Let
$\displaystyle (x,y)\cdot Du = au,\ a>0$.
I computed the characteristics to be
$\begin{align}
\dot{\vec{x}}(s) &= \vec{x}(s)\\
\dot{z}(s)&=\vec{x}(s)\cdot\vec{p}(s)=az(s)
\end{align}$
where I used the notation from Evans – PDE, that is
$\vec x(s)=(x_1(s),x_2(s))$ being the initial curve and $z(s):=u(\vec{x}(s))$ with $\vec p(s):=Du(\vec{x}(s))$, where $D$ is the gradient.
With my characteristics I furthermore get,
$\displaystyle
\begin{align}
\frac{\mathrm{d} x_1}{\mathrm{d}s}&=x_1 \Rightarrow x_1(s)=e^{s}c_1 \\
\frac{\mathrm{d} x_2}{\mathrm{d}s}&=x_2 \Rightarrow x_2(s)=e^{s}c_2\\
\frac{\mathrm{d}\, z(s)}{\mathrm{d}s}&=az(s) \Rightarrow z(s)=e^{as}c_3
\end{align}$
I am given the condition that along the curve $\{x,1\},\, x\in\mathbb{R}$ the initial values $\phi(x)$ are given.
From here I am unsure of how to proceed. I somehow have to recover $u(x,y)$ from my characteristic equations using the initial values.
Edit: My best attempt:
I read from Evans, that I can construct a (local) solution from
$\displaystyle\begin{align}
x_1(r,s)&=e^sc_1(r) \\
x_2(r,s)&=e^sc_2(r) \\
z(r,s)&=e^{as}c_3(r)
\end{align}$
With the initial conditions I get
$\displaystyle\begin{align}
x_1(r,1)&=r \\
x_2(r,1)&=1 \\
z(r,1)&=\phi(r)\end{align}$
yielding
$\displaystyle\begin{align}
x_1&=e^{s-1}r\\
x_2&=e^{s-1} \\
z&=e^{a(s-1)}\phi(r)\end{align}$
finally leading to
$\displaystyle u=x_2^a\phi(x_1/x_2)$
Best Answer
We are almost done. We can make $c_1=1$, as it represents a simple displacementfor the parameter driving the characteristics, $s$. Then, get rid of $s$ because we are interested in the relation among $x,y$ and $u$ for the characteristics.
$y/x=c_2$ or $x/y=c'_2$
$z(s)=u(x(s),y(s))=c_3x(s)^a$ or for easy reading $u=c_3x^a$
But the integration constants must be related: $c_3=f(c'_2)$
Substituting, $u=x^af(x/y)$, being the general solution.
Now $u(x,1)=\phi(x)$
$\phi(x)=x^af(x)\to f(x/y)=(x/y)^{-a}\phi(x/y)$
leading to $u(x,y)=x^a(y/x)^a\phi(x/y)$ or
$u(x,y)=y^a\phi(x/y)$