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\begin{document}

{\bf Question}

Confirm the solution of Q6 by showing that it is not possible to
calculate the second order derivatives in the Taylor expansion of
$u(x,y)$ at the point corresponding to $\alpha$. Then calculate
the second order Taylor expansion of the solution about the point
$(x,y)=(1,0)$.

\medskip

{\bf Answer}

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

$\left.\begin{array} {crcl} (1) & u(\cos\theta,\sin\theta) & = &
\theta\\ (2) & u_x(\cos\theta,\sin\theta) & = & 0 \end{array}
\right\} C: (x,y)=(X(\theta),Y(\theta))=(\cos\theta,\sin\theta)$

Seek $u$ (from $(1)$),\ $u_x$ (from $(2)$),\ $u_y,\ u_{xx},\
u_{xy},\ u_{yy}$ etc. on curve$C$.

Then use Taylor expansion to find $u(x,y)$ \un{off} $C$ as in
notes.

First:

$\begin{array} {rclcrcl} \dot{X}(\theta) & = & -\sin\theta & , &
\dot{Y}(\theta) & = & \cos\theta\\ \ddot{X}(\theta) & = &
-\cos\theta & , & \ddot{Y}(\theta) & = & -\sin\theta \end{array}$

Differentiate boundary conditions

$\ds\frac{d(1)}{d\theta}:=$

$\left.u_{x}\right|_C\dot{X}+\left.u_{y}\right|_C\dot{Y}=1$

$\Rightarrow \undb{0} \times
(-\sin\theta)+\left.u_{y}\right|_C\cos\theta=1$

\ \ \ from $(2)$

$$\Rightarrow \un{\left.u_{y}\right|_C=\sec\theta}\ \ \ (3)$$

Thus we now know $u,\ u_x,\ u_y$ on $C$. To find $u_{xx},\
u_{xy},\ u_{yy}$ on $C$ we need 3 equations, the PDE itself is the
first:

$(4) 2\left.u_{xx}\right|_C-\left.u_{xy}\right|_C+
2\left.u_{yy}\right|_C+\left.u_{x}\right|_C-3\left.u\right|_C=0$

$\Rightarrow 2\left.u_{xx}\right|_C-\left.u_{xy}\right|_C+
2\left.u_{yy}\right|_C+0-3\theta=0$

or $\un{2\left.u_{xx}\right|_C-\left.u_{xy}\right|_C+
2\left.u_{yy}\right|_C=3\theta}$ on $C\ \ \ (5)$

Second equation comes from:

$\ds\frac{d(2)}{d\theta}:=$

$\left.u_{xx}\right|_C\dot{X}+\left.u_{xy}\right|_C\dot{Y}=0$

$\Rightarrow
\un{-\left.u_{xx}\right|_C\sin\theta+\left.u_{xy}\right|_C\cos\theta=0}\
\ \ (6)$

\newpage
Third equation comes from

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

\begin{eqnarray*} & & \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_{y}\right|_C\dot{Y}\dot{X}+\left.u_{yy}\right|_C\dot{Y}^2
+\left.u_{y}\right|_C\ddot{Y}=0 \end{eqnarray*}

$\Rightarrow$
\begin{eqnarray*} & & \left.u_{xx}\right|_C\sin^2\theta-2\sin\theta\cos\theta\left.u_{xy}\right|_C+0
\times(-\cos\theta)\\ & + &
\left.u_{yy}\right|_C\cos^2\theta+\sec\theta\times(-\sin\theta)=0
\end{eqnarray*}

$\Rightarrow
\un{\sin^2\theta\left.u_{xx}\right|_C-2\sin\theta\cos\theta\left.u_{xy}\right|_C
+\cos^2\theta\left.u_{yy}\right|_C=\tan\theta}$\ \ \ $(7)$

Thus we have from $(5),\ (6)$ and $97)$

$\begin{array}{c} (7)\\(6)\\(5) \end{array} \begin{array}{c}
\rightarrow\\\rightarrow\\\rightarrow \end{array}
\left(\undb{\begin{array} {ccc} \sin^2\theta &
-2\sin\theta\cos\theta & \cos^2\theta\\ -\sin\theta & \cos\theta &
0\\ 2 & -5 & 2 \end{array}}
\right)\left(\begin{array}{c}\left.u_{xx}\right|_C\\\left.u_{xy}\right|_C\\
\left.u_{yy}\right|_C\end{array}\right)=\left(\begin{array}{c}
\tan\theta\\0\\3\theta \end{array}\right)$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$\bigtriangleup=det({})$

To solve this we need $\bigtriangleup \ne 0$.

\begin{eqnarray*} & & \bigtriangleup\\ & = &
\sin^2\theta\left|\begin{array}{cc} \cos\theta & 0\\ -5 & 2
\end{array} \right|\\ & & +2\sin\theta\cos\theta\left|\begin{array}{cc} -\sin\theta & 0\\ 2 & 2
\end{array} \right|+\cos^2\theta\left|\begin{array}{cc} -\sin\theta & \cos\theta\\ 2 &
-5 \end{array} \right|\\ & = &
\cos\theta(5\sin\theta\cos\theta-2)\ \rm{after\ algebra}
\end{eqnarray*}

Thus $\bigtriangleup =0$ when $\cos\theta=0 \Rightarrow
\theta=\ds\frac{\pi}{2},\ -\ds\frac{\pi}{2}$,\ etc.

\un{or} $5\sin\theta\cos\theta=2$.

