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\begin{document}

{\bf Question}

Show that the equation $x^2u_{xx}=y^2u_{yy}$ is hyperbolic in the
positive quadrant and find the equations of the characteristics
there, expressing them in the form $f(x,y)=const$,\
$g(x,y)=const$.

If $u,\ u_x,\ u_y$ are given on the quarter circle $x^2+y^2=1,\
x>0,\ y>0$, sketch the domain of the dependence, i.e., the region
in the ($x,y$)-plane for which a unique solution of the equation
may be determined.

By changing the variables $\xi=f(x,y),\ \eta=g(x,y)$ where $f$ and
$g$ are defined above, show that the equation transforms to $2\eta
u_{\xi\eta}=u_{\xi}$.  Verify that this has the general solution
$u=\alpha(\xi)\sqrt{\eta}+\beta(\eta)$ for arbitrary $\alpha,\
\beta$. Hence deduce the general solution of the original equation
in the region $x>0,\ y>0$.

\medskip

{\bf Answer}

$a=x^2,\ b=0,\ c=-y^2$ so

$b^2-ac=x^2y^2$ which is $>0$ everywhere except when $x=0$ and/or
$y=c$

so hyperbolic in positive quadrant.

Characteristics:

$$\ds\frac{dy}{dx}=\pm \ds\frac{xy}{x^2}=\pm \ds\frac{y}{x}$$

$\Rightarrow y=\alpha x,\ \ \ \ \ \ \ \ \ \ y=\ds\frac{\beta}{x}$

\ \ straight lines,\ hyperbolae

i.e., $\ds\frac{y}{x}=const,\ xy=const$

Let $f=\ds\frac{y}{x},\ g=xy$ as per question

Boundary conditions are given on $x^2+y^2=1$

From lectures we can calculate Taylor series everywhere within a
domain bounded by the characteristic.

PICTURE  \vspace{2in}

\newpage
From Q7 we suspect $A$ will be important. $A$ is point
$\left(\ds\frac{1}{\sqrt{2}}, \ds\frac{1}{\sqrt{2}}\right)$. Why
is it important? The characteristics propagate the boundary
condition information from the circle

The domain of dependence can only be defined where there are two
characteristics which have come from the boundary condition
defining curve, bringing information. If we assume that the
boundary conditions are suitably defined on the arc then we have
the domain of dependence shown in the following diagram:

PICTURE \vspace{2in}

Domain of dependence: $g<1\ \ x,y>0$

There is a possible discontinuity along $y=x$.

Since hyperbolic, changing to $\xi=\ds\frac{y}{x},\ \eta=xy$ we
expect to reduce to:

$u_x=-\ds\frac{y}{x^2}u_{\xi}+yu_{\eta}$

$u_y=-\ds\frac{1}{x}u_{\xi}+xu_{\eta}$

$\xi_x=-\ds\frac{Y}{x},\ \xi_y=\ds\frac{1}{x},\ \eta_x=y,\
\eta_y=x$

\begin{eqnarray*} u_{xx} & = &
\ds\frac{+2y}{x^3}u_{\xi}+y(u_{\eta\xi}\xi_x+u_{\eta\eta}\eta_x)
-\ds\frac{y}{x}(u_{\xi\xi}\xi_x+u_{\xi\eta}\eta_x)\\ & = &
\ds\frac{2y}{x^3}u_{\xi}-\ds\frac{y^2}{x^2}u_{\eta\xi}+
y^2u_{\eta\eta}+\ds\frac{y^2}{x^4}u_{\xi\xi}-\ds\frac{y^2}{x^2}u_{\xi\eta}\\
& = &
\ds\frac{y^2}{x^4}u_{\xi\xi}-2\ds\frac{y^2}{x^2}u_{\xi\eta}+y^2u_{\eta\eta}
+\ds\frac{2y}{x^3}u_{\xi} \end{eqnarray*}

\begin{eqnarray*} u_{yy} & = &
\ds\frac{1}{x}(u_{\xi\xi}\xi_y+u_{\xi\eta}\eta_y)+x(u_{\eta\xi}\xi_y+u_{\eta\eta}\eta_y)\\
& = & \ds\frac{1}{x^2}u_{\xi\xi}+2u_{\xi\eta}+x^2u_{\eta\eta}
\end{eqnarray*}

\newpage
Thus equation becomes:

$\ \
\ds\frac{y^2}{x^2}u_{\xi\xi}-2y^2u-{\xi\eta}+x^2y^2u_{\eta\eta}+2\ds\frac{y}{x}u_{\xi}$

$=\ds\frac{y^2}{x^2}u_{\xi\xi}+2y^2u_{\xi\eta}+x^2y^2u_{\eta\eta}$

i.e., $2\ds\frac{y}{x}u_{\xi}=4y^2u_{\xi\eta}$

or $2\xi u_{\xi}=4\xi\eta u_{\xi\eta}$

or $2\eta u_{\xi\eta}=u_{\xi}$ as required.

Substitute in suggested solution
$u=\alpha(\xi)\sqrt{\eta}+\beta(\eta)$ to confirm it satisfies
equation.

Hence original equation has general solution

$$\un{u(x,y)=\alpha\left(\ds\frac{y}{x}\right)\sqrt{xy}+\beta(xy)}$$

\end{document}
