$ \newcommand{\RR}{\mathbb{R}} \newcommand{\NN}{\mathbb{N}} \newcommand{\OO}{\mathcal{O}} \newcommand{\mathcow}{\OO} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\CC}{\mathbb{C}} \newcommand{\KK}{\mathbb{K}} \newcommand{\PP}{\mathcal{P}} \newcommand{\TT}{\mathcal{T}} \newcommand{\BB}{\mathcal{B}} \newcommand{\LL}{\mathcal{L}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\veca}{\vec{a}} \newcommand{\vecb}{\vec{b}} \newcommand{\vecd}{\vec{d}} \newcommand{\vece}{\vec{e}} \newcommand{\vecf}{\vec{f}} \newcommand{\vecn}{\vec{n}} \newcommand{\vecp}{\vec{p}} \newcommand{\vecr}{\vec{r}} \newcommand{\vecu}{\vec{u}} \newcommand{\vecv}{\vec{v}} \newcommand{\vecx}{\vec{x}} \newcommand{\vecy}{\vec{y}} \newcommand{\vecz}{\vec{z}} \renewcommand{\vec}[1]{\mathbf{#1}} $ Shaw Research Notes

August 19th, 2021

We examine the difference-of-exponentials weight function $w(x,y) = M_1e^{-|x-y|/\sigma_1} - M_2e^{-|x-y|/\sigma_2}$. We derive the traveling pulse solution and the asymptotic approximation to the wave response function for this weight function in the case of the Heaviside firing-rate. The results do not match simulation, but we believe the simulation is not accurate enough to properly compare, and needs to be improved by using a semi-analytic version. Another error was corrected from the 2021-06-10 report.

To-Do List Snapshot

Inhibitory Weight Kernel - Difference of Exponentials

We are considering the adaptive model $$\begin{align*} \mu u_t &= -u + w * f(u - a) \\ \alpha a_t &= -a + \gamma f(u - a) \end{align*}$$ where $f(y) = H(y - \theta)$, but now with the inhibitory weight kernel $$ w(x, y) = M_1 e^{-|x-y|/\sigma_1} - M_2 e^{-|x-y|/\sigma_2}. $$

