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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
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\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
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\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
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\begin{document}


\begin{center}
Geometric Approach
\end{center}

\bigskip

Given 
\[
au_{x}+bu_{y}=0
\]

a,b constants. $au_{x}+bu_{y}$ is a directional derivative in the direction

\[
V=(a,b)=a\mathit{i}+b\mathit{j}
\]

Hence, $u$ is constant in the direction $V$. Lines parallel to $V$ have the
formula $bx-ay=$constant \ (i.e. slope $\frac{b}{a}$) = characteristic
lines. Hence, $u$ is a function of \ $bx-ay$ .

\[
u(x,y)=f(bx-ay)
\]

\bigskip Hence, on the characteristic lines $c=bx-ay$ is constant and $u=f(c)
$. The vector $(b,-a)$ is orthogonal to $V$. Hence, $u$ varies along these
lines.

\bigskip

Method \#2.

Let $x^{^{\prime }}=ax+by$ $y^{^{\prime }}=bx-ay$ \newline
By the chain rule \quad $u_{x}=au_{x^{^{\prime }}}+bu_{y^{^{\prime }}}$ $%
u_{y}=bu_{x^{^{\prime }}}-au_{y^{^{\prime }}}$.

\noindent So

$au_{x}+bu_{y}=(a^{2}+b^{2})u_{x^{^{\prime }}}$. Hence, thr original
equation is equivalent to $u_{x^{^{\prime }}}=0$.

Conclusion:%
\[
u(x,y)=f(y^{^{\prime }})=f(bx-ay)
\]

\newpage

\begin{center}
general geometric approach
\end{center}

\[
(\ast )\qquad a(x,y,u)\frac{\partial u}{\partial x}+b(x,y,u)\frac{\partial u%
}{\partial y}=c(x,y,u)
\]

initial condition: curve $\Gamma $ in (x,y,u).

Parameterize: $x=x(s)\qquad y=y(s)\qquad u=u(s)$

Solution: a surface $u=u(x,y)$ that passes through the initial curve.

\bigskip 

Given a solution surface $F(x,y,u)=0.$

$\nabla F=(F_{x},F_{y},F_{z})$ is normal to this surface.

Consider $z=u(x,y)$ \ \ \ \ or equivalently \ \ $F(x,y,u)=u(x,y)-z$

Then $\nabla F=(u_{x},u_{y},-1).$

We rewrite $(\ast )$ as \ \ \ \ $(a,b,c)\cdot (u_{x},u_{y},-1)=0.$

So $(a,b,c)$ is normal to the gradient which itself is normal to the surface.

Hence, $(a,b,c)$ lies in the tangent plane to the solution surface$.$

\end{document}
