User:Mim.cis/sandbox2

Sobolev Space and Reproducing Kernel Hilbert Space
The Sobolev embedding theorem dictates how much differentiation is required so that the space of vector field is continuously embedded in 1-times differentiable vector fields vanishing at infinity $$ V \rightarrow C_0^1({\mathbb R}^3). $$ The Hilbert space of vector field $$ (V,\| \cdot \|_V ) $$ is constructed with inner-product defined via a one-to-one differential operator $$ A : V \rightarrow V^* $$, with $$ V^* $$ the dual space. The dual space contains generalized vector functions or distributions $$ \sigma \in V^* $$, for $$ X \subset {\mathbb R}^3 $$, then $$ \sigma= (\sigma_1,\sigma_2,\sigma_3) \in V^*,$$ with $$ ( \sigma |f ) \doteq \int_X \sum_{i=1}^3 f_i(x) \sigma_i (dx) \, \ for \ f \in V \. $$ The norm on $$ V $$ becomes


 * $$ \langle v, v^\prime \rangle_V \doteq (Av | v^\prime ), \ v,v^\prime \in V, Av \in V^* \.

$$

The operator $$ A $$ is selected to dominate a Sobolev norm of sufficient order required for the smooth embedding condition. Sobolev Space The necessary smoothness in derivative is determined by backround space dimension; for $$ X \subset {\mathbb R}^3$$, the Sobolev norm is given by



\| v_t \|_{H_0^m({\mathbb R}^3)}^2 \doteq \sum_{i=1}^3 \int_X |\sum_{\alpha:\alpha_1+\alpha_2+\alpha_3 \leq m} \frac{\partial^{\alpha_1+\alpha_2+\alpha_3}}{\partial x_1^{\alpha_1} \partial x_2^{\alpha_2} \partial x_3^{\alpha_3}} v_{it}(x) |^2 dx \. $$
 * As Dupuis et al. demonstrated, the condition for the number of m-derivatives in the Sobolev norm requires for volumes the number of integral-squared derivatives $$m \geq \frac{d}{2}+s$$ for s-continuous derivatives in $${\mathbb R}^d$$. For $$d=3, {\mathbb R}^3$$ with $$s=1$$ continuouos derivative, we construct the norm of $$V$$ to dominate the Sobolev space norm with 3-derivatives $$\|v_i\|_V \geq \| v_i \|_{H_0^3(R^3)}$$, with Sobolev norm standardly defined $$\|v_i\|_{H_0^m({\mathbb R}^3)}^2

\doteq \sum_{\alpha+\beta+\gamma \leq 3} \int_X |\frac{\partial^{\alpha+\beta+\gamma} v_i(x)}{\partial x^\alpha \partial x_2^\beta \partial x_3^\gamma}|^2dx$$. For$$X={\mathbb R}^2, d=2,$$ for $$s=1$$ continuous derivative requires $$m=2$$; then for $$L=(\nabla^2-c)$$ is sufficient with $$\| v \|_V^2 \doteq \int_X |Lv|^2 dx$$. For$$X={\mathbb R}^3, d=3,$$ for $$s=1$$ continuous derivative requires $$m\geq 2.5$$; then for $$L=(\nabla^2-c)^{1.5}$$ is sufficient with $$\| v \|_V^2 \doteq \int_X |Lv|^2 dx$$.

We choose our Hilbert space $$ (V,\| \cdot \|_V) $$ with norm so that it dominates the Sobolev norm of proper order, for $$ {\mathbb R}^3 $$ then $$ \| v_t \|_V^2 \doteq (Av_t|v_t) \geq \| v_t \|_{H_0^m({\mathbb R}^3)}^2 ; $$ finiteness of $$ (Av_t | v_t) < \infty $$ implies the Sobolev norm is finite. For d-dimensional backround space $$ X \subset {\mathbb R}^d$$, the Sobolev norm associated to the d-components $$ v_{it},i=1,\dots, d $$, the necessary condition for smooth embedding with k-derivatives, $$V \subset C_0^k({\mathbb R}^d,{\mathbb R}^d)$$ must satisfy $$ m \geq s+d/2 $$

For 1-continuous derivative, the backround space $$ X \subset {\mathbb R}^2 $$, then $$ m\geq 2 $$; for $$ X \subset {\mathbb R}^3 $$, then $$ m \geq 2.5 $$.

For modelling purposes, $$ (V,\| \cdot \|_V) $$ is modelled as a reproducing Kernel Hilbert space with the reproducing kernel $$ K = A^{-1} $$ which smooths the distributions mapping $$ K: V^* \rightarrow V $$, with $$ (\sigma | K \sigma^\prime) = \langle K \sigma, K \sigma^\prime \rangle_V \. $$ In particular this gives the inverse property $$ (Av|w) = \langle K(Av), w \rangle_V = \langle v, w \rangle_V $$ As an operator $$K$$ is matrix-valued, defined by the matrix with kernel  $$ (K\sigma (\cdot))_i \doteq \sum_j \int_X k_{ij}(\cdot, x) \sigma_j(dx) \. $$

Our smoothness condition for smooth flows of the inverse requires control of the first derive $$ v_t \rightarrow Dv_t(x), x \in X $$ which is true for smooth kernel $$ k_{ij}(x,y) $$ in both variables $$ x,y $$.

