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The quadratic stress identity and the local angular-balance identity

Statement

Let $U \subseteq R^{3}$ be open and let $F : U \to R^{3}$ be of class $C^{1}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), with coordinates $x_{1}, x_{2}, x_{3}$ , partial derivatives $\partial_{j} = \partial / \partial x_{j}$ , Euclidean inner product $⟨ \cdot, \cdot ⟩$ and norm $∣ \cdot ∣_{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ), divergence and curl as in Divergence and curl of a $C^{1}$ vector field, and cross product as in The cross product in $R^{3}$ . Put

$Q_{i j} : = F_{i} F_{j} - \frac{1}{2} δ_{i j} ∣ F ∣_{2}^{2}, δ_{i j} : = {\begin{cases} 1, & i = j, \\ 0, & i \neq j, \end{cases} (div Q)_{i} : = \sum_{j = 1}^{3} \partial_{j} Q_{i j} .$

Then $Q$ is a $C^{1}$ tensor field on $U$ and

$div Q = F div F - F \times curl F .$

Moreover, writing $ϵ_{i j k}$ for the Levi-Civita symbol ( $ϵ_{i j k} = + 1$ if $(i, j, k)$ is an even permutation of $(1, 2, 3)$ , $- 1$ if it is an odd permutation, and $0$ if two indices coincide), every symmetric tensor $Q$ (that is, $Q_{k j} = Q_{j k}$ for all $j, k$ ) satisfies

$ϵ_{i j k} Q_{k j} = 0 (i \in {1, 2, 3}),$

the sum over $j, k$ being the nine-term sum with each pair counted separately. Finally, let $I \subseteq R$ be open. A tensor field here is a matrix-valued map $Π : U \times I \to R^{3 \times 3}$ ; assume its entries and all spatial first partial derivatives are continuous and $Π_{k j} = Π_{j k}$ . Let $g, f : U \times I \to R^{3}$ be continuous, assume the componentwise time derivative $\partial_{t} g$ exists, and suppose $\partial_{t} g + div Π = - f$ on $U \times I$ , where $(div Π)_{i} = \sum_{ℓ = 1}^{3} \partial_{ℓ} Π_{i ℓ}$ and $(x \times Π)_{i ℓ} : = ϵ_{i j k} x_{j} Π_{k ℓ}$ with the sum over $j, k$ . Then the vector field $x \times g$ and the tensor field $x \times Π$ satisfy

$\partial_{t} (x \times g) + div (x \times Π) = - x \times f .$

No symmetry of $Q$ beyond the displayed form is assumed, and no boundary integral is asserted: the first two identities are pointwise on $U$ and the angular identity is pointwise on $U \times I$ .

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ ; the $C^{1}$ field $F : U \to R^{3}$ with $Q_{i j} = F_{i} F_{j} - \frac{1}{2} δ_{i j} ∣ F ∣_{2}^{2}$ ; a symmetric tensor $Q$ in the second claim and a spatially $C^{1}$ symmetric matrix field $Π$ on $U \times I$ , continuous $g, f$ there, and the existing time derivative $\partial_{t} g$ satisfying the stated balance.

[F1]

Cross product: for $u, v \in R^{3}$ , $u \times v = (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1})$ (The cross product in $R^{3}$ ).

[F2]

Divergence and curl: $div F = \sum_{i = 1}^{3} \partial_{i} F_{i}$ and $curl F = (\partial_{2} F_{3} - \partial_{3} F_{2}, \partial_{3} F_{1} - \partial_{1} F_{3}, \partial_{1} F_{2} - \partial_{2} F_{1})$ (Divergence and curl of a $C^{1}$ vector field).

[F3]

Inner product and norm: $⟨ u, v ⟩ = \sum_{i = 1}^{3} u_{i} v_{i}$ and $∣ F ∣_{2}^{2} = ⟨ F, F ⟩$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F4]

Product rule and differentiability of products of $C^{1}$ functions, with the partial-derivative convention for maps on open Euclidean sets (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, $C^{k}$ Euclidean maps and diffeomorphisms).

Proof

technique · direct; expand the divergence of the quadratic tensor in coordinates, match the resulting expression with $-F\times\operatorname{curl}F$, cancel symmetric terms in the Levi-Civita contraction, and differentiate the cross-product products in the angular-balance identity

1.1givenF3F4algebra

The entries $Q_{i j} = F_{i} F_{j} - \frac{1}{2} δ_{i j} ⟨ F, F ⟩$ are sums of products of $C^{1}$ functions, hence $C^{1}$ on $U$ by [F3] and [F4], so both sides of the first identity are defined; moreover $\partial_{j} ⟨ F, F ⟩ = \partial_{j} \sum_{l = 1}^{3} F_{l} F_{l} = 2 \sum_{l = 1}^{3} F_{l} \partial_{j} F_{l}$ by the product rule [F4].

