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\begin{document}

{\bf Question}

Prove the standard forms for elliptic, parabolic and hyperbolic
second order linear PDEs with constant coefficients of type

$$au_{xx}+2bu_{xy}+cu_{yy}=0$$

as derived in lectures. Hence check the form of their general
solutions given in lectures.


\medskip

{\bf Answer}

Use lecture notes with the transformations

$$\left\{\begin{array} {l} \xi=\alpha y+\beta x\\ \eta=\gamma
y+\delta x \end{array} \right.$$

where $\alpha,\ \beta,\ \gamma,\ \delta=const$

These give rise to

$\lq\lq \pl_x=\ds\frac{\pl}{\pl x}" \rightarrow \pl_x=\ds\frac{\pl
\xi}{\pl x}\pl_\xi+\ds\frac{\pl \eta}{\pl
x}\pl_\eta=\beta\pl_\xi+\delta \pl_\eta$

$\lq\lq \pl_y=\ds\frac{\pl}{\pl y}"\ \rm{(shorthand)}\ \rightarrow
\pl_y=\ds\frac{\pl \xi}{\pl y}\pl_\xi+\ds\frac{\pl \eta}{\pl
y}\pl_\eta=\alpha\pl_\xi+\gamma \pl_\eta$

$\Rightarrow
\pl^2_x=(\beta\pl_\xi+\delta\pl_\eta)^2=\beta^2\pl^2_\xi
+2\beta\delta\pl^2_{\xi\eta}+\delta^2\pl^2_\eta$

Note: Would have to be more careful if $\beta$ and $\delta$ were
\un{not} constants

$\ \
\pl^2_{xy}=(\beta\pl_\xi+\delta\pl_{\eta})(\alpha\pl_\xi+\gamma\pl_{\eta})
=\alpha\beta\pl^2_{\xi}+(\alpha\delta+\gamma\beta)\pl^2_{\xi\eta}$

$\Rightarrow
\pl^2_y=(\alpha\pl_\xi+\gamma\pl_\eta)^2=\alpha^2\pl^2_\xi
+2\alpha\gamma\pl^2_{\xi\eta}+\gamma^2\pl^2_\eta$

$\Rightarrow au_{xx}+2bu_{xy}+cu_{yy}=0$ becomes

\begin{eqnarray*} (a\beta^2+2b\alpha\beta+c\alpha^2) &amp; + &amp;
(2a\beta\delta+2b(\alpha\delta+\gamma\beta)+2c\alpha\delta)u_{\xi\eta}\\
&amp; + &amp; (a\pl^2 +2b\gamma\delta+c\gamma^2)u_{\eta\eta}=0
\end{eqnarray*}

3 cases:

$\left. \begin{array} {crcl} (i) &amp; \rm{hyperbolic} &amp; : &amp; b^2&gt;ac\\
(ii) &amp; \rm{parabolic} &amp; : &amp; b^2=ac\\ (iii) &amp; \rm{elliptic} &amp; : &amp;
b^2&lt;ac \end{array} \right\}$ see lecture notes.

\newpage
These give $(\star)$ as:

\begin{description}
\item[(i)]
$\ds\frac{\pl^2u}{\pl\xi\pl\eta}=0 \Rightarrow \rm{(integrate)}
\ds\frac{\pl u}{\pl \xi}=p'(\xi) \Rightarrow \rm{(integrate)}
u=p(\xi)+q(\eta)$

\item[(ii)]
$\ds\frac{\pl^2u}{\pl\eta^2}=0 \Rightarrow \rm{(integrate)}
\ds\frac{\pl u}{\pl \eta}=q(\xi) \Rightarrow \rm{(integrate)}
u=p(\xi)+\eta q(\xi)$

\item[(iii)]
$\ds\frac{\pl^2u}{\pl\eta^2}+\ds\frac{\pl^2u}{\pl\eta^2}=0
\Rightarrow$ solutions outlines later in the problem sheet
\end{description}

Remember can treat $p(\xi),\ q(\eta)$ etc. as arbitrary functions
of $\xi,\ \eta$ (cf. constants in ODEs)

\end{document}
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