Physlib.Mathematics.VariationalCalculus.HasVarAdjDeriv

Let $F: (X \to U) \to (X \to V)$ be an operator between function spaces that possesses a variational adjoint derivative $F'$ at $u$. If $v: X \to U$ is a smooth function (of class $C^\infty$), then its image under the operator $F$, given by $F(v): X \to V$, is also a smooth function.

Let $F: (X \to U) \to (X \to V)$ be an operator between function spaces. Suppose $F$ possesses a variational adjoint derivative $F'$ at a function $u: X \to U$. Then the function $F(u): X \to V$ is infinitely differentiable (of class $C^\infty$).

Suppose an operator $F: (X \to U) \to (X \to V)$ has a variational adjoint derivative at a function $u: X \to U$. Let $\phi: \mathbb{R} \times X \to U$ be an infinitely differentiable ($C^\infty$) family of functions, where the mapping $(s, x) \mapsto \phi(s, x)$ is smooth. Then, for any fixed point $x \in X$, the function mapping the parameter $s \in \mathbb{R}$ to the value of the operator $F$ evaluated at the function $\phi(s, \cdot)$ and then evaluated at $x$, given by $$s \mapsto F(\phi(s, \cdot))(x),$$ is infinitely differentiable ($C^\infty$) with respect to $s$.

Suppose an operator $F: (X \to U) \to (X \to V)$ has a variational adjoint derivative at $u: X \to U$. Let $\phi: \mathbb{R} \times X \to U$ be an infinitely differentiable ($C^\infty$) family of functions. Let $u_0: X \to U$ be the function $u_0(x) = \phi(0, x)$ and let $\delta u: X \to U$ be the partial derivative of $\phi$ with respect to its first argument at $0$, given by $\delta u(x) = \left. \frac{\partial \phi(s, x)}{\partial s} \right|_{s=0}$. Then, for any fixed point $x \in X$, the function mapping $s' \in \mathbb{R}$ to the value of the operator $F$ evaluated at the linear variation $u_0 + s' \delta u$ and then evaluated at $x$, given by $$s' \mapsto F(u_0 + s' \delta u)(x),$$ is infinitely differentiable ($C^\infty$) with respect to $s'$.

Suppose an operator $F: (X \to U) \to (X \to V)$ has a variational adjoint derivative at the function $u: X \to U$. Then, for any infinitely differentiable (smooth) function $\delta u: X \to U$ and any fixed point $x \in X$, the function mapping $s \in \mathbb{R}$ to the value of the operator at the perturbed function $u + s \delta u$ evaluated at $x$, given by $$s \mapsto F(u + s \delta u)(x),$$ is infinitely differentiable ($C^\infty$) with respect to $s$.

Suppose an operator $F: (X \to U) \to (X \to V)$ has a variational adjoint derivative at $u: X \to U$. Let $\phi: \mathbb{R} \times X \to U$ be an infinitely differentiable ($C^\infty$) family of functions. Let $u_0: X \to U$ be the function $u_0(x) = \phi(0, x)$ and let $\delta u: X \to U$ be the partial derivative of $\phi$ with respect to its first argument at $s=0$, given by $\delta u(x) = \left. \frac{\partial \phi(s, x)}{\partial s} \right|_{s=0}$. Then, for any fixed point $x \in X$, the function mapping $s' \in \mathbb{R}$ to the value of the operator $F$ evaluated at the linear variation $u_0 + s' \delta u$ and then evaluated at $x$, given by $$s' \mapsto F(u_0 + s' \delta u)(x),$$ is differentiable with respect to $s'$.

Let $F: (X \to U) \to (X \to V)$ be an operator mapping between function spaces. Suppose $F$ is linear on smooth functions, meaning that for all smooth $\phi_1, \phi_2: X \to U$ and $c \in \mathbb{R}$, $F(\phi_1 + \phi_2) = F(\phi_1) + F(\phi_2)$ and $F(c \cdot \phi_1) = c \cdot F(\phi_1)$. Additionally, assume that for any smooth family of functions $\phi: \mathbb{R} \times X \to U$, the derivative at $s=0$ commutes with $F$ at any point $x \in X$: $$\left. \frac{d}{ds} \right|_{s=0} F(\phi(s, \cdot))(x) = F\left( \left. \frac{\partial \phi}{\partial s} \right|_{s=0} \right)(x).$$ Then, for any such smooth $\phi$ and any $x \in X$, the derivative of $F$ applied to the path $\phi(s, \cdot)$ is equal to the derivative of $F$ applied to the first-order linear approximation of $\phi$ at $s=0$: $$\left. \frac{d}{ds} \right|_{s=0} F(\phi(s, \cdot))(x) = \left. \frac{d}{ds} \right|_{s=0} F\left( y \mapsto \phi(0, y) + s \cdot \left. \frac{\partial \phi}{\partial s} \right|_{s=0}(y) \right)(x).$$