2.1step 1.1F2F4algebra

For fixed $i$ , $\sum_{j = 1}^{3} \partial_{j} Q_{i j} = \sum_{j = 1}^{3} \partial_{j} (F_{i} F_{j}) - \frac{1}{2} \sum_{j = 1}^{3} δ_{i j} \partial_{j} ⟨ F, F ⟩ = \sum_{j = 1}^{3} (F_{j} \partial_{j} F_{i} + F_{i} \partial_{j} F_{j}) - \sum_{l = 1}^{3} F_{l} \partial_{i} F_{l} = F_{i} div F + \sum_{j = 1}^{3} (F_{j} \partial_{j} F_{i} - F_{j} \partial_{i} F_{j})$ by step 1.1, the product rule [F4] and the divergence formula [F2].

2.2step 1.1algebra

For a symmetric tensor and fixed $i$ , the nine-term sum $ϵ_{i j k} Q_{k j}$ vanishes: terms with $j = k$ have $ϵ_{i j k} = 0$ , and for $j \neq k$ the pair $(j, k)$ and the pair $(k, j)$ contribute $ϵ_{i j k} Q_{k j} + ϵ_{i k j} Q_{j k} = ϵ_{i j k} Q_{k j} - ϵ_{i j k} Q_{k j} = 0$ , because $ϵ$ changes sign under exchange of its last two indices and $Q$ is symmetric. This is the second claim.

3.1step 2.1F1F2algebra

The first coordinate of $F \times curl F$ is, by [F1] and [F2], $F_{2} (\partial_{1} F_{2} - \partial_{2} F_{1}) - F_{3} (\partial_{3} F_{1} - \partial_{1} F_{3}) = F_{2} \partial_{1} F_{2} + F_{3} \partial_{1} F_{3} - F_{2} \partial_{2} F_{1} - F_{3} \partial_{3} F_{1}$ ; adding and subtracting $F_{1} \partial_{1} F_{1}$ shows that $- (F \times curl F)_{1} = \sum_{j = 1}^{3} (F_{j} \partial_{j} F_{1} - F_{j} \partial_{1} F_{j})$ , and the same expansion in the second and third coordinates gives $- (F \times curl F)_{i} = \sum_{j = 1}^{3} (F_{j} \partial_{j} F_{i} - F_{j} \partial_{i} F_{j})$ for every $i$ . Substituting this into step 2.1 gives $div Q = F div F - F \times curl F$ , which is the first identity.

3.2step 2.2F4algebra

At each time the field $(x \times Π)_{i ℓ} = ϵ_{i j k} x_{j} Π_{k ℓ}$ is spatially $C^{1}$ , and differentiating the product in $x_{ℓ}$ and summing over $ℓ$ gives $(div (x \times Π))_{i} = \sum_{ℓ = 1}^{3} \partial_{ℓ} (ϵ_{i j k} x_{j} Π_{k ℓ}) = ϵ_{i j k} \sum_{ℓ = 1}^{3} (δ_{j ℓ} Π_{k ℓ} + x_{j} \partial_{ℓ} Π_{k ℓ}) = ϵ_{i j k} Π_{k j} + ϵ_{i j k} x_{j} \sum_{ℓ = 1}^{3} \partial_{ℓ} Π_{k ℓ} = ϵ_{i j k} x_{j} (div Π)_{k}$ , where the term $ϵ_{i j k} Π_{k j}$ vanishes by step 2.2 applied to the symmetric field $Π$ ; here $δ_{j ℓ}$ is the Kronecker symbol and the product rule [F4] was used in each summand.

4.1step 3.2F4algebra∎

Substituting $\partial_{t} g_{k} = - f_{k} - (div Π)_{k}$ and differentiating the product with the time-independent factor $x_{j}$ gives $\partial_{t} (x \times g)_{i} = ϵ_{i j k} x_{j} \partial_{t} g_{k} = - ϵ_{i j k} x_{j} f_{k} - ϵ_{i j k} x_{j} (div Π)_{k} = - (x \times f)_{i} - (div (x \times Π))_{i}$ by step 3.2; hence $\partial_{t} (x \times g) + div (x \times Π) = - x \times f$ , which is the third claim.

Remarks

The identity $div Q = F div F - F \times curl F$ requires only first derivatives and no interchange of mixed partials, so the $C^{1}$ hypothesis is the natural one; the symmetry claim for $ϵ_{i j k} Q_{k j}$ is a purely algebraic statement about symmetric tensors.
The angular-balance identity is local; passing to an integral balance over a region requires a divergence theorem and boundary terms, which are not asserted here. For any time-dependent data meeting the displayed balance, the angular identity follows; this item adopts no evolution equation for $F$ .

The quadratic stress identity and the local angular-balance identity

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources