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\begin{document}

{\bf Question}

Classify each of the following differential equations below as
either hyperbolic, parabolic or elliptic. Find the
characteristics, transform the equation to characteristic
coordinates and express the general solution in the original
variables.

\begin{description}
\item[(a)]
$u_{xx}+3u_{xy}+2u_{yy}=0$

\item[(b)]
$u_{xx}+2u_{xy}+2u_{yy}=0$

\item[(c)]
$u_{xx}+2u_{xy}+u_{yy}=0$

\item[(d)]
$u_{xx}+4u_{xy}-5u_{yy}=0$

\item[(e)]
$u_{xx}+u_{yy}=0$

\item[(f)]
$u_{xx}-u_{yy}=0$

\end{description}

\medskip

{\bf Answer}

All are second order linear homogeneous with constant
coefficients.

\begin{description}
\item[(a)]
$a=1,\ b=\ds\frac{3}{2},\ c=2;\ b^2-ac=\ds\frac{1}{4}&gt;0$,
hyperbolic.

Characteristics

$$\left\{\begin{array} {l} \xi =
y+\left(-\ds\frac{3}{2}+\sqrt{\ds\frac{9}{4}-2}\right)x\\
\eta=y+\left(-\ds\frac{3}{2}-\sqrt{\ds\frac{9}{4}-2}\right)x
\end{array} \right.$$

i.e., $$\left\{\begin{array} {l} \xi = y-x\\ \eta=y-2x
\end{array}\right.$$

i.e., $$\left\{\begin{array} {l} y = x+const\\ y=2x+const
\end{array} \right.$$

Transforms to $\ds\frac{\pl^2u}{\pl\xi}{\pl\eta}=0$ in
characteristic coordinates

whence $u=f(\xi)+g(\eta)$ in general solution

so \un{$u(x,y)=f(y-x)+g(y-2x)$}

\newpage
CHECK:

$u_x=-f'-2g'$\ \ $u_{xx}=f''+4g''$

$u_y=f'+g'$\ \ \ \ \ \ $u_{yy}=f''+g''$

\ \ \ \ \ \ \ \ \ \ \ $u_{xy}=-f''-2g''$

where $f'$ denotes $\ds\frac{df(\xi)}{d\xi}$ and $g'$ denotes
$\ds\frac{dg(\eta)}{d\eta}$ etc.

so

\begin{eqnarray*} u_{xx}+3u_{xy}+2u_{yy} &amp; = &amp;
+f''+4g''+3(-f''-2g'')+2f''+2g''\\ &amp; = &amp; 3f''+6g''-3f''-6g''\\ &amp; =
&amp; \un{0} \surd \end{eqnarray*}

\item[(b)]
$a=1,\ b=1,\ c=2;\ b^2-ac=-1&lt;0$ elliptic

No \un{real} characteristics. \un{However} try the solution

$$\left\{\begin{array} {l} \xi =
ay+\left(-b+\sqrt{b^2-ac}\right)x\\
\eta=ay+\left(-b+\sqrt{b^2-ac}\right)x \end{array} \right.$$

i.e., $$\left\{\begin{array} {l} \xi = y+(i-1)x\\ \eta=y+(-i-1)x
\end{array} \right.$$

NB complex conjugate coefficients of $x$

Although these are complex characteristics (which we normally
reject) we can use these formally as in the hyperbolic case to
eliminate the $\ds\frac{\pl^2u}{\pl\xi^2}$ and
$\ds\frac{\pl^2u}{\pl\eta^2}$ terms in the transformed (2.1),
equation (2.4) of the lecture notes. Instead of obtaining
Laplace's equation as in the lecture notes, we obtain
$\ds\frac{\pl^2u}{\pl\xi\pl\eta}=0$ with solutions
$u=p(\xi)+q(\eta)$, i.e.,

$$\un{u(x,y)=p(y-(1-i)x)+q(y-(1+i)x)}$$

This formula could only become real with appropriate boundary
conditions (see later). However, we can check that it \un{does}
satisfy the original equation:

\newpage
CHECK:

$u_x=p'(i-1)+q'(-i-1)$\ \ $u_{xx}=p''(i-1)^2+q''(-i-1)^2$

$u_y=p'+q'$\ \ \ \ \ \ $u_{yy}=p''+q''$

\ \ \ \ \ \ \ \ \ \ \ $u_{xy}=p''(i-1)+q''(-i-1)$

\begin{eqnarray*} &amp; &amp; u_{xx}+2u_{xy}+2u_{yy}\\ &amp; = &amp;
p''[(i-1)^2+2(i-1)+2]+q''[(-i-1)^2+2(-i-1)+2]\\ &amp; = &amp;
p''[-1-2i+1+2i-2+2]+q''[-1+x+1-2i-2+2]\\ &amp; = &amp; \un{0} \surd
\end{eqnarray*}

