From: Bjørn Rustad Date: Fri, 24 Oct 2014 14:55:01 +0000 (+0200) Subject: THEOREMS X-Git-Url: http://git.rustad.me/?a=commitdiff_plain;h=8e7c288351deee9437873b185277c80d4a3fc471;p=master THEOREMS --- diff --git a/theory.tex b/theory.tex index fbd2aa7..104068d 100644 --- a/theory.tex +++ b/theory.tex @@ -335,155 +335,230 @@ We are now left with finding $u \in BV(\Omega)$ that minimizes If we parametrize straight lines as shown in Figure \fixme{ref}, we can define the set of all staight lines as $\mathcal{L} = \{ (\phi, \rho) : -\phi \in [0, 2\pi), \rho \geq 0 \}$. By defining the Lebesgue measure on -this set $d\mathcal{L} = \dpdr$ we are ready to introduce the +\phi \in [0, 2\pi), \rho \geq 0 \}$. We will write a line $l_{\phi, +\rho} = l_{\nu, \rho}$ where $\nu$ is a unit vector along the line, +i.e.\ $\nu = (\sin \phi, -\cos \phi)^T$. By defining the +measure on this set $d\mathcal{L} = \dpdr$ we are ready to introduce the Cauchy--Crofton formula, which gives us a way to calculate the length of a curve by looking at the measure of the set of lines that intersect the curve. Note that the measure $d\mathcal{L}$ is invariant under rigid motions, meaning combinations of translations and rotations. \begin{theorem}[The Euclidean Cauchy--Crofton formula] - Given a curve $C$ in $\mathbb{R}^2$, the length of this curve - $\abs{C}$ is related to the set of lines $\mathcal{L}$ as follows + Given a curve differentiable $C$ in $\mathbb{R}^2$, the length of + this curve $\abs{C}$ is related to the set of lines $\mathcal{L}$ as + follows \begin{equation} - \int_\mathcal{L} n(L \cap C) \, d\mathcal{L}(L) = 2 \abs{C}, + \int_\mathcal{L} \#(l_{\phi, \rho} \cap C) \, + d\mathcal{L}(l_{\phi, \rho}) = 2 \abs{C}, \end{equation} - where $n(L \cap C)$ is the number of times the line $L$ intersects - with the curve $C$. + where $\#(l_{\phi, \rho} \cap C)$ is the number of times the line + $l_{\phi, \rho}$ intersects the curve $C$. \end{theorem} This elegant formula is very useful when we later will discretize our energy function. The set of lines $\mathcal{L}$ is then discretized depending on our choice of neighborhood, and the length of the curve $C$ can be approximated by a sum. +Our space is equipped with a metric tensor $M(p)$ such that the inner +product of two vectors in a point $p$ is calculated as $\langle a, +b\rangle_M = \langle a, M(p) b \rangle$. The length of a curve $\gamma$ +parametrized by some parameter $t$ then becomes +\begin{equation} + \abs{\gamma}_M = \int_\gamma \langle \dot{\gamma}, + M\big(\gamma(t)\big) \, \dot{\gamma} \rangle \, dt +\end{equation} +We will now prove a Cauchy--Crofton formula in this case where we have a +metric tensor in each point. \begin{theorem}[The Riemannian Cauchy--Crofton formula] - In the case of a Riemannian geometry, where the scalar product in - each point depends on a metric tensor $M(p)$ varying continuously over - our