Partial derivative in two dimensions

Question

I am struggling with section 3.3 of the following thesis https://smartech.gatech.edu/xmlui/bitstream/handle/1853/29610/grigo_alexander_200908_phd.pdf. Page 21 is fine, then the problems occur in pages 22 and 23 which I believe are mainly down to notation.

"For $\epsilon_0 > 0$ consider a family of $C^5$ function $L_{\epsilon}: U \times U \rightarrow \mathbb{R^2}$ for $|\epsilon|< \epsilon_0$ which satisfy

$ \displaystyle \partial_{\epsilon}\Big|_{\epsilon=0}L_{\epsilon}(s,s_1)=C \frac{s^4+s_1^4}{24}+O_5(s,s_1)$ and $\partial_s \partial_{s_1}L_0(0,0) \neq 0$ for some $C \neq 0$."

What do the $\partial_{\epsilon}$ and $\partial_s \partial_{s_1}$ mean? What do they act on?

Then $L_{ij} := \partial^i_s \partial^j_{s_1}L_0(s,s_1)$. What are $\partial^i_s$ and $\partial^j_{s_1}$, what do $i$ and $j$ correspond to?

In particular on page 23 I cannot see how $\partial_s L_\epsilon(s, S_{\epsilon}(s,y))=y$ and $\partial_{s_1}L_{\epsilon}(s,S_{\epsilon}(s,y))=Y_{\epsilon}(s,y)$ are derived? I seem to be missing some identities used.

Finally, on page 24, I cannot see how $\partial_{\epsilon}\Big|_{\epsilon=0}A_{\epsilon}$ only the third order term Im c_{21}? Why is it third order, why arent the other terms third order?

Answer 1

It seems that there is a typo. It should be $L_\epsilon : U \times U \to \mathbb{R}$.

Reasons for that: It is written that $T_\epsilon(s, y) \equiv (S_\epsilon(s, y), Y_\epsilon(s,y))$ is the area-preserving maps. And after that: $T_\epsilon(s_*, y_*) = (s_*, y_*)$. This means that $T_\epsilon : \mathbb{R}^2 \to \mathbb{R}^2$. And this means that $S_\epsilon: \mathbb{R}^2 \to \mathbb{R}$ and $Y_\epsilon: \mathbb{R}^2 \to \mathbb{R}$. I.e. $T_\epsilon$ is a real map.

At the same time, $L_\epsilon$ is a generating function (this implicitly says that $L_\epsilon$ maps to $\mathbb{R}$).

Moreover, for maps this guy uses Jacobian as derivatives (and it is ok for vector-valued functions) (see Proposition 3.3.2. there).

The same story is for the $\partial_\epsilon$: for $T_\epsilon$ this derivative is a vector (since $T_\epsilon$ is a vector), but for $S_\epsilon$ it is a scalar (see p.23).

These are just my observations. I'm not completely sure that they are correct.

Answer 2

Since $L_\epsilon$ is a vector-valued function (it maps $U\times U$ to $\mathbb{R}^2$), you can write $L_\epsilon = (L_\epsilon^1, L_\epsilon^2)$, where each $L_\epsilon^i: U\times U \to \mathbb{R}$ for $i=1,2$.

For each $L_\epsilon^i$ you can find the partial derivatives in the standard way. Therefore, $$ \partial_\epsilon L_\epsilon(s, s_1) = (\partial_\epsilon L_\epsilon^1(s, s_1), \partial_\epsilon L_\epsilon^2(s, s_1)), $$ and the same for $\partial_s \partial_{s_1} L_\epsilon(s, s_1)$.

This means that $C$ should be a vector $C = (C_1, C_2)$, and $C \neq 0$ means $(C_1, C_2) \neq (0, 0)$.