Application of Quasi-Steady-State Methods to Nonlinear Models of Intracellular Transport by Molecular Motors

Abstract

Molecular motors such as kinesin and dynein are responsible for transporting material along microtubule networks in cells. In many contexts, motor dynamics can be modelled by a system of reaction–advection–diffusion partial differential equations (PDEs). Recently, quasi-steady-state (QSS) methods have been applied to models with linear reactions to approximate the behaviour of the full PDE system. Here, we extend this QSS reduction methodology to certain nonlinear reaction models. The QSS method relies on the assumption that the nonlinear binding and unbinding interactions of the cellular motors occur on a faster timescale than the spatial diffusion and advection processes. The full system dynamics are shown to be well approximated by the dynamics on the slow manifold. The slow manifold is parametrized by a single scalar quantity that satisfies a scalar nonlinear PDE, called the QSS PDE. We apply the QSS method to several specific nonlinear models for the binding and unbinding of molecular motors, and we use the resulting approximations to draw conclusions regarding the parameter dependence of the spatial distribution of motors for these models.

Appendices

Appendix A: Microtubule Density and Binding by Motor Complexes

1.1 Kinesin Model with Non-uniform MT Density

To explicitly incorporate the possibility that MT density, m(x) (as well as fraction of MT pointing to the right, P(x)) varies across the cell, we can write the kinesin model equations as

$$\begin{aligned} \frac{\partial p^{\text {R}}}{\partial t}&= -v \frac{\partial p^{\text {R}}}{\partial x} + P(x) k_{\text {bm}}m(x) g(p^{\text {U}}) - k_{\text {u}}p^{\text {R}},\end{aligned}$$

(74a)

$$\begin{aligned} \frac{\partial p^{\text {L}}}{\partial t}&= v \frac{\partial p^{\text {L}}}{\partial x} + (1-P(x)) k_{\text {bm}}m(x) g(p^{\text {U}}) - k_{\text {u}}p^{\text {L}},\end{aligned}$$

(74b)

$$\begin{aligned} \frac{\partial p^{\text {U}}}{\partial t}&= D_0 \frac{\partial ^2 p^{\text {U}}}{\partial x^2} - k_{\text {bm}}m(x) g(p^{\text {U}}) + k_{\text {u}}p^{\text {R}}+ k_{\text {u}}p^{\text {L}}. \end{aligned}$$

(74c)

This modification of the model introduces another factor into coefficients that are already spatially dependent, but otherwise leaves the model structure unchanged. Hence, the techniques in the paper apply as before with $k_{\text {bm}}m(x)$ replacing the parameter $k_{\text {b}}$.

For the purposes of our proof-of-concept analysis, we now restrict attention to uniform MT density so that $m(x)\equiv m_0$ is a constant. Then the model for kinesin is given by (75) as below, with the assignment

$$\begin{aligned} k_{\text {b}}= k_{\text {bm}}m_0. \end{aligned}$$

That is, the binding constant $k_{\text {b}}$ is understood to represent the net rate of binding, which includes both the per-MT-binding rate and the MT density.

1.2 Kinesin–Dynein Model and the Function Q(x)

The kinesin–dynein model simplifies the binding of free motor complexes into states that move right with probability Q(x) and left with probability $1-Q(x)$. We consider the case of motor complexes that all have $n_k$ kinesin and $n_d$ dynein components. (The case of complexes with a variety of motor numbers can be handled by considering the mean composition of a complex or the mean ratio between the two motor types.) Let us also define the parameters $k_{bd}$ and $k_{bk}$ as the binding rates for a (single) dynein and for a (single) kinesin to a MT, and consider m(x) as the local MT density. Then we can decompose the quantity $k_{\text {b}}Q$ in the model as follows:

$$\begin{aligned} k_{\text {b}}Q(x) = m(x) \left[ P(x)n_k k_{bk} + (1-P(x)) n_d k_{bd} \right] . \end{aligned}$$

This related the aggregate binding rate to the probability that a kinesin binds to right-pointing MT and that dynein binds to left-pointing MT. Similarly,

$$\begin{aligned} k_{\text {b}}(1-Q(x)) = m(x) \left[ (1-P(x))n_k k_{bk} + P(x) n_d k_{bd} \right] . \end{aligned}$$

Since such details merely substitute one spatially dependent function for another, the analysis we have described carries over as before.

Appendix B: Scaling the Models and the QSS Reduction

1.1 The Kinesin Model

We consider the kinesin model with uniform MT density and demonstrate its scaling here. This system is

$$\begin{aligned} \frac{\partial p^{\text {R}}}{\partial t}&= -v \frac{\partial p^{\text {R}}}{\partial x} + P k_{\text {b}}g(p^{\text {U}}) - k_{\text {u}}p^{\text {R}}, \end{aligned}$$

(75a)

$$\begin{aligned} \frac{\partial p^{\text {L}}}{\partial t}&= v \frac{\partial p^{\text {L}}}{\partial x} + (1-P) k_{\text {b}}g(p^{\text {U}}) - k_{\text {u}}p^{\text {L}}, \end{aligned}$$

(75b)

$$\begin{aligned} \frac{\partial p^{\text {U}}}{\partial t}&= D_0 \frac{\partial ^2 p^{\text {U}}}{\partial x^2} - k_{\text {b}}g(p^{\text {U}}) + k_{\text {u}}p^{\text {R}}+ k_{\text {u}}p^{\text {L}}. \end{aligned}$$

(75c)

We define T by

$$\begin{aligned} T \equiv \int _0^{L_0} \left( p^{\text {R}}(x)+p^{\text {L}}(x)+p^{\text {U}}(x)\right) \, \mathrm{d}x \equiv \int _0^{L_0} y(x) \, \mathrm{d}x. \end{aligned}$$

Then T is the total amount of motors inside the cell, and ${\rho =T/L_0}$ is the average density of motors in the cell.

Scale space, time, and densities as follows:

$$\begin{aligned} x^{\star } = \frac{x}{L_0}, \quad t^{\star }=\frac{tv}{L_0}, \quad p^{J\star }=\frac{p^J}{\rho },\quad y^{\star }=\frac{y}{\rho }, \end{aligned}$$

where $y^{\star }={p^{\text {R}}}^{\star }+{p^{\text {L}}}^{\star }+{p^{\text {U}}}^{\star }$ is the total scaled density. We have scaled distance by the cell length and time by the time that a motor takes to walk across the cell. The densities of each state are scaled by the average motor density across the cell.

Then we can recast the total amount as

$$\begin{aligned} T= \int _0^{1} \left( \rho {p^{\text {R}}}^{\star }(x^{\star })+\rho {p^{\text {L}}}^{\star }(x^{\star }) +\rho {p^{\text {U}}}^{\star }(x^{\star }) \right) d(L_0 x^{\star }). \end{aligned}$$

Taking out the constant factor of $\rho L_0\equiv T$ from the integral results in

$$\begin{aligned} T= \rho L_0\int _0^{1} \left( {p^{\text {R}}}^{\star }(x^{\star })+{p^{\text {L}}}^{\star }(x^{\star }) +{p^{\text {U}}}^{\star }(x^{\star }) \right) \mathrm{d}x^{\star }, \end{aligned}$$

which leads to

$$\begin{aligned} \int _0^{1}y^{\star } \mathrm{d}x^{\star }= \int _0^{1} \left( {p^{\text {R}}}^{\star }(x^{\star })+ {p^{\text {L}}}^{\star }(x^{\star })+{p^{\text {U}}}^{\star }(x^{\star }) \right) \mathrm{d}x^{\star }=1. \end{aligned}$$

With this scaling, the integral of the total scaled density is unity, which we assume throughout our numerical computations.

