Throughout we work in the setting of Second-Order Equations on Euclidean Open Sets and of Bounded Open Domain in Euclidean Space , whose notation, including that of Euclidean Space and Real Matrices: Standing Notation on which the former rests, is in force in the dimensions n n n and m m m , natural numbers with 1 ≤ n 1\le n 1 ≤ n and 1 ≤ m 1\le m 1 ≤ m . In addition tr \operatorname{tr} tr denotes the trace of a square real matrix; for A ∈ M m × n ( R ) A\in\mathcal{M}_{m\times n}(\mathbb{R}) A ∈ M m × n ( R ) and k ∈ [ m ] k\in[m] k ∈ [ m ] the k k k th row of A A A is the point of R n \mathbb{R}^{n} R n whose j j j th coordinate is A k j A_{kj} A kj ; T = { t ∈ R : 0 ≤ t } T=\{t\in\mathbb{R}:0\le t\} T = { t ∈ R : 0 ≤ t } ; s 2 = s s s^{2}=ss s 2 = ss for s ∈ R s\in\mathbb{R} s ∈ R ; and 3 = 1 + 1 + 1 3=1+1+1 3 = 1 + 1 + 1 , so that 0 < 3 0<3 0 < 3 , since 0 < 2 0<2 0 < 2 by claim 8 of Elementary Order Arithmetic in an Ordered Field and 0 ≤ 1 0\le1 0 ≤ 1 by claim 6 there, whence 0 < 3 0<3 0 < 3 by claim 3 there.
Two elementary consequences of the ordered field axioms are used freely below. First, multiplication by a nonnegative real number preserves ≤ \le ≤ : if a ≤ b a\le b a ≤ b and 0 ≤ λ 0\le\lambda 0 ≤ λ then either a = b a=b a = b , and the products are equal, or a < b a<b a < b , and then λ a ≤ λ b \lambda a\le\lambda b λa ≤ λb by claim 10 of Elementary Order Arithmetic in an Ordered Field when 0 < λ 0<\lambda 0 < λ , while λ = 0 \lambda=0 λ = 0 makes both products 0 0 0 . Second, if a ≤ b a\le b a ≤ b and 0 ≤ e 0\le e 0 ≤ e then a ≤ b + e a\le b+e a ≤ b + e , since b = b + 0 ≤ b + e b=b+0\le b+e b = b + 0 ≤ b + e by the compatibility of ≤ \le ≤ with addition and ≤ \le ≤ is transitive.
The data. Let Σ : Ω → M m × n ( R ) \Sigma:\Omega\to\mathcal{M}_{m\times n}(\mathbb{R}) Σ : Ω → M m × n ( R ) be a function, and for x ∈ Ω x\in\Omega x ∈ Ω and k ∈ [ m ] k\in[m] k ∈ [ m ] write σ k ( x ) ∈ R n \sigma_{k}(x)\in\mathbb{R}^{n} σ k ( x ) ∈ R n for the k k k th row of Σ ( x ) \Sigma(x) Σ ( x ) . Let L ∈ R L\in\mathbb{R} L ∈ R be nonnegative, and assume that
tr ( ( Σ ( x ) − Σ ( y ) ) ⊤ ( Σ ( x ) − Σ ( y ) ) ) ≤ L 2 ∥ x − y ∥ 2 for all x , y ∈ Ω . \operatorname{tr}\Bigl(\bigl(\Sigma(x)-\Sigma(y)\bigr)^{\top}\bigl(\Sigma(x)-\Sigma(y)\bigr)\Bigr)\ \le\ L^{2}\lVert x-y\rVert^{2}\qquad\text{for all }x,y\in\Omega . tr ( ( Σ ( x ) − Σ ( y ) ) ⊤ ( Σ ( x ) − Σ ( y ) ) ) ≤ L 2 ∥ x − y ∥ 2 for all x , y ∈ Ω.
By the entry formula for a difference of matrices and the coordinate formula for a difference of points , the k k k th row of Σ ( x ) − Σ ( y ) \Sigma(x)-\Sigma(y) Σ ( x ) − Σ ( y ) is σ k ( x ) − σ k ( y ) \sigma_{k}(x)-\sigma_{k}(y) σ k ( x ) − σ k ( y ) , so The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound (ii) rewrites the left-hand side as ∑ k = 1 m ∥ σ k ( x ) − σ k ( y ) ∥ 2 \sum_{k=1}^{m}\lVert\sigma_{k}(x)-\sigma_{k}(y)\rVert^{2} ∑ k = 1 m ∥ σ k ( x ) − σ k ( y ) ∥ 2 ; the hypothesis is thus a Lipschitz condition on Σ \Sigma Σ with constant L L L , the size of a matrix being measured by the sum over its rows of the squared Euclidean norms.
