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\begin{slide}
\begin{center}
\LARGE \textcolor{highlight}{Solving $1^{\mbox{st}}$-Order ODEs using Symmetry Methods} 

\vspace{0.5in}

\normalsize

\begin{tabular}{rcl}
Hannah Gilmore &amp; $\Longleftrightarrow$ &amp;
\texttt{gilmorehf@appstate.edu}\\
Michael Kelley &amp; $\Longleftrightarrow$ &amp;
\texttt{kelleyma1@appstate.edu}\\
Hadi Morrow &amp; $\Longleftrightarrow$ &amp;
\texttt{alsoqih@appstate.edu}\\
Huy Tu &amp; $\Longleftrightarrow$ &amp;
\texttt{tuhq@appstate.edu}
\end{tabular}

\vspace{0.25in}       
       
\textcolor{highlight}{\large Thursday, April 26, 2012}
\end{center}
\end{slide}


\begin{slide}
\begin{center}
\LARGE How Do You Solve\\ \textcolor{highlight}{``First-Order ODEs''} \\ Using Symmetry?\\ 
\end{center}
\end{slide}

\begin{slide}

An \textcolor{highlight}{ordinary differential equation} (ODE) is an equation involving (ordinary) derivatives.

A \textcolor{highlight}{solution} of an ODE is a function such that the ODE is satisfied when the solution and its derivatives are plugged into the equation.

The \textcolor{highlight}{order} of an ODE is the highest order derivative appearing in the equation.

\textcolor{highlight}{\bf Example:} $y''+y=0$ is a \textcolor{highlight}{second order} ODE. $y=\sin(x)$ is a solution and so is $y=\cos(x)$. In fact, for any choice of real numbers $C_1$ and $C_2$, we have that $y=C_1\sin(x)+C_2\cos(x)$ is a solution.

We looked at \textcolor{highlight}{symmetry methods} for solving \textcolor{highlight}{first order} ODEs. These same techniques can be extended to solve higher order ODEs and PDEs (partial differential equations).

\end{slide}

\begin{slide}

\begin{center}\textcolor{highlight}{An Equation with Rotational Symmetry}\end{center}

\begin{minipage}[b]{5in}
\includegraphics[width=5in]{ODE_Symmetry-figure01.jpg}
\end{minipage}
\begin{minipage}[b]{4in}
{\bf Example:}\\ \hspace*{0.25in} \textcolor{highlight}{\Large $y' = \frac{y^3+x^2 y-y-x}{x^3+x y^2+y-x}$}

The figure on the left shows several solutions of this equation. We can see that they have an obvious rotational symmetry. 
It turns out that this \textcolor{highlight}{can be detected by looking at the form of the equation}.

Once this is known, we can \textcolor{highlight}{easily} cook up an integrating factor and \textcolor{highlight}{solve} the equation.
\end{minipage}

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{The Set Up}\end{center}

We start with the first order ODE: \textcolor{highlight}{$\frac{dy}{dx} = \frac{B(x,y)}{A(x,y)}$}.

First, we convert this equation to its \textcolor{highlight}{Pfaffian form}:\\ $\mbox{\;}$ \hfill \textcolor{highlight}{$-Bdx + Ady = 0$} \hfill \hfill

If $\displaystyle -B_y=-\frac{\partial B}{\partial y}=\frac{\partial A}{\partial x}=A_x$, then this would be an \textcolor{highlight}{exact} differential equation which could easily be solved. Our goal is to find an \textcolor{highlight}{integrating factor} to make this equation exact so we can solve it.

\end{slide}

\begin{slide}

\begin{center}\Large\textcolor{highlight}{Finding an Integrating Factor}\end{center}

Let $\Psi(x,y) = \psi$ be an implicitly defined solution (so $\psi$ is a constant depending on some initial condition).

So $\Psi_x \,dx + \Psi_y \,dy = \color{highlight}\cancelto{0}{\color{regcolor}d\psi}$ \quad or equivalently $\Psi_x+\Psi_y\; \frac{dy}{dx}=0$.

Suppose $y(x)$ is an explicit solution: $\Psi(x,y(x)) = \Psi_x\, \color{highlight}\cancelto{1}{\color{regcolor}\frac{dx}{dx}} \color{regcolor}+ \Psi_y\, \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{B(x,y)}{A(x,y)}$ from original equation thus,

$\Psi_x(x,y) + \Psi_y(x,y) \frac{B(x,y)}{A(x,y)} = 0$ and so

 \mbox{\;}\hfill \fbox{\textcolor{highlight}{$A(x,y) \Psi_x(x,y) + B(x,y) \Psi_y(x,y) = 0$}} \hfill \hfill

\end{slide}

\begin{slide}

\begin{center}\Large\textcolor{highlight}{Symmetries of Solutions}\end{center}

In general, if $\mathcal{F}$ is a (structure preserving) map or operation such that $\mathcal{F}$ sends some object X to itself, then we say $\mathcal{F}$ is a \textcolor{highlight}{symmetry} of X. 

A \textcolor{highlight}{symmetry of a differential equation} is a mapping from a collection of functions to itself which \textcolor{highlight}{sends solutions} (of that ODE) \textcolor{highlight}{to solutions}.

Let \textcolor{highlight}{$\tilde{x} = F(s,x,y)$} and \textcolor{highlight}{$\tilde{y} = G (s,x,y)$} such that for each real number $s$, $\tilde{y}(\tilde{x})$ is a solution whenever $y(x)$ is a solution. In other words, for each choice of $s$, $F$ and $G$ define a symmetry of our ODE. {\small {}[Secretly we are defining a 1-parameter Lie group.]}

Moreover, we assume that \textcolor{highlight}{$F(0,x,y) = x$} and \textcolor{highlight}{$G(0, x, y) = y$}, so $s=0$ corresponds to the \textcolor{highlight}{identity} map (this sends each solution to itself).

\end{slide}

\begin{slide}

\begin{center}\Large\textcolor{highlight}{Tangents of Symmetries}\end{center}

Since $\Psi(x,y)=\psi$ is a solution, then $\Psi(\tilde{x},\tilde{y})=\psi(s)$ is a solution (for possibly some other choice of constant $\psi(s)$). Since $s=0$ corresponds to the identity, $\Psi(x,y) = \psi = \psi(0)$. In general, \textcolor{highlight}{$\Psi \big(F(s,x,y), G(s,x,y)\big) = \psi(s)$}.

Expand this equation in terms of its Maclaurin series (in $s$): 
{\small
\[ \begin{array}{rcr}
     \Psi \big(F(s,x,y), G(s,x,y) \big)\Bigg|_{s=0} \;\, &amp; \begin{array}{c} \Huge\strut\\ \Huge\strut \end{array} &amp; \psi(0) \\  
 + \Big[ \Psi_x (\dots) F_s(s,x,y) + \Psi_y (\dots) G_s (s,x,y)\Big]\Bigg|_{s=0}s &amp; \quad = \quad &amp; \psi'(0) \\
 + O(s^2) \Huge\strut &amp; &amp; +O(s^2)
\end{array} \]
}

\vspace{-0.2in}

\small
$\quad\, \Psi \color{highlight}\cancelto{x}{\color{regcolor}\big(F(0,x,y)} \color{regcolor}, \color{highlight}\cancelto{y}{\color{regcolor}G(0,x,y)} \color{regcolor}\big)$

$+\big[ \Psi_x \color{highlight}\cancelto{x}{\color{regcolor}\big(F(0,x,y)} \color{regcolor}, \color{highlight}\cancelto{y}{\color{regcolor}G(0,x,y)} \color{regcolor}\big) \cdot F_s(0,x,y) + \Psi_y\color{highlight}\cancelto{x}{\color{regcolor}\big(F(0,x,y)} \color{regcolor}, \color{highlight}\cancelto{y}{\color{regcolor}G(0,x,y)} \color{regcolor}\big)\cdot G_s(0,x,y) \big] s$