space, the Cauchy--Crofton formula becomes + In the case where the scalar product in each point depends on a + metric tensor $M(x)$ varying continuously over our space, the + Cauchy--Crofton formula becomes \begin{equation} - \abs{C}_R = \int_\mathcal{L} n(L \cap C) \, \frac{\det - M(p)}{2 \left( u_L^T \cdot M(p) \cdot u_L \right)} \, - d\mathcal{L}(L). + \abs{C}_M = \int_\mathcal{L} \sum_{p \in l_{\nu, \rho} \cap C} + \, \frac{\det M(x)}{2 \left( \nu^T \cdot M(x) \cdot \nu \right)} + \, d\mathcal{L}(l_{\nu, \rho}). \end{equation} - Here, $u_L$ is a unit vector in the point $p$ along the line - currently being integrated over. \end{theorem} \begin{proof} Assume first that our space is equipped with at constant metric - tensor $M$. The length of our curve using this Riemannian metric can - then be calculated by transforming the curve back into euclidean - space and applying the Euclidean Cauchy--Crofton formula + tensor $M$. The length of our curve using this tensor can be + calculated by transforming the curve and applying the Euclidean + Cauchy--Crofton formula \begin{align} - \abs{C}_M &= \abs{M^{\sfrac{1}{2}}C} \\ - &= \int n(L \cap M^{\sfrac{1}{2}}C) \, - d\mathcal{L}(L) \\ - &= \int n(M^{-\sfrac{1}{2}}L \cap C) \, - d\mathcal{L}(L) \\ - &= \int n(P \cap C) \abs{J(\Mhalf)}\, - d\mathcal{L}(P) + \abs{C}_M &= \int_C \sqrt{\langle \dot{C}, M \, \dot{C} \rangle} \, dt + = \int_C \sqrt{\langle \Mhalf \dot{C}, \Mhalf + \dot{C} \rangle} = \abs{M^{\sfrac{1}{2}}C} \\ + &= \int_\mathcal{L} \#(l_{\phi, \rho} \cap M^{\sfrac{1}{2}}C) \, + d\mathcal{L}(l_{\phi, \rho}) \\ + &= \int_\mathcal{L} \#(M^{-\sfrac{1}{2}}l_{\phi, \rho} \cap C) \, + d\mathcal{L}(l_{\phi, \rho}) \\ + &= \int_\mathcal{L} \#(m_{\phi, \rho} \cap C) \abs{J_M(l_{\phi, \rho})}\, + d\mathcal{L}(m_{\phi, \rho}) \end{align} - where $J(M^{\sfrac{1}{2}})$ is the Jacobian of our coordinate - transformation $P = M^{-\sfrac{1}{2}} L$, which is what we need to - find next. + where $J_M(l_{\phi, \rho})$ is the Jacobian of our coordinate + transformation $F : \mathcal{L} \to \mathcal{L}$, which maps + $l_{\phi, \rho} \mapsto \Mhalf l_{\phi, \rho}$. As $M\in \mathbb{R}^{2\times2}$ is symmetric, so is - $M^{\sfrac{1}{2}}$, and it emits an eigendecomposition $\Mhalf - = U\Lambda U^T$ where the components correspond to the following + $M^{\sfrac{1}{2}}$, and it admits an eigendecomposition $\Mhalf + = U\Sigma U^T$ where the components correspond to the following coordinate transformations \begin{align} - U(\alpha,\rho) &= (\alpha + \xi, \rho), \\ - U^T(\alpha,\rho) &= (\alpha - \xi, \rho), \\ - \Lambda &= \begin{pmatrix} + %U(\phi,\rho) &= (\phi + \xi, \rho), \\ + %U^T(\phi,\rho) &= (\phi - \xi, \rho), \\ + U(l_{\nu,\rho}) &= l_{\phi + \xi, \rho} = l_{U\nu, \rho} \\ + U^T(l_{\nu,\rho}) &= l_{\phi - \xi, \rho} = l_{U^T\nu, \rho} \\ + \Sigma &= \begin{pmatrix} \sigma_1 & 0 \\ 0 & \sigma_2 \end{pmatrix}. \end{align} As $U$ and $U^T$ correspond to rotations they do not contribute to - the Jacobian, except for changing the input angle of the $\Lambda$ - operator. Given a line + the Jacobian, except for changing the input angle of the operator + $\Lambda$ such that $J_M(l_{\phi, \rho}) = J_{\Sigma^2}(U^T l_{\phi, + \rho})$. Given a line \begin{equation} - L_{\alpha, \rho} = \begin{pmatrix} - \rho\cdot \cos \alpha \\ - \rho\cdot \sin \alpha + l_{\phi, \rho} = \begin{pmatrix} + \rho\cdot \cos \phi \\ + \rho\cdot \sin \phi \end{pmatrix} + \mathbb{R} \begin{pmatrix} - -\sin \alpha \\ - \cos \alpha + -\sin \phi \\ + \cos \phi \end{pmatrix}, \end{equation} - the $\Lambda$ operator transforms it into + the operator $\Sigma$ transforms it into \begin{equation} - \Lambda L_{\alpha, \rho} = \begin{pmatrix} - \sigma_1 \rho\cdot \cos \alpha \\ - \sigma_2 \rho\cdot \sin \alpha + \Sigma l_{\phi, \rho} = \begin{pmatrix} + \sigma_1 \rho\cdot \cos \phi \\ + \sigma_2 \rho\cdot \sin \phi \end{pmatrix} + \mathbb{R} \begin{pmatrix} - -\sigma_1 \sin \alpha \\ - \sigma_2 \cos \alpha + -\sigma_1 \sin \phi \\ + \sigma_2 \cos \phi \end{pmatrix}, \end{equation} - which is has parameters + which is the line $l_{\theta, \eta}$ with \begin{align} - \beta &= \arctan \left( \frac{\sigma_1}{\sigma_2} \tan \alpha + \theta &= \arctan \left( \frac{\sigma_1}{\sigma_2} \tan \phi \right) \\ - \nu &= \left\langle \begin{pmatrix} - \sigma_1 \rho\cdot \cos \alpha \\ - \sigma_2 \rho\cdot \sin \alpha + \eta &= \left\langle \begin{pmatrix} + \sigma_1 \rho\cdot \cos \phi \\ + \sigma_2 \rho\cdot \sin \phi \end{pmatrix}, \begin{pmatrix} - \cos \beta \\ - \sin \beta + \cos \theta \\ + \sin \theta \end{pmatrix} - \right\rangle. + \right\rangle = \sigma_1 \rho \cdot \cos \phi \cdot \cos \theta + + \sigma_2 \rho \cdot \sin \phi \cdot \sin \theta. \end{align} - As $\partial_\rho \beta = 0$, the Jacobian becomes - $\abs{J} = \partial_\alpha\beta \cdot \partial_\rho \nu$. Differentiation - yields + As $\partial_\rho \theta = 0$, the Jacobian becomes + $\abs{J_{\Sigma^2}(l_{\phi, \rho})} = \partial_\phi\theta \cdot + \partial_\rho \eta$. Differentiation yields \begin{align} - \partial_\alpha \beta &= \frac{ \frac{\sigma_1}{\sigma_2} - \sec^2 \alpha}{1 + \frac{\sigma_1^2}{\sigma_2^2} - \tan^2 \alpha}, \\ - \partial_\rho \nu &= \sigma_1 \cos \alpha \cdot \cos \beta + \sigma_2 - \sin \alpha \cdot \sin \beta - = \frac{\sigma_1 \cos \alpha \left( 1 + \tan^2 \alpha \right) - }{\sqrt{1 + \frac{\sigma_1^2}{\sigma_2^2} \tan^2 \alpha}} + \partial_\phi \theta &= \frac{ \frac{\sigma_1}{\sigma_2} + \sec^2 \phi}{1 + \frac{\sigma_1^2}{\sigma_2^2} + \tan^2 \phi} + = \frac{\sigma_1 \sigma_2}{\sigma_1^2 \sin^2 \phi + \sigma_2^2 + \cos^2 \phi}, \\ + \partial_\rho \eta &= \sigma_1 \cos \phi \cdot \cos \theta + + \sigma_2 \sin \phi \cdot \sin \theta. \end{align} - and assuming that $v = (v_x, v_y)^T$ is a unit vector along the line - parametrized by the angle $\alpha$ we obtain + In the expression for $\partial_\rho \eta$ we use that $\sin + (\arctan( x )) = x / \sqrt{1 + x^2}$ and that $\cos ( \arctan ( x )) + = 1 / \sqrt{1 + x^2}$ to obtain + \begin{equation} + \partial_\rho \eta = \frac{\sigma_1 \cos \phi + \sigma_2 \sin + \phi \, \frac{\sigma_1}{\sigma_2} \tan \phi} + {\sqrt{1 + \frac{\sigma_1^2}{\sigma_2^2} + \tan^2 \phi}} + = \frac{\sigma_1 \sigma_2 \cos^2 \phi + \sigma_1 \sigma_2 + \sin^2 \phi}{\sqrt{\sigma_1^2 \sin^2 \phi + \sigma_2^2 + \cos^2 \phi}}. + \end{equation} + If $\nu = (\nu_x, \nu_y)^T$ is a unit vector along the line + $l_{\phi, \rho} = l_{\nu, \rho}$ then + \begin{equation} + \abs{J_{\Sigma^2}(l_{\nu, \rho})} + = \frac{\sigma_1^2 \sigma_2^2} + {\left(\sigma_1^2 \sin^2 \phi + \sigma_2^2 \cos^2 \phi + \right)^{\sfrac{3}{2}}} + = \frac{\sigma_1^2 \sigma_2^2 + }{\left(\sigma_1^2 \nu_x^2 + \sigma_2^2 + \nu_y^2\right)^{\sfrac{3}{2}}} + = \frac{\det \Sigma^2}{(\nu^T \, \Sigma^2 \, + \nu)^{\sfrac{3}{2}}}. + \end{equation} + We are interested in the Jacobian of the whole transformation, so + all that is left to do is insert $U^T l_{\nu, \rho}$ to obtain + \begin{equation} + \abs{J_M(l_{\nu, \rho})} = \abs{J_{\Sigma^2}(U^T l_{\nu, \rho})} + = \frac{\det M}{\left(\nu^T U \Sigma^2 U^T \nu\right)^{\sfrac{3}{2}}} + = \frac{\det M}{\left(\nu^T M \nu\right)^{\sfrac{3}{2}}} + \end{equation} + We have now proven that for a constant metric tensor $M$, the length + of the differentiable curve $C$ with regards to this tensor can be + calculated as \begin{equation} - \abs{J} = \frac{\frac{\sigma_1}{\sigma_2} \cdot \sigma_1 \left( - 1 + \tan^2 \alpha \right)^{\sfrac{3}{2}}}{\left( 1 + - \frac{\sigma_1^2}{\sigma_2^2} \tan^2 - \alpha \right)^{\sfrac{3}{2}}} - = \frac{\sigma_1^2 \sigma_2^2 \left( v_x^2 + - v_y^2\right)^{\sfrac{3}{2}}}{\left(\sigma_1^2 v_x^2 + \sigma_2^2 - v_y^2\right)^{\sfrac{3}{2}}}. + \abs{C}_M = \int_C \langle \dot{C}, M \dot{C} \rangle \, dt + = \int_\mathcal{L} \# (l_{\nu, \rho} \cap C) + \frac{\det M}{\left(\nu^T M \nu \right)^{\sfrac{3}{2}}} \, + d\mathcal{L} (l_{\nu, \rho}) \end{equation} - The angle of the line that $\Lambda$ takes as input has already been - rotated by the $U^T$ operator, such that if $u$ is a unit vector - along the line being integrated over, then $v = U^Tu$ and + We now argue that this also holds for a non-constant but continuous + metric tensor $M(x)$. By partitioning the domain into disjoint sets + $U_i$ such that $\Omega = \cup U_i$, we make a piecewise constant + approximation $M_\pi(x)$ such that if $x \in U_i$ then $M_\pi(x) = + M(x_i)$ for some fixed $x_i \in U_i$. We then approximate the right + side of \fixme{ref} by \begin{equation} - \abs{J} = \frac{\det M}{\left(u^T M u\right)^{\sfrac{3}{2}}} + \abs{C}_{M_\pi} = \sum_i \int_\mathcal{L} \#(l_{\nu, \rho} \cap + C \cap U_i) \, w_i(\nu) \, d\mathcal{L}(l_{\nu, \rho}) + \label{eq:mpi_approx} \end{equation} - And