$ \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{\vecw}{\vec{w}} \newcommand{\vecx}{\vec{x}} \newcommand{\vecy}{\vec{y}} \newcommand{\vecz}{\vec{z}} \renewcommand{\vec}[1]{\mathbf{#1}} $ Shaw Research Notes

October 5th, 2021

We have begun exploring stability of perturbations to the adaptive model. We are stuck showing inconsistency in the case where the eigenvalues have positive real component.

To-Do List Snapshot

Stability Analysis

We begin with the model \begin{align*} \mu u_t &= -u + \int w(x,y) f(u(y,t) - a(y,t)) \ dy \\ \alpha a_t &= -a + \gamma f(u - a) \end{align*} and make the substitution to characteristic coordinates $\xi = x - c t$ $$\begin{align*} -c\mu u_\xi + \mu u_t &= -u + \int w(\xi,y) f(u(y,t) - a(y,t)) \ dy \\ -c\alpha a_\xi + \alpha a_t &= -a + \gamma f(u - a). \end{align*}$$ The traveling pulse solution $U(\xi), A(\xi)$ has been found previously (report 2021-06-03). Assume a small perterbation $$\begin{align*} u &= U(\xi) + \varepsilon \varphi(\xi, t) + \OO(\varepsilon^2) \\ a &= A(\xi) + \varepsilon \psi(\xi, t) + \OO(\varepsilon^2). \end{align*}$$ Substituting into our characteristic coordinate equations gives $$\begin{align*} -c \mu U' - c\mu \varepsilon \varphi_\xi + \mu \varepsilon \varphi_t &= -U - \varepsilon \varphi + \int w(\xi, y) f\big(U-A + \varepsilon(\varphi - \psi) + \OO(\varepsilon^2) \big) \ dy + \OO(\varepsilon^2) \\ -c\alpha A' - c\alpha\varepsilon\psi_\xi + \alpha \varepsilon \psi_t &= -A - \varepsilon \psi + \gamma f\big(U - A + \varepsilon(\varphi - \psi) + \OO(\varepsilon^2)\big) + \OO(\varepsilon^2). \end{align*}$$

Next we linearize must linearize $f$. We intend to take $f(\cdot) = H(\cdot - \theta)$ so we will consider a two-sided linearization: $$ f\big(U-A + \varepsilon(\varphi - \psi) + \OO(\varepsilon^2)\big) = f(U-A) + \OO(\varepsilon^2) + \varepsilon(\varphi - \psi) \underbrace{\begin{cases} \partial_+f(U-A) ,& \varphi > \psi \\ \partial_-f(U-A) ,& \varphi < \psi \end{cases}}_{f'(U-A)} $$ where $\partial_+$ and $\partial_-$ denote the right and left derivatives respectively. We will use the compact notation $f'(U-A)$ which is consistent if the left and right derivatives match.

This gives $$\begin{align*} -c \mu U' - c\mu \varepsilon \varphi_\xi + \mu \varepsilon \varphi_t &= -U - \epsilon \varphi + \int w(\xi, y) f(U-A) \ dy + \varepsilon f'(U-A)\int w(\xi, y)(\varphi - \psi) \ dy + \OO(\varepsilon^2) \\ -c\alpha A' - c\alpha\varepsilon\psi_\xi + \alpha \epsilon \psi_t &= -A - \varepsilon \psi + \gamma f(U - A) + \gamma f'(U-A)\varepsilon(\varphi - \psi) + \OO(\varepsilon^2). \end{align*}$$ Collecting $\OO(1)$ terms $$\begin{align*} -c \mu U' &= -U + \int w(\xi, y) f(U-A) \ dy \\ -c\alpha A' &= -A + \gamma f(U - A) \end{align*}$$ we see that $U$ and $A$ must indeed be the traveling pulse solution. Collecting the $\OO(\varepsilon)$ terms we have $$\begin{align*} - c\mu \varphi_\xi + \mu \varphi_t &= - \varphi + f'(U-A)\int w(\xi, y)(\varphi - \psi) \ dy \\ - c\alpha\psi_\xi + \alpha \psi_t &= - \psi + \gamma f'(U-A)(\varphi - \psi). \end{align*}$$

In the case of $f(\cdot) = H(\cdot - \theta)$ we have $$\begin{align*} f'(U - A) &= \frac{\delta(\xi)}{U'(0) - A'(0)} + \frac{\delta(\xi+\Delta)}{U'(-\Delta) - A'(-\Delta)} \\ &= c_1\delta(\xi) + c_2\delta(\xi+\Delta). \end{align*}$$ where $U'(\cdot) - A'(\cdot)$ is interpreted as a right/left derivative if $\varphi \gtrless \psi$. Thus the constants $c_1$ and $c_2$ depend on the sign of $\varphi - \psi$.