Let $F: (X \to U) \to (X \to V)$ be an operator between function spaces that is linear on the space of smooth functions, meaning $F(\phi_1 + \phi_2) = F(\phi_1) + F(\phi_2)$ and $F(c \cdot \phi) = c \cdot F(\phi)$ for any $c \in \mathbb{R}$ and smooth functions $\phi, \phi_1, \phi_2: X \to U$. Suppose $u: X \to U$ is a smooth function and $F$ has a variational adjoint $F'$. Then the linear operator that maps a variation $\delta u: X \to U$ to the directional derivative of $F$ at $u$, defined by $$\delta u \mapsto \left( x \mapsto \left. \frac{d}{ds} \right|_{s=0} F(u + s \delta u)(x) \right),$$ has $F'$ as its variational adjoint.

Let $F: (X \to U) \to (X \to V)$ be an operator between function spaces. Suppose $F$ satisfies the following conditions: 1. **Linearity:** $F$ is linear on the space of smooth functions, i.e., $F(\phi_1 + \phi_2) = F(\phi_1) + F(\phi_2)$ and $F(c \cdot \phi) = c \cdot F(\phi)$ for any $c \in \mathbb{R}$ and smooth functions $\phi, \phi_1, \phi_2: X \to U$. 2. **Smoothness:** $F$ maps any smooth family of functions $\phi: \mathbb{R} \times X \to U$ to a smooth function on $\mathbb{R} \times X$ defined by $(s, x) \mapsto F(\phi(s, \cdot))(x)$. 3. **Commutativity of Differentiation:** For any smooth family $\phi: \mathbb{R} \times X \to U$, the derivative at $s=0$ commutes with $F$ at any point $x \in X$: $$\left. \frac{d}{ds} \right|_{s=0} F(\phi(s, \cdot))(x) = F\left( \left. \frac{\partial \phi}{\partial s} \right|_{s=0} \right)(x).$$ If $u: X \to U$ is a smooth function and $F$ has a variational adjoint $F'$, then the variational adjoint derivative of $F$ at $u$ is $F'$.

Let $u: X \to U$ be a $C^\infty$ smooth function. The identity operator $F : (X \to U) \to (X \to U)$ defined by $F(\phi) = \phi$ has a variational adjoint derivative at $u$ given by the identity operator $F'(\psi) = \psi$.

Let $u: X \to U$ and $v: X \to V$ be $C^\infty$ smooth functions. The constant functional $F: (X \to U) \to (X \to V)$ defined by $F(\phi) = v$ for all $\phi$ has a variational adjoint derivative at $u$ given by the zero operator $F'(\psi) = 0$.

Let $F: (Y \to V) \to (Z \to W)$ and $G: (X \to U) \to (Y \to V)$ be operators between function spaces, and let $u: X \to U$ be a function. If $F$ has a variational adjoint derivative $F'$ at the point $G(u)$ and $G$ has a variational adjoint derivative $G'$ at the point $u$, then the composition $F \circ G$ has a variational adjoint derivative at $u$ given by the composition of the adjoint derivatives in reverse order, $G' \circ F'$.

Let $F, G: (X \to U) \to (Y \to V)$ be functionals and $u: X \to U$ be a function. Suppose that $F$ has a variational adjoint derivative $F'$ at $u$ (denoted by $\text{HasVarAdjDerivAt } F \ F' \ u$). If $F$ and $G$ agree on all smooth functions, i.e., $F(\phi) = G(\phi)$ for every smooth function $\phi: X \to U$ ($\phi \in C^\infty$), then $G$ also has $F'$ as its variational adjoint derivative at $u$.

Let $X$ and $Y$ be spaces, where $X$ is equipped with a measure that is finite on compact sets and strictly positive on non-empty open sets. Let $F: (X \to U) \to (Y \to V)$ be a functional. If $F'$ and $G'$ are both variational adjoint derivatives of $F$ at $u: X \to U$ (satisfying the `HasVarAdjDerivAt` condition), then for any test function $\phi: Y \to V$ (i.e., a smooth function with compact support), it holds that $F' \phi = G' \phi$.

Let $X$ and $Y$ be normed real inner product spaces equipped with measures, where $X$ has a measure that is finite on compact sets and strictly positive on non-empty open sets, and $Y$ is finite-dimensional. Let $F: (X \to U) \to (Y \to V)$ be a functional. If $F$ has two variational adjoint derivatives $F'$ and $G'$ at a point $u: X \to U$ (denoted by $\text{HasVarAdjDerivAt } F \ F' \ u$ and $\text{HasVarAdjDerivAt } F \ G' \ u$), then for every smooth function $\phi: Y \to V$ ($\phi \in C^\infty$), it holds that $F' \phi = G' \phi$.