$+ O(s^2) \qquad = \qquad \psi(0) +\psi' (0)s + O(s^2) $

\end{slide}


\begin{slide}

\begin{center}\Large\textcolor{highlight}{Tangents of Symmetries}\end{center}

{\bf Definition:} \textcolor{highlight}{$\xi(x,y) = F_s (0,x,y)$} and \textcolor{highlight}{$\eta(x,y) = G_s (0,x,y)$}\\
{\small {}[Secretly we have the Lie algebra of our 1-parameter Lie group.]}

For all points, $(x,y)$, lying on the solution curve we have $\Psi(x,y) = \psi(0)$.

So we get $\psi(0) + \big[\Psi_x \xi + \Psi_y \eta \big]s + O(s^2) = \psi(0)+\psi'(0)s + O(s^2)$.

Therefore, $\Psi_x \xi + \Psi_y \eta = \psi'(0)$ and finally after normalizing $\psi'(0)=1$, we get


\mbox{\;} \hfill \fbox{\large\textcolor{highlight}{$\Psi_x \, \xi + \Psi_y \, \eta = 1$}} \hfill \hfill


\end{slide}

\begin{slide}
Using Cramer's rule, we can now solve the system of equations: $A\Psi_x+B\Psi_y=0$ and $\xi\Psi_x+\eta\Psi_y=1$.

$$\textcolor{highlight}{\Psi_x} = \det \begin{bmatrix} 0 &amp; B \\ 1 &amp; \eta \end{bmatrix} / \det \begin{bmatrix} A &amp; B \\ \xi &amp; \eta \end{bmatrix}  \textcolor{highlight}{= \frac{-B}{A\eta - B\xi}}$$
$$\textcolor{highlight}{\Psi_y} =\det \begin{bmatrix} A &amp; 0 \\ \xi &amp; 1 \end{bmatrix} / \det \begin{bmatrix} A &amp; B \\ \xi &amp; \eta \end{bmatrix} \textcolor{highlight}{= \frac{A}{A\eta - B\xi}}$$

{\bf Define:} \textcolor{highlight}{$\displaystyle M=\frac{1}{A\eta - B\xi}$} is an \textcolor{highlight}{integrating factor} which makes the Pfaffian form of our original equation exact.

In particular, $M(-Bdx + Ady) = M\cdot 0$ yields $\Psi_x dx + \Psi_y dy = 0$.

Therefore, we can solve our original equation:\\
\mbox{\;} \hfill $\Psi(x,y) = $``$\int  \Psi_x dx + \Psi_y dy $'' $=$ Constant. \hfill \hfill

\end{slide}

\begin{slide}

\begin{center}\Large\textcolor{highlight}{Reality Sets In}\end{center}

Unfortunately, there is \textcolor{highlight}{no known way} of finding symmetries of a random differential equation from the equation itself. If we could, we could solve \textcolor{highlight}{all} ODEs. In practice, one guesses a possible form that the symmetry could take and then sees if that works.

However, given a symmetry it is possible to determine what the most general ODE with that symmetry looks like. This has been done for some simple symmetries on the next slide.

