Math Insight

Two-dimensional linear problems

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Total points: 10
  1. Let the response, $C$, of a cone and the response, $H$, a horizontal cell to a light stimulus $L$ be given by \begin{align*} \diff{C}{t} &= \frac{1}{\tau_C}(-C - kH +L)\\ \diff{H}{t} &= \frac{1}{\tau_H}(-H + C), \end{align*} where the parameter $k$ is a positive constant describing the strength of the inhibition from the horizontal cell onto the cone. Let the time constants describing how quickly $C$ and $H$ change be $\tau_C = 22$ ms and $\tau_H =84$ ms. (Model is from Spikes, decision, and action by Hugh Wilson.)

    In all your responses, be sure to keep at least 4 significant digits.

    1. Calculate the steady state of the model.

      $C_{ss} = $

      $H_{ss} = $

    2. Find the values of $k$ for which the system is each of the following types.

      For each type, enter your response in the following manner:

      • If the type occurs for just one value of $k$, enter your response as k = blah, where blah represents an expression.
      • If the type occurs for a range of values of $k$, enter your response as an inequality, such as k > blah or even as blah1 < k < blah2.
      • If the type cannot occur for any valid value of $k$, then enter none.

      Conditions on $k$ for each type of system:

      1. Saddle:
      2. Stable node:
      3. Unstable node:
      4. Stable spiral:
      5. Unstable spiral:
      6. Center:
    3. Let $k=0.46$ and $L=3$, which means the system is a
      .

      Solve the system to determine $C(t)$ and $H(t)$ for the initial condition $C(0)=0$ and $H(0)=0$.
      $C(t)=$

      $H(t) = $

    4. Plot the solution $C(t)$ and $H(t)$ as a function of time. Also plot the solution $(C(t),H(t))$ on the phase plane.

      (Online, we don't have a way to grade these plots, so they don't count toward your score.)

  2. Consider the system of differential equations \begin{align*} \diff{x_1}{t} &= - 4 x_{1} - 2 x_{2} + 1\\ \diff{x_2}{t} &= - 4 x_{1} + 11 x_{2} - 1 \end{align*} with initial conditions \begin{align*} x_1(0) &= 5\\ x_2(0) &= -5 \end{align*}
    1. As a first step, rewrite the system in vector form as \begin{align*} \diff{\vc{x}}{t} &= A\vc{x} + \vc{b}\\ \vc{x}(0) &= \vc{x}_0 \end{align*} where:
      $\vc{x} = $

      $\vc{A} = $

      $\vc{b} = $

      $\vc{x}_0 = $

    2. Calculate the equilibrium of the system.

      $\vc{x}_{eq} = $

      (If rounding, include at least 4 significant digits.)

    3. Let $\vc{y} = \vc{x}-\vc{x}_{eq}$. Rewrite the system in terms of $\vc{y}$.


      The initial condition is $\vc{y}(0) = \vc{y}_0$, where $y_0=$
      .

    4. Calculate the eigenvalues of $A$.
      $\lambda = $

      (Separate answers by commas. Include at least 4 significant digits.)
    5. Write down the general solution for $\vc{y}(t)$. The general solution will have two arbitrary vectors. Let $\vc{u}$ be an arbitrary multiple of the eigenvector corresponding to the smaller eigenvalue. Let $\vc{v}$ be an arbitrary multiple of the eigenvector corresponding to the larger eigenvalue.

      $\vc{y}(t) = $

    6. The actual multiple of $\vc{u}$ and $\vc{v}$ used for the solution $\vc{y}$ depend on the initial condition $y(0)=y_0$. Given the value of the initial conditions that you calculated above, determine the solution $\vc{y}(t)$.

      First, write the solution in vector form.

      $\vc{y}(t) =$
      $\cdot$
      $+$
      $\cdot$

      The first blank should be $\vc{u}$, the appropriate multiple of the eigenvector corresponding to the smaller eigenvalue, and the second blank should contain an exponential function involving the smaller eigenvalue. The third blank should be $\vc{v}$, the appropriate multiple of the eigenvector corresponding to the larger eigenvalue, and the fourth blank should contain an exponential function involving the larger eigenvalue.

