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\begin{document}

{\bf Question}

Let $f(x,y)$ satisfy

$$\left\{\begin{array}{rl} (x+1)\ds\frac{\pl f}{\pl
x}+\ds\frac{\pl f}{\pl y}=x &amp; \rm{for}\ x&gt;0,\ y&gt;0\\ f(0,y)=q(y), &amp;
f(x,0)=r(x) \end{array} \right.$$

Show that the Laplace transform in $y,\ \overline{f}(x)$ say,
satisfies

$$\left\{\begin{array}{rcl}
(x+1)\ds\frac{d\overline{f}}{dx}+p\overline{f} &amp; = &amp;
p^{-1}x+r(x)\\ \overline{f}(0) &amp; = &amp; \overline{q} \end{array}
\right.$$

Solve this for $\overline{f}$ in the special case

$$q(y)=y,\ r(x)=0.$$

Use the inversion integral to calculate $f(x,y)$.

\medskip

{\bf Answer}

Transform the equation and boundary conditions

$$\ds\int_0^{\infty} dy\ (x+1) \ds\frac{\pl f}{\pl
x}e^{-py}+\ds\frac[-^{\infty}\ds\frac{\pl f}{\pl y} e^{-py} dy =
\ds\int_0^{\infty}dy\ x e{-py}$$

becomes, by standard methods:

$$(x+1)\ds\frac{\pl \bar{f}}{\pl
x}+p\bar{f}-\undb{f(x,0)}=\ds\frac{x}{p}$$

\hspace{2.8in} $r(x)$

Also the boundary conditions:

$$\ds\int_0^{\infty}f(0,y)e^{-py}dy
=\ds\int_0^{\infty}q(y)e^{-py}dy=\bar{q}=\bar{f}(0)$$

as required.

Must now solve this ODE and transformed boundary condition. ODE is
linear and has an integrating factor: special case given is:

$$(x+1)\ds\frac{\pl \bar{f}}{\pl{x}}+p\bar{f}=\ds\frac{x}{p},\
\bar{f}(0)=\ds\int_0^{\infty}y e^{-p}y\ dy=\ds\frac{1}{p^2}$$

Integrating factor $=e^{\int\frac{p}{x+1}}\
dx=e^{p\ln(1+x)}=(1+x)^p$

$\begin{array}{crcl} \Rightarrow &amp; \ds\frac{\pl \bar{f}}{\pl
x}(1+x)^p+p(1+x)^{p-1}\bar{f} &amp; = &amp; \ds\frac{x}{(1+x)p}(1+x)^p\\
\Rightarrow &amp; \ds\frac{\pl}{\pl x}[(1+x)^p \bar{f}] &amp; = &amp;
\ds\frac{x}{(1+x)p}(1+x)^p\\ \Rightarrow &amp; (1+x)^p\bar{f} &amp; = &amp;
\ds\int\ds\frac{x}{p}(1+x)^{p-1}\ dx\ +c\\ &amp; &amp; = &amp;
\undb{\ds\frac{x(1+x)^p}{p^2}-\ds\frac{(1+x){p+1}}{p^2(p+1)}}+c\\
&amp; &amp; &amp; \rm{by\ integration\ by\ parts}\\ \Rightarrow &amp; \bar{f} &amp; =
&amp;
\ds\frac{x}{p^2}-\ds\frac{(1+x)}{p^2(p+1)}+\ds\frac{c}{(1+x)^p}\\
&amp; \bar{f}(0) &amp; = &amp; \ds\frac{1}{p^2}\\ \Rightarrow &amp;
\ds\frac{1}{p^2} &amp; = &amp; -\ds\frac{1}{p^2(p+1)}+c\\ &amp; &amp; &amp;
\Rightarrow c=\ds\frac{1}{p^2}+\ds\frac{1}{p^2(p+1)} \end{array}$

\begin{eqnarray*} \Rightarrow \bar{f} &amp; = &amp;
\ds\frac{x}{p^2}-\ds\frac{(1+x)}{p^2(p+1)}+\ds\frac{(p+2)}{(p+1)p^2}\ds\frac{1}{(1+x)^p}\\
&amp; = &amp;
\ds\frac{xp+x-1-x}{p^(p+1)}+\left[\ds\frac{2}{p^2}-\ds\frac{1}{p(p+1)}\right]
\ds\frac{1}{(1+x)^p}\\ &amp; = &amp;
\ds\frac{xp+p-1-p}{p^2(p+1)}+\left[\ds\frac{2}{p^2}-\ds\frac{1}{p(p+1)}\right]
\ds\frac{1}{(1+x)^p}\\ &amp; = &amp;
\ds\frac{(x+1)}{p(p+1)}-\ds\frac{1}{p^2}+\left[\ds\frac{2}{p^2}-
\ds\frac{1}{p(p+1)}\right]\ds\frac{1}{(1+x)^p}\end{eqnarray*}

