Results 1 to 7 of 7

Math Help - Differential inequation!

  1. #1
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Differential inequation!

    \frac{dr}{dt}>\frac{k}{\sqrt{r}} and r(0)=R.

    Why does it follow that r(t)\rightarrow \infty as t\rightarrow \infty?

    My attempt:

    We can find k'>k so that \frac{dr}{dt}=\frac{k'}{\sqrt{r}} and solve the differential equation to obtain

    r=\left(\frac{3k't}{2}+R^{3/2}\right)^{2/3}

    Clearly, r(t)\rightarrow \infty as t\rightarrow \infty.

    But, how do we know that \sqrt{r}\frac{dr}{dt} is a constant? It could well be a variable that is always greater than k.
    Last edited by alexmahone; September 22nd 2011 at 02:48 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2

    Re: Differential inequation!

    I don't think your working is correct, because I don't think you can assume the equality holds. Is k positive? If so, why not do this:

    \dot{r}>\frac{k}{\sqrt{r}}

    \sqrt{r}\,\dot{r}>k

    \int\sqrt{r}\,\dot{r}\,dt>k\int dt

    \int r^{1/2}\,dr>kt+C

    and so on. The end result is very like what you got, except that all the steps here are valid (I think), whereas, like I said, I don't think you can assume that the equality holds which gives you your differential equation.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Differential inequation!

    Quote Originally Posted by Ackbeet View Post
    I don't think your working is correct, because I don't think you can assume the equality holds. Is k positive? If so, why not do this:

    \dot{r}>\frac{k}{\sqrt{r}}

    \sqrt{r}\,\dot{r}>k

    \int\sqrt{r}\,\dot{r}\,dt>k\int dt

    \int r^{1/2}\,dr>kt+C

    and so on. The end result is very like what you got, except that all the steps here are valid (I think), whereas, like I said, I don't think you can assume that the equality holds which gives you your differential equation.
    Why does k have to be positive for your working to be valid?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Differential inequation!

    Quote Originally Posted by Ackbeet View Post
    I don't think your working is correct, because I don't think you can assume the equality holds. Is k positive? If so, why not do this:

    \dot{r}>\frac{k}{\sqrt{r}}

    \sqrt{r}\,\dot{r}>k

    \int\sqrt{r}\,\dot{r}\,dt>k\int dt

    \int r^{1/2}\,dr>kt+C

    and so on. The end result is very like what you got, except that all the steps here are valid (I think), whereas, like I said, I don't think you can assume that the equality holds which gives you your differential equation.
    Note that integrating doesn't always preserve inequality. Consider f(x)=e^{3x} on [0,\infty). Evidently f(x)>\frac{1}{2} on that interval, what happens if you integrate it?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Differential inequation!

    An easier way would be to consider the graphs of dr/dt=k√r, r(0)=R and dr/dt>k√r, r(0)=R. Clearly, the graph of dr/dt>k√r, r(0)=R lies above the graph of dr/dt=k√r, r(0)=R. Since r(t)-->∞ as t-->∞ in the lower graph, r(t)-->∞ as t-->∞ in the upper graph as well.
    Attached Thumbnails Attached Thumbnails Differential inequation!-differential.png  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Differential inequation!

    Quote Originally Posted by alexmahone View Post
    An easier way would be to consider the graphs of dr/dt=k√r, r(0)=R and dr/dt>k√r, r(0)=R. Clearly, the graph of dr/dt>k√r, r(0)=R lies above the graph of dr/dt=k√r, r(0)=R. Since r(t)-->∞ as t-->∞ in the lower graph, r(t)-->∞ as t-->∞ in the upper graph as well.
    On second thoughts, this may not be correct.

    If ds/dt = k/sqrt(s), then we find s^{3/2} = \dfrac{3}{2}kt + R^{3/2} and it's obvious s\to \infty.

    The problem is that if r > s, then k / sqrt(r) < k / sqrt(s), (remembering both r and s are functions of t), so it might not actually be true that \dfrac{dr}{dt} > \dfrac{k}{\sqrt{s}} = \dfrac{ds}{dt}.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2

    Re: Differential inequation!

    Quote Originally Posted by Drexel28 View Post
    Note that integrating doesn't always preserve inequality. Consider f(x)=e^{3x} on [0,\infty). Evidently f(x)>\frac{1}{2} on that interval, what happens if you integrate it?
    I'm not seeing where you're going with this. While it's true that the integral on the entire interval doesn't converge for either integrand on either side of the inequality (they are both infinity), that would be true for something as simple as 1 > 1/2, as well. It is true that

    \int_{0}^{b}e^{3x}\,dx>\int_{0}^{b}\frac{1}{2}\,dx for all b\in(0,\infty).

    So any finite interval integral that exists will preserve inequalities.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help on this inequation
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 17th 2010, 11:22 AM
  2. Why this inequation?
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: July 12th 2010, 09:55 PM
  3. Help with an inequation.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 28th 2010, 02:21 AM
  4. Inequation
    Posted in the Algebra Forum
    Replies: 6
    Last Post: January 7th 2010, 12:40 AM
  5. inequation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 7th 2008, 08:10 PM

Search Tags


/mathhelpforum @mathhelpforum