For x ∈ Ω x\in\Omega x ∈ Ω and X ∈ S ( n ) X\in\mathcal{S}(n) X ∈ S ( n ) the product Σ ( x ) ⊤ Σ ( x ) X \Sigma(x)^{\top}\Sigma(x)X Σ ( x ) ⊤ Σ ( x ) X is an unambiguously defined element of M n ( R ) \mathcal{M}_{n}(\mathbb{R}) M n ( R ) , as recorded in The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound . Hence
F ( x , r , p , X ) = − tr ( Σ ( x ) ⊤ Σ ( x ) X ) F(x,r,p,X)=-\operatorname{tr}\bigl(\Sigma(x)^{\top}\Sigma(x)X\bigr) F ( x , r , p , X ) = − tr ( Σ ( x ) ⊤ Σ ( x ) X )
defines a function F : Ω × R × R n × S ( n ) → R F:\Omega\times\mathbb{R}\times\mathbb{R}^{n}\times\mathcal{S}(n)\to\mathbb{R} F : Ω × R × R n × S ( n ) → R , that is, a second-order equation operator on Ω \Omega Ω .
Let ω : T → R \omega:T\to\mathbb{R} ω : T → R be given by ω ( t ) = 3 L 2 t \omega(t)=3L^{2}t ω ( t ) = 3 L 2 t . Since 0 ≤ L 0\le L 0 ≤ L , multiplying by the nonnegative L L L gives 0 ≤ L 2 0\le L^{2} 0 ≤ L 2 , and multiplying by the nonnegative 3 3 3 gives 0 ≤ 3 L 2 0\le3L^{2} 0 ≤ 3 L 2 ; so ω \omega ω is a modulus of continuity by Linear Moduli of Continuity (i) .
The claim. (i) F F F and ω \omega ω satisfy the structure condition of the comparison principle for the Dirichlet problem .
Justification. Let x , y ∈ Ω x,y\in\Omega x , y ∈ Ω , let r ∈ R r\in\mathbb{R} r ∈ R , let α ∈ R \alpha\in\mathbb{R} α ∈ R be positive and let X , Y ∈ S ( n ) X,Y\in\mathcal{S}(n) X , Y ∈ S ( n ) satisfy
− 3 α ( ∥ ξ ∥ 2 + ∥ η ∥ 2 ) ≤ ξ ⋅ ( X ξ ) − η ⋅ ( Y η ) ≤ 3 α ∥ ξ − η ∥ 2 for all ξ , η ∈ R n ; -3\alpha\bigl(\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}\bigr)\ \le\ \xi\cdot(X\xi)-\eta\cdot(Y\eta)\ \le\ 3\alpha\lVert\xi-\eta\rVert^{2}\qquad\text{for all }\xi,\eta\in\mathbb{R}^{n}; − 3 α ( ∥ ξ ∥ 2 + ∥ η ∥ 2 ) ≤ ξ ⋅ ( X ξ ) − η ⋅ ( Y η ) ≤ 3 α ∥ ξ − η ∥ 2 for all ξ , η ∈ R n ;
write p = α ( x − y ) p=\alpha(x-y) p = α ( x − y ) , and abbreviate A = Σ ( x ) A=\Sigma(x) A = Σ ( x ) and B = Σ ( y ) B=\Sigma(y) B = Σ ( y ) , with rows a k = σ k ( x ) a_{k}=\sigma_{k}(x) a k = σ k ( x ) and b k = σ k ( y ) b_{k}=\sigma_{k}(y) b k = σ k ( y ) for k ∈ [ m ] k\in[m] k ∈ [ m ] .
Step 1: the difference as a sum of quadratic forms. By The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound (i) , applied to A A A with X X X and to B B B with Y Y Y , and by claims 2 and 3 of Properties of Finite Sums , which together give ∑ k = 1 m ( u k − v k ) = ∑ k = 1 m u k − ∑ k = 1 m v k \sum_{k=1}^{m}(u_{k}-v_{k})=\sum_{k=1}^{m}u_{k}-\sum_{k=1}^{m}v_{k} ∑ k = 1 m ( u k − v k ) = ∑ k = 1 m u k − ∑ k = 1 m v k ,
F ( y , r , p , Y ) − F ( x , r , p , X ) = tr ( A ⊤ A X ) − tr ( B ⊤ B Y ) = ∑ k = 1 m ( a k ⋅ ( X a k ) − b k ⋅ ( Y b k ) ) . F(y,r,p,Y)-F(x,r,p,X)=\operatorname{tr}\bigl(A^{\top}AX\bigr)-\operatorname{tr}\bigl(B^{\top}BY\bigr)=\sum_{k=1}^{m}\Bigl(a_{k}\cdot(Xa_{k})-b_{k}\cdot(Yb_{k})\Bigr). F ( y , r , p , Y ) − F ( x , r , p , X ) = tr ( A ⊤ A X ) − tr ( B ⊤ B Y ) = k = 1 ∑ m ( a k ⋅ ( X a k ) − b k ⋅ ( Y b k ) ) .
