Zero-data bounded classical parabolic errors are weak energy solutions

Use the domains $U_n$ and divergence-compatible smooth convolution approximants from [F2], with zero initial and lateral data and source $r_n$. Their spatial mollifier support stays inside $U$ at every point of $U_n$; their temporal mollifier is supported at positive past lags, sampling $t-s$ with $s>0$. Thus $a_n$ retains ellipticity and its bound, $b_n,c_n,d_n$ retain their separate bounds, and the expanded ${\beta }_n,q_n$ are the convolutions of the bounded original expanded coefficients, retaining their bounds and $q_n\le 0$. No endpoint limit of the original $b,c,d$ at $T$ is used. Each fixed smooth approximant may be extended locally past $T$ for its construction; compactness is used only at times below $T$. The resulting weak solutions $e_n$ are smooth in the interior, uniformly bounded, and have a locally classical subsequential limit with zero continuous initial/lateral data at every time below $T$ by [F2]. Classical comparison on cylinders ending below $T$ identifies the limit with $e$.

Test the weak equation $e_{n,t}-L_n{e}_n=-r_n$ with $e_n$. The divergence terms have no boundary contribution because $e_n\in H_0^1(U_n)$ in time-integrated sense. In particular the lower-order spatial form is ${\int }_{U_n}{a}_{n}D{e}_n\cdot D{e}_n+{\int }_{U_n}(b_n-c_n)\cdot D{e}_n{e}_n-{\int }_{U_n}{d}_n{e}_n^2\mathrm{.}$ The cross term is $b_n-c_n$, since the $b$ term is integrated by parts and the $c$ term is not. Write $B=\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }c{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $D=\mathrm{\parallel }d{\mathrm{\parallel }}_{\mathrm{\infty }}$. With $y_n=\mathrm{\parallel }{e}_n{\mathrm{\parallel }}_2^2$, [F3], ellipticity and Young's inequality give $\frac{1}{2}{y}_n^{\mathrm{\prime }}+\frac{{\lambda }_0}{2}\mathrm{\parallel }D{e}_n{\mathrm{\parallel }}_2^2\le (\frac{B^2}{2{\lambda }_0}+D+\frac{1}{2})y_n+\frac{1}{2}\mathrm{\parallel }{r}_n{\mathrm{\parallel }}_2^2\mathrm{.}$ The constants are uniform in $n$; derivatives of the smoothed coefficients do not occur.

Put $C_0=B^2\mathrm{/}{\lambda }_0+2D+1$. The previous inequality gives $y_n^{\mathrm{\prime }}+{\lambda }_0\mathrm{\parallel }D{e}_n{\mathrm{\parallel }}_2^2\le C_0{y}_n+\mathrm{\parallel }{r}_n{\mathrm{\parallel }}_2^2$. Multiply by $e^{-C_{0}t}$ and integrate, using $y_n(0)=0$. Then ${\sup }_t{y}_n\le e^{C_{0}T}\mathrm{\parallel }{r}_n{\mathrm{\parallel }}_{L^2}^2$. Integrating the same differential inequality without the factor bounds ${\lambda }_0\int \mathrm{\parallel }D{e}_n{\mathrm{\parallel }}_2^2$ by $C_{0}T\sup y_n+\mathrm{\parallel }{r}_n{\mathrm{\parallel }}_{L^2}^2$. The boundedness of $r$ and the finite cylinder give a uniform right side.

Extend each $e_n$ by zero from $U_n$ to $U$. For almost every time this lies in $H_0^1(U)$: extend its compactly supported smooth $H^1(U_n)$ approximants by zero, which preserves their norms, and use the closure definition in [F1]. The bounds in step 3.1 therefore give a bounded sequence in $L^2(0,T;H_0^1(U))$. Apply [F4] and pass to a weakly convergent subsequence. The local uniform convergence from step 1.1 identifies this weak limit with $e$ almost everywhere, by testing on compact interior cylinders. Such cylinders exhaust $U\times (0,T)$; a locally zero $L^2$ function is zero almost everywhere on their union. Hence $e\in L^2(0,T;H_0^1(U))$.