Substituting the scaled variables into the PDE system (75) leads to

$$\begin{aligned} \frac{v}{L_0} \frac{\partial (\rho {p^{\text {R}}}^{\star })}{\partial t^{\star }}&= \frac{-v}{L_0}\frac{\partial (\rho {p^{\text {R}}}^{\star })}{\partial x^{\star }} + k_{\text {u}}\left( P(x) \frac{k_{\text {b}}}{k_{\text {u}}} g(\rho {p^{\text {U}}}^{\star }) - (\rho {p^{\text {R}}}^{\star }) \right) , \end{aligned}$$

(76a)

$$\begin{aligned} \frac{v}{L_0}\frac{\partial (\rho {p^{\text {L}}}^{\star })}{\partial t^{\star }}&= \frac{v}{L_0} \frac{\partial (\rho {p^{\text {L}}}^{\star })}{\partial x^{\star }} + k_{\text {u}}\left( (1-P(x)) \frac{k_{\text {b}}}{k_{\text {u}}} g(\rho {p^{\text {U}}}^{\star }) - (\rho {p^{\text {L}}}^{\star })\right) , \end{aligned}$$

(76b)

$$\begin{aligned} \frac{v}{L_0}\frac{\partial (\rho {p^{\text {U}}}^{\star })}{\partial t^{\star }}&= \frac{D_0}{L_0^2} \frac{\partial ^2(\rho {p^{\text {U}}}^{\star })}{\partial {x^{\star }}^2} + k_{\text {u}}\left( \rho {p^{\text {R}}}^{\star } + \rho {p^{\text {L}}}^{\star } - \frac{k_{\text {b}}}{k_{\text {u}}} g(\rho {p^{\text {U}}}^{\star }) \right) . \end{aligned}$$

(76c)

Then we can consider two cases, depending on whether the function g is linear or not.

Case I: g is linear In this case, we can eliminate the factor $\rho$ from every term. Dividing each term in the equations by $v \rho /L_0$ and dropping the stars leads to

$$\begin{aligned} \frac{\partial {p^{\text {R}}} }{\partial t }&= -\frac{\partial {p^{\text {R}}} }{\partial x } + \frac{1}{\varepsilon } \left( P(x) k_{\text {a}}{p^{\text {U}}} - {p^{\text {R}}} \right) , \end{aligned}$$

(77a)

$$\begin{aligned} \frac{\partial {p^{\text {L}}} }{\partial t }&= \frac{\partial {p^{\text {L}}} }{\partial x } + \frac{1}{\varepsilon } \left( (1-P(x)) k_{\text {a}}{p^{\text {U}}} - {p^{\text {L}}} \right) , \end{aligned}$$

(77b)

$$\begin{aligned} \frac{\partial {p^{\text {U}}} }{\partial t }&= D \frac{\partial ^2 {p^{\text {U}}} }{\partial {x }^2} + \frac{1}{\varepsilon } \left( {p^{\text {R}}} + {p^{\text {L}}} - k_{\text {a}}{p^{\text {U}}} \right) , \end{aligned}$$

(77c)

where D, $\varepsilon$, and $k_{\text {a}}$ are defined by

$$\begin{aligned} D \equiv \frac{D_0}{v L_0}, \qquad \varepsilon \equiv \frac{v}{L_0 k_{\text {u}}}, \qquad k_{\text {a}}\equiv \frac{k_{\text {b}}}{k_{\text {u}}}. \end{aligned}$$

(78)

In this case, these dimensionless parameters represent, respectively, the ratio of (time to be transported:time to diffuse) across the cell (D), the ratio of (time spent unbound:time to walk) across the cell ($\varepsilon$), and the ratio of (time spent unbound:time spent bound) ($k_{\text {a}}$).

Case II: g is Michaelian or Hill

$$\begin{aligned} g(p) = g_{\text {m}}\frac{p^n}{K^n+p^n}, \quad n = 1, 2, \ldots . \end{aligned}$$

Then, (76) becomes

$$\begin{aligned} \frac{v}{L_0} \frac{\partial (\rho {p^{\text {R}}}^{\star })}{\partial t^{\star }}&= \frac{-v}{L_0}\frac{\partial (\rho {p^{\text {R}}}^{\star })}{\partial x^{\star }} + k_{\text {u}}\left( P(x) \frac{k_{\text {b}}}{k_{\text {u}}} \frac{g_{\text {m}}(\rho {p^{\text {U}}}^{\star })^n}{[K^n+(\rho {p^{\text {U}}}^{\star })^n]}- (\rho {p^{\text {R}}}^{\star }) \right) , \end{aligned}$$

(79a)

$$\begin{aligned} \frac{v}{L_0}\frac{\partial (\rho {p^{\text {L}}}^{\star })}{\partial t^{\star }}&= \frac{v}{L_0} \frac{\partial (\rho {p^{\text {L}}}^{\star })}{\partial x^{\star }} + k_{\text {u}}\left( (1-P(x)) \frac{k_{\text {b}}}{k_{\text {u}}} \frac{ g_{\text {m}}(\rho {p^{\text {U}}}^{\star })^n}{[K^n+(\rho {p^{\text {U}}}^{\star })^n]}- (\rho {p^{\text {L}}}^{\star })\right) , \end{aligned}$$

(79b)

$$\begin{aligned} \frac{v}{L_0}\frac{\partial (\rho {p^{\text {U}}}^{\star })}{\partial t^{\star }}&= \frac{D_0}{L_0^2}\frac{\partial ^2(\rho {p^{\text {U}}}^{\star })}{\partial {x^{\star }}^2} + k_{\text {u}}\left( \rho {p^{\text {R}}}^{\star } + \rho {p^{\text {L}}}^{\star } - \frac{k_{\text {b}}}{k_{\text {u}}} \frac{g_{\text {m}}(\rho {p^{\text {U}}}^{\star })^n}{[K^n+(\rho {p^{\text {U}}}^{\star })^n]}\right) . \end{aligned}$$

(79c)

Define a new constant $A\equiv {K/\rho }$. This constant is the ratio of the motor concentration at which the binding rate is half-maximal to the average motor density in the cell. Divide numerator and denominator of the Hill function by $\rho ^n$. Further, divide every term in the equations by $v \rho /L_0$ as before. Then we obtain after rearranging and dropping the starred notation is

$$\begin{aligned} \frac{\partial {p^{\text {R}}} }{\partial t }&= -\frac{\partial {p^{\text {R}}} }{\partial x } + \frac{1}{\varepsilon } \left( P(x) k_{\text {a}}\frac{ ({p^{\text {U}}} )^n}{[A^n+ ({p^{\text {U}}} )^n]}- {p^{\text {R}}} \right) , \end{aligned}$$

(80a)

$$\begin{aligned} \frac{\partial {p^{\text {L}}} }{\partial t }&= \frac{\partial {p^{\text {L}}} }{\partial x } + \frac{1}{\varepsilon } \left( (1-P(x)) k_{\text {a}}\frac{ ( {p^{\text {U}}} )^n}{[A^n+ ({p^{\text {U}}})^n]} - {p^{\text {L}}} \right) , \end{aligned}$$

(80b)

$$\begin{aligned} \frac{\partial {p^{\text {U}}} }{\partial t }&= D \frac{\partial ^2 {p^{\text {U}}} }{\partial {x }^2} + \frac{1}{\varepsilon } \left( {p^{\text {R}}} + {p^{\text {L}}} - k_{\text {a}}\frac{ ({p^{\text {U}}} )^n}{[A^n+ ({p^{\text {U}}} )^n]}\right) , \end{aligned}$$