Traveling Pulse Solution

When the choice of parameters allows for traveling pulses, the pulses are given by the equations $$\begin{align*} U{\left(\xi \right)} &= \begin{cases} \frac{M_{1} \sigma_{1}^{2} \left(e^{\frac{\xi}{\sigma_{1}}} - e^{\frac{\Delta + \xi}{\sigma_{1}}}\right)}{\mu c - \sigma_{1}} + \frac{M_{2} \sigma_{2}^{2} \left(- e^{\frac{\xi}{\sigma_{2}}} + e^{\frac{\Delta + \xi}{\sigma_{2}}}\right)}{\mu c - \sigma_{2}} + \left(\frac{M_{1} \sigma_{1}^{2} \left(1 - e^{- \frac{\Delta}{\sigma_{1}}}\right)}{\mu c - \sigma_{1}} + \frac{M_{2} \sigma_{2}^{2} \left(-1 + e^{- \frac{\Delta}{\sigma_{2}}}\right)}{\mu c - \sigma_{2}} + \gamma - \gamma e^{- \frac{\Delta}{\alpha c}} + \theta\right) e^{\frac{\Delta}{\mu c}} e^{\frac{\xi}{\mu c}} & \text{for}\: \Delta < - \xi \\- \frac{M_{1} \sigma_{1}^{2} e^{\frac{- \Delta - \xi}{\sigma_{1}}}}{\mu c + \sigma_{1}} + \frac{M_{1} \sigma_{1}^{2} e^{\frac{\xi}{\sigma_{1}}}}{\mu c - \sigma_{1}} + 2 M_{1} \sigma_{1} + \frac{M_{2} \sigma_{2}^{2} e^{\frac{- \Delta - \xi}{\sigma_{2}}}}{\mu c + \sigma_{2}} - \frac{M_{2} \sigma_{2}^{2} e^{\frac{\xi}{\sigma_{2}}}}{\mu c - \sigma_{2}} - 2 M_{2} \sigma_{2} + \left(\frac{M_{1} \sigma_{1}^{2} e^{- \frac{\Delta}{\sigma_{1}}}}{\mu c + \sigma_{1}} - \frac{M_{1} \sigma_{1}^{2}}{\mu c - \sigma_{1}} - 2 M_{1} \sigma_{1} - \frac{M_{2} \sigma_{2}^{2} e^{- \frac{\Delta}{\sigma_{2}}}}{\mu c + \sigma_{2}} + \frac{M_{2} \sigma_{2}^{2}}{\mu c - \sigma_{2}} + 2 M_{2} \sigma_{2} + \theta\right) e^{\frac{\xi}{\mu c}} & \text{for}\: \xi < 0 \\\frac{M_{1} \sigma_{1}^{2} \left(- e^{\frac{- \Delta - \xi}{\sigma_{1}}} + e^{- \frac{\xi}{\sigma_{1}}}\right)}{\mu c + \sigma_{1}} + \frac{M_{2} \sigma_{2}^{2} \left(e^{\frac{- \Delta - \xi}{\sigma_{2}}} - e^{- \frac{\xi}{\sigma_{2}}}\right)}{\mu c + \sigma_{2}} & \text{otherwise} \end{cases}\\ A{\left(\xi \right)} &= \begin{cases} \gamma \left(e^{\frac{\Delta}{\alpha c}} - 1\right) e^{\frac{\xi}{\alpha c}} & \text{for}\: \Delta < - \xi \\\gamma \left(1 - e^{\frac{\xi}{\alpha c}}\right) & \text{for}\: \xi < 0 \\0 & \text{otherwise} \end{cases}. \end{align*}$$ with speed c and pulse-width Δ chosen such that $$\begin{align*} 0 &= \frac{M_{1} \sigma_{1}^{2} \left(1 - e^{- \frac{\Delta}{\sigma_{1}}}\right)}{\mu c + \sigma_{1}} + \frac{M_{2} \sigma_{2}^{2} \left(-1 + e^{- \frac{\Delta}{\sigma_{2}}}\right)}{\mu c + \sigma_{2}} - \theta\\ 0 &= - \frac{M_{1} \sigma_{1}^{2}}{\mu c + \sigma_{1}} + \frac{M_{1} \sigma_{1}^{2} e^{- \frac{\Delta}{\sigma_{1}}}}{\mu c - \sigma_{1}} + 2 M_{1} \sigma_{1} + \frac{M_{2} \sigma_{2}^{2}}{\mu c + \sigma_{2}} - \frac{M_{2} \sigma_{2}^{2} e^{- \frac{\Delta}{\sigma_{2}}}}{\mu c - \sigma_{2}} - 2 M_{2} \sigma_{2} - \gamma \left(1 - e^{- \frac{\Delta}{\alpha c}}\right) - \theta + \left(\frac{M_{1} \sigma_{1}^{2} e^{- \frac{\Delta}{\sigma_{1}}}}{\mu c + \sigma_{1}} - \frac{M_{1} \sigma_{1}^{2}}{\mu c - \sigma_{1}} - 2 M_{1} \sigma_{1} - \frac{M_{2} \sigma_{2}^{2} e^{- \frac{\Delta}{\sigma_{2}}}}{\mu c + \sigma_{2}} + \frac{M_{2} \sigma_{2}^{2}}{\mu c - \sigma_{2}} + 2 M_{2} \sigma_{2} + \theta\right) e^{- \frac{\Delta}{\mu c}}. \end{align*}$$

Note that if we take $M_1 = \tfrac{1}{2}, \sigma_1 = 1, M_2 = 0$ then $w$ reduces to the exponential weight kernel and these results are consistent with report 2021-06-03.

Fig 1. The traveling pulse solution for the adaptive model, with Heaviside firing-rate, and a difference-of-exponetials weight kernel.

