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Linear isotropic symmetric stress classification

Statement

Let $Sym (3) = {D \in R^{3 \times 3} : D^{T} = D}$ , let $I$ be the $3 \times 3$ identity matrix, and define $SO (3) = {Q \in R^{3 \times 3} : Q^{T} Q = I, \det Q = 1} .$ An orthogonal matrix is one with $Q^{T} Q = I$ ; define $O (3)$ by this condition without the determinant restriction. For $D \in Sym (3)$ set $tr D = \sum_{i = 1}^{3} D_{i i}$ . Suppose $L : Sym (3) \to Sym (3)$ is real-linear, meaning $L (a D + b E) = a L (D) + b L (E)$ for all $a, b \in R$ and $D, E \in Sym (3)$ , and satisfies $L (Q D Q^{T}) = Q L (D) Q^{T} (Q \in SO (3), D \in Sym (3)) .$ Then there are unique $λ, μ \in R$ such that $L (D) = 2 μ D + λ (tr D) I (D \in Sym (3)) .$ Conversely, each map of this form is real-linear and satisfies the displayed equivariance. Linearity and rotational equivariance are hypotheses of this mathematical classification; it does not assert that a material response is linear or isotropic.

Statement references

Kelly, Solid Mechanics Part III, §4.A “Isotropic Functions,” pp. 406–411

Krzysztof Wilmanski, Continuum Thermodynamics, §4.2 “Isotropy,” pp. 46–50

Facts & Assumptions

Given: The real-linear map $L$ , its domain and codomain, and the stated $SO (3)$ equivariance.

[A1]

All products and transposes are ordinary real matrix operations. For $x, y \in R^{3}$ , $x^{T} y = \sum_{i = 1}^{3} x_{i} y_{i}$ .

[F1]

For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent: a nonempty closed bounded subset of $R^{n}$ is compact, and every continuous real-valued function on a nonempty compact subset of $R^{n}$ attains its maximum and minimum. The supplier states this theorem in ZF.

Proof

1.1A1F1givenconstruct

Fix $D \in Sym (3)$ and define $q_{D} (x) = x^{T} D x$ on $S^{2} = {x \in R^{3} : x^{T} x = 1}$ . The sphere is nonempty, closed because $x \mapsto x^{T} x$ is continuous, and bounded because each $∣ x_{i} ∣ \leq 1$ there. The function $q_{D}$ is continuous because it is a finite sum of coordinate products. By [F1], $q_{D}$ has a maximum at some $e \in S^{2}$ .

1.2A1givenalgebra

Let $X = diag (d_{1}, d_{2}, d_{3})$ . The proper half-turns $Q_{1} = diag (1, - 1, - 1)$ , $Q_{2} = diag (- 1, 1, - 1)$ , and $Q_{3} = diag (- 1, - 1, 1)$ each leave $X$ fixed. Equivariance gives $L (X) = Q_{i} L (X) Q_{i}^{T}$ for each $i$ ; because $L (X)$ is symmetric, these three equalities force every off-diagonal entry to vanish.

2.1A1step 1.1algebra

Let $y \in S^{2}$ satisfy $e^{T} y = 0$ and put $c (t) = (e + t y) / \sqrt{1 + t^{2}}$ . Then $c (t)^{T} c (t) = 1$ and differentiating at $0$ gives $c^{'} (0) = y$ . The differentiable function $t \mapsto q_{D} (c (t))$ has a maximum at $0$ , so its derivative there is $0$ : its right difference quotients are nonpositive and its left difference quotients are nonnegative. Differentiating gives $0 = y^{T} D e + e^{T} D y = 2 y^{T} D e$ , since $D^{T} = D$ . By scaling, this holds for every $y ⊥ e$ , so $D e$ is a scalar multiple of $e$ .

2.2A1step 1.2givenalgebra

Let $E_{j j}$ be the diagonal matrix with a $1$ in position $(j, j)$ and $0$ elsewhere, and let the $j$ th column of a matrix $M$ be the diagonal vector $(L (E_{j j})_{11}, L (E_{j j})_{22}, L (E_{j j})_{33})^{T}$ . Linearity and step 1.2 give $diag L (X) = M (d_{1}, d_{2}, d_{3})^{T}$ , where $diag A = (A_{11}, A_{22}, A_{33})^{T}$ . A permutation matrix $P$ permutes the coordinate basis; if $\det P = - 1$ , then $- P \in SO (3)$ and $(- P) X (- P)^{T} = P X P^{T}$ . Equivariance says $diag L (P X P^{T}) = diag (P L (X) P^{T})$ , hence $M P d = P M d$ for every vector $d$ and $M P = P M$ . Equivalently $M_{i j} = M_{π (i) π (j)}$ for every permutation $π$ , so its diagonal entries have a common value $c$ and its off-diagonal entries have a common value $d_{0}$ . Thus $L (X) = (c - d_{0}) X + d_{0} (tr X) I$ .

3.1A1F1step 1.1step 2.1construct

If $z ⊥ e$ , then $e^{T} D z = (D e)^{T} z = 0$ , so $e^{⊥}$ is invariant under $D$ , and the restriction of $D$ to this two-dimensional plane is symmetric. Its unit circle is a nonempty closed bounded subset of $R^{3}$ ; repeating the argument of steps 1.1 and 2.1 on this invariant plane gives a unit eigenvector in $e^{⊥}$ . The remaining one-dimensional orthogonal complement is invariant and consists of eigenvectors. Hence $D$ has an orthonormal eigenbasis. Only finitely many vectors are chosen in this construction; no arbitrary choice principle is used.

4.1A1step 3.1step 2.2givenalgebra

By step 3.1 write an arbitrary $D \in Sym (3)$ as $D = O X O^{T}$ with $O$ orthogonal and $X$ diagonal. If $\det O = - 1$ , change the sign of one eigenvector; this changes $\det O$ to $1$ and leaves $O X O^{T}$ unchanged. We may therefore take $O \in SO (3)$ . Equivariance and step 2.2 yield $L (D) = (c - d_{0}) D + d_{0} (tr D) I$ .

5.1A1step 4.1algebra∎

Set $2 μ = c - d_{0}$ and $λ = d_{0}$ . Conversely, for any real $λ, μ$ , the map $D \mapsto 2 μ D + λ (tr D) I$ is linear and equivariant: $Q I Q^{T} = I$ , and $tr (Q D Q^{T}) = \sum_{j, k = 1}^{3} D_{j k} \sum_{i = 1}^{3} Q_{i j} Q_{i k} = \sum_{j = 1}^{3} D_{j j} = tr D$ because the columns of $Q$ are orthonormal. If two coefficient pairs give the same map, evaluation at $diag (1, - 1, 0)$ makes their $2 μ$ coefficients equal, and evaluation at $I$ then makes their $λ$ coefficients equal. This proves existence, converse, and uniqueness.

Remarks

The proof uses only proper rotations; the replacement $P \mapsto - P$ handles improper coordinate permutations specifically because the dimension is three. This classification supplies the linear isotropic form when those assumptions are imposed. It does not derive linearity, isotropy, objectivity, a fluid constitutive law, or empirical coefficient values.

Depends on

For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent