Differential inequation!

**alexmahone** · September 22nd 2011, 02:22 AM

$\frac{dr}{dt}>\frac{k}{\sqrt{r}}$ 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 $k$ .

**Ackbeet** · September 22nd 2011, 10:31 AM

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.

**alexmahone** · September 23rd 2011, 05:12 PM

Originally Posted by Ackbeet

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?

**Drexel28** · September 23rd 2011, 09:47 PM

Originally Posted by Ackbeet

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?

**alexmahone** · September 24th 2011, 12:18 AM

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.

**alexmahone** · September 24th 2011, 07:23 AM

Originally Posted by alexmahone

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 $\dfrac{dr}{dt} > \dfrac{k}{\sqrt{s}} = \dfrac{ds}{dt}$ .

**Ackbeet** · September 24th 2011, 11:37 AM

Originally Posted by Drexel28

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.

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