Can a vector-function $v:\mathbb{R}^n\to \mathbb{R}^n$ be an eigenvector of its own Jacobian matrix? Good morning,
I've came across this question, which has been puzzling me for some days. Suppose we are given a vector-valued function $v:\mathbb{R}^n\to \mathbb{R}^n$, $v(x)=\left( v_1(x),\dots, v_n(x) \right)$ for every $x\in \mathbb{R}^n$.
Given a real function $\lambda:\mathbb{R}^n\to \mathbb{R}$, we are interested into the pde $\nabla v(x).v(x)=\lambda(x) v(x)$, in other question we ask that $v$ is an eigenvector of its own Jacobian matrix. Apart from simple cases (e.g. constant functions, or cases where the component $v_i$ depends just on $x_i$) I have not been able to find an exhaustive answer, nor I was able to find references on Google. It is not even clear to me how to treat the case of separable variables up to now.
However, I feel that something general should be known on such equations. Can anybody provide me with some references, or give a hint on how to attack the general problem?
Thank you very much!
 A: I am surprised that Ramesh's answer was voted up. Its first paragraph does not convince me. Here is my analysis:
Consider an integral curve of $v$, that is a curve $t\mapsto X(t)$ so that $\dot X=v(X)$. Then
$$\frac{d}{dt}v=(\dot X\cdot \nabla)v=\lambda v.$$
This shows that $t\mapsto v(X(t))$ keeps a constant direction. Therefore the curve is a straight line (or a segment of).
Conversely, let $\Omega$ be a domain, fibered by straight lines. Choose a vector field $v$ parallel to the fibers. Then differentiation along a fiber gives $\dot v\parallel v$, that is $(v\cdot\nabla)v\parallel v$.
A: Since you asked also about the case $\frac12 \nabla |v|^2 = \lambda v$:
When $\lambda \equiv 0$ on a domain, the solution on that domain is any arbitrary $v$ with constant norm, and is not very interesting. 
There's also always the trivial solution where $v \equiv 0$, regardless of $\lambda$. 
So, assume we are on a domain where $\lambda$ never vanish, and $v$ doesn't vanish. Then we can write $v = \sigma n$ where $\sigma = |v|$ and $n$ is a unit normal vector. Our equation becomes
$$ \nabla \sigma = \lambda n $$ 
which implies
$$\tag{*} |\nabla \sigma| = |\lambda|.$$
Notice that given a solution of this equation, we can simply take $v = \sigma \frac{\nabla \sigma}{\lambda}$ and this gives a solution of the original equation. 
Equations of the form (*) are called eikonal equations. Similar to the situation discussed in Denis Serre's answer, these equations can be treated using the method of characteristics. However, unlike in the case treated by Serre, there is no guarantee that the characteristics are straight lines. (You can see this by taking any $\sigma$ with no critical points and defining $\lambda$ by (*).) These equations can admit locally well-posed initial value problems, and because of the freedom in choosing initial data these local solutions are non-unique. In general there is no guarantee that a local solution can be extended to a global one, due to the possibility of forming caustics. 
For more information, one can start on Wikipedia. 
