x_i^2= & {} \frac{1}{2} \mathbf {E}\left( \mathbf {E}\left\{ \left( t_i^2 \sin ^2\theta _i\right) | \theta _i \right\} \right) \\= & {} \mathbf {E}\frac{\sin ^2 \theta _i}{\lambda ^2(\theta _i)} \\= & {} \int _0^\pi \frac{\phi (\theta ) \sin ^2 \theta }{\lambda ^2(\theta )} ~d\theta \\= & {} \frac{\cosh 2 \beta }{2\beta ^2} - \frac{\sinh 2 \beta }{4 \beta ^3} \end{aligned}$$

Correspondingly, for the z-direction:

$$\begin{aligned} \mathbf {E}z_i^2= & {} \mathbf {E}\left( \mathbf {E}\left\{ t_i^2 \cos ^2\theta _i | \theta _i \right\} \right) \\= & {} \mathbf {E}\frac{2 \cos ^2\theta _i }{\lambda ^2(\theta _i)} \\= & {} \int _0^\pi \frac{2 \phi (\theta ) ~\cos ^2\theta }{\lambda ^2(\theta )}~d\theta \\= & {} \frac{2\beta ^2 +1}{2\beta ^3} \sinh 2 \beta - \frac{\cosh 2 \beta }{\beta ^2} \end{aligned}$$

which yields the variance

$$\begin{aligned} \mathbf {V}z_i = \mathbf {E}z_i^2 - (\mathbf {E}z_i)^2 = \frac{\left( 4 \beta ^3 + 6 \beta \right) \sinh 2 \beta - \left( 6\beta ^2 + 2\right) \cosh 2 \beta + 2 - 2\beta ^2}{4\beta ^4}. \end{aligned}$$

Rotational symmetry dictates that the run displacement $(x_i,y_i,z_i)$ has diagonal covariance matrix ${\varSigma }_x = \hbox {diag}(\mathbf {V}x_i,\mathbf {V}y_i,\mathbf {V}z_i )$ and that the covariance between run duration and displacement is 0 in the two directions $e_x$ and $e_y$, so

$$\begin{aligned} \rho = (0,0,\mathbf {Cov}(z_i,t_i))'. \end{aligned}$$

Here, the covariance between run duration $t_i$ and displacement $z_i$ in z-direction is obtained with similar calculations:

$$\begin{aligned} \mathbf {E}t_i z_i= & {} \mathbf {E}\left( \mathbf {E}\left\{ t_iz_i | \theta _i \right\} \right) \\= & {} \mathbf {E}\left( \mathbf {E}\left\{ t_i^2 \cos \theta _i | \theta _i \right\} \right) \\= & {} \mathbf {E}\frac{2\cos \theta _i}{\lambda ^2(\theta _i)} \\= & {} \int _0^\pi \frac{2 \phi (\theta ) \cos \theta }{ \lambda ^2(\theta ) } ~d\theta \\= & {} \frac{\cosh \beta }{\beta } - \frac{\sinh 2 \beta }{2 \beta ^2} \end{aligned}$$

yielding

$$\begin{aligned} \mathbf {Cov}(t_i,z_i) = \mathbf {E}t_i z_i - \tau \mu _z = \frac{\left( 2\beta ^2 + 1\right) \cosh 2\beta - 2 \beta \sinh 2 \beta -1}{2\beta ^3} ~~. \end{aligned}$$

We now have the statistics which are required to apply Proposition 1. In the x- and y-directions, we get

$$\begin{aligned} D_x =D_y = \frac{\mathbf {E}x_i^2}{2 \tau } = \frac{ 2 \beta \cosh 2 \beta - \sinh 2 \beta }{ 8 \beta ^2 \sinh \beta } \end{aligned}$$

while in the z direction, we get

$$\begin{aligned} D_z = \frac{\mathbf {V}z_i + \frac{(\mathbf {E}z_i)^2}{\tau ^2}\mathbf {V}t_i -2\frac{\mathbf {E}z_i}{\tau } \mathbf {Cov}(t_i,z_i)}{2\tau } = \frac{1}{4} \frac{\sinh 2\beta - 2 \beta }{\beta ^2 \sinh \beta }. \end{aligned}$$

