Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a differential equations question. The only thing I know about differential equations is that there can be many subtleties.

Suppose that I have a function $g:\mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. I have a solution $x_0$ to $g(x_0, t=0)=0$. I desire to show the existence of a solution to $g(x_1, t=1)=0$. My strategy is to take the known solution $x_0$ and vary it from $t=0$ to $t=1$.

Let $J$ be the Jacobian $J_{ij} = \partial g_i / \partial x_j$. Define the function $f(x,t) = - J^{-1} \partial g/\partial t$. Then, setting $dx/dt = f(x,t)$ gives \begin{equation} \frac{dg(x,t)}{dt} = J \frac{dx}{dt} + \frac{\partial g}{\partial t} = 0. \end{equation} So it would seem that integrating this differential equation from $t=0$ to $t=1$ will lead to the desired solution $g(x_1, t=1)=0$.

What I need then is a theorem of the following form:

Suppose $g(x_0, t=0)=0$, and suppose that $\exists f$ such that \begin{equation} \frac{dx}{dt}=f(x,t) \implies \frac{dg(x,t)}{dt}=0. \end{equation} If $f$ and $g$ are well behaved enough in the vicinity $g(x,t) \approx 0$, then $\exists x$ such that $g(x,1)=0$.

When $g(x,t)=0$, I have a lower bound on the smallest singular value of $J$, and $\partial g/\partial t$ is bounded as well. So, in this case $f$ is bounded. In fact, I think that I can pretty much prove any needed type of "well-behavedness" in the region $\left| g(x,t) \right| < \epsilon$ for some $\epsilon$. So, what properties of $f$ and $g$ are needed, and what theorem will help me here? It looks like Picard-Lindelof may help, but it seems to only give the existence of a unique solution to the differential equation, and I need to show that that solution satisfies $g(x_1, t=1)=0$. Furthermore, $f$ is not well behaved when $g$ is far from zero, and so it seems I cannot use Picard-Lindelof without prior assumption that $g(x,t)$ stays small (which is kind of assumes the fact that I am trying to prove).

NOTE: since I didn't get any answers here, I also asked on mathoverflow.

share|cite|improve this question

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Misha below.

The right tag for this question is topology and the answer is degree theory. You could start by reading, say,


for a quick introduction.

Read also the book Differential forms in algebraic topology by Bott and Tu for in depth discussion. (Actually, read this book in any case!)

Warning: Wikipedia article confuses local diffeomorphisms and covering maps, but you do not need to worry about this. It also unnecessarily restricts the discussion to the case of bounded domains, while all you need to assume is that the homotopy is proper, see below.

Here is the upshot: For general continuous (or even smooth) functions $g$, the existence of solution for $t=0$ does not imply existence of solution for $t=1$. However, if you assume that $g(x, 0)$ has nonzero degree over its value $0$ and the family $g(\cdot , t)$ is a proper homotopy, then $g(x,1)$ also has nonzero degree over $0$, in particular, the equation $g(x,1)=0$ also has (at least one) solution. The key principles are:

i. proper homotopy preserves the degree


ii. map $h$ has nonzero degree $\Rightarrow$ existence of solution of the equation $h(x)=0$ (the function $h$ is surjective).

Here, every continuous map $g(x,t)$ defines a homotopy of the function $h_0=g(x,0)$ to the function $h_1=g(x,1)$. This homotopy is proper if the map $g: {\mathbb R}^n \times [0,1]\to {\mathbb R}^n$ is a proper map (inverse image of compact is compact). In calculus terms: $$ \lim_{|x|\to\infty, t\to t_0} g(x,t)=\infty $$

Below are two examples to think about ($n=1$):

  1. $g(x,t)=x^2 + t - \frac{1}{2}$. Then the equation $g(x,t)=0$ has solution for $t=0$ and all $t\le 1/2$ but no solutions for $t>1/2$. In this case, $g(x,t)$ (as a function of $x$) has zero degree at $0$ for every $t$.

  2. $g(x,t)= (t-1)x +1$. In this case $g(x,t)=0$ again has a solution for all $t\ne 1$, but the equation $g(x,1)=0$ has no solutions. In this case, the map $g(x,t)$ (as a function of $x$) has nonzero degree for all $t\ne 1$, but the homotopy is not proper (the map $g(x,1)$ is not a proper map).

In most places, you will read about degree of maps between compact manifolds, while you are interested in maps of ${\mathbb R}^n$. However, the 1-point compactification of ${\mathbb R}^n$ is the sphere $S^n$. Properness allows you to extend your homotopy $g(\cdot, t)$ to $S^n$, so now you can appeal to the usual degree theory.

Interestingly, the degree theory generalizes (with some difficulty) to the case of maps of Banach spaces which, in turn, is very useful for proving existence of solutions of differential equations. (Google "Degree Theory" + "Banach Spaces".)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.