\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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\begin{document}

{\bf Question}

Consider the two dimensional wave equation,

$$c^2\ds\frac{\pl^2 u}{\pl x^2}-\ds\frac{\pl^2 u}{\pl y^2}=0.$$

with Cauchy boundary conditions:

$$u(x,0)=f(x),\ u_y(x,0)=g(x0.$$

Recall from lectures that the general solution is given by

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

Assuming that $f$ and $g$ are suitably differentiable, calculate
the Taylor expansion of this solution about $9x,y)=(0,0)$ up to
second order,

\begin{description}
\item[(i)]
directly from the exact solution

\item[(ii)]
by parametrising the given boundary conditions on given curve
$C:(x,y)=(X(\tau),Y(\tau))=(\tau,0)$.

Confirm that the two agree.
\end{description}

\medskip

{\bf Answer}

$c^2u_{xx}-u_{yy}=0$

$(1)\ \ u(x,0)\ =f(x)$

$(2)\ \ u_y(x,0)=g(x)$

$\Rightarrow X(\tau)=\tau,\ Y(\tau)=0$

Data is given on $x$-axis. Hence $C$ is parametrised as:

On $C: (x,y)=(X(\tau),Y(\tau))$


$(3) \Rightarrow \left\{\begin{array}{ll} \dot{X}=1,\ \dot{Y}=0\\
\ddot{X}=0,\ \ddot{Y}=0 \end{array}\right.$

\begin{description}
\item[(i)]
Must expand exact solution as Taylor about $(0,0)$

$$u(x,y)=\ds\frac{1}{2}[f'(x+cy)+f(x-cy)]+\ds\frac{1}{2c}\ds\int_{x-cy}{x+cy}
g(s) \,ds$$

$\left\{ \begin{array} {rcl} \ds\frac{\pl u}{\pl x} & = &
\ds\frac{1}{2}[f'+f']+\ds\frac{1}{2c}[g(x+cy)-g(x-cy)]\\
\Rightarrow \left.\ds\frac{\pl u}{\pl x}\right|_{(0,0)} & = &
\ds\frac{1}{2}(f'(0)+f'(0)]+\ds\frac{1}{2c}[g(0)-g(0)]=\un{f'(0)}
\end{array} \right.$

$\left\{ \begin{array} {rcl} \ds\frac{\pl u}{\pl y} & = &
\ds\frac{1}{2}[cf'-cf']+\ds\frac{c}{2c}[g(x+cy){\bf{+}}g(x-cy)]\\
\Rightarrow \left.\ds\frac{\pl u}{\pl y}\right|_{(0,0)} & = &
\ds\frac{1}{2}(cf'(0)-cf'(0)]+\ds\frac{1}{2}[g(0)+g(0)]=\un{g(0)}
\end{array} \right.$

Also
$u(0,0)=\ds\frac{1}{2}[f(0)+f(0)]+\ds\frac{1}{2c}\ds\int_0^0g(s)\,ds=\un{f(0)}$

$\ds\frac{\pl^2 u}{\pl x^2} =
\ds\frac{1}{2}[f"+f"]+\ds\frac{1}{2c}[g'-g'] \Rightarrow
\left.\ds\frac{\pl^2 u}{\pl x^2}\right|_{(0,0)}=f"(0)$

$\ds\frac{\pl^2 u}{\pl x\pl y} =
\ds\frac{1}{2}[cf"-cf"]+\ds\frac{1}{2c}[cg'+cg'] \Rightarrow
\left.\ds\frac{\pl^2 u}{\pl x\pl y}\right|_{(0,0)}=g'(0)$

$\ds\frac{\pl^2 u}{\pl y^2} =
\ds\frac{1}{2}[c^2f"+c^2f"]+\ds\frac{1}{2c}[c^2g'-c^2g']
\Rightarrow \left.\ds\frac{\pl^2 u}{\pl
y^2}\right|_{(0,0)}=c^2f"(0)$