Assume separability: $$ \begin{bmatrix} \varphi(\xi,t) \\ \psi(\xi, t) \end{bmatrix} = g(t)\begin{bmatrix} \varphi_0(\xi) \\ \psi_0(\xi) \end{bmatrix}. $$ Substituting and solving we have \begin{align*} \frac{g'}{g} &= \lambda = \frac{c\mu \varphi_0' - \varphi_0 + f'(U-A)\int w(\xi, y)(\varphi_0 - \psi_0) \ dy }{\mu \varphi_0} \\ \frac{g'}{g} &= \lambda = \frac{c\alpha \psi_0' - \psi_0 + \gamma f'(U-A)(\varphi_0 - \psi_0) }{\alpha \psi_0}. \end{align*} This gives $g(t) = e^{\lambda t}$ and the following system of ODEs $$\begin{align*} \varphi_0' - \frac{1+\lambda \mu}{c\mu} \varphi_0 &= -\frac{1}{c\mu}f'(U-A) \int w(\xi, y) \big(\varphi_0(y) - \psi_0(y) \big) \ dy \\ \psi_0' - \frac{1+\lambda \alpha}{c\alpha} \psi_0 &= -\frac{1}{c\alpha}\gamma f'(U-A) (\varphi_0 - \psi_0 ) \\ \end{align*}$$

Using integrating factors we have $$\begin{align*} \frac{d}{d\xi} \big[ \varphi_0 e^{-\frac{1+\lambda \mu}{c\mu}\xi} \big] &= -e^{-\frac{1+\lambda \mu}{c\mu}\xi}\frac{1}{c\mu}f'(U-A) \int w(\xi, y) \big(\varphi_0(y) - \psi_0(y) \big) \ dy \\ \varphi_0 e^{-\frac{1+\lambda \mu}{c\mu}\xi} &= A + \int -e^{-\frac{1+\lambda \mu}{c\mu}\xi}\frac{1}{c\mu}f'(U-A) \int w(\xi, y) \big(\varphi_0(y) - \psi_0(y) \big) \ dy \ d\xi\\ &= A + \int -e^{-\frac{1+\lambda \mu}{c\mu}\xi}\frac{1}{c\mu}\big(c_1 \delta(\xi) + c_2 \delta(\xi + \Delta)\big) \int w(\xi, y) \big(\varphi_0(y) - \psi_0(y) \big) \ dy \ d\xi\\ &= A - c_1 H(\xi) \underbrace{\frac{1}{c\mu}\int w(0, y) \big(\varphi_0(y) - \psi_0(y) \big) \ dy }_{w_0}\\ &\phantom{===} - c_2 H(\xi+\Delta) e^{-\frac{1+\lambda \mu}{c\mu}\Delta} \underbrace{\frac{1}{c\mu}\int w(\Delta, y) \big(\varphi_0(y) - \psi_0(y) \big) \ dy}_{w_\Delta} \\ \varphi_0(\xi) &= e^{\frac{1+\lambda \mu}{c\mu} \xi} \bigg(A - w_0 c_1 H(\xi) - w_\Delta c_2 e^{-\frac{1+\lambda \mu}{c\mu}\Delta} H(\xi + \Delta) \bigg) \end{align*}$$ where these $w_0$ and $w_\Delta$ depend linearly on $\varphi_0 - \psi_0$.

Similarly, we have $$\begin{align*} \frac{d}{d\xi} \big[ \psi_0 e^{-\frac{1 + \lambda\alpha}{c\alpha} \xi} \big] &= -\frac{1}{c\alpha}e^{-\frac{1 + \lambda\alpha}{c\alpha} \xi} \gamma \big( c_1 \delta(\xi) + c_2 \delta(\xi + \Delta) \big)(\varphi_0 - \psi_0) \\ \psi_0 e^{-\frac{1 + \lambda\alpha}{c\alpha} \xi} &= B - c_1 \underbrace{\frac{\gamma}{c\alpha} \big(\varphi_0(0) - \psi_0(0) \big)}_{v_0}H(\xi) \\ &\phantom{===} -c_2 e^{-\frac{1 + \lambda\alpha}{c\alpha} \Delta} \underbrace{\frac{\gamma}{c\alpha} \big(\varphi_0(\Delta) - \psi_0(\Delta)\big)}_{v_\Delta} H(\xi + \Delta) \\ \psi_0(\xi) &= e^{\frac{1+ \lambda \alpha}{c \alpha} \xi} \big(B - c_1v_0 H(\xi) - c_2 v_\Delta e^{-\frac{1 + \lambda\alpha}{c\alpha} \Delta} H(\xi + \Delta) \big) \end{align*}$$ where these $v_0$ and $v_\Delta$ depend linearly on $\varphi_0 - \psi_0$.

We next consider the values of $\lambda$ for which the initial solution $[\varphi_0, \ \psi_0]^T$ remains bounded. We expect adaptation to be on a slower timescale than excitation and thus restrict ourselves to the more relevant case $\mu < \alpha$. For $\lambda >0 > -\tfrac{1}{\alpha} > -\tfrac{1}{\mu}$ we require $$\begin{align*} 0 &= A - c_1 w_0 - c_2 e^{-\frac{1 + \lambda\mu}{c\mu} \Delta} w_\Delta \\ 0 &= B - c_1 v_0 - c_2 e^{-\frac{1 + \lambda\alpha}{c\alpha} \Delta} v_\Delta. \end{align*}$$ We want to show that this is a contradiction and that $\lambda \le 0$.