(80c)

where D and $\varepsilon$ are as before, but $k_{\text {a}}$ now depends on whether g is a Michaelis–Menten or a Hill function. This holds for any Hill coefficient n. Note that, in particular, for the case $n=1$, which is the Michaelian case considered, we have that

$$\begin{aligned} \frac{\partial {p^{\text {R}}} }{\partial t }&= -\frac{\partial {p^{\text {R}}} }{\partial x } + \frac{1}{\varepsilon } \left( P(x) k_{\text {a}}\frac{ {p^{\text {U}}} }{[1+ c{p^{\text {U}}} ]}- {p^{\text {R}}} \right) , \end{aligned}$$

(81a)

$$\begin{aligned} \frac{\partial {p^{\text {L}}} }{\partial t }&= \frac{\partial {p^{\text {L}}} }{\partial x } + \frac{1}{\varepsilon } \left( (1-P(x)) k_{\text {a}}\frac{ {p^{\text {U}}} }{[1+ c{p^{\text {U}}} ]} - {p^{\text {L}}} \right) , \end{aligned}$$

(81b)

$$\begin{aligned} \frac{\partial {p^{\text {U}}} }{\partial t }&= D \frac{\partial ^2 {p^{\text {U}}} }{\partial {x }^2} + \frac{1}{\varepsilon } \left( {p^{\text {R}}} + {p^{\text {L}}} - k_{\text {a}}\frac{ {p^{\text {U}}} }{[1+ c{p^{\text {U}}} ]}\right) , \end{aligned}$$

(81c)

where $c\equiv {1/A}={\rho /K}$. In (80) and (81) $k_{\text {a}}$ is defined by

$$\begin{aligned} k_{\text {a}}\equiv \frac{k_{\text {b}}g_{\text {m}}}{k_{\text {u}}\rho } \, \quad \hbox {(Hill)}, \qquad k_{\text {a}}\equiv \frac{k_{\text {b}}g_{\text {m}}}{k_{\text {u}}K}, \quad \hbox {(Michaelis--Menten)}. \end{aligned}$$

(82)

In either case, the parameter $k_{\text {a}}$ describes the ratio of time spent bound to the time spent unbound, mediated by the nonlinear binding kinetics.

Finally, we scale the boundary conditions in (5) to get

$$\begin{aligned} \left. \left( p^{\text {R}}- p^{\text {L}}- D \frac{\partial p^{\text {U}}}{\partial x} \right) \right| _{x= 0,1} = 0, \end{aligned}$$

(83)

together with

$$\begin{aligned} p^{\text {R}}(0,t) = 0 \qquad \text {and} \qquad p^{\text {L}}(1,t) = 0. \end{aligned}$$

(84)

1.2 Kinesin–Dynein Model Scaling

Define $k_{\text {c}}\equiv k_{\text {rl}}- k_{\text {lr}}$. Then the model can be written as

$$\begin{aligned} \frac{\partial p^{\text {R}}}{\partial t}&= -v_{\text {r}}\frac{\partial p^{\text {R}}}{\partial x} + k_{\text {b}}Q p^{\text {U}}- k_{\text {u}}p^{\text {R}}- k_{\text {c}}p^{\text {R}}p^{\text {L}}, \end{aligned}$$

(85a)

$$\begin{aligned} \frac{\partial p^{\text {L}}}{\partial t}&= v_{\text {l}}\frac{\partial p^{\text {L}}}{\partial x} + k_{\text {b}}(1-Q) p^{\text {U}}- k_{\text {u}}p^{\text {L}}+ k_{\text {c}}p^{\text {R}}p^{\text {L}}, \end{aligned}$$

(85b)

$$\begin{aligned} \frac{\partial p^{\text {U}}}{\partial t}&= D_0 \frac{\partial ^2 p^{\text {U}}}{\partial x^2} - k_{\text {b}}p^{\text {U}}+ k_{\text {u}}(p^{\text {R}}+ p^{\text {L}}). \end{aligned}$$

(85c)

Scale all variables as before. Then terms of the form $(k_{\text {c}}/k_{\text {u}}) p^{\text {R}}p^{\text {L}}$ will lead to the form $(k_{\text {c}}/k_{\text {u}}) \rho {p^{\text {R}}}^{\star } \rho {p^{\text {L}}}^{\star }$, so that what remains, after cancelling out a factor of $v_{\text {r}}\rho /L_0$ from every term in each equation, and dropping the starred quantities, is

$$\begin{aligned} \frac{\partial p^{\text {R}}}{\partial t }&= - \frac{\partial p^{\text {R}}}{\partial x } + \frac{1}{\varepsilon } \left( k_{\text {a}}Q p^{\text {U}}- p^{\text {R}}- k p^{\text {R}}p^{\text {L}}\right) , \end{aligned}$$

(86a)

$$\begin{aligned} \frac{\partial p^{\text {L}}}{\partial t }&= v \frac{\partial p^{\text {L}}}{\partial x } + \frac{1}{\varepsilon } \left( k_{\text {a}}(1-Q) p^{\text {U}}- p^{\text {L}}+k p^{\text {R}}p^{\text {L}}\right) , \end{aligned}$$

(86b)

$$\begin{aligned} \frac{\partial p^{\text {U}}}{\partial t }&= D \frac{\partial ^2 p^{\text {U}}}{\partial {x }^2} + \frac{1}{\varepsilon } \left( p^{\text {R}}+ p^{\text {L}}- k_{\text {a}}p^{\text {U}}\right) , \end{aligned}$$

(86c)

where the parameters are

$$\begin{aligned} v\equiv \frac{v_{\text {l}}}{v_{\text {r}}}, \qquad D \equiv \frac{D_0}{v_{\text {r}}L_0}, \qquad \varepsilon \equiv \frac{v_{\text {r}}}{k_{\text {u}}L_0}, \qquad k_{\text {a}}\equiv \frac{k_{\text {b}}}{k_{\text {u}}}, \qquad k \equiv \frac{k_{\text {c}}\rho }{k_{\text {u}}}=\frac{(k_{\text {rl}}-k_{\text {lr}})\rho }{k_{\text {u}}}.\nonumber \\ \end{aligned}$$

(87)

Here $\rho$ is the average density of motors inside the cell. These dimensionless parameters represent, respectively, the (left:right) walking speed ratio (v), the ratio of (time to be transported:time to diffuse) across the cell (D), the ratio of (time spent unbound:time to walk) across the cell ($\varepsilon$), the ratio of (time spent unbound:time spent bound) ($k_{\text {a}}$), and the turning parameter k, which represents the ratio of (net right–left direction switches:unbinding rate). We comment that the average density of motors $\rho$ enters into the turning rate parameter due to the nonlinearity of the model with respect to the turning of motors when they collide on a MT.

1.2.1 Details of QSS Reduction of Kinesin–Dynein Model

Next, we provide some details of the QSS reduction of the kinesin–dynein model. Upon setting $f_2=f_3=0$ in (49) we get the two equations

$$\begin{aligned} k p^{\text {R}}p^{\text {L}}= p^{\text {L}}- k_{\text {a}}(1-Q) p^{\text {U}}, \qquad -k_{\text {a}}p^{\text {U}}+ p^{\text {R}}+ p^{\text {L}}=0. \end{aligned}$$

(88)

It is convenient to let $p^{\text {L}}$ be the free variable and parameterize the quasi-steady-state in terms of $p^{\text {L}}=\alpha$. By solving (88) for $p^{\text {R}}$ and $p^{\text {U}}$, we get the quasi-steady-state solution $\mathbf {p}^0$ as given in (51). We then readily show that the nonzero eigenvalues $\lambda _{\pm }$ of the Jacobian of the kinetics satisfy the quadratic equation given in (52) and (53). A necessary and sufficient condition for $\hbox {Re}(\lambda _{\pm })<0$ is that $\sigma _1<0$ and $\sigma _2>0$ in (53). To establish this result we need some properties of H(Q) defined in (53). We first observe that $H(0)=1$, so that trivially $\sigma _1<0$ and $\sigma _2>0$ when $Q=0$. Then, since $H^{\prime }(Q)=-{(1+k\alpha )/(1+k\alpha -Q)^2}<0$, it follows that $\sigma _1<0$ and $\sigma _2>0$ on $0\le Q\le 1$ provided that we can show that $\sigma _1<0$ and $\sigma _2>0$ when $Q=1$. These inequalities do hold at $Q=1$, since by using $H(1)={(k\alpha -1)/(k\alpha )}$ we readily obtain that $\sigma _1=-1-k_{\text {a}}-k\alpha$ and $\sigma _2=k\alpha (1+k_{\text {a}})>0$ when $Q=1$. This proves that $\hbox {Re}(\lambda _{\pm })<0$ for any Q in $0\le Q\le 1$. As a result, $\mathbf {p}^0$ defined in (51) is a slow manifold in the sense of Definition (3.1) for any Q in $0\le Q\le 1$. Finally, by using $\mathbf {p}^0$ and the operator M, as defined in (49), in the solvability condition (24), we readily derive the QSS PDE model (54).

