I am interested in the following problem.
Let $D = \mathrm{diag}(d_1, d_2, \ldots, d_n) \in \mathbb{R}^n$ be positive definite, let $B, K \in \mathbb{R}^n$, and let $G\in L^\infty((0, T)\times (0, L); \mathbb{R}^{n \times n})$ be such that $\|G\|_{L^\infty((0, T)\times (0, L); \mathbb{R}^{n \times n})} \leq R$ with $R>0$ as small as we want. Show that there exists $M>0$ (independent of $y, \overline{y}$) such that: if $y \colon [0, T] \times [0, \ell] \rightarrow \mathbb{R}^n$ is solution to \begin{align*} \begin{cases} \partial_t y + D \partial_x y + B y = G(t,x) \overline{y}(x,t) & \text{in }(0, T)\times(0, \ell)\\ y(t, 0) = K \overline{y}(t, \ell) &\text{for }t \in (0, T)\\ y(0, x) = 0 & \text{for }x \in (0, \ell) \end{cases} \end{align*} for some $\overline{y}$ which is at least in $L^\infty(0, T; H_x^1)$, then $y$ necessarily fulfills \begin{align*} \|y\|_{L^\infty(0, T; L_x^2)} + M \|y(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)} \leq \frac{1}{2} \left( \|\overline{y}\|_{L^\infty(0, T; L_x^2)} + M \|\overline{y}(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)} \right). \end{align*}
Here I used the shortened notation $L_x^2 = L^2(0, \ell; \mathbb{R}^n)$ and $H_x^1 = H^1(0, \ell; \mathbb{R}^n)$.
I tried the following (below I do not write the argument $t$ for $y$ and $\overline{y}$ in the integrands in order to lighten the notation). Using the governing equations and that $2\left\langle y, D \partial_x y \right\rangle = \partial_x \left( \langle y, Dy \rangle \right)$ we get \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \left( \|y(t, \cdot)\|_{L_x^2}^2 \right) &= 2 \int_0^\ell \left \langle y, \partial_t y \right \rangle dx \\ &= - 2 \int_0^\ell \left \langle y, D \partial_x y \right \rangle dx + 2 \int_0^\ell \left \langle y,- B y \right \rangle dx + 2 \int_0^\ell \left \langle y, G(t,x)\overline{y} \right \rangle dx\\ &= - \left \langle y(t, \ell), D y(t, \ell) \right \rangle + \left \langle y(t, 0), D y(t, 0) \right \rangle + 2 \int_0^\ell \left \langle y,- B y \right \rangle dx\\ &\quad + 2 \int_0^\ell \left \langle y, G(t,x)\overline{y} \right \rangle dx, \end{align*} where $\langle \cdot, \cdot \rangle$ denotes the inner product in $\mathbb{R}^n$. Then, using the boundary conditions we get \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \left( \|y(t, \cdot)\|_{L_x^2}^2 \right) &= - \left \langle y(t, \ell), D y(t, \ell) \right \rangle + \left \langle K \overline{y}(t, \ell), D K \overline{y}(t, \ell) \right \rangle + 2 \int_0^\ell \left \langle y,- B y \right \rangle dx\\ &\quad + 2 \int_0^\ell \left \langle y, G(t,x)\overline{y} \right \rangle dx. \end{align*} On the other hand, \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\left( \int_0^t \left \langle y(\tau, \ell), D y(\tau, \ell) \right \rangle d\tau \right) &= \left \langle y(t, \ell), D y(t, \ell) \right \rangle. \end{align*} Hence, denoting $\eta(t) := \|y(t, \cdot)\|_{L_x^2}^2 + \int_0^t |D^\frac{1}{2} y(\tau, \ell)|^2 d\tau$ (and using the Cauchy-Schwarz and Young inequalities), we obtain that \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \eta(t) &=|D^\frac{1}{2}K\overline{y}(t, \ell)|^2 + 2 \int_0^\ell \left \langle y,- B y \right \rangle dx + 2 \int_0^\ell \left \langle y, G(t,x)\overline{y} \right \rangle dx\\ %&\leq C_1 \left(|\overline{y}(t,\ell)|^2 + \|y(t, \cdot)\|_{L_x^2}^2 + \|G\|_{L^\infty}\|\overline{y}(t)\|_{L_x^2}^2\right)\\ &\leq |D^\frac{1}{2} K \overline{y}(t,\ell)|^2 + R \|\overline{y}(t)\|_{L_x^2}^2 + (2\|B\| + R) \eta(t). \end{align*} Thus, by Gronwall's inequality and the fact that $y(0, \cdot)\equiv 0$, we get \begin{align*} \eta(t) &\leq e^{(2\|B\|+R) t} \int_0^t \left(|D^\frac{1}{2}K\overline{y}(\tau,\ell)|^2 + R\|\overline{y}(\tau)\|_{L_x^2}^2 \right)d\tau\\ &\leq e^{(2\|B\|+R) T} \left(\|D^\frac{1}{2}K\overline{y}(\cdot,\ell)\|_{L^2(0, T; \mathbb{R}^n)}^2 + TR\|\overline{y}\|_{L^\infty(0, T; L_x^2)}^2 \right)d\tau. \end{align*} Consequently, \begin{align*} \|y\|_{L^\infty(0, T; L_x^2)}^2 + \|D^\frac{1}{2}y(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)}^2 \leq e^{(2\|B\|+R) T} \left(\|D^\frac{1}{2}K\overline{y}(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)}^2 + T R \|\overline{y}\|_{L^\infty(0, T; L_x^2)}^2 \right). \end{align*} So, we have obtained that \begin{align*} \|y\|_{L^\infty(0, T; L_x^2)} + \|D^\frac{1}{2}y(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)} \leq 2 e^{(\|B\|+\tfrac{1}{2}R) T} \left(\|D^\frac{1}{2}K\overline{y}(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)} + \sqrt{TR} \|\overline{y}\|_{L^\infty(0, T; L_x^2)} \right) \end{align*} which also implies that \begin{align*} \|y\|_{L^\infty(0, T; L_x^2)} + \|y(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)} \leq C e^{(\|B\|+\tfrac{1}{2}R) T} \left(\|\overline{y}(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)} + \sqrt{TR} \|\overline{y}\|_{L^\infty(0, T; L_x^2)} \right), \end{align*} with \begin{align*} C= 2\frac{\max\{1, \|D^\frac{1}{2}K\|\}}{\min\{1, \min_{1\leq i\leq n}d_i\}}. \end{align*}
However, I don't see how to proceed further to obtain the desired inequality. It seems that the term $\sqrt{TR} \|\overline{y}\|_{L^\infty(0, T; L_x^2)}$ can be made small via a certain choice of $T$ or $R$, but I do not know how to deal with the other term.
Is there another way to do the estimations for the term $\|y(\cdot, \ell)\|_{L^2(0, T; \mathbb{R}^n)}$?
Any help, hint, solution, comment is appreciated. Thank you!
Also asked on MO: https://mathoverflow.net/questions/393297/contraction-like-inequality-how-to-deal-with-the-boundary-term