Proposition 1 now establishes the limiting Gaussian distribution of the position at tumbles, and it remains only to show that the same limiting distribution applies during a run: The displacement during a run from the position at last tumble will be bounded in probability. Therefore, $(\bar{X}_t - X_t)/\sqrt{t}$ converges to 0 in probability, so that $(X_t - v t)/\sqrt{t}$ and $(\bar{X}_t - vt)/\sqrt{t}$ has the same limiting distribution, i.e., the limiting Gaussian distribution applies also to the position at any point in time, whether at a tumble or during a run. $\square$

Appendix 3: Monte Carlo Simulation

Here, we describe the Monte Carlo simulation performed to verify the results of Theorem 2, i.e., the estimates and associated confidence intervals displayed in Fig. 1. The run–tumble model with constant speed $U=1$, base tumble rate $\lambda _0=1$, and different values of the parameter $\beta$ is simulated stochastically. The initial distribution for the velocity is the stationary distribution which scales with $1/\lambda (v)$; this is a truncated exponential distribution for $v_z=v\cdot e_z$ and conditional on this, a uniform distribution of the two other components on the circle $\{ (v_x,v_y) : v_x^2 + v_y^2 = 1- v_z^2\}$. The motion is simulated one tumble at a time until the time of last tumble exceeds the simulation time $T=10000$; after this, the position at time T is obtained by interpolation in the last run and reported. This simulation time is orders of magnitude larger than the decorrelation time of the velocity process (compare Fig. 2) which ensures that the diffusion approximation is reasonable. $N=1000$ independent replicates of the motion are simulated for each value of $\beta$. Estimates of the drift and diffusivity are obtained as sample means computed from mean displacement and its variance, respectively. 95 % confidence intervals for the mean are obtained using $\pm 1.96$ times the standard error between replicates; these intervals are smaller than the plotting symbols so practically invisible in the plot. 95 % confidence intervals for the diffusivity are obtained from percentiles in $M=1,000$ bootstrap replicates.

Appendix 4: Proof of Theorem 3

We first nondimensionalize the model; this amounts to taking $U=1$ and $\kappa =1$. We aim to apply Proposition 1 and therefore examine the statistics of time and displacement during a single run. A run starts at time 0 with a random velocity $V_0$ sampled uniformly on the half-sphere $B_+$ with respect to solid angle and ends when the Brownian motion on the half-sphere hits the plane $\{ v: e_z v = 0\}$.

We first transform the problem to a one-dimensional one by focusing on the velocity in the z-direction. During a run, the instantaneous velocity vector $V_t$ is Brownian motion on the surface of the unit sphere, which can be written as the solution to an Itô stochastic differential equation (Øksendal 2010)

$$\begin{aligned} {\text {d}}V_t = - V_t ~{\text {d}}t + \left( I - V_tV_t' \right) ~ {\text {d}}B_t \end{aligned}$$

(5)

where $B_t$ is three-dimensional Brownian motion. Let $U_t = e_z V_t$ denote the velocity in the direction of the gradient. Using Itô’s lemma, we obtain

$$\begin{aligned} {\text {d}}U_t = - U_t ~{\text {d}}t + \sqrt{ 1 - |U_t|^2 } ~ {\text {d}}W_t \end{aligned}$$

