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A Single Cell

Non-Linear and Linear DEsno_title

Example (Non-Linear and Linear DEs)  

Earlier, we briefly explored the role of a cellular action potential plays in short term memory. Cells such as neuronal, cardiac and muscle cells are excitable, i.e. when stimulated with a subthreshold stimulus, the cell's electrical potential remains more or less constant. On the other hand, when the cell is stimulated with a suprathreshold stimulus, the cell's electrical potential will change dramatically, and over time return to its rest value.

The action potential is a mechanism for cells communicating with each other. Cellular communication happens when a cell releases a packet of neurotransmitter that binds to a receptor on a nearby cell, or when the cell changes its transmembrane potential and the change is sensed by an adjoining cell. An example of the first method is a nerve cell talking to another nerve cell using synaptic coupling. The latter method could take place in heart cells that are electrically coupled by gap junctions.

What is the minimum complexity of a cell capable of talking (an excitable cell)? In order for it to be useful in signaling, it must be able to have two stable equilibria, rest and excited. By equilibria, we mean points where the derivative, $ dU/dt$, is zero. By stable, we mean that when you push the solution to either side of the equilibrium, the process described by the ode moves the solution back to the equilibrium. In order to have two stable states, the derivative of the current voltage relationship (if we are considering membrane potential as our means of communication) must equal zero at three different conditions, two of these zeros will be stable and one unstable. (See Examples, and for illustrations of stable and unstable equilibria.) Therefore a model, driven by a cubic equation of the form:

$\displaystyle \frac{\textrm{d}U_B}{\textrm{d}t} = f(U) = U(a-U)(U-1),$ (2.8.14)

where we assume $ 0 < a < 1$, is required to meet the above conditions. Equilibria exist when this derivative is zero, and thus has equilibria, at $ U = 0$, $ U = a$ and $ U = 1$. A graph of $ f(U)$, current, is shown in Figure 2.8.1 and is called the nullcline. From this, we can graphically explore the behavior of any ode. An equilibrium exists at each point where the nullcline crosses the $ dU/dt$ (current) $ = 0$ axis.

Figure: Cubic nullcline for Equation 2.8.14. Here $ a = 0.25$.

Notice that for $ U < 0$, the nullcline $ (dU/dt)$ is positive and for $ 0 < U <
a$, the derivative is negative. If the solution is sitting at $ U = 0$ and you push it to the left (say $ U = -0.9$) then the value of $ f(U)$ is positive so that $ dU/dt > 0$ and the solution moves back to $ U = 0$. Similarly, when you push it to the right (say $ U = 0.1$), then the value of $ f(U)$ is negative so that $ dU/dt < 0$ and the solution moves to the left, back to $ U = 0$. Thus, $ U = 0$ is a stable equilibrium. When $ a < U < 1$, the derivative is again positive and thus $ U = a$ is an unstable equilibrium. Finally, when $ U > 1$, the derivative is negative making $ U = 1$ a stable equilibrium.

The switching nature of this model can be readily demonstrated. Assume that we are resting at $ U = 0$. Now as you move $ U$ to the right, the derivative is $ < 0$ so that if we turn the solution loose, it will migrate back to the stable equilibrium at $ U = 0$. However, if we continue to push so that $ U > a$, then now $ dU/dt > 0$ and the solution will continue to the right until it reaches the point, $ U = 1$. The point, $ a$ is called is called the threshold and with this switch, we have a mechanism for ``talking'', switching from a stable rest ($ U = 0$) state to a stable excitable ($ U = 1$) state.

We can add a stimulus function, $ S(t)$ to $ f(U)$ that has magnitude $ m$ by simply creating a function that is equal to $ S$ for a given interval of time. That is:

$\displaystyle S(t) = \left\{ \begin{array}{rl} 0, & t \le 0 m, & 0 < t \le t_s 0, & t_s < t \end{array} \right.$ (2.8.15)

where $ t_s$ is the duration of the stimulus. This function could be a reflection of coupling from other cells. For example, a neurotransmitter opening an excitatory channel transiently. Adding S(t) to $ f(U)$ gives us:

$\displaystyle \frac{\textrm{d}U_B}{\textrm{d}t} = f(U) + S(t).$ (2.8.16)

Depending on the size of $ m$, if we initially start at $ f(U) = 0$, we can switch from the rest state to the excitable state as shown in 2.8.2

Here, the stimulus amplitude is 0.25 and the duration of the stimulus is altered. Starting at the stable equilibrium at 0,0, the potential increases linearly until the stimulus value returns to zero. By varying the duration of the stimulus, we can achieve sub-threshold, threshold and suprathreshold values. Shown are the durations of the stimuli. Note that for a duration of 1.4, the value of U exceeds the threshold (0.25), at the end of the stimulus, and thus, rapidly moves toward the higher stable equilibrium at 1. As the stimulus duration is reduced, the transition time to the equilibrium at U = 1 is progressively longer until the duration is 1.02. Now, the value of U at the end of the stimulus is no longer suprathreshold and the potential decays back to the stable equilibrium at U = 0.

