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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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\begin{document}

{\bf Question}

State which of the following PDEs are elliptic, which are
hyperbolic and which are parabolic. Try to find the equations of
the characteristics where these are real.

\begin{description}
\item[(a)]
$u_{xx}+3u_{xy}+u_{yy}=2u+x^2y^2$

\item[(b)]
$yu_{xx}+(x+y)u_{xy}+xu_{yy}=0$

\item[(c)]
$x^2u_{xx}-y^2u_{yy}=xy$

\item[(d)]
$r^2u_{rr}+u_{\theta\theta}$

\item[(e)]
$\bigtriangledown \cdot [F(r)\bigtriangledown\phi]$

\item[(f)]
$x(\pi-x)u_{xx}+xu_x+u_t=0$

\end{description}


\medskip

{\bf Answer}

\begin{description}
\item[(a)]
This is an \un{inhomogeneous} type with $d$ of th electure notes
given as $d=2u+x^2-y^2$.

However $a,\ b,\ c$ are given by

$a=1,\ b=\ds\frac{3}{2},\ c=1 \Rightarrow $ hyperbolic everywhere

Characteristics are given by constant coefficient results,

$$\left\{\begin{array} {l} \xi =
y+\left[\ds\frac{-3+\sqrt{5}}{2}\right]x\\
\eta=y+\left(\ds\frac{-3-\sqrt{5}}{2}\right)x
\end{array} \right.$$

or

$$\left\{\begin{array} {l} y =
\left(\ds\frac{3-\sqrt{5}}{2}\right)x+const\\
y=\left(\ds\frac{3+\sqrt{5}}{2}\right)x+const \end{array}
\right.$$

Note we can solve the homogeneous equation simply as

$$u_{homogeneous}(x,y)=p(\xi)+q(\eta)$$

with $\xi,\ \eta$ above, but complete solution is:

$$u=u_{homogeneous}+u_{particular\ integral}$$

\item[(b)]
$a=y,\ b=\ds\frac{(x+y)}{2},\ c=x$

$b^2-ac=\ds\frac{(x+y)^2}{4}-yx=\left(\ds\frac{x-y}{2}\right)^2
\geq 0$

So hyperbolic except where $x=y$ where it's parabolic.

For $y \ne x$ characteristics are given by:

$$adx^2+2b\,dx\,dy+cdy^2=0$$

$\Rightarrow y\,dx^2+(x+y)\,dx\,dy+xdy^2=0$

$\Rightarrow$

\begin{eqnarray*} \ds\frac{dy}{dx} &amp; = &amp; \ds\frac{b \pm
\sqrt{b^2-ac}}{a}\\ &amp; = &amp; \ds\frac{(x+y)}{2y} \pm
\ds\frac{\sqrt{(x-y)^2}}{2y}\\ &amp; = &amp; \ds\frac{1}{2y}[x+y \pm
(x-y)]\\ &amp; = &amp; \ds\frac{x}{y}\ \rm{or}\ \un{1} \end{eqnarray*}

Therefore $\ds\frac{dy}{dx}=1$ or
$\ds\frac{dy}{dx}=\ds\frac{x}{y}$

$\Rightarrow y=x+const$ or $y^2-x^2=const$

PICTURE \vspace{2in}

Note that the two families of characteristics become parallel on
\un{$y=x$} where equation is parabolic.

\item[(c)]
$a=x^2,\ b=0,\ c=-y^2$, inhomogeneous

$b^2-ac=0+x^2y^2&gt;0$ for all $x \ne 0,\ y \ne 0$

Characteristics are:

$$\ds\frac{dy}{dx}=\ds\frac{b \pm \sqrt{b^2-ac}}{a}=\pm
\ds\frac{xy}{x^2}=\pm \ds\frac{y}{x}$$

$\Rightarrow \ds\frac{dy}{dx}=\ds\frac{y}{x},\
\ds\frac{dy}{dx}=-\ds\frac{y}{x}$

$\Rightarrow \ln y=\ln kx,\ \ln y=-\ln(x\bar k)$

$\Rightarrow y=kx,\ \ \ds\frac{\bar{\bar {k}}}{x}$ where $k$ and
${\bar{\bar {k}}}$ are constants

PICTURE \vspace {2in}

parallel when $y=0$ or $x=0$ i.e., when parabolic

\item[(d)]
$r^2u_{rr}+u_{\theta\theta}=\lq\lq a"u_r$

The \lq\lq d" is irrelevant to classification

$a=r^2,\ b=0,\ c=1$

$b^2-ac=-r^2 \leq 0 \rightarrow$ elliptic $r\ne 0$, parabolic
$r=0$

No real characteristics for elliptic case

Parabolic case: $\ds\frac{dr}{d\theta}=0 \Rightarrow \un{r=const}$

\newpage
\item[(e)]
\begin{eqnarray*} {\bf{\bigtriangledown}} \cdot
[F({\bf{r}}){\bf{\bigtriangledown}}\phi] &amp; = &amp;
f({\bf{r}}){\bf{\bigtriangledown}}^2\phi\\ &amp; = &amp;
F({\bf{r}})(\phi_{xx}+\phi_{yy}) \end{eqnarray*}

so if $F({\bf{r}})(\phi_{xx}+\phi_{yy})=0$ it's elliptic
everywhere except where $F({\bf{r}})=0$

($a=F({\bf{r}}),\ b=0,\ c=F({\bf{r}}),\ b^2-ac=-F({\bf{r}})^2&lt;0$)

\item[(f)]
$a=x(\pi-x),\ b=0,\ c=0$

(only second order derivatives count)

so $b^2-ac=0$ for all $x,\ t$

Thus parabolic with $\ds\frac{dt}{dx}=0 \Rightarrow t=const$

\end{description}
\end{document}
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