Prove of the shape-derivative identity relating the shape and material derivative of a shape-dependent function I've started reading about shape optimization. Most of the concepts I've encountered so far (such as the shape derivatives of domain and boundary integrals and the corresponding) seem to be complex, but turned out to be quite simple. However, I really struggle to wrap my head around the different notions of "derivatives" for a "shape-dependent" function $y$.
The setting is as follows: Let

*

*$d\in\mathbb N$;

*$D\subseteq\mathbb R^d$ be open and $\mathcal A\subseteq 2^D$ with $D\in\mathcal A$;

*$E_\Omega\subseteq\mathbb R^{\Omega}$ be a $\mathbb R$-Banach space for $\Omega\in\mathcal A$ and $E:=\bigcup_{\Omega\in\mathcal A}E_\Omega$;

*$y:\mathcal A\to E$ with $$y(\Omega)\in E_\Omega\;\;\;\text{for all }\Omega\in\mathcal A;\tag1$$

*$\tau>0$, $T_t$ be a $C^1$-diffeomorphism from $U$ onto an open subset of $\mathbb R^d$ for $t\in[0,\infty)$ and $$V:=\bigcup_{t\in[0,\:\tau)}T_t(D);$$

*$v:[0,\tau)\times V\to\mathbb R^d$ be differentiable in the second argument with $$v\left(t,T_t(x)\right)=\frac{\partial T}{\partial t}(t,x)\;\;\;\text{for all }(t,x)\in[0,\tau)\times D;\tag2$$

*$\Omega\in\mathcal A$ and $\Omega_t:=T_t(\Omega)$ for $t\in[0,\tau)$.

Now the "shape derivative* is defined as follows:

Definition 1 (shape derivative) Let $Y:[0,\tau)\to E_d$ with $$\left.Y(t)\right|_{\Omega_t}=y(\Omega_t)\;\;\;\text{for all }t\in[0,\tau).$$ Then $y$ is called shape differentiable at $\Omega$ in direction $v$ if $Y$ is Fréchet differentiable at $0$. In that case, $$y'(\Omega;v):=\left.Y'(0)\right|_{\Omega}\tag4.$$
(Please note that we most probably need to assume a certain regularity (at least continuity) of the time-dependence of $Y$ (and most probably of $T$ as well). I've omitted them, cause it's part of my question what we need to assume precisely.)

The second definition is given by the "material derivative*:

Definition 2 (material derivative) $\dot y(\Omega;v)\in E_\Omega$ is called **material derivative of $y$ at $\Omega$ in direciont$ v$ if $$y(\Omega_t)\circ\left.T_t\right|_{\Omega}\in E_\Omega\;\;\;\text{for all }t\in[0,\tau)\tag5$$ and $$[0,\tau)\to E_\Omega\;,\;\;\;t\mapsto y(\Omega_t)\circ\left.T_t\right|_{\Omega}\tag6$$ is Fréchet differentiable at $0$ with derivative equal to $\dot y(\Omega;v)\in E_\Omega$, i.e. $$\frac{y(\Omega_t)\circ\left.T_t\right|_{\Omega}-y(\Omega)}t\xrightarrow{t\to0+}\dot y(\Omega;v)\tag7.$$


Question 1: What do we need to assume in order to show that $(3)$ is well-defined, i.e. independent of the choice of $Y$?
Question 2: How can we relate the shape and material derivative?

I guess we need to assume that there is a continuous linear $$\iota_A:E_A\to E_D$$ for all $A\in\mathcal A$. Let $$y_t:=y(\Omega_t)\circ\left.T_t\right|_{\Omega}\;\;\;\text{for }t\in[0,\tau).$$ Then we could write $$\frac{Y(t)-Y(0)}t=\frac{Y(t)-\iota_\Omega y_t}t+\frac{\iota_\Omega y_t-Y(0)}t\;\;\;\text{for all }t\in(0,\tau)\tag8.$$ If the answer to question 1 is positive, then we could assume $$Y(t)=\iota_{\Omega_t}y(\Omega_t)\;\;\;\text{for all }t\in[0,\tau)\tag9$$ and, assuming $y$ has a material derivative at $\Omega$ in direction $v$, we could conclude $$\frac{\iota_\Omega y_t-Y(0)}t=\iota_\Omega\frac{y_t-y(\Omega)}t\xrightarrow{t\to0+}\iota_\Omega\dot y(\Omega;v)\tag{10}.$$

Note that there is a proof of the "shape-derivative identity" in this paper, but I think their proof is missing rigor and they seem to assume $E_\Omega$ is a closed subspace of $L^1(\Omega)$:



 A: The classical material derivative $D\varphi/Dt$ of a test function $\varphi \in C_c^\infty(\mathbb{R}_+\times D)$ is obtained by setting
$$
\dfrac{D\varphi}{Dt}(x) := \dfrac{\partial\tilde\varphi}{\partial t}(0,x)\;\;\mbox{with}\;\; \tilde\varphi(t,x) = \varphi(t,T_t(x))
$$
for $x\in D$. Expanding out using the chain rule, we have
$$
\dfrac{D\varphi}{Dt}(x) = \dfrac{\partial \phi}{\partial t}(0,x)+ \sum_{i=1}^d v^i(x)\dfrac{\partial \phi}{\partial x^i}(0,x)\;\;(x\in D).
$$
with $v(x) = \lim_{t\to 0}t^{-1}(T_t(x)-x)$. I'm imagining here that it's $(T_t)_{t>0}$ that has been prescribed, but one can go in the other way too (i.e. go from a vector field to a flow rather than a flow to a vector field).
The 'shape identity' is the natural generalisation of the formula above to distributions $y(\Omega_t)$ of the form
$$
\langle y(\Omega_t),\varphi\rangle=\int_{\Omega_t} y_{\Omega_t}(x)\varphi(x)\mathrm{d}x\;\;(\varphi\in C^\infty_c(D))
$$
with $y_{\Omega_t}\in L^1_\mathrm{loc}(\Omega_t)$ and $\Omega_t = T_t(\Omega)$. In this formulation $y_{\Omega_t}(x)$ is trying to be $\varphi(t,x)$ from the smooth formulation while the 'shape derivative' is trying to be $x\mapsto (t\mapsto y_{\Omega_t}(x))'(0)$.
Regarding your first question, it seems like the easiest ways to make everything work are:

*

*Regard everything as a distibution on $D$; or

*Make sure that $E_\Omega = \{f_{|\Omega}: f\in E_D\}$ for all $\Omega\in \mathcal{A}$, do what you need in $E_D$, then restrict back to $\Omega$.

These two approaches both let you form linear combinations and take limits 'normally', so remove the problems associated with everything living in different spaces. It's important to check with the second approach that the behavior of the limit in $\Omega$ doesn't depend on the extensions chosen, but there's a result in the reference which shows how to do that (i.e. by testing against a smooth bump supported in $\Omega$).
Regarding your comment about the reference assuming that $E_\Omega$ is closed in $L^1(\Omega)$; I'm not sure I agree - isn't the fact that quotients converge to something in $L^1$ just part of their definition?
As for minimum requirements, I think you at least want the quotients
$$
\dfrac{\langle y(\Omega_t)\circ T_t, \varphi\rangle - \langle y(\Omega),\varphi\rangle}{t}\;\;\mbox{and}\;\;\dfrac{
\langle y(\Omega_t),\varphi\rangle - \langle y(\Omega),\varphi\rangle}{t}
$$
to converge as $t\to 0$ for all test functions $\varphi$, since these are what give you the distributional 'material' and 'shape' derivatives.