Here,

$$\begin{aligned} {\text {d}}W_t = \frac{1}{\sqrt{1-|U_t|^2}} e_z ( I - V_t V_t') ~ {\text {d}}B_t \end{aligned}$$

and, noting that

$$\begin{aligned} \left| \frac{1}{\sqrt{1-|U_t|^2}} e_z ( I - V_t V_t') \right| ^2 = 1 \end{aligned}$$

we recognize $W_t$ as standard Brownian motion (compare Øksendal (2010)). Thus, the process $U_t$ is itself a Markov process with generator L given by

$$\begin{aligned} (L f)(u) = - u \frac{\partial f}{\partial u} + \frac{1}{2} \left( 1-u^2\right) \frac{\partial ^2f}{\partial u^2} = \frac{\partial }{\partial u} \left( \frac{1}{2} \left( 1-u^2\right) \frac{\partial f}{\partial u} \right) \end{aligned}$$

Up to the factor 1 / 2, this is Legendre’s operator of order 0.

From the point of view of the individual run, the boundary $u=0$ is absorbing in that the run terminates when $U_t$ reaches this boundary. Let $T=\inf \{ t : U_t \le 0 \}$ denote this stopping time. We seek the statistics of $T$ and $Z_T$. During the run, the time t and up-gradient displacement $Z_t$ are additive processes, studied in an oceanographic context by Thygesen et al. (2007). Define the moment functions

$$\begin{aligned} \psi _{ij}(u) = \mathbf {E}^u T^i Z_T^j \end{aligned}$$

then it follows that these functions $\psi _{ij}$ for $i,j\in \mathbf {N}_0$, $(i,j)\ne (0,0)$ are governed by boundary value problems

$$\begin{aligned} L\psi _{10} + 1= & {} 0 \\ L\psi _{01} + u= & {} 0 \\ L\psi _{ij} + i \psi _{(i-1)j} + j u \psi _{i(j-1)}= & {} 0 \hbox { for } i,j \in \mathbf {N}\end{aligned}$$

on the domain $u\in (0,1)$. The boundary condition at $u=0$ is $\psi _{ij}(0)=0$ while the boundary condition at $u=1$ is that the limit $\lim _{u\rightarrow 1} \psi _{ij}(u)$ exists, since the operator L is singular at that point (recall that $u=1$ does not correspond to a boundary point for the original Brownian motion on the half-sphere $B_+$).

For the low-order moments, we find the solutions:

$$\begin{aligned} \psi _{10}(u)= & {} \mathbf {E}^u T= 2 \log (1+u) \\ \psi _{01}(u)= & {} \mathbf {E}^u Z_T= u \\ \psi _{20}(u)= & {} \mathbf {E}^u T^2 = 8(\log 2 -1 ) \log (u + 1) - 4 (\log (2))^2 - 8 \hbox {dilog}\left( \frac{u}{2} + \frac{1}{2}\right) + 2 \frac{\pi ^2}{3} \\ \psi _{11}(u)= & {} \mathbf {E}^u TZ_T= (4+2z)\log (u+1) -2u \\ \psi _{02}(u)= & {} \mathbf {E}^u Z_T^2 = \frac{2}{3} u^2 +\frac{4}{3} \log (u+1) \end{aligned}$$

Here, the dilog function is $\hbox {dilog} (x) = \int _1^x (1-t)^{-1} \log t ~{\text {d}}t$.

These statistics are conditional on the up-gradient velocity $u=U_0$ at the beginning of the run. To find the unconditional statistics, we take expectation with respect to the distribution of $U_0$. Since $V_0$ is uniformly distributed on the half-sphere $B_+$ with respect to solid angle, $U_0$ is uniformly distributed on [0, 1]. Hence,

$$\begin{aligned} \mathbf {E}T^i Z_T^j = \int _0^1 \psi _{ij}(u)~{\text {d}}u \end{aligned}$$

We find:

$$\begin{aligned} \mathbf {E}T= & {} 4 \log 2 - 2 \approx 0.77 \\ \mathbf {E}Z_T= & {} \frac{1}{2} \\ \mathbf {E}T^2= & {} 16 (1- \log 2 )^2 \approx 1.51 \\ \mathbf {E}TZ_T= & {} -9/2 + 8 \log 2 \approx 1.05 \\ \mathbf {E}Z^2_T= & {} -\frac{10}{9} + \frac{8}{3} \log 2 \approx 0.74 \end{aligned}$$

