Locally integrable conservative weak and initial forms

Adopt the Axiom of Countable Choice, meaning that every sequence of nonempty sets admits a choice function. For $n\ge 1$, use the complete Euclidean Lebesgue measure space $({\mathbb{R}}^n,{\mathcal{L}}_n,{\lambda }_n)$, where ${\mathcal{L}}_n=\mathcal{L}({\mathbb{R}}^n)$ is the Caratheodory-measurable family for Lebesgue outer measure and ${\lambda }_n$ is that outer measure restricted to ${\mathcal{L}}_n$. The sigma-algebra and complete-measure assertions are the Lebesgue measure theorem under this axiom. For a Lebesgue-measurable set $E\subseteq {\mathbb{R}}^n$, use the trace sigma-algebra ${\mathcal{L}}_E:=\{E\cap B:B\in {\mathcal{L}}_n\}$ and the measure ${\lambda }_E(A):={\lambda }_n(A)$ for $A\in {\mathcal{L}}_E$; its countable additivity is inherited from ${\lambda }_n$. A map $f:E\to {\mathbb{R}}^k$ is measurable when $f^{-1}(B)\in {\mathcal{L}}_E$ for every Borel set $B\subseteq {\mathbb{R}}^k$, where the Borel sigma-algebra is generated by the open sets. For matrix-valued maps use their finite coordinate arrays. A property holds almost everywhere on $E$ when its exceptional set is contained in a measurable ${\lambda }_E$-null set. Nonnegative integrals are suprema of nonnegative simple minorants; signed real integrals are the difference of the positive- and negative-part integrals when those parts are finite. Finite-dimensional integrals are taken componentwise.

Let $T>0$, let $\mathrm{\Omega }\subseteq {\mathbb{R}}^3$ be open, and let $m\ge 1$ be an integer. Put $D:=\mathrm{\Omega }\times [0,T)\subseteq {\mathbb{R}}^4$ in spatial-first coordinates $(x,t)$. Both $D$ and $\mathrm{\Omega }$ are Borel and hence Lebesgue measurable. Spatial integrals use ${\lambda }_{\mathrm{\Omega }}$ from ${\lambda }_3$, and space-time integrals use ${\lambda }_D$ from ${\lambda }_4$. For a finite coordinate array $v$, set $\mathrm{\parallel }v{\mathrm{\parallel }}_1:={\sum }_i\mathrm{\mid }{v}_i\mathrm{\mid }$, summing all entries for a matrix. A measurable field $f:E\to {\mathbb{R}}^k$ is locally integrable when ${\int }_K\mathrm{\parallel }f{\mathrm{\parallel }}_{1}d{\lambda }_E<\mathrm{\infty }$ for every compact $K\subseteq E$. For $E=D$ the compact sets are taken in the relative topology, including those meeting $t=0$. Fields on $D$ are identified up to ${\lambda }_D$-almost-everywhere equality, and initial fields on $\mathrm{\Omega }$ up to ${\lambda }_{\mathrm{\Omega }}$-almost-everywhere equality. The slice $\mathrm{\Omega }\times \{0\}$ is contained in a Lebesgue-null coordinate hyperplane in ${\mathbb{R}}^4$; values of a space-time representative on that slice do not determine the separately prescribed initial field.

In the displayed weak identities, ${\int }_0^T{\int }_{\mathrm{\Omega }}fdxdt$ denotes the joint integral of the zero extension of $f$ from $D$ with respect to ${\lambda }_4$. For each test, its compact support bounds the test and its first derivatives, so every component of the tested products is measurable and is bounded in absolute value by a fixed multiple of the norm of a locally integrable field on that compact set. Thus all tested products and their finite sums have finite signed integrals. Euclidean Lebesgue measure on ${\mathbb{R}}^3\times \mathbb{R}$ is the completed product of ${\lambda }_3$ and ${\lambda }_1$; completed-product Fubini therefore also gives the displayed iterated notation, with section integrals taken for almost every time and set to zero on a measurable exceptional null set. No integrability of every individual time slice is required. The integral is unchanged by the stated almost-everywhere identifications. The same conventions apply to the scalar inequality and to the initial integrals on $\mathrm{\Omega }$.

