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!



Can a vector-function $v:\mathbb{R}^n\to \mathbb{R}^n$ be an eigenvector of its own Jacobian matrix?
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 $(v\cdot\nabla)v\parallel v$.


