\chapter{Mathematics Examples} This appendix provides an example of \LaTeX's typesetting capabilities. Most of text was obtained from the University of Wisconsin-Madison Math Department's example thesis file. \section{Matrices} The equations for the {\em dq}-model of an induction machine in the synchronous reference frame are \begin{eqnarray} \left[\begin{array}{c} v_{qs}^e\\v_{ds}^e\\v_{qr}^e\\v_{dr}^e \end{array}\right] &=& \left[ \begin{array}{cccc} r_s + x_s\frac{\rho}{\omega_b} & \frac{\omega_e}{\omega_b}x_s & x_m\frac{\rho}{\omega_b} & \frac{\omega_e}{\omega_b}x_m \\ -\frac{\omega_e}{\omega_b}x_s & r_s + x_s\frac{\rho}{\omega_b} & -\frac{\omega_e}{\omega_b}x_m & x_m\frac{\rho}{\omega_b} \\ x_m\frac{\rho}{\omega_b} & \frac{\omega_e -\omega_r}{\omega_b}x_m & r_r'+x_r'\frac{\rho}{\omega_b} & \frac{\omega_e - \omega_r}{\omega_b}x_r' \\ -\frac{\omega_e -\omega_r}{\omega_b}x_m & x_m\frac{\rho}{\omega_b} & -\frac{\omega_e - \omega_r}{\omega_b}x_r' & r_r' + x_r'\frac{\rho}{\omega_b} \end{array} \right] \left[\begin{array}{c} i_{qs}^e\\i_{ds}^e\\i_{qr}^e\\i_{dr}^e\end{array} \right] \label{volteq}\\ T_e&=&\frac{3}{2}\frac{P}{2}\frac{x_m}{\omega_b}\left(i_{qs}^ei_{dr}^e - i_{ds}^ei_{qr}^e\right) \label{torqueeq}\\ T_e-T_l&=&\frac{2J\omega_b}{P}\frac{d}{dt}\left(\frac{\omega_r}{\omega_b}\right) \label{mecheq}. \end{eqnarray} \section{Multi-line Equations} \LaTeX{} has a built-in equation array feature, however the equation numbers must be on the same line as an equation. For example: \begin{eqnarray} \Delta u + \lambda e^u &= 0&u\in \Omega, \nonumber \\ u&=0&u\in\partial\Omega. \end{eqnarray} Alternatively, the number can be centered in the equation using the following method. % % The equation-array feature in LaTeX is a bad idea. For centered % numbers you should set your own equations and arrays as follows: % \def\dd{\displaystyle} \begin{equation}\label{gelfand} \begin{array}{rl} \dd \Delta u + \lambda e^u = 0, & \dd u\in \Omega,\\[8pt] % add 8pt extra vertical space. 1 line=23pt \dd u=0, & \dd u\in\partial\Omega. \end{array} \end{equation} The previous equation had a label. It may be referenced as equation~(\ref{gelfand}). \section{More Complicated Equations} \section*{Rellich's identity}\label{rellich.section} \setcounter{theorem}{0} % % Standard developments of Pohozaev's identity used an identity by Rellich~\cite{rellich:der40}, reproduced here. \begin{lemma}[Rellich] Given $L$ in divergence form and $a,d$ defined above, $u\in C^2 (\Omega )$, we have \begin{equation}\label{rellich} \int_{\Omega}(-Lu)\nabla u\cdot (x-\overline{x})\, dx = (1-\frac{n}{2}) \int_{\Omega} a(\nabla u,\nabla u) \, dx - \frac{1}{2} \int_{\Omega} d(\nabla u, \nabla u) \, dx \end{equation} $$ + \frac{1}{2} \int_{\partial\Omega} a(\nabla u,\nabla u)(x-\overline{x}) \cdot \nu \, dS - \int_{\partial\Omega} a(\nabla u,\nu )\nabla u\cdot (x-\overline{x}) \, dS. $$ \end{lemma} {\bf Proof:}\\ There is no loss in generality to take $\overline{x} = 0$. First rewrite $L$: $$Lu = \frac{1}{2}\left[ \sum_{i}\sum_{j} \frac{\partial}{\partial x_i} \left( a_{ij} \frac{\partial u}{\partial x_j} \right) + \sum_{i}\sum_{j} \frac{\partial}{\partial x_i} \left( a_{ij} \frac{\partial u}{\partial x_j} \right) \right]$$ Switching the order of summation on the second term and relabeling subscripts, $j \rightarrow i$ and $i \rightarrow j$, then using the fact that $a_{ij}(x)$ is a symmetric matrix, gives the symmetric form needed to derive Rellich's identity. \begin{equation} Lu = \frac{1}{2} \sum_{i,j}\left[ \frac{\partial}{\partial