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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:

1. Regard everything as a distibution on $D$; or
2. 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?

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:

1. Regard everything as a distibution on $D$; or
2. 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.

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 $x\mapsto (t\mapsto y_{\Omega_t}(x))'(0)$.

The proof of the shape identity relies on the fact that

$$ \int_\Omega f_t(T_t(x))\varphi(x)\mathrm{d}x = \int_{\Omega_t} f_t(x)\varphi(T_t^{-1}(x))|\mathrm{det}(T_t'(x)^{-1})|\mathrm{d}x $$

for $f_t\in L^1_\mathrm{loc}(\Omega_t)$ and $\varphi\in C^\infty_c(D)$. This lets us transfer the material derivative to $\varphi$ via

$$ \left\langle \dfrac{1}{t}(f_t-f_0),\varphi \right\rangle= \int_{\Omega_t} f_t(x)\left(\dfrac{\varphi(T_t^{-1}(x))|\mathrm{det}(T_t'(x)^{-1})|-\varphi(x)}{t}\right)\mathrm{d}x $$

The easiest case is when $T_t$ is volume-preserving, so that the determinant term disappears, we don't have to worry about the spatial regularity of $T_t$, and the result follows by dominated convergence using the identity for test functions. The next easiest case is when $T_t$ is smooth and orientation preserving, which seems to require a 'partition of zero' before taking limits.

Regarding your first question, it seems like the easiest ways to make everything work are:

1. Regard everything as a distibution on $D$ and proceed as above; or
2. 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 the linear combinations and take the limits, 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?

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:

1. Regard everything as a distibution on $D$; or
2. 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?

Just for my own sanity, I'm going to write $y_\Omega$ for what you call $y(\Omega)$, and keep $y(\Omega)$ for the associated distribution, so that

$$ \langle y(\Omega),\varphi\rangle=\int_\Omega y_\Omega(x)\varphi(x)\mathrm{d}x $$

for $\varphi\in C^\infty_c(D)$ and $\Omega\in \mathcal{A}$. I've tacitly assumed here that $E_\Omega\subset L^1_\mathrm{loc}(\Omega)$, but that's not a major restriction.

As for other asumptions, I think we probably want each $T_t$ to be $C^1$ with a $C^1$ inverse, and for the limit $v(x) = \lim_{t\to 0}t^{-1}(T_t(x)-x)$ to exist for all (or maybe just almost all) $x\in \Omega$.

The shape derivative $y'(\Omega)(v)$ of $y$ relative to $v$ is then defined by the requirement that

$$ \langle y'(\Omega)(v), \varphi\rangle =\lim_{t\to 0} \dfrac{ \langle y(\Omega_t),\varphi\rangle - \langle y(\Omega),\varphi\rangle}{t} $$

for all test functions $\varphi$.

On the other hand, the material derivative $\dot y(\Omega)(v)$ of $y$ relative to $v$ is defined by the requirement that

$$ \langle \dot y(\Omega)(v),\varphi\rangle = \lim_{t\to 0}\dfrac{\langle y(\Omega_t)\circ T_t, \varphi\rangle - \langle y(\Omega),\varphi\rangle}{t} $$

for all test functions $\varphi$. The pullback here is to be understood in the distributional sense, i.e.

$$ \langle y(\Omega_t)\circ T_t, \varphi\rangle=\langle y(\Omega_t), |\mathrm{det}(T_t')|^{-1}\varphi\circ T_t^{-1}\rangle $$

Regarding your first question, it seems like the easiest ways to make everything work are:

1. Regard everything as distributions on $D$; or
2. 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 the linear combinations and take the limits, 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 'weak' and 'strong' formulations, I do think you're going to have to consider 'weak' formulations eventually, at which point you'll probably want to use the fact that

$$ \int_\Omega f(T_t(x))\varphi(x)\mathrm{d}x = \int_{\Omega_t} f(x)\varphi(T_t^{-1}(x))|\mathrm{det}(T_t'(x)^{-1})|\mathrm{d}x $$

to transfer the internal differentiation from non-smoooth $f$ to a test function $\varphi$. With weak formulations, it seems like everything gets a bit nicer if you assume that $T_t$ is volume preserving (since then you don't have the Jacobian hanging around complicating your integrals).

All we're really trying to generalise here is the identity

$$ \dfrac{\partial \tilde \phi}{\partial t}(0,x) = \dfrac{\partial \phi}{\partial t}(0,x)+ \sum_{i=1}^d v^i(x)\dfrac{\partial \phi}{\partial x^i}(0,x) $$

satisfied by $\phi\in C^1(\mathbb{R}\times \mathbb{R}^d)$ when $\tilde \phi(t,x)=\phi(t,T_t(x))$. The thing on the left is the 'material derivative' (sometimes written $D\phi/Dt$) of $\phi$. The 'shape derivative' is trying to be the first term on the right.

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 $x\mapsto (t\mapsto y_{\Omega_t}(x))'(0)$.

The proof of the shape identity relies on the fact that

$$ \int_\Omega f_t(T_t(x))\varphi(x)\mathrm{d}x = \int_{\Omega_t} f_t(x)\varphi(T_t^{-1}(x))|\mathrm{det}(T_t'(x)^{-1})|\mathrm{d}x $$

for $f_t\in L^1_\mathrm{loc}(\Omega_t)$ and $\varphi\in C^\infty_c(D)$. This lets us transfer the material derivative to $\varphi$ via

$$ \left\langle \dfrac{1}{t}(f_t-f_0),\varphi \right\rangle= \int_{\Omega_t} f_t(x)\left(\dfrac{\varphi(T_t^{-1}(x))|\mathrm{det}(T_t'(x)^{-1})|-\varphi(x)}{t}\right)\mathrm{d}x $$

The easiest case is when $T_t$ is volume-preserving, so that the determinant term disappears, we don't have to worry about the spatial regularity of $T_t$, and the result follows by dominated convergence using the identity for test functions. The next easiest case is when $T_t$ is smooth and orientation preserving, which seems to require a 'partition of zero' before taking limits.

Regarding your first question, it seems like the easiest ways to make everything work are:

1. Regard everything as a distibution on $D$ and proceed as above; or
2. 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 the linear combinations and take the limits, 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?