In the context of continuum mechanics, given a tensor $A$, the second invariant $i_2(A) = \frac{\text{tr}(A)^2 – \text{tr}(A^2)}{2} $. I need to compute its derivative $Di_2(A)[S]$ (I'm using the notation from Gurtin's book – An introduction to continuum mechanics), where $S$ is a tensor.
(see here for better comprehension)
I will analyize one term at a time:
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$D\text{tr}(A)^2[S] = 2\text{tr}(A)[D \text{tr}(A)[S]] = 2 \text{tr}(A)\text{tr}(S)$
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$D\text{tr}(A^2)[S] = \text{tr}[DA^2[S]] = \text{tr}(AS+ SA)$
where I used the fact that $D \text{tr}(A)[S] = \text{tr}(S)$ because the trace is a linear operator.
So I would obtain $$D i_2(A)[S] = \text{tr}(A)\text{tr}(S) – \frac{1}{2} \text{tr}(AS+SA)$$
Is this correct?
Best Answer
Yes, your answer is right, though for the second term, your notation is misleading. Taken literally, $D\text{tr}(A^2)[S]$ means the derivative of trace at the point $A^2$ evaluated on $S$. This would just equal $\text{tr}(S)$, since trace is linear. Though, what you meant is that \begin{align} D\left(A\mapsto \text{tr}(A^2)\right)_A[S] &= [(D\text{tr}_{A^2})\circ \left(D(A\mapsto A^2)_A\right)][S]\tag{chain rule}\\ &= \text{tr}(AS + SA). \end{align} Here, I put the point of evaluation of derivatives in subscript to avoid excessive bracketing.
This is just one of those issues which comes up over and over in differential calculus, about functions vs function values.