On the transport equation $x \cdot \nabla u = |x|^2$ I have no clue about how to  solve the following system
\begin{equation*}
 \begin{cases}
 x\cdot\nabla u=|x|^2, \quad x\in\mathbb{R}^n, \\
 u|_{x_1=1}=3x_n.
 \end{cases}
 \end{equation*}
Study the domain where the function is defined.
Can you help me with some hints?
 A: Since 
$$
x \cdot \nabla u = |x|^2,
$$
but also
$$
x \cdot x = |x|^2,
$$
I would try
$$
\nabla u = x,
$$
which leads to 
$$
u(x) = {1 \over 2} x \cdot x + {\tt const}.
$$
In general, I would recommend looking into the method of characteristics.
A: Use the method of characteristics to decompose the problem. Observe, we have the string of equations: 
$$\frac{dx_1}{x_1} = \ldots = \frac{dx_n}{x_n} = \frac{dz}{x_1^2 + \ldots x_n^2}$$
Solving for the constant characteristics for the first n-1 equations, we get: 
$$\phi_i = \frac{x_{i-1}}{x_i}$$
So the last equation becomes: 
$$\frac{dx_n}{x_n} = \frac{dz}{x_n^2 \phi_{n-1}^2(1 + \phi_{n-2}^2(\ldots (1 + \phi_1^2) \ldots )))}$$
Which, if we allow $\psi := \phi_{n-1}^2(1 + \phi_{n-2}^2(\ldots (1 + \phi_1^2) \ldots )))$, then this boils down to: 
$$\frac{dz}{\psi} = x_n dx_n$$
$$\implies \frac{z}{\psi} = \frac{1}{2}x_n^2 + \eta(\phi_1, \ldots, \phi_{n-1})$$
$$\implies z = \psi(\frac{1}{2}x_n^2 + \eta)$$
where we defined $\psi$ already, which we know equals: 
$$\psi = \frac{|x|^2}{x_n^2}$$
and $\eta$ is a function on the constants $\phi_1, \ldots, \phi_{n-1}$.
This leads to: 
$$u = \frac{1}{2} x \cdot x + \eta(\phi_1, \ldots, \phi_{n-1})\frac{|x|^2}{x_n^2}$$
Now considering the initial conditions: 
$$u(x_1 = 1, \cdot ) = 3x_n = \frac{1}{2}\sum_{i=1}^n x_i^2 + \eta(1, \phi_2, \ldots, \phi_{n-1}) \frac{1}{x_n^2} \sum_{i=1}^n x_i^2$$
$$\implies 3x_n^3 - \frac{x_n^2}{2}\sum_{i=1}^n x_i^2 = \eta(1, \phi_2, \ldots, \phi_{n-1}) \sum_{i=1}^n x_i^2$$
$$\implies \frac{3x_n^3}{|x|^2} - \frac{x_n^2}{2} = \eta(1, \phi_2, \ldots, \phi_{n-1}) $$
Of course we can easily translate the variables $\phi_2 = x_2, \phi_3 = \frac{x_3}{x_2}, \ldots$ which instantly makes $\eta$ a function of $x_2, \ldots, x_n$. Therefore, 
$$ \implies \eta = \frac{3x_n^3}{1 + \sum_{i=2}^n x_i^2} - \frac{x_n^2}{2}$$
Thus, we get: 
$$u = \frac{1}{2} x \cdot x + |x|^2 \left ( \frac{3x_n}{1 + \sum_{i=2}^n x_i^2} - \frac{1}{2} \right )$$
Which simplifies to: 
$$u = x \cdot x  \left ( \frac{3x_n}{1 + \sum_{i=2}^n x_i^2} \right )$$