Nullspace of the Adjoint

From report 2021-06-10 we know that the first order asymptotic approximation of the wave response function can be calculated using the null-space of the following adjoint operator $$ \LL^* \vecv = \vecv + c \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix} \vecv_\xi - f'\big(U(\xi) - A(\xi)\big)\int_\RR w(y, \xi) \begin{bmatrix}1&\gamma \\ -1 &-\gamma\end{bmatrix} \vecv(y) \ dy. $$ The report goes on to calculate a basis for the null-space in the case of the Heaviside firing rate and the exponential weight kernel. However, when the Heaviside firing-rate is used, the result can be shown to be independent of the weight kernel. Thus, we find that the adjoint has a one-dimensional null-space that is spanned by the vector $\vecv = [v_1 \ v_2]^T$ where $$\begin{align*} v_1(\xi) &= \alpha H(\xi) e^{-\xi/c\mu} \\ v_2(\xi) &= -\mu H(\xi) e^{-\xi/c\alpha}. \end{align*}$$

We can compute the null-space as follows: $$\begin{align*} \vec{0} &= \LL^* \vecv \\ &= \vecv + c \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix} \vecv_\xi - \frac{d}{d \xi}\bigg[ H\big( U(\xi) - A(\xi) \big) \bigg] \begin{bmatrix}1&\gamma \\ -1 &-\gamma \end{bmatrix} \int_\RR w(y, \xi) \vecv(y) \ dy \\ &= \vecv + c \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix} \vecv_\xi - \frac{\delta(\xi)}{U'(0) - A'(0)} \begin{bmatrix}1&\gamma \\ -1 &-\gamma \end{bmatrix} \int_\RR w(y, 0)\vecv(y) \ dy \\ \vecv + c\begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix} \vecv_\xi &= \frac{\delta(\xi)}{U'(0) - A'(0)} \begin{bmatrix}1&\gamma \\ -1 &-\gamma \end{bmatrix} \int_\RR w(y, 0)\vecv(y) \ dy \\ \vecv_\xi + \frac{1}{c} \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix}^{-1} \vecv &= \frac{\delta(\xi)}{U'(0) - A'(0)} \begin{bmatrix}\frac{1}{\mu} &\frac{\gamma}{\mu} \\ -\frac{1}{\alpha} &-\frac{\gamma}{\alpha} \end{bmatrix} \int_\RR w(y, 0)\vecv(y) \ dy \\ \left[ \exp\left(\frac{\xi}{c} \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix}^{-1} \right) \vecv \right]_\xi &= \exp\left(\frac{\xi}{c} \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix}^{-1} \right) \frac{\delta(\xi)}{U'(0) - A'(0)} \begin{bmatrix}\frac{1}{\mu} &\frac{\gamma}{\mu} \\ -\frac{1}{\alpha} &-\frac{\gamma}{\alpha} \end{bmatrix} \int_\RR w(y, 0)\vecv(y) \ dy \\ \exp\left(\frac{\xi}{c} \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix}^{-1} \right) \vecv &= H(\xi) \left[ \exp\left(\frac{\xi}{c} \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix}^{-1} \right) \frac{1}{U'(0) - A'(0)} \begin{bmatrix}\frac{1}{\mu} &\frac{\gamma}{\mu} \\ -\frac{1}{\alpha} &-\frac{\gamma}{\alpha} \end{bmatrix} \int_\RR w(y, 0)\vecv(y) \ dy \right]_{\xi = 0} \\ \exp\left(\frac{\xi}{c} \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix}^{-1} \right) \vecv &= H(\xi) \frac{1}{U'(0) - A'(0)} \begin{bmatrix}\frac{1}{\mu} &\frac{\gamma}{\mu} \\ -\frac{1}{\alpha} &-\frac{\gamma}{\alpha} \end{bmatrix} \int_\RR w(y, 0)\vecv(y) \ dy. \end{align*}$$

The integral on the right-hand-side is some constant vector $$ \frac{1}{U'(0) - A'(0)} \int_\RR w(y, 0)\vecv(y) \ dy = \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} $$ so we have \begin{align*} \vecv(\xi) &= H(\xi) \exp\left(-\frac{\xi}{c} \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix}^{-1} \right) \begin{bmatrix}\frac{1}{\mu} &\frac{\gamma}{\mu} \\ -\frac{1}{\alpha} &-\frac{\gamma}{\alpha} \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix}. \end{align*}

This constant matrix is rank 1 and thus the null-space of $\LL^*$ is one-dimensional. Fixing the arbitrary scaling of the null-vector, we arrive at $$\begin{align*} \vecv(\xi) &= H(\xi) \exp\left(-\frac{\xi}{c} \begin{bmatrix}\mu&0\\0&\alpha\end{bmatrix}^{-1} \right) \begin{bmatrix} \alpha \\ -\mu \end{bmatrix} \end{align*}$$ or equivalently $$\begin{align*} v_1(\xi) &= \alpha H(\xi) e^{-\xi/c\mu} \\ v_2(\xi) &= -\mu H(\xi) e^{-\xi/c\alpha}. \end{align*}$$

