# Math Insight

### Oscillations in higher-dimensional linear systems

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Total points: 10
1. The following system depends on a parameter $r$. \begin{align*} \diff{ z_1}{t} &= r z_{3} - 8 z_{1} + 8 z_{2}\\ \diff{ z_2}{t} &= r z_{1} - 6 z_{2} - z_{3}\\ \diff{ z_3}{t} &= r z_{2} + 4 z_{1} - 8 z_{3} \end{align*}
1. Find a value of the parameter $r$ where the following system will oscillate.
$r=$

(Include at least 4 significant digits. Enter none if the system never oscillates for any value of $r$.)
2. For this value of $r$, calculate the eigenvalues of $A$.
$\lambda_1=$
, $\lambda_2=$
, $\lambda_3=$
3. Write the general form of the solution to the system of equations in vector form, where $\vc{ z } = \left ( z_{1}, \quad z_{2}, \quad z_{3}\right )$. You don't need to calculate the eigenvectors of $A$. Instead, your general solution should involve three arbitrary vectors $\vc{c}_1$, $\vc{c}_2$, and $\vc{c}_3$ (which would depend on the eigenvectors and the initial conditions). Rather than writing the solution in terms of complex exponentials, it should be in terms real-valued quantities with sines and cosines.

$\vc{ z }(t) =$

2. Let $E$ be the firing rate of an excitatory neuron and $I$ be the firing rate of an inhibitory neuron. If we model the neurons as linear systems and add a “poor man's delay” in the connection from $I$ to $E$, we could model the system as \begin{align*} \diff{E}{t} &= \frac{1}{\tau_e}(-E - a I_d + P)\\ \diff{I}{t} &= \frac{1}{\tau_i}(-I + bE+Q)\\ \diff{I_d}{t} &= \frac{1}{\delta}(-I_d + I), \end{align*} where $a$ is the strength of inhibitory connection and $b$ is the strength of the excitatory connection. We set the time constant of the excitatory neuron to $\tau_e=10$ ms, the time constant of the inhibitory neuron to $\tau_i=90$ ms, and set $b=6$. The parameter $P$ is the input to the excitatory neuron, and $Q$ is the input to the inhibitory neuron.

The variable $I_d$ is an approximation to the value of $I$ delayed by $\delta$, where $\delta$ is measured in ms. In the following, we will vary the strength $a$ of the inhibitory feedback and examine how the delay required for an oscillation changes.

For all numerical answers, include at least 4 significant digits.

1. A preliminary check, let's review the system without a delay to make sure we cannot get an oscillation without a delay. If we eliminate the delay, setting $I_d$ to $I$ so that the system becomes \begin{align*} \diff{E}{t} &= \frac{1}{\tau_e}(-E - a I + P)\\ \diff{I}{t} &= \frac{1}{\tau_i}(-I + bE+Q),\\ \end{align*} can the system oscillate for any value of $a$?
, because no what the value of $a$ is, the trace of the matrix underlying the system is
and the determinant is
. The equilibrium is always
and can never be a
.
2. Restoring the delay, so that we are back to the system of three equations involving $I_d$, let's first examine the case when $a=4$.

For $a=4$, find two values of the delay $\delta$ for which the system oscillates.
$\delta_1 =$
ms, $\delta_2 =$
ms

You should have found one value less than 100 ms and the other value greater than 100 ms. We will focus on the shorter delay, as the longer delay is much too long to be realistic for delays in a local neuronal circuit. The shorter delay is $\delta=$
ms.

When the delay is tuned to this value, what is the frequency of the oscillations?
Hz.

3. For $a=8$, find the value of the delay $\delta$ that is less than 100 ms for which the system oscillates.
$\delta =$
ms

When the delay is tuned to this value, what is the frequency of the oscillations?
Hz.

4. For $a=15$, find the value of the delay $\delta$ that is less than 100 ms for which the system oscillates.
$\delta =$
ms

When the delay is tuned to this value, what is the frequency of the oscillations?
Hz.

5. As the inhibitory feedback becomes stronger, we observe the following trends. The length of the delay required for oscillations becomes
and the oscillation frequency becomes
.