$ \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

September 16th, 2021

Zack found an error in the derivation for report 2021-06-10. We start over here, and derive the adjoint and begin computation of the nullspace.

To-Do List Snapshot

Expansion

Begin with the model:

$$\begin{align*} \mu u_t &= -u + \int_\RR w(x,y) f\big[u(y,t) - a(y,t) \big] \ dy + \epsilon I(x, t) \\ \alpha a_t &= -a + \gamma f\big[u(x,t) - a(x,t)\big]. \end{align*}$$ Convert to $\xi = x - ct$ coordinates \begin{align*} -c\mu u_\xi + \mu u_t &= -u + \int_\RR w(\xi,y) f\big[u(y,t) - a(y,t) \big] \ dy + \epsilon I(\xi, t) \\ -c\alpha a_\xi + \alpha a_t &= -a + \gamma f\big[u(\xi,t) - a(\xi,t)\big]. \end{align*} Make the expansion $$\begin{align*} u(\xi, t) &= U\big(\xi - \epsilon \nu(t) \big) + \epsilon u_1(\xi, t) + \OO(\epsilon^2) \\ a(\xi, t) &= A\big(\xi - \epsilon \nu(t) \big) + \epsilon a_1(\xi, t) + \OO(\epsilon^2) \\ \end{align*}$$ where $$\begin{align*} U &= U\big(\xi - \epsilon \nu(t) \big) A &= A\big(\xi - \epsilon \nu(t) \big) \end{align*}$$ satisfy the traveling wave solution. Substitute and linearize to find $$\begin{align*} -c\mu U' - c\mu \epsilon \partial_\xi u_1 - \epsilon \mu U' \nu_t + \epsilon \mu \partial_t u_1 &= -U - \epsilon u_1 + \int_\RR w(\xi, y)\bigg[ f(U-A) + \epsilon (u_1 - a_1) f'(U-A) + \OO(\epsilon^2) \bigg] \ dy + \epsilon I(\xi, t) \\ -c\alpha A' - c\alpha \epsilon \partial_\xi a_1 - \epsilon \alpha A' \nu_t +\alpha \epsilon \partial_t a_1 &= -A -\epsilon a_1 + \gamma f(U-A) + \epsilon \gamma (u_1 - a_1) f'(U-A) + \OO(\epsilon^2) \end{align*}$$ Equating the $\OO(1)$ terms gives us that $U$ and $A$ are the traveling wave solution. Equating the $\OO(\epsilon)$ terms gives us $$\begin{align*} -c\mu \partial_\xi u_1 - \mu U' \nu_t + \mu \partial_t u_1 &= -u_1 + \int_\RR f'(U-A)w(\xi, y)\bigg[ (u_1 - a_1) \bigg] \ dy + I(\xi, t) \\ -c\alpha \partial_\xi a_1 - \alpha A' \nu_t +\alpha \partial_t a_1 &= -a_1 + \gamma (u_1 - a_1) f'(U-A) \end{align*}$$ or $$\begin{align*} \begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecu_t + \underbrace{\vecu - c\begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecu_\xi - \begin{bmatrix} (f'(U-A)w) \ast \cdot & -(f'(U-A)w) \ast \cdot \\ \gamma f'(U-A) & - \gamma f'(U-A) \end{bmatrix} \vecu}_{\LL \vecu} &= \begin{bmatrix} \mu U' \nu_t + I \\ \alpha A' \nu_t \end{bmatrix} \end{align*}$$ where $w \ast \cdot$ denotes applying the operator rather than the usual multiplication and $\vecu = [u_1 \ a_1]^T$. A bounded solution exists if the right-hand-side is orthogonal to the nullspace of $\LL^*$. We find this nullspace to be