x_i} \left( a_{ij} \frac{\partial u}{\partial x_j} \right) + \frac{\partial}{\partial x_j} \left( a_{ij} \frac{\partial u}{\partial x_i} \right) \right]. \end{equation} Multiplying $-Lu$ by $\frac{\partial u}{\partial x_k} x_k$ and integrating over $\Omega$, yields $$\int_{\Omega}(-Lu)\frac{\partial u}{\partial x_k} x_k \, dx= -\frac{1}{2} \int_{\Omega} \sum_{i,j}\left[ \frac{\partial}{\partial x_i} \left( a_{ij} \frac{\partial u}{\partial x_j} \right) + \frac{\partial}{\partial x_j} \left( a_{ij} \frac{\partial u}{\partial x_i} \right) \right] \frac{\partial u}{\partial x_k} x_k \, dx$$ Integrating by parts (for integral theorems see~\cite[p. 20] {zeidler:nfa88IIa}) gives $$= \frac{1}{2} \int_{\Omega} \sum_{i,j} a_{ij} \left[ \frac{\partial u}{\partial x_j} \frac{\partial^2 u}{\partial x_k\partial x_i} + \frac{\partial u}{\partial x_i} \frac{\partial^2 u}{\partial x_k\partial x_j} \right] x_k \, dx $$ $$ + \frac{1}{2} \int_{\Omega} \sum_{i,j} a_{ij} \left[ \frac{\partial u}{\partial x_j} \delta_{ik} + \frac{\partial u}{\partial x_i} \delta_{jk} \right] \frac{\partial u}{\partial x_k} \, dx $$ $$- \frac{1}{2} \int_{\partial\Omega} \sum_{i,j} a_{ij} \left[ \frac{\partial u}{\partial x_j} \nu_{i} + \frac{\partial u}{\partial x_i} \nu_{j} \right] \frac{\partial u}{\partial x_k} x_k \, dx $$ = $I_1 + I_2 + I_3$, where the unit normal vector is $\nu$. One may rewrite $I_1$ as $$I_1 = \frac{1}{2} \int_{\Omega} \sum_{i,j} a_{ij} \frac{\partial}{\partial x_k}\left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) x_k \, dx $$ Integrating the first term by parts again yields $$I_1 = -\frac{1}{2} \int_{\Omega} \sum_{i,j} a_{ij} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) \, dx + \frac{1}{2} \int_{\partial\Omega} \sum_{i,j} a_{ij} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) x_k \nu_k \, dS $$ $$ - \frac{1}{2} \int_{\Omega} \sum_{i,j} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) x_k \frac{\partial a_{ij}}{\partial x_k}\, dx. $$ Summing over $k$ gives $$\int_{\Omega}(-Lu)(\nabla u\cdot x)\, dx = -\frac{n}{2} \int_{\Omega} \sum_{i,j} a_{ij} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) \, dx $$ $$ + \frac{1}{2} \int_{\partial\Omega} \sum_{i,j} a_{ij} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) (x\cdot \nu ) \, dS - \frac{1}{2} \int_{\Omega} \sum_{i,j} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) (x\cdot \nabla a_{ij}) \, dx $$ $$ + \frac{1}{2} \int_{\Omega} \sum_{i,j,k} a_{ij} \left[ \frac{\partial u}{\partial x_j} \frac{\partial u}{\partial x_k} \delta_{ik} + \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_k} \delta_{jk} \right] \, dx $$ $$- \frac{1}{2} \int_{\partial\Omega} \sum_{i,j} a_{ij} \left[ \frac{\partial u}{\partial x_j} \nu_{i} + \frac{\partial u}{\partial x_i} \nu_{j} \right] (\nabla u\cdot x) \, dS. $$ Combining the first and fourth term on the right-hand side simplifies the expression $$\int_{\Omega}(-Lu)(\nabla u\cdot x)\, dx = (1-\frac{n}{2}) \int_{\Omega} \sum_{i,j} a_{ij} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) \, dx $$ $$ + \frac{1}{2} \int_{\partial\Omega} \sum_{i,j} a_{ij} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) (x\cdot \nu ) \, dS - \frac{1}{2} \int_{\Omega} \sum_{i,j} \left( \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \right) (x\cdot \nabla a_{ij}) \, dx $$ $$ - \frac{1}{2} \int_{\partial\Omega} \sum_{i,j} a_{ij} \left[ \frac{\partial u}{\partial x_j} \nu_{i} + \frac{\partial u}{\partial x_i} \nu_{j} \right] (\nabla u\cdot x) \, dS. $$ Using the notation defined above, the result follows.