NB

Say we carried out the method of the lecture notes and transformed
as

$$\left\{\begin{array} {lclcl} \bar\xi &amp; = &amp;
\ds\frac{cx-by}{\sqrt{ac-b^2}} &amp; = &amp; 2x-y\\ \bar\eta &amp; = &amp; y &amp; = &amp;
y \end{array} \right.$$

The we would obtain Laplace's equation.

$$u_{\bar\xi\bar\xi}+u_{\bar\eta\bar\eta}=0$$

What is the general solution of this? Try the solution

$u=f(\bar\xi+i\bar\eta)+g(\bar\xi-i\bar\eta)$ $f,\ g$ arbitrary

It \un{works}, so the general solution must be

\begin{eqnarray*} u(x,y) &amp; = &amp; f(2x-y+iy)+g(2x-y-i-y)\\ &amp; = &amp;
f(2x+(i-1)y)+g(2x+(-i-1)y) \end{eqnarray*}

How does this relate to the solution above?

Well $f,\ g,\ p,\ q$ are arbitrary functions.

We can scale the arguments:

\begin{eqnarray*} 2x+(i-1)y &amp; = &amp;
(i-1)\left[\ds\frac{2x}{(i-1)}+y\right]\\ &amp; = &amp;
(i-1)\left[\ds\frac{2x}{2}(-1-i)+y\right]\\ &amp; = &amp; (i-1)[y-(1+i)x]
\end{eqnarray*}

Similarly \begin{eqnarray*} 2x-(1+i)y &amp; = &amp;
-(1+i)\left[y-\ds\frac{2x}{(1+i)}\right]\\ &amp; = &amp; -(1+i)[y-(1-i)x]
\end{eqnarray*}

So if we let

$$\left\{\begin{array} {rcl} f((i-1)\bar X) &amp; = &amp; p(\bar X)\\
g(-(1+i)\bar Y) &amp; = &amp; p(\bar X) \end{array} \right.$$

we obtain

$$\un{u(x,y)=p(y-(1-i)x)+q(y-(1+i)x)}$$

In $x,\ y$ coords, the same general solution as \un{above}
$\surd$.

MORAL:

General solution to Laplace equation

$u_{\bar\xi\bar\xi}+u_{\bar\eta\bar\eta}=0$ is
$u=f(\bar\xi+i\bar\eta)+g(\bar\xi-i\bar\eta)$

\item[(c)]
$a=1,\ b=1,\ c=1;\ b^2-ac=0$, parabolic.

Characteristics: only one set

$$\left\{\begin{array} {l} \xi = ay-bx \leftarrow\
\rm{characteristic}\\ \eta=y \leftarrow\ \rm{not\ important}
\end{array} \right.$$

i.e., $$\left\{\begin{array} {l} \xi = y-x\\ \eta=y
\end{array}\right.$$

Transforms equation to $\ds\frac{\pl^2 u}{\pl \eta^2}=0$

whence $u(\xi,\eta)=p(\xi)+\eta q(\xi)$ is general solution

$$\Rightarrow \un{u(x,y)=p(y-x)+yq(y-x)}$$

\newpage
\item[(d)]
$a=1,\ b=2,\ c=-5;\ b^2-ac=9&gt;0$, hyperbolic.

Characteristics

$$\left\{\begin{array} {l} \xi = y+x\\ \eta=y-5x
\end{array} \right.$$

i.e., $$\left\{\begin{array} {l} y = -x+const\\ y=5x+const
\end{array}\right.$$

Gives general solution to transformed equation $u_{\xi\eta}=0$ as
$u=p(\xi)+q(\eta)$, so

$$\un{u(x,y)=p(y+x)+q(y-5x)}$$

\item[(e)]
$u_{xx}+u_{yy}=0$,\ $a=1,\ b=0,\ c=1;\ b^2-ac=-1&lt;0$ \un{elliptic}

For general solution see (B) above.

For solution with boundary conditions given see Q7

\item[(f)]
$u_{xx}-u_{yy}=0$,\ $a=1,\ b=0,\ c=-1;\ b^2-ac=1&gt;0$
\un{hyperbolic}

Characteristics

$$\left\{\begin{array} {l} \xi = y+x\\ \eta=y-x
\end{array} \right.$$

So general solution is

$$\un{u(x,y)=p(y+x)+q(y-x)}$$

\end{description}

\end{document}
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