According to Proposition 1, we can now compute the advective velocity as

$$\begin{aligned} \frac{\mathbf {E}Z_T}{\mathbf {E}T} = \frac{1}{8 \log 2 - 4} \approx 0.64 \end{aligned}$$

To determine the diffusivity, we need the variance of the run duration

$$\begin{aligned} \mathbf {V}T= \mathbf {E}T^2 - (\mathbf {E}T)^2 = 12 - 16 \log 2 \approx 0.91 \quad , \end{aligned}$$

the variance of the up-gradient displacement during a run

$$\begin{aligned} \mathbf {V}Z_T= \mathbf {E}Z_T^2 - (\mathbf {E}Z_T)^2 = - \frac{49}{36} + \frac{8}{3} \log 2 \approx 0.49 \quad , \end{aligned}$$

and the covariance

$$\begin{aligned} \mathbf {Cov}(T, Z_T) = \mathbf {E}TZ_T- (\mathbf {E}Z_T)(\mathbf {E}T) = -7/2 + 6\log 2 \approx 0.66 \quad . \end{aligned}$$

With this, we can employ Proposition 1 to obtain the diffusivity in the z-direction:

$$\begin{aligned} D_z= & {} \frac{1}{2} \frac{\mathbf {V}Z_T+ \frac{(\mathbf {E}Z_T)^2}{(\mathbf {E}T)^2} \mathbf {V}T- 2 \mathbf {E}Z_T\frac{\rho }{\mathbf {E}T} }{\mathbf {E}T}\\= & {} {\frac{1}{144}}\,{\frac{-85+490\,\ln \left( 2 \right) -796\, \left( \ln \left( 2 \right) \right) ^{2}+384\, \left( \ln \left( 2 \right) \right) ^{3}}{ \left( -1+2\,\ln \left( 2 \right) \right) ^{3}}}\\\approx & {} 1.00 \cdot 10 ^{-2} \end{aligned}$$

By rotational symmetry, the chemotactic drift in the x- (or y)-direction is zero, and the diffusivity matrix is diagonal. It remains to find the diffusivity $D_x$ in the x-direction, and to this end, we must find $\mathbf {V}X_T$. Conditional on $V_0$, this is governed by the two boundary value problems on the half-sphere $B_+$:

$$\begin{aligned} \nabla ^2 \gamma _1 + x = 0,\quad \nabla ^2 \gamma _2 + 2x\gamma _1 = 0 \end{aligned}$$

with Dirichlet boundary condition on the boundary $\{ (x,y,z): z=0, x^2+y^2 = 1 \}$. These equations do not appear to admit analytical solution in closed form. Instead, two numerical approaches were used: first, numerical solution of the two equations using the finite element software Comsol and next, a Monte Carlo method where the stochastic differential Eq. (5) is simulated using a modified Euler method until absorption. The two numerical methods both estimate

$$\begin{aligned} \mathbf {E}X_T^2 = 0.131 \pm 0.001 \end{aligned}$$

With this, we find

$$\begin{aligned} D_x = \frac{\mathbf {E}X_T^2}{2 \mathbf {E}T} \approx 8.54 \cdot 10^{-2} \quad . \end{aligned}$$

Finally, according to the general result (3), the expected age in stationarity is

$$\begin{aligned} \mathbf {E}A_t = \frac{1}{2} \frac{\mathbf {V}T}{\mathbf {E}T} + \frac{1}{2} \mathbf {E}T= \frac{3 - 4 \log 2}{2 \log 2 - 1} + 2 \log 2 - 1 \approx 0.975 \quad . \end{aligned}$$

$\square$