Autonomous differential equation

[jellicorse]

Homework Statement

A pond forms as water collects in a conical depression of radius a and depth h. Suppose water flows in at a constant rate, k and is lost through evaporation at a rate proportional to the surface area.


I was wondering whether anyone could give me some guidance on this part of the question:

Find the equilibrium depth of water in the pool. Is the equilibrium asymptotically stable?

Homework Equations

$$\frac{dV}{dt}=k-\alpha \pi(\frac{3aV}{\pi h})^{\frac{2}{3}}$$

Where $$\alpha$$ = coefficient of evaporation.

radius at any time r(t) and depth at any time l(t).

$$\frac{a}{h}=\frac{r}{l}$$

The Attempt at a Solution

$$\frac{dV}{dt}=0=k-\alpha \pi(\frac{3aV}{\pi h})^{\frac{2}{3}}$$


Putting the differential equation only in terms of radius and alpha:
$$\alpha \pi r^2 =k$$

$$r=\sqrt{(\frac{k}{\alpha \pi})}$$

given that $$l=r\cdot\frac{h}{a}$$,

$$l=\frac{h}{a}\sqrt{(\frac{k}{\alpha \pi})}$$

But I am not sure how to find whether or not this is asymptotically stable.

 

[maajdl]

Consider initial conditions near the equilibrium point.
If initially the volume is a little bit larger than the equilibrium volume, does the volume come back to equilibrium, or does it increase without end?
And the reverse, if the initial volume is a little bit smaller than the equilibrium volume?
You could also linearize the equation around the equilibrium point and answer the same questions, easily.

 

[jellicorse]

Thanks Maajdl.

I can see that solving for V, the equilibrium volume is $$V=\frac{\pi h}{3a}(\frac{k}{\alpha \pi})^{\frac{3}{2}}$$

But without picking some arbitrary values and plotting a direction field (or plotting Dv against V) I am not sure how to tell whether the volume diverges or converges to equilibrium.

(I haven't yet studied linearisation so am not quite ready to use this option)

 

[maajdl]

Assume Vo is the equilibrium volume.
If V>Vo, will the evaporation overcome the water supply and bring V back to Vo?
If V<Vo, will the water supply overcome the evaporation to bring V back to Vo?

This site may inspire you:

http://tutorial.math.lamar.edu/Classes/DE/EquilibriumSolutions.aspx

The situations described there are a little bit more complicated, but the same reasoning will apply.
The pictures shown there are useful to visualize what will happen.
Such "phase space flows" are more general than this one variable problem, and are at the hearth of the stability theory of differential equations. (and are even related to top mathematical physics, like the KAM theorem)

 

[jellicorse]

Thanks a lot maajdl... I'll have a good look through that webpage too. I need to get to grips with this topic!

 

[pasmith]

Homework Statement

A pond forms as water collects in a conical depression of radius a and depth h. Suppose water flows in at a constant rate, k and is lost through evaporation at a rate proportional to the surface area.


I was wondering whether anyone could give me some guidance on this part of the question:

Find the equilibrium depth of water in the pool. Is the equilibrium asymptotically stable?

Homework Equations

$$\frac{dV}{dt}=k-\alpha \pi(\frac{3aV}{\pi h})^{\frac{2}{3}}$$

Where $$\alpha$$ = coefficient of evaporation.

radius at any time r(t) and depth at any time l(t).

It is much, much easier to look at $dl/dt$ (as you are strongly hinted to do by the instruction to find the equilibrium depth, not the equilibrium volume). Since
$$r = \frac{al}{h}$$
we have
$$V = \frac{\pi}{3}r^2 l = \frac{\pi a^2}{3h^2} l^3$$
and hence
$$\frac{dV}{dt} = \frac{\pi a^2}{h^2} l^2 \frac{dl}{dt}.$$
Writing $k = \frac{\pi a^2}{h^2}\alpha C^2$ and using $r^2 = a^2l^2/h^2$ in the ODE gives
$$\frac{dV}{dt} = \frac{\pi a^2}{h^2}\alpha C^2 - \alpha \frac{\pi a^2}{h^2} l^2 = \frac{\pi a^2}{h^2} \alpha\left(C^2 - l^2\right)$$
so that
$$\frac{dl}{dt} = \alpha\left(\frac {C^2}{l^2} - 1\right)$$
where, of course,
$$C = \frac{h}{a} \sqrt{\frac{k}{\pi \alpha}}.$$

No linearization is necessary here; just sketch a graph of $dl/dt$ against $l$, and see what the sign of $dl/dt$ is immediately above and below the fixed point.

 

[maajdl]

It is also possible to sketch the graph of dV/dt against V .

 

[jellicorse]

Writing $k = \frac{\pi a^2}{h^2}\alpha C^2$ and using $r^2 = a^2l^2/h^2$ in the ODE gives
$$\frac{dV}{dt} = \frac{\pi a^2}{h^2}\alpha C^2 - \alpha \frac{\pi a^2}{h^2} l^2 = \frac{\pi a^2}{h^2} \alpha\left(C^2 - l^2\right)$$
so that
$$\frac{dl}{dt} = \alpha\left(\frac {C^2}{l^2} - 1\right)$$

Thanks, Pasmith. After studying it, I can see what you have done. The thing that concerns me is that it wouldn't have occurred to me to make these steps. You must be thinking several steps ahead, visualizing how it will all fit together...

 

[jellicorse]

It is also possible to sketch the graph of dV/dt against V .

That's true... I would quite like to try to see if I can get it symbolically though.

 

[maajdl]

Thanks, Pasmith. After studying it, I can see what you have done. The thing that concerns me is that it wouldn't have occurred to me to make these steps. You must be thinking several steps ahead, visualizing how it will all fit together...

All you need is exercise.
These steps are very standard and will see them again and again, specially in classical mechanics.
Choosing l instead of V as variable would be highly natural if you would like to study if overflow might occur!