Now remember that from Q6 $\alpha=\tan^{-1}\ds\frac{1}{2}$

PICTURE \vspace{1in}

$\Rightarrow \sin\alpha=\ds\frac{1}{\sqrt{5}},\
\cos\alpha=\ds\frac{2}{\sqrt{5}}$

$\Rightarrow 5\sin\alpha\cos\alpha=5\times\ds\frac{1}{\sqrt{5}}
\times \ds\frac{2}{\sqrt{5}}=2(!)$

Thus it is not possible to find

$$\left.u_{xx}\right|_C,\ \left.u_{xy}\right|_C,\
\left.u_{yy}\right|_C$$

when $\theta=\alpha=\tan^{-1}\frac!{}{2}$ as found in Q6.

OK, now assume that $0 \leq \theta < \alpha$

Then solutions of $(8)$ are given by Cramer's rule (see lecture
notes)

$\left\{\begin{array} {rcl} \left.u_{xx}\right|_C & = &
\ds\frac{1}{\bigtriangleup} \left|\begin{array}{ccc} \tan\theta &
-2\sin\theta\cos\theta & \cos^2\theta\\ 0 & \cos\theta & 0\\
3\theta & -5 & 2 \end{array} \right|\\ & & \\
\left.u_{xy}\right|_C & = &\ds\frac{1}{\bigtriangleup}
\left|\begin{array}{ccc} \sin^2\theta & \tan\theta &
\cos^2\theta\\ -\sin\theta & 0 & 0\\ 2 & 3\theta & 2\end{array}
\right|\\ & & \\ \left.u_{yy}\right|_C & = &
\ds\frac{1}{\bigtriangleup} \left|\begin{array}{ccc} \sin^3\theta
& -2\sin\theta\cos\theta & \tan\theta\\ -\sin\theta & \cos\theta &
0\\ 2 & -5 & 3\theta
\end{array} \right| \end{array}\right.$

$\Rightarrow$

$\left\{\begin{array} {rcl} \left.u_{xx}\right|_C & = &
\ds\frac{1}{\bigtriangleup}
\left\{\tan\theta\left|\begin{array}{cc} \cos\theta & 0\\ -5 & 2
\end{array}\right|+2\sin\theta\cos\theta\left|\begin{array}{cc} 0 & 0\\ 3\theta & 2
\end{array}\right|\right.\\ & & \left.+\cos^2\theta\left|\begin{array}{cc} 0 & \cos\theta\\ 3\theta &
-5 \end{array}\right|\right\}
\\ \left.u_{xy}\right|_C & = & \ds\frac{1}{\bigtriangleup}
\left\{\sin^2\theta\left|\begin{array}{cc} 0 & 0\\ 3\theta & 2
\end{array}\right|-2\tan\theta\left|\begin{array}{cc} -\sin\theta & 0\\ 2 & 2
\end{array}\right|\right.\\ & & \left.+\cos^2\theta\left|\begin{array}{cc} -\sin\theta & 0\\ 2 &
3\theta \end{array}\right|\right\}\\ \left.u_{yy}\right|_C & = &
\ds\frac{1}{\bigtriangleup}
\left\{\sin^2\theta\left|\begin{array}{cc} \cos\theta & 0\\ -5 &
3\theta
\end{array}\right|+2\sin\theta\cos\theta\left|\begin{array}{cc}
-\sin\theta & 0\\ 2 & 3\theta
\end{array}\right|\right.\\ & & \left.+\tan\theta\left|\begin{array}{cc} -\sin\theta & \cos\theta\\ 2 &
-5 \end{array}\right|\right\}
\end{array}\right.$

$\left\{\begin{array} {rcl} \left.u_{xx}\right|_C & = &
\ds\frac{1}{\bigtriangleup} \{2\sin\theta-3\theta\cos^3\theta\}\\
& &
\\ \left.u_{xy}\right|_C & = &\ds\frac{\tan\theta}{\bigtriangleup}
\{2\sin\theta-3\theta\cos^3\theta\}
\\ & &
\\ \left.u_{yy}\right|_C & = &
\ds\frac{\tan\theta}{\bigtriangleup}\{5\sin\theta-2\cos\theta-3\theta\sin\theta\cos^2\theta\}
\end{array}\right.\ \ \ (9)$

Hence when $\theta=0 \Longleftrightarrow (x,y)=(1,0)$

we have $\bigtriangleup=\cos 0 (5\times 0 \times 1-2)=-2$

So

$(1) \Rightarrow u(1,0)=0$

$(2) \Rightarrow u_x(1,0)=0$

$(3) \Rightarrow u_y(1,0)=\sec 0=1$

$(9) \Rightarrow \left\{ \begin{array} {rclcl} u_{xx}(1,0) & = &
-\ds\frac{1}{2}(0-0) & = & 0\\u_{xy}(1,0) & = &
\ds\frac{0}{-2}(0-0) & = & 0\\u_{yy}(1,0) & = &
\ds\frac{0}{-2}(0-2-0) & = & 0
\end{array} \right.$

Thus, to second order

\begin{eqnarray*} u(x,y) & = & u(1,0)+(x-1)\left.\ds\frac{\pl u}{\pl
x}\right|_{(1,0)}\\ & & +y\left.\ds\frac{\pl u}{\pl
x}\right|_{(1,0)}+\ds\frac{(x-1)^2}{2}\left.\ds\frac{\pl^2 u}{\pl
x^2}\right|_{(1,0)}\\ & & +(x-1)y\left.\ds\frac{\pl^2 u}{\pl x\pl
y }\right|_{(1,0)}+y^2\left.\ds\frac{\pl^2 u}{\pl
y^2}\right|_{(1,0)}+\cdots\\ & = & 0+(x-1)\times 0+y\times
1+0+0+0+\cdots\\ & = & \un{y+\rm{3rd\ order\ terms!}}
\end{eqnarray*}

\end{document}