Thus the Taylor expansion around $(0,0)$ is

\begin{eqnarray*} u(x,y) & = & u(0,0)_+x\left.\ds\frac{\pl u}{\pl
x}\right|_{(0,0)}+y\left.\ds\frac{\pl u}{\pl y}\right|_{(0,0)}\\ &
& +\ds\frac{x^2}{2}\left.\ds\frac{\pl^2 u}{\pl
x^2}\right|_{(0,0)}+xy\left.\ds\frac{\pl^2 u}{\pl x\pl y
}\right|_{(0,0)}+\ds\frac{y^2}{2}\left.\ds\frac{\pl^2 u}{\pl
y^2}\right|_{(0,0)}+\cdots \end{eqnarray*}

\begin{eqnarray*} u(x,y) & = & f(0)+xf'(0)+yg(0)\\ & &
+\ds\frac{x^2}{2}f"(0)+xy g'(0)+\ds\frac{y^2c^2}{2}f"(0)+\cdots
\end{eqnarray*}

\item[(ii)]
Seek $\left.u\right|_{(0,0)},\ \left.u_{x}\right|_{(0,0)},\
\left.u_{y}\right|_{(0,0)},\ \left.u_{xx}\right|_{(0,0)},\
\left.u_{xy}\right|_{(0,0)},\ \left.u_{yy}\right|_{(0,0)}$ from
equation and boundary data only.

From $(1)\ \ \ u(0,0)=f(0)$

From $(2)\ \ \ u_y(0,0)=g(0)$

Get $u_x$ by differentiating $(1)$ with respect to $\tau$:

$\ds\frac{d(1)}{d\tau}:=
\left.u_{x}\right|_C\dot{X}+\left.u_{y}\right|_C\dot{Y}=f_x\dot{X}
\Rightarrow \un{\left.u_{x}\right|_C=f'(x)}$

Now seek $\left.u_{xx}\right|_{(0,0)},\
\left.u_{xy}\right|_{(0,0)},\ \left.u_{yy}\right|_{(0,0)}$ etc.
from 3 equations.

First is PDE itself:

$(4)$ \un{$x^2\left.u_{xx}\right|_C-\left.u_{yy}\right|_C=0$}

Second is

$\ds\frac{d(2)}{d\tau}:=
\left.u_{yx}\right|_C\dot{X}+\left.u_{yy}\right|_C\dot{Y}=g_x\dot{X}
\Rightarrow \un{\left.u_{xy}\right|_C=g'(x)}$ from $(3)$

Third is from

$\ds\frac{d^2(1)}{d\tau^2}:=$

$\left.u_{xx}\right|_C\dot{X}^2+\left.u_{xy}\right|_C\dot{X}\dot{Y}
+\left.u_{x}\right|_C\ddot{X}$

$+\left.u_{yx}\right|_C\dot{Y}\dot{X}+\left.u_{yy}\right|_C\dot{Y}^2
+\left.u_{y}\right|_C\ddot{Y}=f_{xx}\dot{X}^2+f_x\ddot{X}$

$\Rightarrow \un{\left.u_{xx}\right|_C=f_{xx}=f"(x)}$ from $(3)$

Thus we have

$\begin{array}{c} (7)\\(6)\\(5) \end{array} \begin{array}{c}
\rightarrow\\\rightarrow\\\rightarrow \end{array}
\left(\undb{\begin{array} {ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ c^2 & 0 &
-1 \end{array}} \right)\left(\begin{array}{c}u_{xx}\\u_{xy}\\
u_{yy}\end{array}\right)=\left(\begin{array}{c} f"\\g'(x)\\0
\end{array}\right)$

So $\bigtriangleup = det({})=-1 \ne$ anywhere on $C$ so can find
solution

Not difficult to solve

$\Rightarrow \left\{\begin{array} {rcl} u_{xx} & = & f''(x)\\
u_{xy} & = & g'(x)\\ u_{yy} & = & c^2f''(x) \end{array} \right\}\
(B)$

Hence expanding about $(0,0)$ we have:

\begin{eqnarray*} u(x,y) & = & f(0)\ \rm{(from\ (1)}\ +xf'(0)\
\rm{(from\ (A)}\ +yg(0)\ \rm{(from\ (2)}\\ & &
+\undb{\ds\frac{x^2}{2}f''(0)+xyg'(0)+\ds\frac{y^2c^2}{2}f''(0)}+\cdots
\end{eqnarray*}

which is exactly the same as the result of part (i) as required.
\end{description}
\end{document}