Again, $A_\xi$ and $v_2(\xi)$ are orthogonal. This will always be the case for a Heaviside firing rate and a traveling pulse solution. The wave response is then given by $$\begin{align*} \nu(t) &= -\frac{ \int_0^t \int_\RR \alpha H(\xi) e^{-\xi/c\mu} I(\xi, \tau) \ d\xi \ d\tau}{\int_\RR \mu U_\xi \alpha H(\xi) e^{-\xi/c\mu} d\xi} \\ &= -\frac{1}{\mu} \frac{\int_0^t \int_0^\infty e^{-\xi/c\mu} I(\xi, t) \ d\xi \ d\tau }{\int_0^\infty U_\xi e^{-\xi/c\mu} d\xi }. \end{align*}$$ This formulation is for arbitrary weight-kernels, however, it does depend on the weight kernel in that $U_\xi$ depends on the weight-kernel.

In the case of the difference-of-exponentials weight kernel, we then have $$ \nu(t) = \left( \frac{M_{1} \mu^{2} \sigma_{1}^{2} c \left(1 - e^{- \frac{\Delta}{\sigma_{1}}}\right)}{\left(\mu c + \sigma_{1}\right)^{2}} + \frac{M_{2} \mu^{2} \sigma_{2}^{2} c \left(-1 + e^{- \frac{\Delta}{\sigma_{2}}}\right)}{\left(\mu c + \sigma_{2}\right)^{2}} \right)^{-1} \int_0^t \int_0^\infty e^{-\xi/c\mu} I(\xi, \tau) \ d\xi \ d\tau. $$

In report 2021-06-10 this was simplified to not explicitly depend on $\Delta$ by use of one of the conditions implicitly determining $c$ and $\Delta$. The corresponding condition in this case is $$ 0 = \frac{M_{1} \sigma_{1}^{2} \left(1 - e^{- \frac{\Delta}{\sigma_{1}}}\right)}{\mu c + \sigma_{1}} + \frac{M_{2} \sigma_{2}^{2} \left(-1 + e^{- \frac{\Delta}{\sigma_{2}}}\right)}{\mu c + \sigma_{2}} - \theta $$ from which, it is not obvious how to simplify our expression for $\nu(t)$.

Spatially Homogeneous Pulse

In the case where $I(x, t) = I_0$, we our asymptotic approximation to the wave response becomes $$ \nu(t) = \nu_\infty = \frac{\mu c}{\frac{M_{1} \mu^{2} \sigma_{1}^{2} c \left(1 - e^{- \frac{\Delta}{\sigma_{1}}}\right)}{\left(\mu c + \sigma_{1}\right)^{2}} + \frac{M_{2} \mu^{2} \sigma_{2}^{2} c \left(-1 + e^{- \frac{\Delta}{\sigma_{2}}}\right)}{\left(\mu c + \sigma_{2}\right)^{2}}} I_0. $$ For the parameters used above ($\mu=1$, $\alpha=5$, $\gamma=3$, $\theta=0.1$, $M_1=3$, $M_2=1$, $\sigma_1=1$, $\sigma_2=2$, $c=1.421$, $\Delta=7.267$) this gives $$ \nu(t) = \nu_\infty \approx 5.5926 I_0. $$

Unfortunately, this result seems to disagree with numerical simulation. Figure 1 shows the measured wave response from a numerical simulation and it does not appear to match the prediction. We believe this to be due to low accuracy of the simulation. In this case, the spatial discretization is roughly on the order of the stimulus ($h = 0.008$). Presently, the simulation is using a quadrature matrix to perform the convolution and thus each time step costs $\mathcal{O}(N^2)$. This can be reduced by using a semi-analytic convolution as with previous simulations. We suspect the asymptotic approximation to be correct as it is consistent with the exponential weight kernel (a special case of the difference of exponentials where $M_2 = 0$).

Fig 1. The asymptotic approximation of the wave response is not agreeing with the numerical simulations.