Solve the given Euler PDE with initial value. The Euler PDE for a homogeneous function $u(x,y,z)$ is
$$xu_x+yu_y+zu_z=\alpha u$$
Show that the initial value problem $u(x,y,1)= h(x,y)$ has a solution $u=z^{\alpha}h(\frac{x}{z},\frac{y}{z}),z\neq 0$ and $u(\lambda x, \lambda y, \lambda z) = \lambda^{\alpha}u (x,y,z)$. 
Here is what I did:
$x(s,0)=s,y(s,0)=s,z(s,0)=1$ and $u(s,0)=h(x,y)$
The characteristic system is 
$$
\begin{cases}
\frac{dx}{dt}=x\\
\frac{dy}{dt}=y\\
\frac{dz}{dt}=z\\
\frac{du}{dt}=\alpha u 
\end{cases}
$$
Using the initial condition we get
$$x(s,t)=se^t,y(s,t)=se^t,z(s,t)=e^t\mbox{ and } u(s,t)=z^{\alpha}h(x,y)$$
I was unable to go further because 
$$J=\frac{\partial(x,y)}{\partial(s,t)}=0$$
 A: $$xu_x+yu_y+zu_z=\alpha u$$
System of differential equations for the characteristic curves :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}=\frac{du}{\alpha u}$$
A first equation of characteristic comes from : $\frac{dx}{x}=\frac{dz}{z} \quad\to\quad \frac{x}{z}=c_1$
A second equation of characteristic comes from : $\frac{dy}{y}=\frac{dz}{z} \quad\to\quad \frac{y}{z}=c_2$
A fird equation of characteristic comes from : $\frac{du}{u}=\frac{dz}{\alpha z} \quad\to\quad \frac{u}{z^{\alpha}}=c_3$
The general solution of the PDE expressed on the form of implicite equation $\Phi(c_1,c_2,c_3)=0$ is :
$$\Phi\left(\frac{x}{z},\frac{y}{z},\frac{u}{z^{\alpha}}\right)=0$$
where $\Phi$ is any differentiable function of three variables. In fact it is a way to express any relationships between the three variables. An equivalent form consists to express any relationship between one of the variables and the two others :
$\frac{u}{z^{\alpha}}=F\left(\frac{x}{z},\frac{y}{z}\right)$ where $F$ is any differentiable function of two variables. 
$$u(x,y,z)=z^{\alpha}F\left(\frac{x}{z},\frac{y}{z}\right)$$
With the condition $u(x,y,1)=h(x,y)=1^{\alpha}F\left(\frac{x}{1},\frac{y}{1}\right) =F(x,y)$ the function $F$ is determined :
$$F(X,Y)=h(X,Y)$$
where $X$ and $Y$ are any dummy variables. Hence, with $X=\frac{x}{z}$ and $Y=\frac{y}{z}$ the particular solution consistent with the given condition is :
$$u(x,y,z)=z^{\alpha}h\left(\frac{x}{z},\frac{y}{z}\right)$$
A: I will start with the second question. It is easy to show that the solution to this PDE is an homogeneous function, to do this, call $x'=\lambda x$, $y' = \lambda y$ and $z' = \lambda z$, and then note that
$$
\frac{d}{d\lambda}(\lambda^\alpha u(x, y, z)) = \alpha \lambda^{\alpha -1}u(x,y,z) = \frac{d}{d\lambda}u(\lambda x, \lambda y, \lambda z) = \frac{\partial (\lambda x)}{\partial \lambda}\frac{\partial u}{\partial x'}
+ \frac{\partial (\lambda y)}{\partial \lambda}\frac{\partial u}{\partial y'} + \frac{\partial (\lambda z)}{\partial \lambda}\frac{\partial u}{\partial z'}
$$
now, replacing $\lambda = 1$ 
$$
\alpha u(x, y, z) = x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}+ z \frac{\partial u}{\partial z}
$$
Now the second problem. Call $\lambda = 1/z$ with $z\ne 0$, so that the expression $u(\lambda x, \lambda y, \lambda z) = \lambda^\alpha u (x, y, z)$ becomes 
$$
u(x/z, y/z, 1) = 1/z^\alpha u (x, y, z) = h(x/z, y/z)
$$
and then you get your solution
$$
u(x, y, z) = z^\alpha h(x/z, y/z)
$$
