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\title{MATH 208 - HW \# 7}
\author{Nathan Marianovsky \\ November 1, 2017}

\begin{document}

\maketitle

\begin{problem}
Let $\alpha = -y\text{d}x+\text{d}z$, a one-form on $\mathbb{R}^3$. Find non-vanishing vector fields $V$ on $\mathbb{R}^3$ s.t. $\mathcal{L}_V\alpha = 0$. Find other vector fields s.t. $\mathcal{L}_V\alpha = f\alpha$, $f \neq 0$ a function. 
\end{problem}

\vspace{.5cm}

\begin{solution}
We take $V = V^1 \partial_x + V^2 \partial_y + V^3 \partial_z$ to be an arbitrary vector field. The Lie derivative can be computed via:
\begin{equation*}
\mathcal{L}_V\alpha = \Bigg( V^j \frac{\partial\alpha_i}{\partial x^j} + \alpha_j \frac{\partial V^j}{\partial x^i} \Bigg) \text{d}x^i
\end{equation*}
With $\alpha_2 = 0$ and the derivative of $\alpha_3$ being zero throws out a lot of terms providing:
\begin{equation*}
\mathcal{L}_V\alpha = \Big( -y \partial_x V^1 - V^2 + \partial_x V^3 \Big) \text{d}x + \Big( -y\partial_y V^1 + \partial_y V^3 \Big) \text{d}y + \Big( -y\partial_z V^1 + \partial_z V^3 \Big) \text{d}z
\end{equation*}
\begin{itemize}

\item
For $\mathcal{L}_V\alpha = 0$ we want the vector field to satisfy:
\begin{equation*}
\begin{split}
\partial_x \Big( -yV^1 + V^3 \Big) &= V^2 \\
\partial_y \Big( -yV^1 + V^3 \Big) &= -V^1 \\
\partial_z \Big( -yV^1 + V^3 \Big) &= 0
\end{split}
\end{equation*}

\item
For $\mathcal{L}_V\alpha = f\alpha$ we want the vector field to satisfy:
\begin{equation*}
\begin{split}
\partial_x \Big( -yV^1 + V^3 \Big) &= V^2 - yf \\
\partial_y \Big( -yV^1 + V^3 \Big) &= -V^1 \\
\partial_z \Big( -yV^1 + V^3 \Big) &= f
\end{split}
\end{equation*}

\end{itemize}
\end{solution}

\vspace{1cm}
\begin{problem}
Repeat the previous problem with the one-form $\alpha = \text{d}z$. 
\end{problem}

\vspace{.5cm}

\begin{solution}
We take $V = V^1 \partial_x + V^2 \partial_y + V^3 \partial_z$ to be an arbitrary vector field. The Lie derivative can be computed via:
\begin{equation*}
\mathcal{L}_V\alpha = \Bigg( V^j \frac{\partial\alpha_i}{\partial x^j} + \alpha_j \frac{\partial V^j}{\partial x^i} \Bigg) \text{d}x^i
\end{equation*}
With $\alpha_1 = \alpha_2 = 0$ and the derivative of $\alpha_3$ being zero throws out a lot of terms providing:
\begin{equation*}
\mathcal{L}_V\alpha = \partial_x V^3 \text{d}x + \partial_y V^3 \text{d}y + \partial_z V^3 \text{d}z
\end{equation*}
\begin{itemize}

\item
For $\mathcal{L}_V\alpha = 0$ we want the vector field to satisfy $\partial_x V^3 = \partial_y V^3 = \partial_z V^3 = 0$. It easily follows that $V^3 = c$ for some $c \in \mathbb{R} \setminus \{0\}$. Therefore, any vector field of the form $V = V^1 \partial_x + V^2 \partial_y + c \partial_z$ ensures $\mathcal{L}_V\alpha = 0$. 

\item
For $\mathcal{L}_V\alpha = f\alpha$ we want the vector field to satisfy $\partial_x V^3 = \partial_y V^3 = 0$ and $\partial_z V^3 = f$. It easily follows that $V^3 = g(z)$ where $g'(z) = f$. Therefore, any vector field of the form $V = V^1 \partial_x + V^2 \partial_y + g(z) \partial_z$ ensures $\mathcal{L}_V\alpha = f\alpha$. 

\end{itemize}
\end{solution}

\end{document}