      Rewrite this solution in terms of the components of $\vc{y}$, which we'll call $y_1$ and $y_2$.

      $y_1(t) = $

      $y_2(t) = $

      On the below phase plane, plot the two eigenvalues (any vectors that are multiplies of $\vc{u}$ and $\vc{v}$). To do so, drag the black points so that the green and blue vectors point in the correct direction, one for each eigenvector. (The applet automatically plots the opposite of each vector, as both directions correspond to the eigenvector.)

      You can click the each vector to toggle its direction; point each vector in the direction that the solution moves along the eigenvector.

      Feedback from applet
      eigenvectors:
      movement direction along eigenvector:
    7. Lastly, solve the system for the original variables $x_1$ and $x_2$.

      $x_1(t) = $

      $x_2(t) = $

      On the below phase plane, move the red point to the equilibrium. Then plot the two eigenvalues so that they are centered at the equilibrium. To do so, drag the black points so that the green and blue vectors point in the correct direction, one for each eigenvector. (The applet automatically plots the opposite of each vector, as both directions correspond to the eigenvector.)

      You can click the each vector to toggle its direction; point each vector in the direction that the solution moves along the eigenvector.

      Feedback from applet
      eigenvectors:
      equilibrium:
      movement direction along eigenvector:
    8. Plot the solution $x_1(t)$ and $x_2(t)$ as a function of time. Also plot the solution $(x_1(t),x_2(t))$ on the phase plane.

      (Online, we don't have a way to grade these plots, so they don't count toward your score.)

    9. e^{- 4.5156097709407 t}

  3. For the dynamical system $\diff{\vc{x}}{t} = A\vc{x}$, where $$A =\left[\begin{matrix}-2 & 8\\-4 & a\end{matrix}\right],$$ find a value of the parameter $a$ where the system is each of the following types. If the system is cannot be of that type for any real value of $a$, enter none.
    1. Saddle:
    2. Stable node:
    3. Unstable node:
    4. Stable spiral:
    5. Unstable spiral:
    6. Center:

  4. Consider the system of differential equations \begin{align*} \diff{x_1}{t} &= 5 x_{1} + 4 x_{2}\\ \diff{x_2}{t} &= - 4 x_{1} + 4 x_{2} + 3 \end{align*} with initial conditions \begin{align*} x_1(0) &= 1\\ x_2(0) &= 1 \end{align*}
    1. As a first step, rewrite the system in vector form as \begin{align*} \diff{\vc{x}}{t} &= A\vc{x} + \vc{b}\\ \vc{x}(0) &= \vc{x}_0 \end{align*} where:
      $\vc{x} = $

      $\vc{A} = $

      $\vc{b} = $

      $\vc{x}_0 = $

    2. Calculate the equilibrium of the system.

      $\vc{x}_{eq} = $

      (If rounding, include at least 4 significant digits.)

    3. Let $\vc{y} = \vc{x}-\vc{x}_{eq}$. Rewrite the system in terms of $\vc{y}$.


      The initial condition is $\vc{y}(0) = \vc{y}_0$, where $y_0=$
      .

    4. Calculate the eigenvalues of $A$.
      $\lambda = $

      (Separate answers by commas. Include at least 4 significant digits.)
    5. Compute the solution $\vc{y}$ and write the result in terms of its component $y_1$ and $y_2$.

      $y_1(t) = $

      $y_2(t) = $

    6. Lastly, solve the system for the original variables $x_1$ and $x_2$.

      $x_1(t) = $

      $x_2(t) = $

    7. Plot the solution $x_1(t)$ and $x_2(t)$ as a function of time. Also plot the solution $(x_1(t),x_2(t))$ on the phase plane.

      (Online, we don't have a way to grade these plots, so they don't count toward your score.)