Continuous variation from solution of easy problem to solution of hard problem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:53:06Z http://mathoverflow.net/feeds/question/89993 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89993/continuous-variation-from-solution-of-easy-problem-to-solution-of-hard-problem Continuous variation from solution of easy problem to solution of hard problem Dan Stahlke 2012-03-01T21:24:59Z 2012-03-02T14:29:17Z <p>I asked this question a week ago over on <a href="http://math.stackexchange.com/q/112543/25589" rel="nofollow">math.stackexchange</a> and got no reply, so I am asking here with slightly different wording. I am trying to prove that there exists a solution to a problem. I can't solve this problem in general, but I can always continuously vary the parameters of the general problem to turn it into a specific case which is easy to solve. I believe that the solution of the easy problem can then be varied continuously to be a solution to the hard problem.</p> <p>It seems to me that this would be a widely used technique, but I haven't seen it before, and I don't know how to prove that it works. To be specific, suppose that $g:\mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. 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 $t=1$. What I need is a theorem of the following form:</p> <hr> <p>Suppose $g(x_0, t=0)=0$, and suppose that $\exists f$ such that $$\frac{dx}{dt}=f(x,t) \implies \frac{dg(x,t)}{dt}=0,$$ with $x$ a function of $t$. If $f$ and $g$ are "well behaved enough" in the vicinity $|g(x,t)| &lt; \epsilon$ for some $\epsilon$, then $\exists x$ such that $g(x,1)=0$.</p> <hr> <p>$f$ and $g$ in my case are "well behaved enough" that I can probably prove just about any sort of continuity condition that is needed. 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).</p> http://mathoverflow.net/questions/89993/continuous-variation-from-solution-of-easy-problem-to-solution-of-hard-problem/90018#90018 Answer by Misha for Continuous variation from solution of easy problem to solution of hard problem Misha 2012-03-02T07:32:12Z 2012-03-02T07:32:12Z <p>The right tag for this question is <em>topology</em> and the answer is <em>degree theory</em>. You could start by reading, say, <a href="http://en.wikipedia.org/wiki/Degree_of_a_map" rel="nofollow">http://en.wikipedia.org/wiki/Degree_of_a_map</a> </p> <p>or/and </p> <p><a href="http://unapologetic.wordpress.com/2011/12/10/calculating-the-degree-of-a-proper-map/" rel="nofollow">http://unapologetic.wordpress.com/2011/12/10/calculating-the-degree-of-a-proper-map/</a></p> <p>for a quick introduction. </p> <p>Read also the book <em>Differential forms in algebraic topology</em> by Bott and Tu for in depth discussion. (Actually, read this book in any case!) </p> <p>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. </p> <p>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 <em>proper homotopy</em>, 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: </p> <p>i. <em>proper homotopy preserves the degree</em> </p> <p>and </p> <p>ii. <em>map $h$ has nonzero degree $\Rightarrow$ existence of solution of the equation $h(x)=0$ (the function $h$ is surjective)</em>. </p> <p>Here, every continuous map $g(x,t)$ defines a <em>homotopy</em> of the function $h_0=g(x,0)$ to the function $h_1=g(x,1)$. This homotopy is <em>proper</em> 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: $$\lim_{|x|\to\infty, t\to t_0} g(x,t)=\infty$$</p> <p>Below are two examples to think about ($n=1$):</p> <ol> <li><p>$g(x,t)=x^2 + t - \frac{1}{2}$. 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 $t$. </p></li> <li><p>$g(x,t)= (t-1)x +1$. 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). </p></li> </ol> <p>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 <em>usual</em> degree theory. </p> <p>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".) </p>