inequalities – Show `contraction-like’ inequality for solution of first-order PDE

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 $Gin L^infty((0, T)times (0, L); mathbb{R}^{n times n})$ 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 $2leftlangle y, D partial_x y rightrangle = partial_x left( langle y, Dy rangle right)$ 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 dtau right)
&= left langle y(t, ell), D y(t, ell) right rangle.
end{align*}

Hence, denoting $eta(t) := |y(t, cdot)|_{L_x^2}^2 + int_0^t |D^frac{1}{2} y(tau, ell)|^2 dtau$ (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}Koverline{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}^2right)\
&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}Koverline{y}(tau,ell)|^2 + R|overline{y}(tau)|_{L_x^2}^2 right)dtau\
&leq e^{(2|B|+R) T} left(|D^frac{1}{2}Koverline{y}(cdot,ell)|_{L^2(0, T; mathbb{R}^n)}^2 + TR|overline{y}|_{L^infty(0, T; L_x^2)}^2 right)dtau.
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}Koverline{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}Koverline{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= 2frac{max{1, |D^frac{1}{2}K|}}{min{1, min_{1leq ileq 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 MSE: https://math.stackexchange.com/questions/4144078/show-contraction-like-inequality-for-solution-of-first-order-pde