1.3 Myosin Model Scaling

We carry out similar scaling for the myosin model characterized by

$$\begin{aligned} \frac{\partial p^{\text {W}}}{\partial t}&= -v_{\text {w}}\frac{\partial p^{\text {W}}}{\partial x} - \hat{k}_{\text {bw}}\left( {p^{\text {B}}}\right) ^2p^{\text {W}}+ \hat{k}_{\text {b}}p^{\text {U}}- k_{\text {u}}p^{\text {W}}, \end{aligned}$$

(89a)

$$\begin{aligned} \frac{\partial p^{\text {B}}}{\partial t}&= v_{\text {b}}\frac{\partial p^{\text {B}}}{\partial x} + \hat{k}_{\text {bw}}\left( {p^{\text {B}}}\right) ^2 p^{\text {W}}- k_{\text {u}}p^{\text {B}}, \end{aligned}$$

(89b)

$$\begin{aligned} \frac{\partial p^{\text {U}}}{\partial t}&= D_{\text {f}}\frac{\partial ^2 p^{\text {U}}}{\partial x^2} - \hat{k}_{\text {b}}p^{\text {U}}+ k_{\text {u}}(p^{\text {B}}+ p^{\text {W}}). \end{aligned}$$

(89c)

When we scale variables just as before, the terms $\left( {p^{\text {B}}}\right) ^2 p^{\text {W}}$ will lead to the forms $\left( \rho {{p^{\text {B}}}}^{\star }\right) ^2 (\rho {p^{\text {W}}}^{\star })$. This will result in a constant factor $\rho ^2$ that remains after cancelling out $\rho$ from all terms in the equation. As a result, we will obtain, upon dropping the starred quantities,

$$\begin{aligned} \frac{\partial p^{\text {W}}}{\partial t}&= - \frac{\partial p^{\text {W}}}{\partial x} + \frac{1}{\varepsilon } \left( -k_{\text {bw}}\left( {p^{\text {B}}}\right) ^2 p^{\text {W}}+ k_{\text {b}}p^{\text {U}}-p^{\text {W}}\right) , \end{aligned}$$

(90a)

$$\begin{aligned} \frac{\partial p^{\text {B}}}{\partial t}&= v \frac{\partial p^{\text {B}}}{\partial x} + \frac{1}{\varepsilon } \left( k_{\text {bw}}\left( {p^{\text {B}}}\right) ^2p^{\text {W}}- p^{\text {B}}\right) , \end{aligned}$$

(90b)

$$\begin{aligned} \frac{\partial p^{\text {U}}}{\partial t}&= D \frac{\partial ^2 p^{\text {U}}}{\partial x^2} + \frac{1}{\varepsilon } \left( p^{\text {B}}+ p^{\text {W}}-k_{\text {b}}p^{\text {U}}\right) ,\quad \end{aligned}$$

(90c)

where the dimensionless parameters v, D, $\varepsilon$, $k_{\text {bw}}$, and $k_{\text {b}}$ are defined by

$$\begin{aligned} v \equiv \frac{v_{\text {b}}}{v_{\text {w}}}, \qquad D \equiv \frac{D_{\text {f}}}{v_{\text {w}}L_0}, \qquad \varepsilon \equiv \frac{v_{\text {w}}}{k_{\text {u}}L_0}, \qquad k_{\text {bw}}\equiv \frac{\hat{k}_{\text {bw}}\rho ^2}{k_{\text {u}}}, \qquad k_{\text {b}}\equiv \frac{\hat{k}_{\text {b}}}{k_{\text {u}}}. \end{aligned}$$

(91)

Recall that $\rho$ is the average density of motors inside the cell. These dimensionless parameters represent, respectively, the bound:walking motor speed ratio (v), the ratio of (time to be transported:time to diffuse) across the cell (D), the ratio of (time spent unbound:time to walk) across the cell ($\varepsilon$), the interaction parameter $k_{\text {bw}}$, which represents the ratio of (net rate of collisions that result in direction change:unbinding rate), and the ratio of (time spent unbound:time spent bound) ($k_{\text {b}}$). Note that the average density of motors $\rho$ enters into the interaction rate parameter due to the nonlinearity of the model with motor–motor interaction.

1.4 Non-Spatial Myosin Model

In Sect. 4.3, we seek to determine whether the Type I or Type II QSS PDE better approximates the behaviour of the full myosin system. To understand the behaviour, we study the non-spatial myosin model kinetics through a phase-plane analysis, where the advection and diffusive processes in (90) are neglected.

The non-spatial myosin model kinetics are described by the following system of ODEs:

$$\begin{aligned} \frac{\mathrm{d} p^{\text {W}}}{\mathrm{d} t}= & {} -k_{\text {bw}}\left( {p^{\text {B}}}\right) ^2 p^{\text {W}}+ k_{\text {b}}p^{\text {U}}-p^{\text {W}}, \qquad \frac{\mathrm{d} p^{\text {B}}}{\mathrm{d}t} = k_{\text {bw}}\left( {p^{\text {B}}}\right) ^2p^{\text {W}}- p^{\text {B}}, \qquad \nonumber \\ \frac{\mathrm{d} p^{\text {U}}}{\mathrm{d} t}= & {} p^{\text {B}}+ p^{\text {W}}-k_{\text {b}}p^{\text {U}}, \end{aligned}$$

(92)

where time has been scaled to remove the $\varepsilon$ dependence. Due to conservation of mass, we can write $p^{\text {U}}= 1 - p^{\text {W}}- p^{\text {B}}$. This facilitates the reduction of this system of three equations to a system of two equations:

$$\begin{aligned} \frac{\mathrm{d} p^{\text {W}}}{\mathrm{d} t}&= -k_{\text {bw}}\left( {p^{\text {B}}}\right) ^2 p^{\text {W}}+ k_{\text {b}}\left( 1 - p^{\text {W}}- p^{\text {B}}\right) -p^{\text {W}}, \end{aligned}$$

(93a)

$$\begin{aligned} \frac{\mathrm{d} p^{\text {B}}}{\mathrm{d}t}&= k_{\text {bw}}\left( {p^{\text {B}}}\right) ^2p^{\text {W}}- p^{\text {B}}. \end{aligned}$$

(93b)

With $k_{\text {bw}}= 25$ and $k_{\text {b}}= 3$, a phase-plane analysis (see Fig. 17) reveals the existence of an unstable manifold which divides the $(p^{\text {W}},p^{\text {B}})$ plane into two regions. For initial conditions below this unstable manifold, the system converges to a steady-state with $p^{\text {B}}= 0$, but $p^{\text {W}}> 0$, as in Type I QSS. For initial conditions above this unstable manifold, the system converges to a steady-state with $p^{\text {B}}> 0$, as in Type II QSS.