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{Some first-order ODEs and Symmetries}\end{center}
\begin{center}
    \begin{tabular}{| l | l | l ||| l | l | l |}
    \hline
    Equation &amp; $\xi$ &amp; $\eta$ &amp; Equation &amp; $\xi$ &amp; $\eta$ \Huge\strut \\ \hline\hline
    $y' = F[y]$ &amp; 1 &amp; 0 &amp; $y' = \frac{y}{x} + xF[\frac{y}{x}]$ &amp; 1 &amp; $\frac{y}{x}$ \Huge\strut \\ \hline
    $y' = F[x]$ &amp; 0 &amp; 1 &amp; $xy' = y + F[\frac{y}{x}]$ &amp; $x^2$ &amp; $xy$ \Huge\strut \\ \hline
    $y' = F[ax +by]$ &amp; $b$ &amp; $-a$ &amp; $y' = \frac{y}{F[\frac{y}{x}]}$ &amp; $xy$ &amp; $y^2$ \Huge\strut \\ \hline
    $y' = \frac{(y+x)F[x^2 + y^2]}{(x-y)F[x^2+y^2]}$ &amp; $y$ &amp; $-x$ &amp; $y' = \frac{y}{x+F[y]}$ &amp; $y$ &amp; 0 \Huge\strut \\ \hline
    $y' = F[\frac{y}{x}]$ &amp; $x$ &amp; $y$ &amp; $xy' = y+F[y]$ &amp; 0 &amp; $x$ \Huge\strut \\ \hline
    $y' = x^{k-1} F[\frac{y}{x^k}]$ &amp; $x$ &amp; $ky$ &amp; $xy' = \frac{y}{\ln[x] +F[y]}$ &amp; $xy$ &amp; 0 \Huge\strut \\ \hline
    $xy' = F[xe^{-y}]$ &amp; $x$ &amp; 1 &amp; $xy' = y(\ln[x]+F[x])$ &amp; 0 &amp; $xy$ \Huge\strut \\ \hline
    $y' = yF[ye^{-x}]$ &amp; 1 &amp; $y$ &amp; $y' = yF[x]$ &amp; 0 &amp; $y$ \Huge\strut \\ \hline
    \end{tabular}
\end{center}

where $F$ is an arbitrary function.
\end{slide}
  
\begin{slide}
\begin{center}\Large\textcolor{highlight}{Example 1:}\end{center}

Consider the equation \textcolor{highlight}{$\displaystyle y' = \frac{xy}{x^2+y}$}.

This equation can be rewritten as $\displaystyle y' = \frac{y}{x+\frac{y}{x}}$.

Looking at the Table from Brian Cantwell's text, we get that \textcolor{highlight}{$\xi=xy$} and \textcolor{highlight}{$\eta=y^2$} are symmetry tangents.

Therefore, \textcolor{highlight}{$M(x,y) =$} $\frac{1}{A\eta-B\xi} = \frac{1}{1\cdot y^2 - \frac{y}{x+\frac{y}{x}} \cdot xy}=$ \textcolor{highlight}{$\frac{x^2+y}{y^3}$} will work as an integrating factor.

The Pfaffian form the equation: $-B\,dx+A\,dy=0$ becomes $-MB\,dx+MA\,dy=0$ which after some algebra simplifies to
\[ \frac{-x^2}{y(x^2+y^2)}\,dx+\left(\frac{-x^3}{y^2(x^2+y^2)}-\frac{1}{y^2}\right)\,dy = 0 \]

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{Example 1:}\end{center}

Using standard integrating techniques (i.e. partial fractions), we get:
\[ \int \frac{-x^2}{y(x^2+y^2)}\,dx = -\frac{x}{y}+\arctan\left(\frac{x}{y}\right) + C_1(y) \]
\[ \int \left(\frac{-x^3}{y^2(x^2+y^2)}-\frac{1}{y^2}\right)\,dy = -\frac{1}{y}-\frac{x}{y}+\arctan\left(\frac{x}{y}\right) + C_2(x) \]

Therefore, the solution is \textcolor{highlight}{$ -\frac{1}{y}-\frac{x}{y}+\arctan\left(\frac{x}{y}\right) = C$}.

