
\section{Introduction to Vector Fields}

\begin{enumerate}[1.]
\item {\it 
Find and sketch the field lines of the vector field 
${\bf F}(x,y)= -x^3 {\bf i}+y  {\bf j} $
in $\mathbb{R}^2$. Indicate the flow direction in your sketch.
}

\medskip

For a field line $\bfr(t)$, we have $\bfr'(t) = c {\bf F}(\bfr(t)) $ with $c>0$, so that
$$
\frac{dx}{dt}=-cx^3,\ \ \frac{dy}{dt}=cy.$$
If $x,y\neq 0$, we have $-\frac{dx}{x^3}=cdt = \frac{dy}{y}$. Integrating this, we get
$$
\ln|y|=\frac{1}{2x^2}+C,\ \ |y| = e^C e^{1/2x^2},\ \ y=C_1 e^{1/2x^2},
$$
where $C_1=\pm e^C$ is a constant. If $x=0$, we get 
$$
\frac{dx}{dt}=0,\ \ \frac{dy}{dt}=cy,$$
which corresponds to the two half-lines on the $y$-axis, oriented away from the origin. 
Similarly, if $y=0$, we get 
$$
\frac{dx}{dt}=-cx^3,\ \ \frac{dy}{dt}=0,$$
which corresponds to the two half-lines on the $x$-axis, oriented towards the origin. 
At $(0,0)$, we have ${\bf F}={\bf 0}$, so that this is a stationary point with no field lines through it.

\pagebreak

\begin{figure}[h!]
\centering\includegraphics[width=\linewidth]{figures/21wpfig1.png}
\end{figure}








\item
{\it 
Let $f(x,y)=x^2+4y^4$.  Find the field line of the vector field ${\bf F}=\nabla f$ which
passes through the point $(2,6)$.} 

\medskip

We are looking for a field line that satisfies 
$$
(x'(t),y'(t))=\nabla f(x,y)=(2x,16y^4)
$$
and that passes through $(2.6)$. We will assume that 
$x(0)=2,y(0)=6$ (for a field line that passes through $(2,6)$ at time $t_0$, replace $t$ by
$t-t_0$).  This means
$$x'=2x,\ x(0)=2,$$
$$y'=16y^3,\ y(0)=6.$$
We solve for $x(t)$ and $y(t)$:
$$x(t)=C_1e^{2t},\ 2=x(0)=C_1,\ x=2e^{2t},$$
$$\frac{y'}{y^3}=16,\ \int\frac{dy}{y^3}=\int 16 dt,\ 
-\frac{1}{2y^2}=16t+C_2,\ C_2=-\frac{1}{2\cdot 6^2}=-\frac{1}{72},$$
$$y^2=-\left(2\left(16t-\frac{1}{72}\right)\right)^{-1}=\left(\frac{1}{36}-32t\right)^{-1},$$
$$y=\left(\frac{1}{36}-32t\right)^{-1/2}$$
(we took the positive square root because $y(0)=6>0$).

\bigskip


\item 
{\it 
Find all field lines of the vector field ${\bf F}=\frac{1-x^2}{x}\ii + y^2 \jj$, using the following procedure.

(a) Assuming that $x$, $1-x^2$, and $y$ are all non-zero, solve the corresponding ordinary differential equation using the Method of Separation of Variables. (You may have to consider multiple cases.)}


\medskip

We have the differential equation
	\begin{equation}
		\frac{dy}{y^{2}} = \frac{x dx}{1-x^{2}} 
	\end{equation}
	If we integrate both sides, we get
	\begin{align*}
		-\frac{1}{y}+C &= \int \frac{x}{1-x^{2}}dx 
	\end{align*}
	We substitute $u = x^2$ to get
	\begin{align*}
	\frac{1}{y}+C &= \frac{1}{2} \int \frac{1}{u-1}du =\frac{1}{2}\ln|u-1| = \frac{1}{2} \ln|x^2-1|,
	\end{align*}
	which gives us	
	\begin{align*}
    \frac{1}{y}&= \frac{1}{2} \ln|x^2-1|+C,
 \ \        
    y = \frac{2}{\ln|x^2-1|+2C},
	\end{align*}
where $C$ is a constant. Alternatively, we can let $D=e^{2C}$, so that $2C=\ln D$. Then the field line equation
can be rewritten as
$y = \frac{2}{\ln(D|x^{2}-1|)}$, where $D$ is a positive constant.





\bigskip 


{\it (b) What happens when $x=0$, $1-x^2=0$, or $y=0$? Are there any field lines corresponding
to these cases?}

\begin{itemize}

\item The vector field ${\bf F}$ is undefined at $x = 0$, so there are no field lines there. 

\item If $1 - x^2 = 0$, then ${\bf F} = y^{2} {\bf j}$. 
This gives two vertical lines at $x = 1$ and $x = -1$, oriented upwards. 

\item If $y = 0$, we have ${\bf F} = \frac{1 - x^{2}}{x} {\bf i }$. The corresponding field lines lie on the $x$-axis, with orientation depending on the sign of $\frac{1 - x^{2}}{x}$. 

\end{itemize}

\pagebreak





{\it (c) Draw a picture of the field lines you found. Make sure to indicate their orientation. (You can check that by evaluating ${\bf F}$ at sample points.) You should have exactly one field line through every point where ${\bf F}\neq {\bf 0}$.}

	\begin{figure}[htbp]
		\centering\includegraphics[width=\linewidth]{figures/21wpfig2.png}
		\caption{The field lines of $\mathbf{F}(x,y) = \frac{1 - x^2}{x} \mathbf{i} + y^2 \mathbf{j}$ overlain on the vector field.  The solutions corresponding to $x = -1, x=1$ and $y=0$ are the heavy red lines.}
	\end{figure}


\end{enumerate}

