Lecture 09: Derivatives of Tensor Functions
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Overview
Steven M. Wise develops derivatives for scalar-valued and second-order-tensor-valued functions of second-order tensors, using directional derivatives and component formulas to identify the resulting second- and fourth-order derivative tensors. He applies the framework to derive the determinant derivative, the time rate of a determinant, and the derivative of the example T(A) = tr(A)A.
Key takeaways
- For a scalar-valued function s(A), the derivative is a second-order tensor characterized by D s(A) : B = d/dα s(A + αB)|₀ for every second-order direction B.
- In a frame, the components of a scalar-function gradient are ∂s/∂aᵢⱼ; for s(A) = ½ A : A, this gives D s(A) = A.
- At an invertible tensor A, the determinant derivative is D det(A) = det(A)A⁻ᵀ, derived by reducing the directional variation to det(A)tr(A⁻¹B).
- For an invertible, differentiable tensor history S(t), the determinant rate is d(det S)/dt = det(S)tr(S⁻¹Ṡ), a result relevant to the Reynolds transport theorem.
- A map from second-order tensors to second-order tensors has a fourth-order derivative with 81 components in three dimensions, and that derivative acts on a direction tensor to produce a second-order tensor.
- For T(A) = tr(A)A, the directional derivative is D T(A)[B] = tr(B)A + tr(A)B, illustrating how the product rule generates a fourth-order derivative.
Chapters
- A scalar-valued function s maps a second-order tensor A to a scalar, so its derivative at A is a second-order tensor.
- The Fréchet definition writes s(A + H) as s(A) plus the tensor inner product of the derivative with H and a remainder of order o(||H||).
- The derivative term and remainder are scalar-valued, matching the type of s(A).
- The Gâteaux formula identifies the derivative D s(A) through its action on any second-order tensor B.
- Compute the derivative of s(A + αB) with respect to scalar α at α = 0, then equate it to D s(A) : B.
- Fréchet differentiability implies Gâteaux differentiability, and the directional formula is the practical tool used throughout the lecture.
- In an arbitrary frame, a second-order tensor has nine components, so s can be treated as a function from R⁹ to R.
- Applying the vector chain rule to A + αB produces the component derivatives ∂s/∂aᵢⱼ multiplied by bᵢⱼ.
- Reassembling those coefficients in the dyadic basis gives the gradient tensor; because B is arbitrary, its inner-product representation uniquely identifies the derivative.
- For sufficiently smooth s, Taylor's theorem approximates s(A + H) by s(A) + D s(A) : H with a higher-order remainder.
- The derivative supplies the tensor-function equivalent of the tangent-line approximation from ordinary calculus.
- The stronger remainder estimate requires additional smoothness beyond the original Fréchet differentiability definition.
- For s(A) = ½ A : A, the tensor inner product expands in a frame as the double sum aₖₗaₖₗ.
- The product rule and component identity ∂aₖₗ/∂aᵢⱼ = δᵢₖδⱼₗ reduce the gradient components to aᵢⱼ.
- Reassembling the components shows D s(A) = A, the tensor analogue of differentiating ½ x·x.
- The determinant is a scalar-valued function of a second-order tensor, so its derivative is a second-order tensor.
- Wise states that for invertible A, the determinant derivative is det(A) A⁻ᵀ.
- The proof uses the directional derivative with A + αB and rewrites the determinant expression to isolate its dependence on α.
- Assuming A is invertible, factor the perturbed tensor using A and A⁻¹ to expose a term involving the identity and αA⁻¹B.
- The determinant product rule separates det(A) from the determinant of the transformed factor.
- In three dimensions, pulling out a scalar factor α from a 3 × 3 determinant contributes α³.
- The characteristic polynomial of A⁻¹B expresses the transformed determinant through its three scalar invariants.
- Differentiating with respect to α at zero eliminates the constant and higher-power terms, leaving the coefficient involving det(A) and tr(A⁻¹B).
- Using tr(ST) = Sᵀ : T converts that directional derivative into det(A)A⁻ᵀ : B, identifying the derivative since B is arbitrary.
- In a frame, expand tr(ST) as the sum of diagonal components, giving SₗₖTₖₗ with the repeated index k summed.
- The components Sₗₖ are those of Sᵀ, so the expression equals Sᵀ : T.
- This identity makes the determinant derivative's trace expression directly comparable to the directional derivative definition.
- For a differentiable, invertible time-dependent tensor S(t), the determinant is a scalar function of time.
- The chain rule and the determinant gradient give d(det S)/dt = det(S) tr(S⁻¹ Ṡ).
- The rate Ṡ is formed by differentiating the components of S with respect to time; the trace identity also writes the result as det(S)S⁻ᵀ : Ṡ.
- A function T mapping second-order tensors to second-order tensors has a fourth-order tensor as its derivative at A.
- The fourth-order derivative acts linearly on a second-order direction B and returns a second-order tensor.
- Its directional form is obtained by differentiating T(A + αB) with respect to α at zero.
- In an arbitrary frame, each output component Tᵢⱼ is treated as a function of the nine input components aₖₗ.
- The chain rule gives directional components as the double sum of (∂Tᵢⱼ/∂aₖₗ)bₖₗ.
- Expressing that action in the dyadic basis lifts the coefficients into the 81 components of a fourth-order tensor; arbitrary B ensures the derivative tensor is uniquely determined.
- The first derivative of a scalar-valued tensor function is second order, and differentiating that tensor-valued function produces a fourth-order second derivative.
- Higher derivatives continue to involve increasingly higher-order tensor objects.
- For sufficiently smooth functions, Taylor's theorem gives a more accurate approximation with a higher-order remainder.
- The output components are Tᵢⱼ = (tr A)aᵢⱼ, or (aₘₘ)aᵢⱼ with a sum over the diagonal index m.
- Applying the product rule and Kronecker deltas gives fourth-order components for the derivative of T.
- Acting on a direction B yields D T(A)[B] = tr(B)A + tr(A)B; Wise leaves additional examples for independent practice.
Transcript excerpt
0:04 Well, welcome back to continue mechanics then and uh we're in part four of chapter 2. This is our last lecture uh from chapter 2 on tensor calculus. Uh we're going to start out with um functions that are functions of second order tensors but have values in the scalers. So these are called scaler valued functions. And we want to define what we mean by a derivative of these objects. Uh again things aren't going to become any more uh complicated I guess from here. It's just we keep generalizing the concept of derivative to other objects. The nice thing about the way we defined it in the first place using this um little O notation that we
0:52 introduced uh in the first of these parts of um chapter 2 lectures is that we can write pretty much everything exactly the same way. So we say a function a scaler valued function of second order tensors is differentiable at a point and a point in the space is whatever whatever your uh pre-image is here whatever your domain is and so a point here is um is um points in in this space of second order tensors is just going to be a second order tensor. So we say it's differentiable at a second order tensor if and only if there exists a second order tensor such that
1:40 s of a + h is equal to s of a plus this is our derivative second order tensor. This is the tensor inner product with H plus some small quantity O of uh norm of H as H goes to zero. So this is probably meaningless to you. Um and I'll show you where uh a more meaningful version of this comes if we consider the Taylor approximation. But the nice thing is it has the same
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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Steven M. Wise.