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{Example 1:}\end{center}

\hspace*{-0.25in}
\begin{minipage}[b]{3.75in}
\includegraphics[width=3.75in]{ODE_Symmetry-figure02.jpg}
\end{minipage}
\begin{minipage}[b]{5.25in}
Maple can solve this equation as well:

{\small
\verb|--- Trying Lie symmetry methods, 1st order ---|\\
\verb|-&gt; Computing symmetries using: way = 2|
\[ \left[ 0,\frac{x^2y^2+y^4}{x^3+y^2+x^2}\right] \]
\verb|&lt;- successful computation of symmetries.|\\
\verb|trying an integrating factor| 
\verb|                 from the invariance group|\\
\verb|&lt;- integrating factor successful|}
\end{minipage}

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{Example 2:}\end{center}

Consider the equation \textcolor{highlight}{$\displaystyle y' = \frac{y}{x}+x\left(1+\frac{y^2}{x^2}\right)$}.

Looking at the Table from Brian Cantwell's text, we get that \textcolor{highlight}{$\xi=1$} and \textcolor{highlight}{$\eta=y/x$} are symmetry tangents.

Therefore, \textcolor{highlight}{$M(x,y) =$} $\frac{1}{A\eta-B\xi} = \frac{1}{1\cdot \frac{y}{x} - \left(\frac{y}{x}++x\left(1+\frac{y^2}{x^2}\right)\right) \cdot 1}=$ \textcolor{highlight}{$-\frac{x}{x^2+y^2}$} will work as an integrating factor.

The Pfaffian form the equation: $-B\,dx+A\,dy=0$ becomes $-MB\,dx+MA\,dy=0$ which after some algebra simplifies to
\[\frac{x-\ln(y)}{x-\ln(y)+1}\,dx+\frac{1}{x-\ln(y)+1}\cdot \frac{1}{y}\,dy = 0 \]

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{Example 2:}\end{center}

After a quick bit of algebra (essentially polynomial division) and a simple $u$-substitution ($u=x-\ln(y)+1$), we get:
\[ \int \frac{x-\ln(y)\textcolor{highlight}{+1-1}}{x-\ln(y)+1}\,dx = \int 1 - \frac{1}{x-\ln(y)+1}\,dx\] \[=x - \ln(x-\ln(y)+1)+C_1(y) \]
\[ \int \frac{1}{x-\ln(y)+1}\cdot \frac{1}{y}\,dy = -\ln(x-\ln(y)+1) + C_2(x) \]

Therefore, the solution is \textcolor{highlight}{$ x - \ln(x-\ln(y)+1) = C$}.

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{Example 2:}\end{center}

Maple can solve this equation as well:

{\small
\verb|looking for linear symmetries 1st order,|\\
\verb|trying the canonical coordinates of the invariance group|\\
\verb|-&gt; Computing canonical coordinates for the symmetry [1, y]|\\
\verb|-&gt; Calling odsolve with the ODE diff(y(x) x) = y(x) y(x)|\\
\verb|      *** Sublevel 2 ***|\\
\verb|      Methods for first order ODEs:|\\
\verb|      --- Trying classification methods ---|\\
\verb|      trying a quadrature|\\
\verb|      trying 1st order linear|\\
\verb|      &lt;- 1st order linear successful|\\
\verb|-&gt; Computing canonical coordinates for the symmetry [0, (x+1-ln(y))*y]|\\
\verb|&lt;- 1st order, canonical coordinates successful|
}

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{References:}\end{center}

\begin{itemize}
    \item {\it Symmetry Methods for Differential Equations: A Beginner's Guide} by Peter E. Hydon, Cambridge Texts in Applied Mathematics, 2000.
    \item {\it Introduction to Symmetry Analysis} by Brian J. Cantwell, Cambridge Texts in Applied Mathematics, 2002.
\end{itemize}

\end{slide}

\begin{slide}
\begin{center}\Large\textcolor{highlight}{Thank You!}\end{center}

\begin{center}
\Large A big thanks to Dr. T,  Dr. Palmer, and Dr. Cook!
\end{center}

\end{slide}



\end{document}

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