Let $U:\mathrm{\Omega }\times [0,T)\to {\mathbb{R}}^m$, $F:\mathrm{\Omega }\times [0,T)\to {\mathbb{R}}^{m\times 3}$, and $S:\mathrm{\Omega }\times [0,T)\to {\mathbb{R}}^m$ be such fields, and let $U_0:\mathrm{\Omega }\to {\mathbb{R}}^m$ be locally integrable. For each row $a$, define $({\mathrm{div}}_{x}F{)}_a={\sum }_{j=1}^3{\mathrm{\partial }}_{x_j}{F}_{aj}$ in the distributional sense, and define $F:{\mathrm{\nabla }}_x\varphi ={\sum }_{a=1}^m{\sum }_{j=1}^3{F}_{aj}{\mathrm{\partial }}_{x_j}{\varphi }_a$. The equation ${\mathrm{\partial }}_{t}U+{\mathrm{div}}_{x}F=S$ with distributional initial datum $U_0$ means that for every $\varphi \in C_c^1(\mathrm{\Omega }\times [0,T);{\mathbb{R}}^m)$, continuously differentiable up to $t=0$ and supported away from $\mathrm{\partial }\mathrm{\Omega }$ and $t=T$,

${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathrm{\Omega }}(U\cdot {\mathrm{\partial }}_t\varphi +F:{\mathrm{\nabla }}_x\varphi +S\cdot \varphi )dxdt+{\int }_{\mathrm{\Omega }}{U}_0(x)\cdot \varphi (x,0)dx=0.$

No boundary condition on $\mathrm{\partial }\mathrm{\Omega }$ is encoded because the test fields are supported away from it. This initial identity specifies the initial datum distributionally; it does not assert $U(\cdot ,t)\to U_0$ in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$ or any other stronger trace topology. Such a trace must be assumed or proved separately. The fields are abstract mathematical variables; no physical unit system or balance postulate is imposed.

For locally integrable scalar fields $H,R:D\to \mathbb{R}$ and $H_0:\mathrm{\Omega }\to \mathbb{R}$, and a locally integrable flux $Q:\mathrm{\Omega }\times [0,T)\to {\mathbb{R}}^3$, the inequality ${\mathrm{\partial }}_{t}H+{\mathrm{div}}_{x}Q\ge R$ with initial datum $H_0$ means that for every nonnegative $\psi \in C_c^1(\mathrm{\Omega }\times [0,T))$ with the same support convention,

${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathrm{\Omega }}(H{\mathrm{\partial }}_t\psi +Q\cdot {\mathrm{\nabla }}_x\psi +R\psi )dxdt+{\int }_{\mathrm{\Omega }}{H}_0(x)\psi (x,0)dx\le 0.$

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)

Lebesgue measurable sets, the family $\mathcal{L}({\mathbb{R}}^n)$, and the restricted set function ${\lambda }_n$

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra containing every elementary set and ${\lambda }_n$ is a complete measure extending elementary volume

Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable

The Borel sigma-algebra of a topological space

A measurable function between measurable spaces

Measure-null sets and almost-everywhere statements relative to a measure

The nonnegative Lebesgue integral

Integrable real and complex functions, and their integrals

Arithmetic and lattice operations preserve measurability whenever they are defined

A continuous map has Borel preimages of Borel sets

Monotonicity and nonnegative homogeneity of the nonnegative integral

The Lebesgue integral is linear on $L^1(\mu )$

Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree

Lebesgue measure is sigma-finite, and every metrically bounded subset of ${\mathbb{R}}^n$ has finite outer measure

The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures

Tonelli and Fubini for the completed product, with only almost-everywhere section measurability

Every affine hyperplane of ${\mathbb{R}}^n$, and hence every proper linear subspace, is Lebesgue null

Silvestre, Partial Differential Equations: Lecture Notes, Chapter 6, §6.2.2.

Guermond, Introduction to Partial Differential Equations for Engineering Applications, §§24.4.2 and 26.2.2.