Let $X$ be a space equipped with a measure that is finite on compact sets. Let $F: (X \to U) \to (X \to V)$ and $G: (X \to U) \to (X \to W)$ be operators. Suppose $F$ has a variational adjoint derivative $F'$ at the function $u: X \to U$, and $G$ has a variational adjoint derivative $G'$ at $u$. Then the product operator $H: (X \to U) \to (X \to V \times W)$, defined by $H(\phi)(x) = (F(\phi)(x), G(\phi)(x))$, has a variational adjoint derivative $H'$ at $u$ given by: $$H'(\psi)(x) = F'(\psi_1)(x) + G'(\psi_2)(x)$$ where $\psi_1: X \to V$ and $\psi_2: X \to W$ are the component functions of $\psi: X \to V \times W$ such that $\psi(x) = (\psi_1(x), \psi_2(x))$.

Suppose an operator $F: (X \to U) \to (X \to W \times V)$ has a variational adjoint derivative $F'$ at a function $u: X \to U$. Then the operator that maps a function $\phi: X \to U$ to the first component of $F(\phi)$, given by $x \mapsto (F(\phi)(x))_1$, also has a variational adjoint derivative at $u$. This variational adjoint derivative is the operator that maps a function $\psi: X \to W$ to the value of $F'$ evaluated at the function $x' \mapsto (\psi(x'), 0)$.

Suppose an operator $F: (X \to U) \to (X \to W \times V)$ has a variational adjoint derivative $F'$ at a function $u: X \to U$. Then the operator that maps a function $\phi: X \to U$ to the second component of $F(\phi)$, given by $x \mapsto (F(\phi)(x))_2$, also has a variational adjoint derivative at $u$. This variational adjoint derivative is the operator that maps a function $\psi: X \to V$ to the value of $F'$ evaluated at the function $x' \mapsto (0, \psi(x'))$.

Let $u: \mathbb{R} \to U$ be a $C^\infty$ smooth function. Let $F$ be the operator that maps a function $\phi: \mathbb{R} \to U$ to its derivative $\frac{d\phi}{dx}$. Then the variational adjoint derivative of $F$ at $u$ is the operator that maps a function $\phi$ to its negative derivative $-\frac{d\phi}{dx}$.

Let $F: (\mathbb{R} \to U) \to (\mathbb{R} \to V)$ be an operator between function spaces. If $F$ has a variational adjoint derivative $F'$ at the function $u: \mathbb{R} \to U$, then the operator mapping $\phi$ to the derivative $\frac{d}{dx}(F(\phi))$ has a variational adjoint derivative at $u$ given by the mapping $\psi \mapsto F'(-\frac{d\psi}{dx})$.

Let $U$ and $V$ be complete generalized inner product spaces over $\mathbb{R}$. Let $u: X \to U$ be a smooth ($C^\infty$) function. Let $f: X \to U \to V$ be a map such that the uncurried function $(x, y) \mapsto f(x, y)$ is $C^\infty$, and for each $x \in X$, the map $f(x, \cdot): U \to V$ has an adjoint Fréchet derivative $f'(x, u(x))$ at the point $u(x)$. Consider the operator $F$ acting on functions $\phi: X \to U$ defined by pointwise application: $$(F\phi)(x) = f(x, \phi(x)).$$ Then, the variational adjoint derivative of $F$ at $u$ is the operator that maps a function $\psi: X \to V$ to the function: $$x \mapsto f'(x, u(x))(\psi(x)).$$

Suppose $F$ is an operator mapping functions from $X \to U$ to functions from $X \to V$. If $F'$ is the variational adjoint derivative of $F$ at $u$, then the operator mapping $\phi$ to $-F(\phi)$ has a variational adjoint derivative at $u$ given by the mapping $\psi \mapsto -F'(\psi)$.

Suppose $F$ and $G$ are operators mapping functions from $X \to U$ to functions from $X \to V$. If $F$ has a variational adjoint derivative $F'$ at $u: X \to U$, and $G$ has a variational adjoint derivative $G'$ at $u$, then the sum operator defined by $\phi \mapsto F(\phi) + G(\phi)$ has a variational adjoint derivative at $u$ given by the operator $\psi \mapsto F'(\psi) + G'(\psi)$.

