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!

  • $\begingroup$ In your title, "eigenvalue" should be "eigenvector". $\endgroup$ – Jean Marie Becker Dec 5 '18 at 9:23
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    $\begingroup$ Can you specify whether you meant a left or a right eigenvector? (In particular, is $\nabla v \cdot v$ meant as $(v \cdot \nabla)v$ or $\frac12(\nabla |v|^2)$?) The Jacobian matrix is in general not symmetric so that the left and right eigenvectors are not the same. I agree that the "natural" interpretation is $v\cdot \nabla v$ which is completely treated by Denis Serre, but just want to make sure. $\endgroup$ – Willie Wong Dec 5 '18 at 15:49
  • $\begingroup$ @WillieWong Wow, I didn't even thought about that, but I agree that if we interpret the vector as a row or as a column things are different! Now I would be very glad to have some references about the other case as well, which seems in fact somewhat different. Thank you! $\endgroup$ – Gil Sanders Dec 5 '18 at 17:26

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 $(v\cdot\nabla)v\parallel v$.

  • $\begingroup$ Dear prof. Serre, indeed I was doubtful when I read his argument, and I was not convinced since the very beginning. Indeed your analysis is much more akin to my intuition, that is we should look for first integrals of the system, i.e. looking to what is conserved during this dynamics. Do you know any reference where such system are treated and/or analysed and/or studied in concrete cases? Many thanks! $\endgroup$ – Gil Sanders Dec 5 '18 at 12:17
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    $\begingroup$ @GilSanders. First of all, remember that you can vote up if you think the answer is valuable ! Now, this analysis is know as the method of characteristics. It is rather classical for those who work on hyperbolic scalar conservation laws. There are several books on this wide subject (Dafermos, Smoller, mine). $\endgroup$ – Denis Serre Dec 5 '18 at 12:52
  • $\begingroup$ Whoops, I had indeed made a very basic error, implicitly conflating having $v$ as an eigenvector at each point with having every vector as an eigenvector at each point. Careless of me, and thank you for the correction! $\endgroup$ – Sridhar Ramesh Dec 6 '18 at 20:46

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.

  • $\begingroup$ Really Willie, "irregardless"? ;-) $\endgroup$ – Igor Khavkine Dec 5 '18 at 21:01
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    $\begingroup$ Happier, Igor? :-) $\endgroup$ – Willie Wong Dec 5 '18 at 21:06
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    $\begingroup$ Probably @IgorKhavkine is unsadless. $\endgroup$ – LSpice Dec 5 '18 at 21:15

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