Fig. 17

Phase-plane analysis of the non-spatial myosin model. A phase-plane analysis of the non-spatial myosin model (93) reveals the existence of an unstable manifold that divides $(p^{\text {W}},p^{\text {B}})$ space into two regions. For initial conditions below the unstable manifold, the system tends to a steady-state with $p^{\text {B}}=0$, but for initial conditions above the unstable manifold, the system tends to a steady-state with $p^{\text {B}}> 0$ (Color figure online)

Appendix C: Numerics for the Steady-State of the QSS PDEs

In this appendix we show how to numerically compute the steady-state solution of the QSS PDEs by recasting the non-local problem into an initial boundary value problem (IBVP), which is amenable to a numerical shooting method.

For the QSS PDE associated with the kinesin model (29) of Sect. 4.1, the steady-state problem is

$$\begin{aligned} \frac{\mathrm{d}\alpha }{\mathrm{d}x} = \frac{k_{\text {a}}}{D}\left[ 2P(x) -1\right] g(\alpha ), \qquad \int _{0}^{1} \left( k_{\text {a}}g(\alpha ) + \alpha \right) \, \mathrm{d}x =1, \end{aligned}$$

(94)

where $g(\alpha )$ is either the saturated binding model (34) or the Hill function (48). To reformulate (94), we define N(x) by

$$\begin{aligned} N(x) \equiv \int _{0}^{x} \left( k_{\text {a}}g[\alpha (\eta )] + \alpha (\eta ) \right) \, \mathrm{d}\eta -1. \end{aligned}$$

(95)

Then, (94) is equivalent to the ODE system

$$\begin{aligned} \frac{\mathrm{d}\alpha }{\mathrm{d}x} = \frac{k_{\text {a}}}{D}\left[ 2P(x) -1\right] g(\alpha ), \qquad \frac{\mathrm{d} N}{\mathrm{d}x} = k_{\text {a}}g(\alpha ) + \alpha , \end{aligned}$$

(96)

with $N(0)=-1$. We then specify $\alpha (0)=\beta$, where $\beta$ is a value to be determined. We solve the IBVPs (96) for various values of $\beta$ and output the quantity $N(1;\beta )$. In this numerical shooting procedure, Newton’s method on $\beta$ is then used to satisfy the required terminal constraint $N(1;\beta )=0$.

A similar approach can be used to compute steady-state solutions of the QSS PDE (54) for the kinesin–dynein model of Sect. 4.2 subject to the total mass constraint $\int _{0}^{1} y(x)\, \mathrm{d}x=1$. In place of (96) we obtain

$$\begin{aligned} \frac{\mathrm{d}\alpha }{\mathrm{d}x}= & {} -\frac{k_{\text {a}}}{D} \frac{ \left[ v (k \alpha + 1 - Q) -Q \right] }{\left( k \alpha + 1 -Q \right) ^2 + Q(1-Q) } \left( k \alpha + 1 -Q \right) \alpha , \qquad \nonumber \\ \frac{\mathrm{d} N}{\mathrm{d}x}= & {} \left( 1 + \frac{1}{k_{\text {a}}} \right) \frac{ (k\alpha +1) \alpha }{k\alpha + 1 -Q}, \end{aligned}$$

(97)

with $N(0)=-1$ and $\alpha (0)=\beta$, where $\beta >0$ is a shooting parameter determined numerically by satisfying the terminal constraint $N(1;\beta )=0$.

Finally, we consider steady-state solutions of the QSS PDE (68) for the myosin model of Sect. 4.3 subject to the total mass constraint $\int _{0}^{1} y(x)\, \mathrm{d}x=1$. In place of (96) we get

$$\begin{aligned} \frac{\mathrm{d}\alpha }{\mathrm{d}x} = -\frac{k_{\text {b}}}{D} \frac{ \left( v k_{\text {bw}}\alpha ^2 -1 \right) }{ k_{\text {bw}}\alpha ^2 -1 } \alpha , \qquad \frac{\mathrm{d} N}{\mathrm{d}x} = \frac{\left( k_{\text {b}}+ 1 \right) }{k_{\text {b}}k_{\text {bw}}} \frac{ (k_{\text {bw}}\alpha ^2 +1)}{\alpha }, \end{aligned}$$

(98)

with $N(0)=-1$ and $\alpha (0)=\beta$, where $\beta >0$ is computed numerically to satisfy the constraint $N(1;\beta )=0$. A steady-state solution exists only when $k_{\text {bw}}\alpha ^2>1$ on $0\le x\le 1$.

To numerically determine the boundary in parameter space where $k_{\text {bw}}\alpha ^2>1$ holds on $0\le x\le 1$ for the steady-state when $0<v<1$, it is convenient to reformulate (98). We define $A(x)\equiv \sqrt{k_{\text {bw}}} \alpha (x)$ to transform (98) to

$$\begin{aligned}&\frac{\mathrm{d}A}{\mathrm{d}x} = -c_1 \frac{ \left( v A^2 -1 \right) }{ A^2 -1 } A, \qquad \frac{\mathrm{d} N}{\mathrm{d}x} = c_2 \frac{\left( A^2 + 1 \right) }{A}, \qquad \nonumber \\&\hbox {where} \quad c_1 \equiv \frac{k_{\text {b}}}{D}, \qquad c_2 \equiv \frac{k_{\text {b}}+1}{k_{\text {b}}\sqrt{k_{\text {bw}}}}. \end{aligned}$$

(99)

A steady-state solution to the QSS PDE exists only when $A(x)>1$ on $0\le x\le 1$. Since (99) implies that A(x) is monotonic in x whenever $A>1$, then it is possible that $A\rightarrow 1^+$ only for $x\rightarrow 0^+$ or $x\rightarrow 1^-$. However, since $A\rightarrow {1/\sqrt{v}}>1$ on the infinite line as $x\rightarrow \infty$, it follows that we can only have $A\rightarrow 1^+$ as $x\rightarrow 0^+$. To determine the local behaviour as $A\rightarrow 1^+$ and $x\rightarrow 0^+$, we calculate from (99) that ${\mathrm{d}A/\mathrm{d}x}\sim c_1{(1-v)/[2(A-1)]}$ and ${\mathrm{d}N/\mathrm{d}x}\sim 2c_2$. This yields the local behaviour

$$\begin{aligned} A \sim 1 + \sqrt{c_1(1-v)x}, \qquad N \sim -1 + 2c_2 x, \qquad \hbox {as} \quad x\rightarrow 0^+. \end{aligned}$$

(100)

For a fixed v and $D>0$, with $0<v<1$, to determine the region in the parameter space $k_{\text {bw}}$ versus $k_{\text {b}}$ where $A(x)>1$ on $0\le x\le 1$, we proceed as follows. We fix $c_1$ in (99), numerically integrate the IBVPs (99) with the local behaviour (100) imposed at some $x=\delta$, with $0<\delta \ll 1$, and numerically shoot on the value of $c_2$ for which $N(1;c_2)=0$. From (99), this determines $k_{\text {b}}$ and $k_{\text {bw}}$ as $k_{\text {b}}=c_1 D$ and $k_{\text {bw}}= \left[ {(k_{\text {b}}+1)/(k_{\text {b}}c_2)}\right] ^2$.