now we argue that this also holds for a non-constant but - continuous metric tensor $M(x)$. By partitioning the domain into - disjoint sets $U_i$ such that $\Omega = \cup U_i$, we make a - piecewise constant approximation such that $M_\pi(x) = M(x_i)$ for - some fixed $x_i \in U_i$. We then approximate the right side of - \fixme{ref} by + where $w_i$ is the weight-function used in the set $U_i$ using the + constant tensor $M_\pi(x_i)$ and can be written + \begin{equation} + w_i(\nu) = \frac{\det M(x_i)}{\left(\nu^T M(x_i) \nu + \right)^{\sfrac{3}{2}}}. + \end{equation} + We further simplify the approximation by introducing the global + weight-function $w_\pi(\nu, x)$ which is equal to + $w_i(\nu)$ when $x \in U_i$. It can be written + \begin{equation} + w_\pi(\nu, x) = \frac{\det M_\pi(x)}{\left(\nu^T M_\pi(x) \nu + \right)^{\sfrac{3}{2}}}. + \end{equation} + Using this weight in \eqref{eq:mpi_approx} we can get rid of the sum + over the partition $i$ and form a sum of all intersection point of + $C$ and the line $l_{\nu, \rho}$ currently being integrated over. + The approximation becomes \begin{align} - \abs{C}_{M_\pi} &= \sum_i \int_\mathcal{L} n_{C\cap U_i} w_i \, - d\mathcal{L} \\ - &= \sum_i \int_\mathcal{L} \sum_{x\in C\cap U_i \cap - \mathcal{L}} w_\pi(\phi, p) \, d\mathcal{L} \\ - &= \int_\mathcal{L} \sum_{x\in C\cap \mathcal{L}} w_\pi(\phi, p) - \, d\mathcal{L}. + \abs{C}_{M_\pi} &= + \sum_i \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C \cap + U_i} w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}) \\ + &= \int_\mathcal{L} \sum_{x \in l_{\nu, \rho} \cap C} + w_\pi(\nu, x) \, d\mathcal{L}(l_{\nu, \rho}). \end{align} As our partition $\pi$ is refined the weight $w_\pi(x)$ converges - pointwise to the continously varying weight $w(x)$. We now want to - bound the expression for $w_\pi$. + pointwise to the continously varying weight $w(x)$. Now it only + remains to show that the integral converges, and that the left side + converges to the left side of \fixme{ref}. + + Recall from \fixme{ref} that the left hand side is calculated as \begin{equation} - w_\pi = \frac{\det M(p)}{2 \left(u_L^T \cdot M(p) \cdot - u_L\right)^{\sfrac{3}{2}}} + \abs{C}_{M_\pi} = \int_C \sqrt{ \dot{C}^T M_\pi \dot{C} } \, dt. \end{equation} + We know that $M_\pi(x)$ converges pointwise to $M(x)$ and using the + fact that the largest eigenvalue of $M$ is equal to 1, we know that + the integrand is bounded and we can apply Lebesgue's dominated + convergence theorem to see that $\abs{C}_{M_\pi} \to + \abs{C}_{M(x)}$. + + We apply the same theorem to show that the right hand side of + \fixme{ref} converges, but showing that $w_\pi(\nu, x)$ is bounded + is a bit more involved. + If we manage to bound the singular values and keep them away from $0$, that would be great. The singular values \fixme{or eigenvalues} of our tensor is