# 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!