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{\bf Question}

Find and sketch the characteristic curves for

$$2u_{xx}-5u_{xx}+2u_{yy}+u_x-3u=0$$

For what values of $\alpha$ does it have a unique solution
satisfying the conditions

$$u(\cos \theta,\sin\theta)=\theta,\ u_x(\cos
\theta,\sin\theta)=0\ \rm{for}\ 0&lt;\theta&lt;\alpha?$$



\medskip

{\bf Answer}

1st order derivatives are irrelevant for classification.

Thus:

$a=2,\ b=-\ds\frac{5}{2},\ c=2$

$b^2-ac=\left(\ds\frac{5}{2}\right)^2-4=\ds\frac{9}{4}&gt;0$
Therefore hyperbolic everywhere

Characteristic equations given by:

$$\ds\frac{Dy}{dx}=\ds\frac{0\frac{5}{2} \pm
\sqrt{\ds\frac{9}{4}}}{2}=\ds\frac{1}{2}\left(-\ds\frac{5}{2} \pm
\ds\frac{3}{2} \right)=-\ds\frac{1}{2}\ \rm{or}\ -2$$

so characteristics or $\left\{\begin{array} {rcl} y &amp; = &amp;
-\ds\frac{1}{2}+const\\ y &amp; = &amp; -2x + const \end{array} \right.$

i.e., 2 sets of straight lines

PICTURE \vspace{1.5in}

Now we're given boundary conditions as:

$u(\cos\theta,\sin\theta)=\theta,\ u_x(\cos\theta,\sin\theta)=0$
for $0&lt;\theta &lt; 2\pi$

i.e., on an arc of a circle of radius 1.

Now remember the example of lectures where characteristics
\un{carried} information from the curve on which boundary
conditions are given \un{into} the domain of a solution.

Draw first the boundary condition curve. Then superimpose a grid
given by the characteristics.

PICTURE \vspace{2.5in}

Now consider the point $B$. This is where $y=-2x+const$ is first
tangential to the circle.

Any point on the circle between $A$ and $B$, $P_1$, say, has two
characteristics passing through it and into the range of
influence.  However these characteristics given by $y=-2x+const$
can be seen to encounter the circle at another point in the first
quadrant, $P_2$, say. If we think of characteristics as
propagating information from the curve in which the boundary
condition is defined into the range of influence (cf. lecture
example), then we have a possible conflict at this second
intersection point.

PICTURE \vspace{1.5in}

In other words the information propagated from the boundary
condition at $P_1$ may be different from the information given by
the boundary condition at $P_2$.

The only way we can be sure things will work out is if we only
define the boundary conditions up to \un{$B$} (which turns out to
be $\left(\ds\frac{2}{\sqrt{5}},\ds\frac{1}{\sqrt{5}}\right)$ .
Thus we have a domain of dependence.

PICTURE \vspace{1.5in}

Thus $\alpha=\angle AOB=tan^{-1}\left(\ds\frac{1}{2}\right).$

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