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\begin{center}
Euler equation
\end{center}

\begin{align*}
x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} &=\alpha u \\
\sum_{k=1}^{n}x_{k}\frac{\partial u}{\partial x_{k}} &=\alpha u
\end{align*}

Initial condition: \ \ 
\begin{align*}
u(x,1) &=h(x) \\
u(x_{1},x_{2},...,x_{n-1},+1) &=h(x_{1},x_{2},...,x_{n-1})
\end{align*}

characteristic equations

\begin{align*}
\frac{\partial x_{k}}{\partial s} &= x_{k} \\
\frac{\partial u}{\partial s} &= \alpha u
\end{align*}

$\frac{dx_{k}}{x_{k}}=ds$ . \qquad At $s=0$

\begin{equation*}
x_{k}(s=0)=\left\{ 
\begin{array}{cc}
t_{k} & k=1,...,n-1 \\ 
1 & k=n%
\end{array}%
\right.
\end{equation*}

The solution is

\begin{equation*}
x_{k}=\left\{ 
\begin{array}{cc}
t_{k}e^{s} & k=1,...,n-1 \\ 
e^{s} & k=n%
\end{array}%
\right.
\end{equation*}

Then $\frac{\partial u}{\partial s}=\alpha u$ along the characteristic has
the solution

\begin{equation*}
u=e^{\alpha s}h(t_{1},t_{2},...t_{n-1})
\end{equation*}

or

\begin{align*}
u(x,y) &= e^{\alpha s}h(\frac{x}{y}) \\
u(x_{1},x_{2},...,x_{n}) &= e^{\alpha s}h(\frac{x_{1}}{x_{n}},\frac{x_{2}}{%
x_{n}},...,\frac{x_{n-1}}{x_{n}})
\end{align*}

This has the property that%
\begin{equation*}
u(\lambda x_{1},\lambda x_{2},...\lambda x_{n})=\lambda
u(x_{1},x_{2},...x_{n})
\end{equation*}

i.e. homogenous of degree $\alpha $ . Note if $\alpha <0$ then the solution
is singular at the origin.

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