How to solve $\nabla(u\cdot u)=\lambda u$ Suppose $C^2(\mathbb R^n)$ vector field $u(x)$, how to solve this pde?
$$\nabla(u\cdot u)=\lambda u$$
for some constant $\lambda$.
The physical intuition is that: The left hand side of the equation is advection acceleration of irrotational flow, while the right hand side is something like viscosity force.
I get two solutions: one is $u=0$ and the other is $u=\frac12\lambda x$. I have no idea how to get other solutions, neither know the existence of other solutions. Can anyone help?
 A: Here are some nontrivial solutions to the problem. Note that the value of $\lambda$ has no intrinsic meaning.
Consider a smooth closed convex hypersurface $S\subset{\mathbb R}^n$ (an egg), and let $\Omega$ be the exterior of $S$. For ${\bf x}\in\Omega$ define
$$f({\bf x}):=d({\bf x},S)=\inf_{{\bf y}\in S}|{\bf x}-{\bf y}|\ .$$
For a fixed ${ \bf p}\in\Omega$ there is a unique ${\bf z}\in S$ with $f({\bf p})=|{\bf p}-{\bf z}|$, and one has
$$f\bigl({\bf p}+t({\bf p}-{\bf z})\bigr)=f({\bf p})+t|{\bf p}-{\bf z}|\qquad(t\geq0)\ .$$
It follows that
$${\bf E}:=\nabla f$$
is a unit vector field on $\Omega$. Now put
$${\bf U}({\bf x}):=f({\bf x})\>{\bf E}({\bf x})\qquad({\bf x}\in\Omega)\ .$$
Then
$$\nabla({\bf U}\cdot{\bf U})=\nabla( f^2)=2 f\>\nabla f=2f\>{\bf E}=2{\bf U}\ .$$
A: Here is my attempt.
\begin{align}
u:\mathbb R^n&\to\mathbb R, &
\nabla\,\big(\,u\cdot u\,\big)&=\lambda \,u &\iff &&
\begin{cases}
\dfrac{\partial}{\partial x_1} \Big(u_1^2 + \dots + u_n^2 \Big) = \lambda u_1 \\
\qquad\qquad\vdots \\
\dfrac{\partial}{\partial x_n} \Big(u_1^2 + \dots + u_n^2 \Big) = \lambda u_n \\
\end{cases}
\end{align}
Denote $\,F := u\cdot u = u_1^2 + \dots + u_n^2 \,$, so that the original equation will take form
\begin{align}
\nabla F &= \lambda \,u\,& \iff & &  
\dfrac{\partial F}{\partial x_i} & = \lambda\,u_i & \forall \; i = 1,\,\dots,\, n
\end{align}
But then
\begin{align}
F = \sum_{i=1}^n u_i^2 = \dfrac{1}{\lambda^2}\sum_{i=1}^n (\dfrac{\partial F}{\partial x_i})^2 
\implies \lambda^2\,F = \big(\,\nabla F\cdot\nabla F\,\big)= \big\| \nabla F\big\|^2
\end{align}
Therefore all you need to do is to solve equation
\begin{align}
\bbox[4pt, border:2pt solid #FF0000]{\left\| \nabla F\right\|^2_{\phantom{y}}  = \lambda^2\,F}
\end{align}
and the solution of the original problem will be $\;u=\dfrac{1}{\lambda}\,\nabla F.\,$

In particular, consider, for example, solution 
\begin{align}
F &= a\,\left\|x\right\|^2 = x_1^2 + \dots + x_n^2 &\implies &&
\nabla F &= 2\,a\,x &\implies&& a&= \dfrac{\lambda^2}{4}
\\
&&\implies
&& u & = \dfrac{1}{2}\,\lambda\,x
\end{align}