So inversion integral is

$f(x,y)$

$=\ds\frac{1}{2\pi i}\ds\int dp\
\left\{\ds\frac{(x+1)}{p(p+1)}-\ds\frac{1}{p^2}+
\left[\ds\frac{2}{p^2}-\ds\frac{1}{p(p+1)}\right]
\ds\frac{1}{(1+x)^p}\right\}e^{py}$

\hspace{.45in} \setlength{\unitlength}{.1in}\begin{picture}(0,2)

\put(0,0){\vector(0,1){1}}

\put(0,1){\line(0,1){1}}

\end{picture} \hspace{.55in} (1) \hspace{.3in} (2) \hspace{.5in} (3)

(1):

$\ds\frac{1}{2\pi i}\ds\int dp\
\ds\frac{(x+1)}{p(p+1)}e^{py}=(x+1)[1\ -\ e^{-y}]$

\hspace{.6in} \setlength{\unitlength}{.1in}\begin{picture}(0,2)

\put(0,0){\vector(0,1){1}}

\put(0,1){\line(0,1){1}}

\end{picture} \hspace{1.4in} $\stackrel {\uparrow}{p=0}\ \stackrel{\uparrow}{p=-1}$

complete to \un{left} since $y&gt;0$. Simple poles at $p=0,\ -1$.

Semicircular contribution $\to 0$
\newpage
(2):

$\ds\frac{1}{2\pi i}\ds\int{dp\ e^{py}}{-p^2}$ complete to
\un{left} since $y&gt;0$. \un{Double} pole at $p=0$. Semicircular
contribution $\to 0$

$=-\lim_{p \to 0} \left\{\ds\frac{1}{1!}\ds\frac{\pl}{\pl
p}e^{py}\right\}$

$= -\lim_{p \to 0}(ye^{py})$

$= -y$

(3):

must decide whether $y=\ln(1+x)&gt;$ or $&lt;0$. This affects which side
you complete on:


$\ds\frac{1}{2\pi i}\ds\int dp\
\left(\ds\frac{2}{p^2}-\ds\frac{1}{p(p+1)}\right)\ds\frac{e^{py}}{(1+x)^p}$

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$=\ds\frac{1}{2\pi i}\ds\int dp
\left(\ds\frac{2}{p^2}-\ds\frac{1}{p(p+1)}\right)e^{p(y-\log(1+x))}$

\hspace{.45in} \setlength{\unitlength}{.1in}\begin{picture}(0,2)

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\end{picture}

\un{$y&gt;\log(1+x)$}: complete to \un{left} enclose poles at 0 and
$-1$. Semicircular contribution vanishes

(3):

$\ds\frac{1}{2\pi i}\ds\int dp\
\left(\ds\frac{2}{p^2}-\ds\frac{1}{p(p+1)}\right)e^{p(y-\log(1+x))}$

\hspace{.3in} \setlength{\unitlength}{.1in}\begin{picture}(0,2)

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\end{picture}

$=\ds\frac{1}{2\pi i} \ds\int dp\ \ds\frac{2}{p^2}
e^p(y-\log(1+x))-\ds\frac{1}{2\pi i}\ds\int \ds\frac{dp\
e^p(y-\log(1+x))}{p(p+1)}$

\hspace{.45in} \setlength{\unitlength}{.1in}\begin{picture}(0,2)

\put(0,0){\vector(0,1){1}}

\put(0,1){\line(0,1){1}}

\end{picture}\ \ \ \ \ double pole at $p=0$ \ \ \ \hspace{.3in}
\setlength{\unitlength}{.1in}\begin{picture}(0,2)

\put(0,0){\vector(0,1){1}}

\put(0,1){\line(0,1){1}}

\end{picture}\ \ \ 2 simple poles at $p=0$

\hspace{3.7in} and $p=-1$

$=2(y-\log(1+x))-1+e^{-y}(1+x)$

\un{$y&lt;\log(1+x)$}: complete to the \un{right}. No poles enclosed.
Semicircular contribution vanishes.

\newpage
Adding (1), (2) and (3) we get:

$$f(x,y)=\left\{\begin{array}{ll} (x+y)-2\log(1+x) &amp; y&gt;\log(1+x)\\
(x+1)(1-e^{-y})-y &amp; y&lt;\log(1+x) \end{array} \right.$$

Note that if you go back and check, $y=\log(1+x)+const$ are the
characteristics of the equation. Hence you expect some funny
behaviour across them. Check that $f(x,y)$ is continuous across
$y=\log(1+x)$ by substituting $y=\log(1+x)$ into both expressions
for $f(x,y)$ and seeing that they both reduce to

$$f(x,\ \log(1+x))=x-\log(1+x)$$

\end{document}
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