Appendix D: Boundary Layer Analysis

In this appendix we determine the appropriate boundary conditions for our QSS PDEs, and we analyse the boundary layers near $x=0,1$. We focus our discussion on general three-component systems on $0\le x\le 1$ of the form

$$\begin{aligned} p_{1t} = - v_1 p_{1x} + \frac{f_1}{\varepsilon }, \qquad p_{2t} = v_2 p_{2x} + \frac{f_2}{\varepsilon }, \qquad p_{3t} = D p_{3xx} + \frac{f_3}{\varepsilon }, \qquad \end{aligned}$$

(101a)

where $v_1$, $v_2$, D are positive ${\mathscr {O}}(1)$ constants, $\varepsilon \ll 1$, and the kinetics $f_j=f_j(p_1,p_2,p_3)$ for $j=1,\ldots ,3$, satisfy the conservation condition

$$\begin{aligned} f_1 + f_2 + f_3 =0. \end{aligned}$$

(101b)

By imposing the mass constraint $\partial _t \int _{0}^{1} \left( p_1+p_2+p_3\right) \, dx =0$, and setting $p_1(0,t)=p_2(1,t)=0$, we obtain the following boundary conditions for (101a):

$$\begin{aligned} D p_{3x} + v_2 p_2 - v_1 p_1 =0, \quad \hbox {at} \quad x=0,1\,; \qquad p_1(0,t)=0, \quad p_2(1,t)=0.\nonumber \\ \end{aligned}$$

(101c)

We assume that there is a unique one-parameter family $\mathbf {p}^0(\alpha )\equiv (p_1^0(\alpha ),p_2^0(\alpha ),p_3^0(\alpha ))^T$ of solutions to the leading-order problem $\mathbf {f}=(f_1,f_2,f_3)^T=\mathbf {0}$, and that $\mathbf {p}^0$ is a slow manifold for (101) in the sense of Definition 3.1. This is the leading-order outer solution, valid away from boundary layers at $x=0,1$. Then, as shown in Sect. 3, $\alpha =\alpha (x,t)$ satisfies the QSS PDE (24a), which can be written as

$$\begin{aligned} \partial _t \left( p_1^0 + p_2^0 + p_3^0 \right) = \partial _{x} \left( -v_1 p_1^0 + v_2 p_2^0 + D \partial _x p_3^0 \right) . \end{aligned}$$

(102)

We now determine an appropriate boundary condition for (102) as $x\rightarrow 0^+$ by analysing the boundary layer structure for (101) near the left endpoint $x=0$. As $x\rightarrow 0^+$, we obtain from the outer solution that

$$\begin{aligned} p_1 = p_{10}^{0} + {\mathscr {O}}(x), \qquad p_2 = p_{20}^{0} + {\mathscr {O}}(x), \qquad p_3\rightarrow p_{30}^{0}+ x \frac{dp_3^0}{dx}{\big \vert _{x=0}} + \cdots ,\nonumber \\ \end{aligned}$$

(103)

where we have defined $p_{j0}^0\equiv p_j^{0}(\alpha (0,t))$ for $j=1,\ldots ,3$.

We will only analyse in detail the region near $x=0$, as a similar analysis can be done near $x=1$. For $t={\mathscr {O}}(1)$ the two possible dominant balances for the spatial derivatives in (101a) near $x=0$ are $x={\mathscr {O}}(\sqrt{\varepsilon })$ and $x={\mathscr {O}}(\varepsilon )$. On the wider such scale, we let $\xi ={x/\sqrt{\varepsilon }}$ to obtain from (101a) that

$$\begin{aligned} p_{1t} = - \frac{v_1}{\sqrt{\varepsilon }} p_{1\xi } + \frac{f_1}{\varepsilon }, \qquad p_{2t} = \frac{v_2}{\sqrt{\varepsilon }} p_{2\xi } + \frac{f_2}{\varepsilon }, \qquad p_{3t} = \frac{D}{\varepsilon } p_{3\xi \xi } + \frac{f_3}{\varepsilon }. \qquad \end{aligned}$$

(104)

To leading-order we obtain that $f_1=f_2=0$, so that from (101b) we must have $f_3=0$. As a result, we obtain to leading-order that $p_1\sim p_{10}^0$, $p_2\sim p_{20}^0$, and $p_3\sim p_{30}^0$. This implies that our QSS approximation is still valid when $x={\mathscr {O}}(\sqrt{\varepsilon })$.

Next, we analyse the region where $x={\mathscr {O}}(\varepsilon )$. Upon introducing $\eta \equiv {x/\varepsilon }$, we obtain from (101a) that

$$\begin{aligned} \varepsilon p_{1t} = - v_1 p_{1\eta } + f_1, \qquad \varepsilon p_{2t} = v_2 p_{2\eta } + f_2, \qquad \varepsilon p_{3t} = \frac{D}{\varepsilon } p_{3\eta \eta } + f_3.\qquad \end{aligned}$$

(105a)

From (101c), the boundary conditions for this system are

$$\begin{aligned} \frac{D}{\varepsilon } p_{3\eta } + v_2 p_2 - v_1 p_1=0, \quad \hbox {at} \quad \eta =0 \,; \qquad p_1(0,t)=0, \end{aligned}$$

(105b)

while the asymptotic matching conditions, as obtained from (103), are that

$$\begin{aligned} p_1 \sim p_{10}^{0}, \qquad p_2 \sim p_{20}^{0}, \qquad p_3\sim p_{30}^{0}+ \varepsilon \eta \frac{dp_3^0}{dx}{\big \vert _{x=0}}, \qquad \hbox {as} \quad \eta \rightarrow \infty .\qquad \end{aligned}$$

(105c)

For $t={\mathscr {O}}(1)$, we neglect the asymptotically negligible left-hand sides of (105a) to obtain

$$\begin{aligned} -v_1 p_{1\eta } = -f_1, \qquad v_2 p_{2\eta } =- f_2, \qquad \frac{D}{\varepsilon } p_{3\eta \eta } = - f_3. \end{aligned}$$

(106)

By adding the equations in (106), and using the conservation condition (101b), we obtain upon integration in $\eta$ that, for all $\eta >0$,

$$\begin{aligned} \frac{D}{\varepsilon } p_{3\eta } - v_1 p_1 + v_2 p_2 = A, \end{aligned}$$

(107)

where A is independent of $\eta$. By evaluating this expression at $\eta =0$, (105b) yields that $A=0$. With $A=0$, we then evaluate (107) as $\eta \rightarrow \infty$ by using the matching condition (105c). This yields

$$\begin{aligned} D \frac{\mathrm{d}p_3^0}{\mathrm{d}x} - v_1 p_{1}^0 + v_2 p_{2}^0 = 0, \quad \hbox {at} \quad x=0. \end{aligned}$$

(108a)

This key result shows that to obtain the boundary condition at $x=0$ for the QSS PDE for $\alpha (x,t)$ we can simply substitute the outer approximation $p_1=p_1^0(\alpha )$, $p_2=p_2^0(\alpha )$, and $p_3=p_3^0(\alpha )$, into the first condition of (101c). In this sense, the QSS PDE inherits the no-flux boundary condition (101c) at $x=0$. We remark that a similar analysis can be done near $x=1$, with the analogous result that

$$\begin{aligned} D \frac{\mathrm{d}p_3^0}{\mathrm{d}x} - v_1 p_{1}^0 + v_2 p_{2}^0 = 0, \quad \hbox {at} \quad x=1. \end{aligned}$$

(108b)

To complete the boundary layer analysis near $x=0$, we expand

$$\begin{aligned} p_3 = p_{30}^{0} + \frac{\varepsilon }{D} {\mathscr {P}}_3 + \cdots , \end{aligned}$$

(109)

and obtain from the first two equations in (106), together with (107) with $A=0$, the following boundary layer problem on $0<\eta <\infty$:

$$\begin{aligned} v_1 p_{1\eta }&= f_1\left( p_1,p_2,p_{30}^0\right) ; \qquad p_1(0)=0, \quad p_1\rightarrow p_{10}^{0} \, \quad \hbox {as}\quad \eta \rightarrow \infty , \end{aligned}$$

(110a)

$$\begin{aligned} v_2 p_{1\eta }&= -f_2\left( p_1,p_2,p_{30}^0\right) ; \qquad p_2\rightarrow p_{20}^{0} \, \quad \hbox {as}\quad \eta \rightarrow \infty , \end{aligned}$$

(110b)

$$\begin{aligned} {\mathscr {P}}_{3\eta }&= v_1 p_1 - v_2 p_2; \qquad {\mathscr {P}}_{3\eta } \sim D \frac{dp_3^0}{dx}\big \vert _{x=0} \, \quad \hbox {as}\quad \eta \rightarrow \infty . \end{aligned}$$

(110c)

Although the first two equations for $p_1$ and $p_2$ are uncoupled from ${\mathscr {P}}_3$, in general it is not possible to calculate $p_1$ and $p_2$ analytically, especially when $f_1$ and $f_2$ are nonlinear in $p_1$ and $p_2$. However, the system for $p_1$ and $p_2$ are readily studied in the phase-plane.

