My Math Forum ODE inequality

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 July 2nd, 2010, 06:42 PM #1 Member   Joined: Nov 2009 Posts: 72 Thanks: 0 ODE inequality If you've got something like $\frac{du}{dt}\geq f(u)$ is it possible to say that if $\frac{dU}{dt}=f(U)$ then $u\geq U$? [Or something similar?]
 July 9th, 2010, 01:00 PM #2 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: ODE inequality Hello martexel. I saw your post a few days ago, but I haven't had a good chance to post a reply, so I hope you still have some interest in the question. To be honest, I am not completely certain that I understand your question myself, since there are possibly more than one way to interpret it, but I will state what I believe to be the most natural interpretation, and then give my thoughts on that. First, when you say $\frac{\mathrm{d}u}{\mathrm{dt}} \geq f(u)$, I assume you mean that $u : \mathbb{R} \rightarrow \mathbb{R}$ is given as a function of the real parameter $t$, that is to say, $u= u(t)$ (if not, it's hard to make sense of the differentiation). Furthermore, I also assume that the equality and inequality are to be interpreted pointwise, that is, $\frac{\mathrm{d}u}{\mathrm{dt}} \geq f(u)$ means that $\forall t \in \mathbb{R}$, we have $\frac{\mathrm{d}u}{\mathrm{dt}}(t) \geq f(u(t))$, and similarly, that $u \geq U$ means that $\forall t \in \mathbb{R}$, we have $u(t) \geq U(t)$. Assuming that this is the correct reading of the question, now I can answer that unfortunately, the statement is not true in general. Let me give a couple really simple examples to illustrate how and why the statement fails. Let's first recall that the derivative of a function naturally gives only a partial description of a function, since, for instance, two functions differing only by a constant have the same derivative. Now, suppose that we simplify first to the case of equality. That is, we assume that $\frac{\mathrm{d}u}{\mathrm{dt}}= f(u)$, and ask whether $\frac{\mathrm{d}U}{\mathrm{dt}}= f(U)$ implies $u= U$. Suppose that $f$ is a constant function, $f(y)= C$. Then, we can let $u(t)= Ct$ and $U(t)= Ct + D$, where $D$ is a non-zero constant, so that clearly $u \neq U$, but $\frac{\mathrm{d}u}{\mathrm{dt}}= \frac{\mathrm{d}U}{\mathrm{dt}} = C = f(u) = f(U)$, so the desired result doesn't hold. Here's another counterexample, this time using the inequalities as originally stated. Let $f(y)= y$, and suppose that $U(t)= e^t$ and $u(t)= -e^{-t}$. Then, we'd have $\frac{\mathrm{d}U}{\mathrm{dt}}= e^t = f(e^t) = f(U)$, and $\frac{\mathrm{d}u}{\mathrm{dt}}= e^{-t}$. The latter gives $f(u)= f(-e^{-t}) = -e^{-t} < e^{-t}= \frac{\mathrm{d}u}{\mathrm{dt}}$, so we have all the hypotheses in place. However, $u= -e^{-t} < e^t= U$ for all values of the argument. There are many other counterexamples for which the conclusion $u \geq U$ is violated on a portion of the domain, but not the entire domain. You can play around graphing functions in Mathematica and get a good feel for why this is so. It might be possible to get the conclusion you're after by strengthening the conditions on the functions. One way to go might be to impose some smoothness conditions on the function $f$. Additionally, you might want to demand that the conclusion hold for absolute values that is, have $|u| \geq |U|$, or something similar. I haven't had time to think through whether and which such restrictions might result in the conclusion you are looking for, but I wish you luck in your explorations. Please let me know if you do come up with something, I'm eager to think about this more some time later, since I don't get to do too much with DE these days!
 July 13th, 2010, 10:44 PM #3 Member   Joined: Nov 2009 Posts: 72 Thanks: 0 Re: ODE inequality Sorry, I forgot a really important part: u and U have the same initial condition, so that there's some $\overline{t}$ such that $u(\overline{t})=U(\overline{t})$, which rules out your counterexamples. [and yes, your interpretation is correct.]
 July 13th, 2010, 10:56 PM #4 Member   Joined: Nov 2009 Posts: 72 Thanks: 0 Re: ODE inequality Hang on a sec -- after thinking about it for a little while, if you separate variables in $\frac{du}{dt}\geq f(u)$ and $\frac{dU}{dt}=f(U)$ and integrate [taking the aforementioned initial value to be $u(t_0)=u_0=U(t_0)$], you get $\displaystyle\int_{u_0}^{u(t)}\frac{d\tau}{f(\tau) }\geq t-t_0=\int_{u_0}^{U(t)}\frac{d\tau}{f(\tau)}$ and so as long as you make the additional assumption that $f$ is everywhere positive, that'd have to imply that $u(t)\geq U(t)$, wouldn't it? -- since $\displaystyle\int_{u_0}^\xi \frac{d\tau}{f(\tau)}$ would be an increasing function of $\xi$.
 July 16th, 2010, 04:27 PM #5 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: ODE inequality Hi again martexel. I think with the added conditions your conclusion is now irrefutable.

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