Step 2: the matrix hypothesis, row by row. For each k ∈ [ m ] k\in[m] k ∈ [ m ] , taking ξ = a k \xi=a_{k} ξ = a k and η = b k \eta=b_{k} η = b k in the second inequality of the hypothesis gives
a k ⋅ ( X a k ) − b k ⋅ ( Y b k ) ≤ 3 α ∥ a k − b k ∥ 2 . a_{k}\cdot(Xa_{k})-b_{k}\cdot(Yb_{k})\ \le\ 3\alpha\lVert a_{k}-b_{k}\rVert^{2}. a k ⋅ ( X a k ) − b k ⋅ ( Y b k ) ≤ 3 α ∥ a k − b k ∥ 2 .
By claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers and claim 3 of Properties of Finite Sums ,
F ( y , r , p , Y ) − F ( x , r , p , X ) ≤ ∑ k = 1 m 3 α ∥ a k − b k ∥ 2 = 3 α ∑ k = 1 m ∥ a k − b k ∥ 2 . F(y,r,p,Y)-F(x,r,p,X)\ \le\ \sum_{k=1}^{m}3\alpha\lVert a_{k}-b_{k}\rVert^{2}=3\alpha\sum_{k=1}^{m}\lVert a_{k}-b_{k}\rVert^{2}. F ( y , r , p , Y ) − F ( x , r , p , X ) ≤ k = 1 ∑ m 3 α ∥ a k − b k ∥ 2 = 3 α k = 1 ∑ m ∥ a k − b k ∥ 2 .
Step 3: the Lipschitz hypothesis. The k k k th row of A − B A-B A − B is a k − b k a_{k}-b_{k} a k − b k , so The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound (ii) gives ∑ k = 1 m ∥ a k − b k ∥ 2 = tr ( ( A − B ) ⊤ ( A − B ) ) \sum_{k=1}^{m}\lVert a_{k}-b_{k}\rVert^{2}=\operatorname{tr}\bigl((A-B)^{\top}(A-B)\bigr) ∑ k = 1 m ∥ a k − b k ∥ 2 = tr ( ( A − B ) ⊤ ( A − B ) ) , which is at most L 2 ∥ x − y ∥ 2 L^{2}\lVert x-y\rVert^{2} L 2 ∥ x − y ∥ 2 by hypothesis. Since 0 < 3 0<3 0 < 3 and 0 < α 0<\alpha 0 < α , claim 5 of Elementary Order Arithmetic in an Ordered Field gives 0 < 3 α 0<3\alpha 0 < 3 α , so multiplying by 3 α 3\alpha 3 α preserves the inequality and
F ( y , r , p , Y ) − F ( x , r , p , X ) ≤ 3 α ( L 2 ∥ x − y ∥ 2 ) = 3 L 2 ( α ∥ x − y ∥ 2 ) , F(y,r,p,Y)-F(x,r,p,X)\ \le\ 3\alpha\bigl(L^{2}\lVert x-y\rVert^{2}\bigr)=3L^{2}\bigl(\alpha\lVert x-y\rVert^{2}\bigr), F ( y , r , p , Y ) − F ( x , r , p , X ) ≤ 3 α ( L 2 ∥ x − y ∥ 2 ) = 3 L 2 ( α ∥ x − y ∥ 2 ) ,
the last equality by the commutativity and associativity of multiplication in R \mathbb{R} R .
Step 4: conclusion. Put t = α ∥ x − y ∥ 2 + ∥ x − y ∥ t=\alpha\lVert x-y\rVert^{2}+\lVert x-y\rVert t = α ∥ x − y ∥ 2 + ∥ x − y ∥ . By claim 1 of Properties of the Euclidean Norm on R n \mathbb{R}^n R n : Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square the number ∥ x − y ∥ \lVert x-y\rVert ∥ x − y ∥ is nonnegative, hence so is ∥ x − y ∥ 2 \lVert x-y\rVert^{2} ∥ x − y ∥ 2 by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field , and therefore 0 ≤ α ∥ x − y ∥ 2 0\le\alpha\lVert x-y\rVert^{2} 0 ≤ α ∥ x − y ∥ 2 ; consequently α ∥ x − y ∥ 2 ≤ t \alpha\lVert x-y\rVert^{2}\le t α ∥ x − y ∥ 2 ≤ t and 0 ≤ t 0\le t 0 ≤ t , so t ∈ T t\in T t ∈ T . Multiplying α ∥ x − y ∥ 2 ≤ t \alpha\lVert x-y\rVert^{2}\le t α ∥ x − y ∥ 2 ≤ t by the nonnegative 3 L 2 3L^{2} 3 L 2 and using transitivity,
F ( y , r , α ( x − y ) , Y ) − F ( x , r , α ( x − y ) , X ) ≤ 3 L 2 t = ω ( α ∥ x − y ∥ 2 + ∥ x − y ∥ ) , F\bigl(y,r,\alpha(x-y),Y\bigr)-F\bigl(x,r,\alpha(x-y),X\bigr)\ \le\ 3L^{2}t=\omega\bigl(\alpha\lVert x-y\rVert^{2}+\lVert x-y\rVert\bigr), F ( y , r , α ( x − y ) , Y ) − F ( x , r , α ( x − y ) , X ) ≤ 3 L 2 t = ω ( α ∥ x − y ∥ 2 + ∥ x − y ∥ ) ,
which is the inequality required by the structure condition.