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\begin{document}

{\bf Question}

Consider the two dimensional Laplace equation.

$$\ds\frac{\pl ^2 u}{\pl x^2}+\ds\frac{\pl ^2 u}{\pl y^2}=0$$

with Cauchy boundary conditions:

$$u(x,0)=f(x),\ u_y(x,0)=g(x).$$

Recognising Laplace's equation as a constant coefficient second
order linear PDE, show that the general solution is given by

$$u(x,y)=F(x+iy)+G(x-iy)$$

What can you deduce about the characteristaics of this equation?
Solve the equation for the given boundary conditions and show that

$$u(x,y)=\ds\frac{1}{2}[f(x+iy)+f(x-iy)]-\ds\frac{i}{2}\ds\int_{x-iy}^{x+iy}
g(s)ds.$$

Consider now the specific problems

\begin{description}
\item[(i)]
$f(x)=\ds\frac{1}{1+x^2},\ g(x)=0$

\item[(ii)]
$f(x)=exp(-x^2),\ g(x)=0$

\end{description}

Where are the singularities of the boundary conditions? Where are
the singularities of the solution $u(x,y)$?

This question demonstrates that although it is possible to
construct a solution of Laplace's equation with Cauchy boundary
conditions, such solutions often pose problems. Real singularities
can appear in the solution that depend on the complex
singularities of the boundary conditions or unbounded behaviour
can occur in one direction even thought the b.c. was well defined.
In fact the only solution that is analytic everywhere is
$u=const$. Cauchy boundary conditions are therefore often
inappropriate for Laplace's equation with Direchlet or Neumann
conditions giving unique bounded solutions in their respective
domains of interest.


\medskip

\newpage
{\bf Answer}

Use above method (Q2) or just substitute

$$u=F(x+iy)+G(x-iy) \ \ \ (0)$$

$\begin{array} {lclclcl} u_x &amp; = &amp; F'+G' &amp; ; &amp; u_y &amp; = &amp; iF'-iG'\\
u_{xx} &amp; = &amp; F"+G" &amp; ; &amp; u_{yy} &amp; = &amp; -F"-G" \end{array}$

$$u_{xx}+u_{yy}=F"+G"-F"-G"=0$$

Characteristics are \un{complex}.

BC: $u(x,0)=f(x) \Rightarrow F(x)+G(x)=f(x)\ \ \ (1)$

BC: $u(x,0)=g(x) \Rightarrow iF'(x)-iG'(x)=g(x)\ \ \ (2)$

$(2)$ gives $F(x)-G(x)=\ds\frac{1}{i}\ds\int_0^x g(s)\,ds$

Thus

$\left\{\begin{array} {rcl} F(x)+G(x) &amp; = &amp; f(x)\\ F(x)-G(x) &amp; = &amp;
-i\ds\int_0^x g(s)ds \end{array} \right.$

$\Rightarrow$

$\left. \begin{array} {rcl} F(x) &amp; = &amp;
\ds\frac{1}{2}f(x)-\ds\frac{i}{2}\ds\int_0^xg(s)\,ds\\ G(x) &amp; = &amp;
\ds\frac{1}{2}f(x)+\ds\frac{i}{2}\ds\int_0^xg(s)\,ds \end{array}
\right\}\ \ (3)$

so inserting $(3)$ in $(0)$
$$\un{u=\ds\frac{1}{2}[f(x+iy)+f(x-iy)]-\ds\frac{i}{2}\ds\int_{x-iy}^{x+iy}g(s)\,ds\
\ \ (4)}$$

as required.

\begin{description}
\item[(i)]
singularities at $x\pm i$ (i.e., nowhere real)

\item[(ii)]
singularities at $x=\pm -\infty$ (i.e., nowhere finite)

substitute (i) into $(4)$; no contribution from integral

$$u=\ds\frac{1}{2}\left[\ds\frac{1}{1+(x+iy)^2}+\ds\frac{1}{1+(x-iy)^2}\right]$$

This is a real and finite function except where $x=0,\ y=\pm$,
where it blows up.

Complex boundary condition singularities here

$\Rightarrow$ \un{real solution singularities}.

\newpage
Substitute (ii) into $(4)$; no contribution from integral,

\begin{eqnarray*} u &amp; = &amp;
\ds\frac{1}{2}[e^{-(x+iy)^2}+e^{-(x-iy)^2}]\\ &amp; = &amp;
e^{-x^2+y^2}\cos 2xy \end{eqnarray*}

This function increases unboundedly for $y&gt;0$. Thus singularity of
boundary condition at infinity leads to bad solution.

\end{description}
\end{document}
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