Here I look at the Tensor Form third-order derivative of a function-of-a-function. This is based on the book by Pavel Grinfeld on Tensor Analysis.

As *promised*, I will try to look at the tensor expression for

in this post.

This is clearly

Before I begin, let’s just take a look at the first-derivative and the second-derivative, from a previous post, *Tensor Analysis – Grinfeld – Chapter 4*. The first-derivative is obtainable, in the final Tensor Form, just by using chain rule for a function-of-a-function:~

We can differentiate this a second time, again using chain-rule, for a function-of-a-function, and also the product-rule – for one differential times another differential. (The details are given in the previous post.) We get:~

This is Eqn. (4.44) in Pavel’s book.

This then begs the question: Can I differentiate this expression, again, to get the tensor form for the triple-derivative that we require? Let us try:~

Carrying out the differentiation, we get:~

You can see that, just as we introduced A^{j} to find the second differential, here we have introduced A^{k}, to find the third-differential.

In the second-differential, there were two terms added together. The first term contained three parts multiplied together. Therefore, by the product rule, we differentiate each part, in turn, holding the other two constant.

The first part, of the first term, is a function of a function, so it throws out another partial differential of A^{k}. This then gives us three terms in the triple-derivative.

The second term, of the second differential, contained two parts. The first part, when differentiated, throws out another partial differential of A^{k}. Finally, the second part of the second term gives us a third-order differential in A^{i}, when differentiated with respect to . So the second term in the second derivative gives us two more terms in our triple-derivative.

## The Proof of the Pudding

The triple derivative, above, has five terms. All five terms contain i. Three of the terms contain j. Two of the terms contain k. We hold i and k constant and sum over j. We then, holding i constant, and changing to the next k, sum over j all over again.

Exercise 35, in Pavel’s book asks us to find

This would be

Fortunately, this is included in the model anwers provided by *David Sulon*. David does his differentiation in the reverse order to myself (the results will be the same). I therefore want to plug the appropriate ones, twos and threes into our tensor form of the triple derivative for

Thus .

To assist us, we re-arrange the triple differential as:~

Here, I have separated out the tensor form of the triple derivative into its five terms. All of the terms contain i, the first three terms contain j and only the first and fifth term contain k. So what we need to do is, for each of the three i-values, sum over the three j-values and then sum over the three k-values. Then start again for the next i-value.

It is worth making a couple of checks before we begin:~

- For each term, the A’s in the numerator are cancelled by the a’s in the denominator
- Each term has F in the numerator and in the denominator

Therefore, taking a deep breath:~

For each i, I will sum over j, for the first three terms, and then sum over k, for the two terms containing k.

For i = 1 :~

For i = 2 :~

For i = 3 :~

For all those folk following my deliberations here, I apologise for the haphazzard way that it seems to be coming together (23rd June 2018) 🙂

For Exercise (35) *David Sulon* has, after correcting for a couple of typos, and reversing the order, of differentiation (as mentioned earlier):~

What I will do now is carry out the differentiation on the first of these three terms and compare the result with my tensor form, for i = 1, above :~

If you convert the A, B and C, in the equation above, to the corresponding indexed terms in the tensor form, for i = 1, or visa versa, you will see that this is indeed the triple derivative term, which I worked out above. I found this more difficult than working with the tensor form, which is one of the reasons for using Tensors. It did take me a while to get the three i-components of the tensor form right. Again, one of the reasions for using tensors is to just get going with whatever maths you’re doing and then sort out the algebra later. Much better than trying to do both at the same time.

I hope that this has been worthwhile, I would really appreciate any comments. Thanks for visiting. John

John,

Thank you so much for these postings! I haven’t done the Jacobian posting yet, but the other two are great. At the end of the first posting, you showed how to differentiate using the tensor-form only. This was profoundly helpful to me, since I’d been laboriously working through the process similar to Sulon’s solution manual method. This was unmanageable and too bloated a process to do for #36 and #38, so your differentiation using the tensor-form was essential. Your amazing efforts (so much LaTeX and attention to detail) in your second posting to justify and confirm this method were truly magnificent.

The Grinfeld book’s approach is great, and I greatly appreciate the Sulon solutions. The book’s errors are truly a ridiculous abomination though — they’ve added so much confusion for me, and the errata even have errors. The thrash to work though the mess has helped me understand things though, but the time drain is bothersome. Your postings have been crucial to help me work through the difficulties of Chapter 4 so far. Thank you so much from Florida!

Tremendous! Thank you very much for your welcome comment Sanjay. Long time since I been to sunny Florida! give it my love. 🙂

I meant #35 and #38. (My error) 🙂

Also for my original comment, if you could modify my name so just “sanjay” is displayed, I’d be grateful. Best wishes!