We require $$ v \in V $$ a Hilbert space which continuously embeds in 1-times differentiable vector fields vanishing at infinity giving the group of diffeomorphisms generated from smooth flows:



Diff_V \doteq \{\varphi=\phi_1: \dot \phi_t = v_t \circ \phi_t, \phi_0 = id, \int_0^1 \|v_t \|_V dt < \infty \}. $$

Eulerian Momentum Density Evolution
For the smooth vector density case, with $$ Av_t = \mu_t dx $$ with
 * $$ (Av_t | w) =\int_X \mu_t \cdot w dx \ ,w \in V$$ then the Euler equation exists pointwise (not just weakly as in ....) as first derived for the density case appear in

. Substituting the vector density $$ \mu_t $$ into the weak equation $$ \int_X \frac{d}{dt} \mu_t \cdot w dx + \int_X \mu_t \cdot (Dv_t \, w - Dw\, v_t ) dx= 0 $$, and integrating by parts gives the pointwise Euler equation for the density case:

Equation ($$) is the pointwise Euler-equation for geodesics.

The original derivation takes a first order perturbation of the flow at point $$ \phi \in Diff_V $$ according to
 * $$ \phi_t^\epsilon \doteq

(id + \epsilon \delta \phi) \circ \phi = \phi+\epsilon \delta \phi \circ \phi $$ fixing the boundary so that $$ \delta\phi_0=\delta \phi_1=0 \. $$ Calculating the first order perturbation on the vector fields $$ v^\epsilon = v + \epsilon \delta v \ $$ gives $$ \delta v $$ in terms of the derivative in time of the group perturbation $$ \frac{\partial}{\partial t} \delta \phi_t $$ adjusted by the Lie-bracket correction term defined as
 * $$ [v,w]\doteq (Dv)w - (Dw)v,  v,w \in V ;$$ the Lie-bracket in this function setting involves the Jacobian matrix,

unlike the matrix group. To calculate $$ \delta v $$ define the solution $$ \phi^\epsilon $$ to satisfy the flow
 * $$ \frac{d}{dt} \phi_t^\epsilon = v_t^\epsilon \circ \phi_t^\epsilon, \phi_0^\epsilon = id ,$$

giving

\frac{d}{dt} \phi_t^\epsilon = \frac{d}{dt} \phi_t + \epsilon \frac{d}{dt}(\delta \phi_t \circ \phi_t ) =v_t \circ \phi_t + (D \delta \phi_t)\circ \phi_t v_t \circ \phi_t \. $$ We also have
 * $$ \dot \phi_t^\epsilon = (v_t + \epsilon \delta v_t) \circ (\phi_t + \epsilon \delta \phi_t \circ \phi_t) \simeq v_t \circ \phi_t +\epsilon (Dv_t)\circ \phi_t \delta \phi_t \circ \phi_t + \delta v_t \circ \phi_t +o(\epsilon) \.

$$ Equating the above two equations gives
 * $$ \delta v_t = \frac{\partial}{\partial t} \delta \phi_t - [v_t,\delta \phi_t] \ . $$

Taking the first variation gives the result
 * $$ \frac{d}{d \epsilon} J(\phi^\epsilon)|_{\epsilon=0} = \int_0^1 (Av_t | \delta v_t ) dt =\int_0^1 (Av_t | \delta \phi_t - [v_t,\delta \phi_t] ) dt \, \text{for all} \delta \phi_t \in V \.

$$ Defining the adjoint operator $$ ad_v:V \mapsto V $$ given by
 * $$ ad_v(w) \doteq [v,w] \doteq (Dv)w-(Dw)v \in V $$,

then the adjoint operator $$ ad_v^*: V^* \mapsto V^* $$ gives the Euler equation on general distributions $$ Av \in V^* $$; for all smooth $$ w \in V ,$$
 * $$ ( \frac{d}{dt} Av_t + ad_{v_t}^* (Av_t) | w ) = 0 \, $$

implying the Euler equation for all $$ t \in [0,1], $$

Equation ($$) is the weak version of Euler-equation for geodesics through the group. This equation has been called EPDiff, Euler-Poincare equation for diffeomorphisms and has been studied in the context of fluid mechanics

For the smooth vector density case, with $$ Av_t = \mu_t dx $$ with
 * $$ (Av_t | w) =\int_X \mu_t \cdot w dx \ ,w \in V$$

then the Euler equation exists pointwise (not just weakly as in ....). The first derivation of the Euler-lagrange equation but for the density case appear in. Taking the weak equation
 * $$ ( \frac{d}{dt} Av_t|w) + (Av_t | (Dv)w-(Dw)v ) = 0 \,  w \in V, $$

and substituing with the vector density $$\mu_t $$ gives
 * $$ \int_X ( \frac{d}{dt} \mu_t + (Dv_t)^T \mu_t +(D\mu_t)v_t + (\nabla \cdot v) \mu_t ) \cdot w dx = 0 \,  w \in V, $$

This gives the pointwise Euler equation for the density case:

Equation ($$) is the pointwise Euler-equation for geodesics.