Math Insight

Spiking voltage worksheet

Math 2241, Spring 2023
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  1. Let $v(t)$ be the (normalized) voltage of a neuron. A highly simplified dynamical system model for the neuron is \begin{align*} \diff{v}{t} &= -v(v-a)(v-1)\\ v(0) &= v_0 \end{align*} where $v_0$ is the initial condition and $a$ is a parameter between zero and one-half.
    1. Find the equilibria of the dynamical system and determine their stability.

      Equilibria: $v=$

      Enter in increasing order, separated by commas.

      Stability:

      Enter the same order as the equilibria, separated by commas. Enter stable for each stable equilibrium and unstable for each unstable equilibrium.

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    2. For the case when $a=0.2$, sketch a phase line for $v$, showing the equilibria as well as the vector field for the four regions of the phase line separated by the equilibria. Use solid circles for stable and open circles for unstable equilibria.
      Feedback from applet
      equilibria:
      number of equilibria:
      stability of equilibria:
      vector field:

      For any other value of $a$ with $0 \lt a \lt 1$, the phase line would look the same except that the middle equilibrium would be in a different location.

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    3. Again with $a=0.2$, sketch solutions for the dynamical system for initial conditions $v(0)=-0.2$, $v(0)=0$, $v(0)=0.9a=0.18$, $v(0)=a=0.2$, $v(0)= 1.1 a=0.22$, and $v(0)=1$. If a solution happens to be an equilibrium, then use a solid line if it is stable and a dashed line if it is unstable.
      Feedback from applet
      Equilibria:
      Final values of curves:
      Initial conditions of curves:
      Speed profiles of curves:
      Stability of equilibria:

      For any other value of $a$ with $0 \lt a \lt 1$, the solution plot would look the same except that the middle equilibrium would be in a different location.

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    4. This model incorporates the positive feedback of the sodium channels, which upon opening drive the voltage upward (to the number one, in the model). Although in some of the cases above, we set $a=0.2$, it can really be any value between zero and one. Describe the meaning of the parameter $a$.

      If the neuron's voltage starts at $v=0$, then the voltage will
      . Imagine that the neuron subsequently receives an input that jumps its voltage up to $v=v_0$. If
      (enter inequality involving $v_0$), then the voltage will decay back down to zero. On the other hand, if
      (enter inequality involving $v_0$), then the voltage will shoot up to one. Therefore, $a$ is the
      for the voltage to shoot up to one.

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  2. In addition to $v(t)$, let $w(t)$ describe the state of potassium channels, which upon opening bring the voltage down. The Fitzhugh-Nagumo model describing the evolution of the neuron's voltage in response to sodium channels and potassium channels is \begin{align*} \diff{v}{t} &= -v(v-a)(v-1) -w\\ \diff{w}{t} &= \varepsilon (v-\gamma w)\\ v(0) &= v_0\\ w(0)&=w_0 \end{align*} where $v_0$ and $w_0$ are the initial conditions, $a$ is a parameter between zero and one-half, $\varepsilon$ is a small positive parameter, and $\gamma$ is a positive parameter. To be concrete, set $a=0.2$, $\varepsilon=0.01$, and $\gamma=3$.
    1. Calculate the nullclines and plot them on the below phase plane.

      Enter the equation for $v$-nullcline after solving for $w$:
      $w =$

      Enter the equation for the $w$-nullcline after solving for $w$:
      $w = $

      Feedback from applet
      equilibria:
      nullclines:
      number of equilibria:
      trajectory 1:
      trajectory 2:
      vector directions in regions:
      vector directions on nullclines:
      vector locations in regions:
      vector locations on nullclines:
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    2. The equilibria are the points where the two nullclines cross. In this case, there is only one equilibrium. What is the equilibrium? The equilibrium is $(v,w)=$

      Draw its location on the phase plane and verify that both $\diff{v}{t}$ and $\diff{w}{t}$ are zero at this equilibrium.

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    3. The nullclines should chop up the phase plane into four regions. We want to determine direction vectors in each of these four regions. For each of the four regions, pick one point in the region and determine the sign of both $\diff{v}{t}$ and $\diff{w}{t}$ at the point. Since the derivatives cannot change sign in the region, the derivatives at this one point determine the rough direction for the whole region. Based on the signs of the two derivatives, draw a vector indicating the general direction the solution $(v(t),w(t))$ will move. (Since $\varepsilon$ is small, $v$ will typically move faster than $w$, so draw the arrow pointing more left or right than up or down.)
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    4. Since there is just one equilibrium, each nullcline is divided into two segments. Draw an arrow on each of the four resulting nullcline segments.
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    5. Imagine that we start with the initial condition $(v_0,w_0)=(0,0)$. What is the solution to the dynamical system with these initial conditions?