We remark that a similar boundary layer analysis can be done near $x=1$. To study this boundary layer, we now define $\eta ={(1-x)/\varepsilon }$. We readily find in place of (110a) and (110b) that

$$\begin{aligned} v_1 p_{1\eta }&= -f_1\left( p_1,p_2,p_{31}^0\right) \,; \qquad \, \quad p_1\rightarrow p_{11}^{0} \, \quad \hbox {as}\quad \eta \rightarrow \infty , \end{aligned}$$

(111a)

$$\begin{aligned} v_2 p_{1\eta }&= f_2\left( p_1,p_2,p_{30}^0\right) \,; \qquad p_2(0)=0, \qquad p_2\rightarrow p_{21}^{0} \,\quad \hbox {as}\quad \eta \rightarrow \infty , . \end{aligned}$$

(111b)

Here $p_{j1}^{0}\equiv p_j^{0}(\alpha (1,t))$, for $j=1,\ldots ,3$.

1.1 The Kinesin Model

For the kinesin model (25) of Sect. 4.1, the boundary layer system (110) can be solved explicitly. With the QSS approximation $\mathbf {p}^0$, as given in (27), we identify $v_1=v_2=1$, $p_1=p^{\text {R}}$, $p_2=p^{\text {L}}$, and $p_3=p^{\text {U}}$. From (27), we calculate that $p_{10}^0=k_{\text {a}}P(0) g(\alpha _0)$, $p_{20}^0=k_{\text {a}}\left[ 1-P(0)\right] g(\alpha _0)$, and $p_{30}^0=\alpha (0)$, where $\alpha _0\equiv \alpha (0,t)$. Therefore, using the reaction kinetics in (25), (110) becomes

$$\begin{aligned} p_{1\eta }=p_{10}^{0} - p_1, \qquad p_{2\eta }=-p_{20}^{0} + p_2, \qquad {\mathscr {P}}_{3\eta } = p_1 - p_2. \end{aligned}$$

(112)

The solution with $p_1(0)=0$ and $p_2\rightarrow p_{20}^0$ as $\eta \rightarrow \infty$, is simply $p_1=p_{10}^{0}(1-e^{-\eta })$, and $p_2=p_{20}^{0}$. Then, ${\mathscr {P}}_3$ is obtained up to a constant by integrating the last equation in (112). In this way, we obtain the boundary layer solution for $x={\mathscr {O}}(\varepsilon )$ that

$$\begin{aligned} p^{\text {R}}\sim & {} p_{10}^{0}\left( 1-e^{-x/\varepsilon }\right) , \qquad p^{\text {L}}\sim p_{20}^0, \qquad \nonumber \\ p^{\text {U}}= & {} p_{30}^{0} + \frac{\varepsilon }{D} \left( \eta D \frac{\mathrm{d}\alpha }{\mathrm{d}x} \big \vert _{x=0} + p_{10}^{0} e^{-\eta } + A_3 \right) , \end{aligned}$$

(113)

where the constant $A_3$ can only be determined from a two-term outer QSS solution, which is intractable analytically. This analysis shows two key features. Firstly, the right-moving motors have a classic boundary layer behaviour when $x={\mathscr {O}}(\varepsilon )$. Secondly, for $x={\mathscr {O}}(\varepsilon )$ the unbound kinesin motor density $p^{\text {U}}$ differs from its outer approximation only by an error ${\mathscr {O}}({\varepsilon /D})$. A similar calculation can be done for the boundary layer near $x=1$ using (111). We leave the details to the reader.

1.2 The Kinesin–Dynein Model

For the kinesin–dynein model (49), the boundary layer equations (110) for the layer near $x=0$ is analysed via the phase-plane. Using $\mathbf {f}$ in (49), and setting $v_1=1$ and $v_2=v$, (110a) and (110b) on $0<\eta <\infty$ become

$$\begin{aligned} p_{1\eta }&= -p_1 - k p_1 p_2 + k_{\text {a}}Q p_{30}^0, \qquad p_1(0)=0, \qquad \nonumber \\ p_1&\rightarrow p_{10}^0\equiv \frac{Q\alpha _0}{k\alpha _0+1-Q} \, \quad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(114a)

$$\begin{aligned} p_{2\eta }&= -\frac{1}{v}\left[ k_{\text {a}}(1-Q)p_{30}^0 - p_2 + k p_1 p_2\right] , \qquad p_2 \rightarrow \alpha _0 \qquad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(114b)

where $p_{30}^0={(k\alpha _0+1)\alpha _0/[k_a(k\alpha _0+1-Q)]}$. To analyse (114) in the phase-plane, it is convenient to introduce new variables $q_1(\eta )$ and $q_2(\eta )$ defined by

$$\begin{aligned} p_1 = \frac{r_2}{k} q_1, \qquad p_2=\frac{r_1}{k} q_2, \qquad \hbox {where} \qquad r_1 = k \alpha _0, \qquad r_2\equiv \frac{Qr_1}{r_1 + 1-Q}.\nonumber \\ \end{aligned}$$

(115)

In terms of $q_1$ and $q_2$, (114) transforms to the two-component dynamical system

$$\begin{aligned} q_{1\eta }&=g_1(q_1,q_2)\equiv (1-q_1) + r_1(1-q_1 q_2), \qquad q_1(0)=0, \qquad \nonumber \\ q_1&\rightarrow 1 \quad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(116a)

$$\begin{aligned} q_{2\eta }&= g_2(q_1,q_2)\equiv - \frac{1}{v}\left[ 1-q_2 + r_2 (q_1 q_2 -1) \right] , \qquad q_2 \rightarrow 1 \qquad \hbox {as} \quad \eta \rightarrow +\infty . \end{aligned}$$

(116b)

As a function of $r_1$, we have $r_2=0$ when $r_1=0$, $r_2\rightarrow Q<1$ as $r_1\rightarrow \infty$, and that $r_2$ is monotone increasing in $r_1$ since ${dr_2/dr_1}={[Q(1-Q)]/(r_1+1-Q)^2}>0$ holds for $0<Q<1$. It follows that $0<r_2<1$ for any $r_1>0$.