Let $\iota$ be a finite index set. For each $i \in \iota$, let $F_i$ be an operator mapping functions from $X \to U$ to functions from $X \to V$. Suppose that for each $i \in \iota$, $F_i$ has a variational adjoint derivative $F'_i$ at a $C^\infty$ smooth function $u: X \to U$. Then the sum operator $S$, defined by $(S\phi)(x) = \sum_{i \in \iota} F_i(\phi)(x)$, has a variational adjoint derivative at $u$ given by the operator $S'$, where for any function $\psi: X \to V$, $$S'(\psi)(x) = \sum_{i \in \iota} F'_i(\psi)(x).$$

Let $F, G: (X \to U) \to (X \to \mathbb{R})$ be operators. Suppose $F$ and $G$ have variational adjoint derivatives $F'$ and $G'$ at a function $u: X \to U$, respectively. Then the pointwise product operator $H$, defined by $(H\phi)(x) = F(\phi)(x) \cdot G(\phi)(x)$, has a variational adjoint derivative at $u$ given by the operator $H'$, where for any function $\psi: X \to \mathbb{R}$, $$H'(\psi)(x) = F'(x \mapsto \psi(x)G(u)(x))(x) + G'(x \mapsto F(u)(x)\psi(x))(x).$$ In more concise notation, the variational adjoint derivative is $H'(\psi) = F'(\psi \cdot G(u)) + G'(\psi \cdot F(u))$, where the products inside the arguments are pointwise products of functions.

Let $F: (X \to U) \to (X \to \mathbb{R})$ be an operator. Suppose $F$ has a variational adjoint derivative $F'$ at a function $u: X \to U$. For any constant scalar $c \in \mathbb{R}$, the operator $H$ defined by the pointwise multiplication $(H\phi)(x) = c \cdot F(\phi)(x)$ has a variational adjoint derivative $H'$ at $u$, given by: $$H'(\psi) = F'(c \cdot \psi)$$ where $c \cdot \psi$ denotes the function $x' \mapsto c \psi(x')$.

Let $X$ and $U$ be spaces. Let $f: X \to \mathbb{R}$ be an infinitely differentiable function ($f \in C^\infty(X, \mathbb{R})$). Suppose an operator $F: (X \to U) \to (X \to \mathbb{R})$ possesses a variational adjoint derivative $F'$ at a function $u: X \to U$. Then the operator $H$, defined by the pointwise product $(H\phi)(x) = f(x)F(\phi)(x)$, has a variational adjoint derivative at $u$ given by the operator $H'$, where for any function $\psi: X \to \mathbb{R}$: $$H'(\psi) = F'(f \cdot \psi)$$ Here, $f \cdot \psi$ denotes the pointwise product $x' \mapsto f(x')\psi(x')$.

Let $X$ be a finite-dimensional real vector space equipped with an additive Haar measure (volume). Let $u: X \to U$ be a smooth function. For a fixed vector $dx \in X$, consider the operator that maps a function $\phi: X \to U$ to its Fréchet derivative at $x$ in the direction $dx$, denoted by $D\phi(x)[dx]$. The variational adjoint derivative of this operator at $u$ is the operator that maps a function $\psi$ to the negative of its directional derivative, $-D\psi(x)[dx]$.

Let $X$ be a finite-dimensional real vector space equipped with an additive Haar measure. Let $F: (X \to U) \to (X \to V)$ be an operator between function spaces, and let $F'$ be its variational adjoint derivative at a point $u$. For a fixed vector $dx \in X$, consider the operator that maps a function $\phi$ to the directional derivative of $F(\phi)$ in the direction $dx$, denoted by $x \mapsto D(F(\phi))(x)[dx]$. The variational adjoint derivative of this operator at $u$ is given by the operator that maps a function $\psi$ to $F'(\psi^*)$, where $\psi^*(x') = -D\psi(x')[dx]$ is the negative directional derivative of $\psi$.

Let $u: \text{Space } d \to \mathbb{R}$ be an infinitely smooth ($C^\infty$) function. Consider the operator that maps a scalar function $\phi: \text{Space } d \to \mathbb{R}$ to its gradient field $\nabla \phi$. The variational adjoint derivative of this gradient operator at $u$ is the operator that maps a vector field $\psi$ to the negative of its divergence, $-\text{div} \psi$.

Let $u: \text{Space } d \to \mathbb{R}$ be an infinitely smooth ($C^\infty$) function. The variational adjoint derivative of the gradient operator $F$, defined by $F(\phi) = \nabla \phi$ for scalar functions $\phi: \text{Space } d \to \mathbb{R}$, at the point $u$ is the negative divergence operator $F'$, defined by $F'(\psi) = -\text{div} \psi$ for vector fields $\psi$.

Let $d$ be a natural number representing the dimension of the space, and let $u: \text{Space } d \to \mathbb{R}^d$ be a smooth ($C^\infty$) function. The variational adjoint derivative of the divergence operator $\phi \mapsto \text{div } \phi$ at $u$ is the operator that maps a scalar field $\psi$ to its negative gradient $-\nabla \psi$.