Parabolic L2 energy estimate with zero initial-lateral trace

For almost every $t$, by [F4], approximate $e(\cdot ,t)$ in $H^1(U)$ by compactly supported smooth functions. For each approximant the distributional definition of the divergence terms gives the spatial integration-by-parts identity; the coefficient bounds make every pairing continuous in $H^1$, so passing to the limit produces no boundary term because $e(\cdot ,t)\in H_0^1(U)$. Pair the weak equation $-e_t+L_t[e]=r$ against $e$ in $V^{\ast },V$ and use [F1] and [F2]. Moving the time derivative to the left gives $\frac{1}{2}{E}^{\mathrm{\prime }}(t)+{\lambda }_0\mathrm{\parallel }\mathrm{\nabla }e(t){\mathrm{\parallel }}_2^2\le (\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }c{\mathrm{\parallel }}_{\mathrm{\infty }})\mathrm{\parallel }e(t){\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }e(t){\mathrm{\parallel }}_2+\mathrm{\parallel }d{\mathrm{\parallel }}_{\mathrm{\infty }}E(t)+\mathrm{\parallel }r(t){\mathrm{\parallel }}_2\mathrm{\parallel }e(t){\mathrm{\parallel }}_2,$ where $E(t)=\mathrm{\parallel }e(t){\mathrm{\parallel }}_2^2$.

By Young's inequality, the first product on the right is at most $\frac{{\lambda }_0}{4}\mathrm{\parallel }\mathrm{\nabla }e{\mathrm{\parallel }}_2^2+C_{1}E(t)$ and the last product is at most $\frac{1}{2}\mathrm{\parallel }r(t){\mathrm{\parallel }}_2^2+\frac{1}{2}E(t)$, for a finite $C_1$ depending only on the coefficient bounds and ${\lambda }_0$. Hence for $C_2=C_1+\mathrm{\parallel }d{\mathrm{\parallel }}_{\mathrm{\infty }}+\frac{1}{2}$, $E^{\mathrm{\prime }}(t)+{\lambda }_0\mathrm{\parallel }\mathrm{\nabla }e(t){\mathrm{\parallel }}_2^2\le 2C_{2}E(t)+\mathrm{\parallel }r(t){\mathrm{\parallel }}_2^2\mathrm{.}$

Dropping the nonnegative gradient term in step 2.1 and multiplying by $e^{-2C_{2}t}$ gives $\frac{d}{dt}(e^{-2C_{2}t}E(t))\le e^{-2C_{2}t}\mathrm{\parallel }r(t){\mathrm{\parallel }}_2^2$. Integrating from $0$ to any $s\le T$ yields ${\sup }_{0\le s\le T}E(s)\le e^{2C_{2}T}(E(0)+\mathrm{\parallel }r{\mathrm{\parallel }}_{L^2(Q)}^2)\mathrm{.}$

Integrating step 2.1 from $0$ to $T$ and using step 3.1 gives ${\lambda }_0{\int }_0^T\mathrm{\parallel }\mathrm{\nabla }e(t){\mathrm{\parallel }}_2^{2}dt\le E(0)+2C_{2}T{\sup }_{0\le t\le T}E(t)+\mathrm{\parallel }r{\mathrm{\parallel }}_{L^2(Q)}^2\mathrm{.}$ Combining this with step 3.1 and dividing by ${\lambda }_0>0$ proves the stated estimate after increasing the constant.