      The solution is $v(t)=$
      and $w(t)=$
      .

    6. Next, imagine that the neuron receives a small input that jumps its voltage upward, moving the initial condition to $(v_0,w_0)=(0.18,0)$. For that initial condition, what is the initial value of $\diff{v}{t}$ and $\diff{w}{t}$? (Just plug the initial conditions into the equation.)

      $\diff{v}{t}=$

      $\diff{w}{t} =$

      The means that with the initial condition $(v,w)=(0.18,0)$, the solution begins to move
      in the phase plane.

      Sketch a plausible solution in the phase plane for that initial condition. By plausible solution, we mean a curve that begins at the point $(0.18,0)$, starts in the direction you calculated from the derivatives, and is consistent with the general direction of the vector field you determined above. (Hint, the solution should go the equilibrium in fairly short order, though it does have to cross at least one nullcline.)

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    7. Sketch this solution versus time on the below axes. You can't really determine what $v(t)$ and $w(t)$ will look like, but you can sketch the rough idea of how they increase or decrease and where they head as $t$ gets large. For this plot, the time scale is arbitrary.
      Feedback from applet
      v trajectory:
      w trajectory:
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    8. Now, imagine that the neuron receives a large input that jumps its voltage upward, moving the initial condition all the way up to to $(v_0,w_0)=(0.3,0)$. For that initial condition, what is the initial value of $\diff{v}{t}$ and $\diff{w}{t}$? (Just plug the initial conditions into the equation.)

      $\diff{v}{t}=$

      $\diff{w}{t} =$

      You should get that $\diff{v}{t}$ is much larger than $\diff{w}{t}$. What does this mean about the initial direction of movement of the trajectory in the phase plane? It is moving
      . Begin to sketch a curve starting at the initial condition $(0.3,0)$ in the phase plane, moving in this direction.

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    9. The trajectory is going to keep moving in this general direction until it get close to the $v$-nullcline. When it crosses the $v$-nullcline, what direction should the trajectory move?
      Sketch a trajectory from the initial condition up to a point where it crosses the $v$-nullcline that is consistent with this information.
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    10. Once the trajectory moves above the $v$-nullcline, what is the general direction of movement?
      . And, as a reminder, what is the direction of movement across this segment of the $v$-nullcline?
      This means the trajectory cannot cross this segment of the $v$-nullcline again. Hence it stays close to the $v$-nullcline until it passes the bend in the $v$-nullcline. With the nullcline out of the way, the movement should be primarily left or right and not as much up or down. The reason for the primarily horizontal movement $\varepsilon$ is small, which makes $w$ move more slowly than $v$. When it hits the $w$-nullcline, what direction should it be moving?
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    11. Once the trajectory moves to the left of the $w$-nullcline, what is the general direction of movement?
      Since $\varepsilon$ is small, the movement should be primarily left or right and not as much up or down. Continue the trajectory in this direction until it hits the $v$-nullcline again. When it hits the $v$-nullcline, what direction should it be moving?
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    12. Once the trajectory moves below the $v$-nullcline, what should the general direction of movement be?
      And, as a reminder, what is the direction of movement across this segment of the $v$-nullcline?
      This means that the trajectory cannot cross this segment of the $v$-nullcline again. What actually happens next is that the trajectory will stay just below the $v$-nullcline, and head back toward the equilibrium. Finish the sketch of the trajectory using this information.
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    13. The trajectory that you drew should have been a fairly large excursion in the phase plane. The voltage should have shot up, after which $w$ increased until the voltage shot back down. The voltage should have then shot, past $v=0$ to lower values of $v$, after which $w$ decreased back toward zero. Eventually both $v$ and $w$ should evolved back to the origin $(v,w)=(0,0)$. Sketch this solution on plots versus time on the axes below. Your plot should show a spike in the voltage $v(t)$, which is called the action potential.
      Feedback from applet
      v trajectory:
      w trajectory:
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Math 2241, Spring 2023