Adjoint Derivation

Let $\vecu = [u_1 \ u_2]^T$ and $\vecv = [v_1 \ v_2]^T$ be arbitrary vectors (functions of $\xi$ and $t$). Then consider $$\begin{align*} \langle \vecv, \LL \vecu \rangle &= \int_\RR \vecv^T \left( \vecu - c \begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecu_\xi - \begin{bmatrix} (f'(U-A)w) \ast \cdot & -(f'(U-A)w) \ast \cdot \\ \gamma f'(U-A) & - \gamma f'(U-A) \end{bmatrix} \vecu \right) \ d\xi \\ &= \int_\RR \vecv^T\vecu \ d\xi - c\int_\RR v_1 \mu \partial_\xi u_1 + v_2 \alpha \partial u_2 \ d\xi - \int_\RR \left( v_1(\xi) \int_\RR f'\big(U(\xi) - A(\xi)\big)w(\xi, y) u_1(y) \ dy - v_2 \int_\RR f'\big(U(\xi) - A(\xi)\big)w(\xi, y) u_2(y) \ dy + v_2 \gamma u_1 - v_2 \gamma u_2 \right) \ d\xi \\ &= \int_\RR \vecu^T \vecv \ d\xi -c \underbrace{\left(v_1\mu u_1 + v_2 \alpha u_2 \right)\bigg|_{\xi = -\infty}^\infty}_{=0} + c\int_\RR u_1 \mu \partial_\xi v_1 + u_2 \alpha \partial v_2 \ d\xi - \int_\RR \left( v_1(\xi) \int_\RR f'\big(U(\xi) - A(\xi)\big)w(\xi, y) u_1(y) \ dy - v_2(\xi) \int_\RR f'\big(U(\xi) - A(\xi)\big)w(\xi, y) u_2(y) \ dy + v_2 \gamma u_1- v_2 \gamma u_2 \right) \ d\xi \\ &= \int_\RR \vecu^T \vecv \ d\xi + c\int_\RR \vecu^T \begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecv_\xi \ d\xi - \int_\RR \int_\RR f'\big(U(\xi) - A(\xi)\big) v_1(\xi) w(\xi, y) u_1(y) - f'\big(U(\xi) - A(\xi)\big) v_2(\xi) w(\xi, y) u_2(y) \ dy \ d\xi + \int_\RR v_2 \gamma u_1- v_2 \gamma u_2 \ d\xi \\ &= \int_\RR \vecu^T \vecv \ d\xi + c\int_\RR \vecu^T \begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecv_\xi \ d\xi - \int_\RR f'\big(U(\xi) - A(\xi)\big) \int_\RR v_1(\xi) w(\xi, y) u_1(y) - v_2(\xi) w(\xi, y) u_2(y) \ d \xi \ dy + \int_\RR v_2 \gamma u_1- v_2 \gamma u_2 \ d\xi \\ &= \int_\RR \vecu^T \vecv \ d\xi + c\int_\RR \vecu^T \begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecv_\xi \ d\xi - \int_\RR \vecu^T f'(U-A) \begin{bmatrix} w\ast \cdot & \gamma \\ -w\ast \cdot & -\gamma \end{bmatrix} \vecv \ d\xi \\ &= \langle \LL^* \vecv, \vecu \rangle \end{align*}$$ where $$\begin{align*} \LL^* \vecv = \vecv + c\begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecv_\xi - f'(U-A) \begin{bmatrix} w\ast \cdot & \gamma \\ -w\ast \cdot & -\gamma \end{bmatrix} \vecv \end{align*}$$

Adjoint Nullspace

Choosing $f(J) = H(J - \theta)$ we have $$\begin{align*} f'(J(\xi)) &= \frac{\delta(\xi)}{J'(0)} + \frac{\delta(\xi + \Delta)}{J'(-\Delta)}. \end{align*}$$ We seek vectors $\vecv$ such that $$\begin{align*} 0 &= \vecv + c\begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecv_\xi - \left(\frac{\delta(\xi)}{J'(0)} + \frac{\delta(\xi + \Delta)}{J'(-\Delta)} \right) \begin{bmatrix} w\ast \cdot & \gamma \\ -w\ast \cdot & -\gamma \end{bmatrix} \vecv \\ &= \vecv + c\begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecv_\xi - \frac{\delta(\xi)}{J'(0)} \begin{bmatrix} \int_\RR(w(0,y) v_1(y) \ dy + \gamma v_2(0) \\ -\int_\RR w(0,y)v_1(y) \ dy - \gamma v_2(0) \end{bmatrix} + \frac{\delta(\xi + \Delta)}{J'(-\Delta)} \begin{bmatrix} \int_\RR(w(-\Delta,y) v_1(y) \ dy + \gamma v_2(-\Delta) \\ -\int_\RR w(-\Delta,y)v_1(y) \ dy - \gamma v_2(-\Delta) \end{bmatrix} \end{align*}$$ and $$\begin{align*} \vecv + c\begin{bmatrix}\mu & 0 \\ 0 & \alpha \end{bmatrix} \vecv_\xi &= \delta(\xi) w_1 \begin{bmatrix}1\\-1\end{bmatrix} + \delta(\xi + \Delta) w_2 \begin{bmatrix}1\\-1\end{bmatrix} \end{align*}$$ for some constants $w_1, w_2$ which depend linearly on $\vecv$.