Structure Condition: a Degenerate Elliptic Operator with a Continuous Inhomogeneity

In the setting of Second-Order Equations on Euclidean Open Sets and of Bounded Open Domain in Euclidean Space, whose notation, including that of Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation on which the former rests, is in force in the dimension $n$, a natural number with $1\le n$, let $T=\{t\in\mathbb{R}:0\le t\}$ and write $3=1+1+1$.

Two elementary consequences of the ordered field axioms are used freely below. First, multiplication by a nonnegative real number preserves $\le$: if $a\le b$ and $0\le\lambda$ then either $a=b$, and the products are equal, or $a<b$, and then $\lambda a\le\lambda b$ by claim 10 of Elementary Order Arithmetic in an Ordered Field when $0<\lambda$, while $\lambda=0$ makes both products $0$. Second, if $a\le b$ and $0\le e$ then $a\le b+e$, since $b=b+0\le b+e$ by the compatibility of $\le$ with addition and $\le$ is transitive.

The data. Let $G:\mathbb{R}\times\mathbb{R}^{n}\times\mathcal{S}(n)\to\mathbb{R}$ be a function, its value at $(r,p,X)$ written $G(r,p,X)$, which is degenerate elliptic in the matrix variable, meaning that

for every $r\in\mathbb{R}$, every $p\in\mathbb{R}^{n}$ and all $X,Y\in\mathcal{S}(n)$ with $X\preceq Y$.

Let $f:\overline{\Omega}\to\mathbb{R}$ be continuous on $\overline{\Omega}$, as a map into the metric space $(\mathbb{R},d_{\mathbb{R}})$ of The Absolute Value Metric on the Real Line. Let $\omega:T\to\mathbb{R}$ be a modulus of continuity that is nondecreasing, meaning that $\omega(s)\le\omega(t)$ whenever $s,t\in T$ satisfy $s\le t$, and that dominates the oscillation of $f$, meaning that $|f(x)-f(y)|\le\omega(t)$ for all $x,y\in\overline{\Omega}$ and every $t\in T$ with $d_{E}(x,y)\le t$. At least one such $\omega$ exists: $\overline{\Omega}$ is nonempty and compact by Bounded Open Domain in Euclidean Space §closure, so A Continuous Function on a Compact Set Admits a Nondecreasing Modulus of Continuity applies with the metric space $(\mathbb{R}^{n},d_{E})$ and $K=\overline{\Omega}$, and A Continuous Function on a Compact Set Admits a Nondecreasing Modulus of Continuity §monotone and A Continuous Function on a Compact Set Admits a Nondecreasing Modulus of Continuity §domination are exactly the two properties just named.

Since $\Omega\subseteq\overline{\Omega}$ by Bounded Open Domain in Euclidean Space §closure, the formula

defines a function $F:\Omega\times\mathbb{R}\times\mathbb{R}^{n}\times\mathcal{S}(n)\to\mathbb{R}$, that is, a second-order equation operator on $\Omega$.

The claim.¶ $F$ and $\omega$ satisfy the structure condition of the comparison principle for the Dirichlet problem.

Justification. Let $x,y\in\Omega$, let $r\in\mathbb{R}$, let $\alpha\in\mathbb{R}$ be positive and let $X,Y\in\mathcal{S}(n)$ satisfy

write $p=\alpha(x-y)$ for the scalar multiple by $\alpha$ of the difference $x-y$.

Step 1: the hypothesis forces $X\preceq Y$. Let $z\in\mathbb{R}^{n}$ and take $\xi=\eta=z$ in the second inequality. The difference $z-z$ is the origin $0_{\mathbb{R}^{n}}$, so $\lVert z-z\rVert=0$ by claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, whence $\lVert z-z\rVert^{2}=0$ and $3\alpha\lVert z-z\rVert^{2}=0$. Thus $z\cdot(Xz)-z\cdot(Yz)\le0$, that is, $z\cdot(Xz)\le z\cdot(Yz)$. As $z\in\mathbb{R}^{n}$ was arbitrary, $X\preceq Y$ by Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §ordering.

Step 2: reduction to the oscillation of $f$. By the hypothesis on $G$ and Step 1, $G(r,p,Y)-G(r,p,X)\le0$. Since

adding $f(x)-f(y)$ to both sides of $G(r,p,Y)-G(r,p,X)\le0$ gives $F(y,r,p,Y)-F(x,r,p,X)\le f(x)-f(y)$.

Step 3: the modulus. Put $t=\alpha\lVert x-y\rVert^{2}+\lVert x-y\rVert$. By claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ the number $\lVert x-y\rVert$ is nonnegative, hence so is $\lVert x-y\rVert^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, and therefore $0\le\alpha\lVert x-y\rVert^{2}$. Consequently $\lVert x-y\rVert\le t$, and $0\le t$ by transitivity, so $t\in T$. Now $x,y\in\overline{\Omega}$ and $d_{E}(x,y)=\lVert x-y\rVert\le t$, so the domination property gives $|f(x)-f(y)|\le\omega(t)$, while $f(x)-f(y)\le|f(x)-f(y)|$ by claim 3 of Properties of the Absolute Value in an Ordered Field. Combining with Step 2 and using transitivity,

which is the inequality required by the structure condition.