Fix a compact positive-time cylinder $K⋐U\times (0,T)$ and a smooth spacetime cutoff $\chi$ equal to one near $K$. Extend each of $\chi b,\chi c,\chi d$ by zero to ${\mathbb{R}}^{d+1}$; these bounded compactly supported functions belong to $L^2$. The nonnegative product convolution kernels used in step 1.1 have mass one, $L^1$ norm one, and support shrinking to zero in both space and time (with positive past time lag). Thus they satisfy the approximate-identity conditions, although the spatial and temporal scales differ. Apply [F5] with $p=2$ to these extensions. For large $n$ their convolutions agree on $K$ with $b_n,c_n,d_n$, respectively, so $b_n\to b$, $c_n\to c$ and $d_n\to d$ strongly in $L^2(K)$, with their common uniform supremum bounds. Only $a_n,{\beta }_n,q_n,r_n$ are asserted to converge uniformly on such compact cylinders, by their local Hölder regularity. For a smooth test $\varphi$ supported in $K$, the $a_{n}D{e}_n\cdot D\varphi$ term passes by uniform coefficient convergence and weak $L^2$ gradient convergence. In the term $c_n\cdot D{e}_n\varphi$, the test vector $c_n\varphi$ converges strongly in $L^2(K)$, so its pairing with the bounded weakly convergent gradients tends to $\int c\cdot De\varphi$. In the $b_n{e}_n\cdot D\varphi$ and $d_n{e}_n\varphi$ terms, strong local coefficient convergence and local uniform convergence of $e_n$ imply strong $L^2$ convergence of the products. The source and time-test terms pass by local uniform convergence. These arguments prove the limiting weak identity on every compact positive-time cylinder without extra spatial $C^1$ assumptions on $b,c,d$.

For a smooth spacetime test with compact spatial support in $U$ but not necessarily vanishing at initial time, first perform the passage above on $[\delta ,T]$, taking its support strictly below $T$ when necessary. The common bounded coefficient and source norms, supremum bounds and energy bounds control the omitted interval $(0,\delta )$ uniformly: terms linear in gradients are at most a fixed constant times $\sqrt{\delta }\mathrm{\parallel }D{e}_n{\mathrm{\parallel }}_{L^2}$ by Cauchy–Schwarz, and the zeroth-order, source and time-test terms are at most a constant times $\delta$. The same bounds hold for the limit by step 4.1. Letting $\delta \downarrow 0$ therefore gives the identity with zero initial boundary term, inherited from the approximants. Density of spatial compact-smooth functions in $H_0^1(U)$ extends the positive-time identity to all variational tests, since the divergence coefficients are bounded and the limit has the membership established in step 4.1. It follows that $e_t=L_{t}e-r\in L^2(0,T;H^{-1}(U))$.

By [F3], this evolution-space function has a continuous $L^2$ trace. Its integration-by-parts identity compared with the weak identity in step 6.1, for tests $\eta (t)\varphi (x)$ with $\eta (0)\mathrm{\ne }0$ and $\varphi \in C_c^{\mathrm{\infty }}(U)$, gives $(e(0),\varphi {)}_{L^2}=0$. Density in $L^2(U)$ gives $e(0)=0$. Thus $e$ is a zero-initial variational solution as claimed. If two such weak solutions exist, their difference has zero source and zero initial value. The same test and integrating-factor argument as steps 2.1–3.1 on $U$ gives zero $L^2$ norm at every time, proving weak uniqueness.

Apply [F3] to the now identified weak solution $e$, repeating step 2.1 on $U$ with its actual source $r$. The integrating-factor and integrated gradient arguments in step 3.1 yield the stated estimate with, for example, $C_E=e^{C_{0}T}+{\lambda }_0^{-1}(C_{0}T{e}^{C_{0}T}+1)$. The exact value is immaterial, but its dependence is only on the declared quantities, not on boundary derivatives of $e$.

For the final convergence assertion, $\mathrm{\parallel }{r}_m{\mathrm{\parallel }}_{L^2}^2\le T\mathrm{\mid }U\mathrm{\mid }\mathrm{\parallel }{r}_m{\mathrm{\parallel }}_{\mathrm{\infty }}^2\to 0$. The estimate in step 8.1 gives $\int \mathrm{\parallel }D{e}_m{\mathrm{\parallel }}_2^2\to 0$, and either that estimate or $\int \mathrm{\parallel }{e}_m{\mathrm{\parallel }}_2^2\le T\mathrm{\mid }U\mathrm{\mid }\mathrm{\parallel }{e}_m{\mathrm{\parallel }}_{\mathrm{\infty }}^2$ gives the zeroth-order term. Their sum is the squared $L^2(0,T;H^1(U))$ norm. This proves strong convergence and completes the result.