Does $f_1(x,y)<f_2(x,y)$ imply $y_1<y_2$ for solutions to the integral equation $y_k'=f_k(x,y_k)$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:20:10Z http://mathoverflow.net/feeds/question/53016 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53016/does-f-1x-yf-2x-y-imply-y-1y-2-for-solutions-to-the-integral-equation Does $f_1(x,y)<f_2(x,y)$ imply $y_1<y_2$ for solutions to the integral equation $y_k'=f_k(x,y_k)$? AppliedSide 2011-01-24T01:11:08Z 2013-02-22T13:22:00Z <p>Suppose for two given functions $f_1,f_2 \colon \mathbb{R}^2 \to \mathbb{R}$ there exist unique solutions $y_1$ and $y_2$ with the intersection of their intervals of existence $[0,\epsilon)$ to the integral equations $$y_k(x)=\int_{0}^{x} f_k(t,y_k(t)) dt$$. Moreover, suppose that $f_1(x,y)\leq f_2(x,y)$ (or strictly if that makes things easier), does it follow that $y_1 \leq y_2$ on $[0,\epsilon)$?</p> <p>It's true if $f_1$ and $f_2$ are Lipschitz in $y$ and continuous in $x$ (i.e., satisfy the hypotheses of the standard uniqueness and existence theorem) or if $f_1$ and $f_2$ are decreasing in $y$ (i.e., satisfy the hypotheses of Peano's uniqueness theorem). The assumption of unique solutions is certainly necessary.</p> http://mathoverflow.net/questions/53016/does-f-1x-yf-2x-y-imply-y-1y-2-for-solutions-to-the-integral-equation/116354#116354 Answer by Gustaf Gripenberg for Does $f_1(x,y)<f_2(x,y)$ imply $y_1<y_2$ for solutions to the integral equation $y_k'=f_k(x,y_k)$? Gustaf Gripenberg 2012-12-14T09:27:33Z 2012-12-14T09:27:33Z <p>In general the answer seems to be no: Let $$f_1(t,y)=\begin{cases} -1, &amp;t\leq 0,\quad y\in \mathbb R\\ 1,&amp; t>0,\quad y>t/2,\\ -1,&amp; t>0,\quad -t/2 \leq y \leq t/2,\\ -e^{-n^2y},&amp; t\in [1/n,1/(n-1)),\quad y&lt; -t/2,\quad n\geq 1, \end{cases}$$ and let $f_2(t,y)=-f_1(t,-y)$ so that $f_1(t,y)&lt; f_2(t,y)$ when $(t,y)\in \mathbb R^2$. </p> <p>For $x\geq 0$ we have the solutions $y_1(x)=x$ and $y_2(x)= -x$ and in order to show that e.g. $y_1$ is unique one must show that $y_1$ cannot be such that $-x/2 \leq y_1(x)\leq x/2$ some $x>0$ and furthermore show that if $-\infty &lt; y_1(x)&lt; -x/2$ when $1/n \leq x \leq 1/(n-1)$ for some $n>1$ then $$1 > e^{n^2y(1/(n-1))}-e^{n^2y(1/n)}= \frac n{n-1} > 1.$$</p>