Fig. 18

Qualitative analysis of boundary layer behaviour of the kinesin–dynein model. Phase portraits of $q_2$ versus $q_1$ for boundary layer solutions of the kinesin–dynein model near $x=0$ (a) and near $x=1$ (b) from (116) and (117), respectively. In a there is a unique value $q_2=q_2^{0}$ at $q_1=0$ for which (116) has a solution with $(q_1,q_2)\rightarrow (1,1)$ as $\eta \rightarrow +\infty$. In b there is a unique value $q_1=q_1^{0}$ at $q_2=0$ for which (117) has a solution with $(q_1,q_2)\rightarrow (1,1)$ as $\eta \rightarrow +\infty$. The parameter values of $r_1$, $r_2$, and v for b are those consistent with Fig. 7. a $r_1 = 2.0$, $r_2 = 0.5$, $v = 0.5$. b $r_1 = 1.69$, $r_2 = 0.85$, $v = 0.5$ (Color figure online)

By calculating the Jacobian $J_g$ of $g_1$ and $g_2$ at the equilibrium state $q_1=q_2=1$, we find that

$$\begin{aligned} \hbox {det}(J_g) = -\frac{1}{v\left( k\alpha _0+1-Q\right) } \left[ (1-Q)(1+2k\alpha _0) + k\alpha _0^2\right] <0, \end{aligned}$$

so that $q_1=q_2=1$ is a saddle point for the dynamics. In Fig. 18a we plot the phase portrait $q_2$ versus $q_1$ and nullclines for (116) for representative values $r_1=2$, $r_2=0.5$, and $v=0.5$. We observe that the $q_2$ nullcline intersects the $q_2$ axis at $q_2=1-r_2\in (0,1)$ since $0<r_2<1$. This plot indicates the existence of a unique value $q_2(0)=q_2^{0}>1-r_2$ for which (116) has a solution with $(q_1,q_2)\rightarrow (1,1)$ as $\eta \rightarrow +\infty$. This qualitative analysis confirms the existence of a boundary layer solution near $x=0$ for the kinesin–dynein model for all range of parameters.

A similar phase-plane analysis can be performed to analyse the boundary layer system (111) near $x=1$. In place of (116), we obtain that

$$\begin{aligned} q_{1\eta }&=-g_1(q_1,q_2)\equiv -\left[ (1-q_1) + r_1(1-q_1 q_2)\right] , \qquad q_1\rightarrow 1 \quad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(117a)

$$\begin{aligned} q_{2\eta }&= -g_2(q_1,q_2)\equiv \frac{1}{v} \left[ 1-q_2 + r_2 (q_1 q_2 -1) \right] , \qquad q_2(0)=0 \qquad \nonumber \\ q_2&\rightarrow 1 \qquad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(117b)

where in place of (115), $r_1$ and $r_2$ are now defined by $r_1 = k \alpha _1$ and $r_2\equiv {Qr_1/(r_1 + 1-Q)}$, where $\alpha _1=\alpha$ at $x=1$. In Fig. 18b we plot the phase portrait and nullclines for (117) for $r_1=1.69$, $r_2=0.85$, and $v=0.5$, which corresponds to the parameter values used in the caption of Fig. 7. This phase portrait shows the existence of a unique value $q_1(0)=q_1^{0}$ for which (117) has a solution with $(q_1,q_2)\rightarrow (1,1)$ as $\eta \rightarrow +\infty$. Our computations yield $q_1^{0}\approx 1.95$, so that from (115) we get $p_1\approx 0.83$ at $x=1$.

1.3 The Myosin Model

For the full myosin transport model (90), the boundary layer equations (110a)–(110b) near $x=0$ can be studied qualitatively in the phase-plane. Upon setting $v_1=1$ and $v_2=v$, (110a) and (110b) on $0<\eta <\infty$ become

$$\begin{aligned} p_{1\eta }&= - k_{\text {bw}}p_1 p_2^2 - p_1 + k_{\text {b}}p_{30}^0, \qquad p_1(0)=0, \qquad \nonumber \\ p_1&\rightarrow p_{10}^0\equiv \frac{1}{k_{\text {bw}}\alpha _0}, \quad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(118a)

$$\begin{aligned} p_{2\eta }&= -\frac{1}{v}\left( k_{\text {bw}}p_1 p_2^2 - p_2 \right) , \qquad p_2 \rightarrow \alpha _0 \qquad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(118b)

where $p_{30}^0={\left( \alpha _0+{1/[k_{\text {bw}}\alpha _0]}\right) /k_{\text {b}}}$ and $\alpha _0=\alpha (0,t)$. We conveniently introduce new variables $q_1$ and $q_2$ defined by

$$\begin{aligned} p_1 = \frac{1}{k_{\text {bw}}\alpha _0} q_1, \qquad p_2=\alpha _0 q_2, \qquad \end{aligned}$$

(119)

so that in terms of $r\equiv k_{\text {bw}}\alpha _0^2$, (118) becomes

$$\begin{aligned} q_{1\eta }&=g_1(q_1,q_2)\nonumber \\&\equiv -r\left( q_1 q_2^2-1\right) + 1 - q_1,\qquad q_1(0)=0,\qquad q_1\rightarrow 1\quad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(120a)

$$\begin{aligned} q_{2\eta }&= g_2(q_1,q_2)\equiv - \frac{1}{v}\left( q_1 q_2^2 - q_2\right) , \qquad q_2 \rightarrow 1 \qquad \hbox {as} \quad \eta \rightarrow +\infty . \end{aligned}$$

(120b)

At the equilibrium state $q_1=q_2=1$, the determinant of the Jacobian $J_g$ of $g_1$ and $g_2$ is $\hbox {det}(J_g) = {(1-r)/v}$. Therefore, $\hbox {det}(J_g) <0$ and $q_1=q_2=1$ is a saddle point if $r\equiv k_{\text {bw}}\alpha _0^2>1$. In Fig. 19a we plot the phase portrait of $q_2$ versus $q_1$ and nullclines for (120) for the representative values $r=5$ and $v=0.5$. We observe that there is a unique value $q_2(0)=q_2^{0}$ for which (120) has a solution with $(q_1,q_2)\rightarrow (1,1)$ as $\eta \rightarrow +\infty$. As such, there is always a boundary layer solution near $x=0$ for the myosin model.

Fig. 19

Qualitative analysis of boundary layer behaviour of the myosin model. Phase portraits of $q_2$ versus $q_1$ for boundary layer solutions of the myosin model near $x=0$ (a) and near $x=1$ (b) from (120) and (121), respectively. In a there is a unique value $q_2=q_2^{0}$ at $q_1=0$ for which (120) has a solution with $(q_1,q_2)\rightarrow (1,1)$ as $\eta \rightarrow +\infty$. However, for the right boundary layer, the phase-plane in b there is no value $q_1=q_1^{0}>0$ at $q_2=0$ for which $(q_1,q_2)\rightarrow (1,1)$ as $\eta \rightarrow \infty$. a $r = 5$, $v = 0.5$. b $r = 5$, $v = 0.5$ (Color figure online)

A similar boundary layer system near $x=1$ can be obtained from (111) for the myosin model. In place of (120), we obtain that

$$\begin{aligned} q_{1\eta }&=-g_1(q_1,q_2)\equiv r\left( q_1 q_2^2-1\right) -1 + q_1, \qquad \, \qquad q_1\rightarrow 1 \quad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(121a)

$$\begin{aligned} q_{2\eta }&= -g_2(q_1,q_2)\equiv \frac{1}{v} \left( q_1 q_2^2 -q_2\right) , \qquad q_2(0)=0 \qquad q_2 \rightarrow 1 \qquad \hbox {as} \quad \eta \rightarrow +\infty , \end{aligned}$$

(121b)

where r is now defined by $r=k_{\text {bw}}\alpha _1^2$ with $\alpha _1=\alpha (1,t)$. Although the equilibrium point $q_1=q_2=1$ is a saddle point of (121) whenever $r>1$, the phase portrait in the $q_2$ versus $q_1$ plane shown in Fig. 19b shows that there is no value $q_1(0)=q_1^{0}>0$ on $q_2=0$ for which $(q_1,q_2)\rightarrow (1,1)$ as $\eta \rightarrow \infty$.

As such, we conclude for the Type II QSS approximation (64) for the myosin model that there is no steady-state boundary layer solution near $x=1$ that allows the extra boundary condition $p^{\text {B}}=0$ at $x=1$ to be satisfied.