Figure: Switching with near threshold stimulation, a = 0.25
\includegraphics[width=3in]{threshold} .

Now we must determine how the switch can return to a rest state when it is in an excitable state. Physically, charge is removed from the cell until it crosses the threshold and then the cell takes over, lowering its charge until the lower equilibrium has been reached. As a first attempt at modeling this removal of charge, we could define a variable $ V$ such that:

$\displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = aU.$ (2.8.17)

The problem, however, with this definition is that $ V$ will not continue to grow when $ U$ hits the upper equilibrium point. As a result, the system will never return to the rest state. Thus, we must add a second term that will allow $ V$ to grow once $ U$ is at the equilibrium point. In this case we will add a term that causes exponential decay in the charge:

$\displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = aU - bV,$ (2.8.18)

where $ a < b$. From this equation we can see that as $ U$ increases, so does $ aU$ and thus, so does $ V$. Once $ U$ plateaus at the higher equilibrium, $ V$ will continue to increase, at slower and slower rates because $ bV$ will continue to subtract from $ aU$ larger and larger amounts, until $ aU <
bV$. Once $ bV$ dominates, Equation 2.8.18 will become negative and $ V$ will decrease.

With Equation 2.8.18 for $ V$, our equation defining the entire cell becomes

$\displaystyle \frac{\textrm{d}U_B}{\textrm{d}t} = f(U) - V + S(t).$ (2.8.19)

By subtracting $ V$ from $ f(U)$ the switch becomes monostable by removing two of the equilibria. We can determine the location of the remaining equilibrium is determined by examining when the derivatives for both equations are zero.

Since $ V$ only plays a significant role after the initial stimulation, we can omit $ S(t)$ in the following derivations. First, we will solve $ \frac{\textrm{d}U_M}{\textrm{d}t}=0$:

$\displaystyle \frac{\textrm{d}U_M}{\textrm{d}t}$ $\displaystyle \stackrel{\mathrm{set}}{=} 0$    
$\displaystyle f(U) - V$ $\displaystyle = 0$    
$\displaystyle V$ $\displaystyle = f(U).$    

Now we will solve $ \textrm{d}V/\textrm{d}t$:

$\displaystyle \frac{\textrm{d}V}{\textrm{d}t}$ $\displaystyle \stackrel{\mathrm{set}}{=} 0$    
$\displaystyle (aU - bV)$ $\displaystyle = 0$    
$\displaystyle bV$ $\displaystyle = aU$    
$\displaystyle V$ $\displaystyle = aU/b.$    

The result is the strait line in Figure 2.8.3 with a slope of $ a/b $ The intersection gives us a single equilibrium.
Figure: Cubic and linear nullcline for the Fitzhugh Nagumo cell model. When U(t = 0) < 0.25, the potential collapses because dU/dt < 0. However, when U(t=0) > 0.25 the potential grows and produces an action potential.

Figure: A response to sub-thresholdstimulation at $ t = 2$. The cell potential, U, collapses after the end of the stimulation pulse because the phase point did not cross the threshold (0.25) marking the unstable equilibrium 2.8.3 marking the transition from dU/dt < 0 to dU/dt > 0.

Figure: A response to supra-threshold stimulation at $ t = 2$. In this case, the potential, U, exceedes the threshold, crossing into the region where dU/dt > 0, and thus accelerates toward the equilibrium at U = 1. The peak of the action potential never reaches the point where U = 1, due to the cooling effect reflecting the slow parameter, V.

In biology, the dynamics of moving from a rest state to an excitable state is fast because Na channels open quickly. The recovery, however, is slow because K channels open slowly. We can incorporate this into our model by including a scaling factor into Equation 2.8.18:

$\displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = \epsilon(aU - bV).$ (2.8.20)

Keep in mind that $ \epsilon$ does nothing to alter the equilibrium point since it simply divides out when solving for it.

The switch defined by Equation 2.8.14, $ \frac{\textrm{d}U_B}{\textrm{d}t}$ is bistable with stable equilibria at $ U = 0$ and $ U = 1$. What we would like to do is incorperate a variable that will remove charge from the cell during the excitable stage until the threshold is crossed and the cell and reset itself to the rest state. $ \vert\boldsymbol{\vert}$

Now - move the slow function linear nullcline to the right where the equilibrium is unstable. Now the FHN system behaves as an oscillator as shown in the figure. Same exact model - only a shift in the intersection of the two nullclines (f($ U$) = 0, V = $ U$ / $ \gamma$

Figure: Shifted equilibrium to the unstable point

The result is spontaneious oscillation, because the nullclines intersect at a singular point that is unstable.

Figure: Oscillation of the FHN system due to the unstable equilibrium

next up previous index
Next: Taylor series and identifying Up: Examples of Models Previous: Microscopic/Probabilistic Behavior   Index

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Frank Starmer 2004-05-19