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Autonomous differential equation

  • #1
372
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I just need some help with this problem. Thank you.

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

(a) Show that the volume [tex]V(t)[/tex] of water in the pond at time [tex]t[/tex] satisfies the differential equation

[tex]\frac{dV}{dt} = k - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}[/tex]

where [tex]\alpha[/tex] is the coefficient of evaporation.

(b) Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically stable?

(c) Find a condition that must be satisfied if the pond is not to overflow."

My work:

(a)

Consider the following

[tex]\frac{dV}{dt} = k - \alpha \pi \underbrace{\left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}} _{r^2}[/tex]

where [tex]r[/tex] is the radius of the pond. Thus, we have

[tex]r^2 = \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}[/tex]

[tex]r^3 = \frac{3aV}{\pi h}[/tex]

[tex]V = \frac{1}{3} \pi r^2 \left( \frac{hr}{a} \right) = \frac{1}{3} \pi r^2 L[/tex]

which satisfies the differential equation.

(b)

[tex]\frac{dV}{dt} = 0[/tex]

[tex]k - \alpha \pi r^2 = 0[/tex]

[tex]r = \sqrt{\frac{k}{\alpha \pi}}[/tex]

Then, the equilibrium depth of water in the pond is

[tex]L = \frac{h}{a} \sqrt{\frac{k}{\alpha \pi}}[/tex]

In this particular case, I don't know how to show whether or not it is asymptotically stable.

(c)

[tex]L = \frac{h}{a} \sqrt{\frac{k}{\alpha \pi}} \leq h[/tex]


Any help is highly appreciated.
 

Answers and Replies

  • #2
372
0
An equilibrium depth of water in the pond implies that there is an equilibrium volume. Thus, a direction field can show whether or not it is asymptotically stable, which is what I have at

http://mygraph.cjb.net/

Here are the (random) values that I used to plot it:

[tex]\left\{ \begin{array}{ll} k = 1 \\ \alpha = 0.6 \\ a = 0.3 \\ h = 0.5 \end{array} \right.[/tex]

Based on this information, I'd say the equilibrium is asymptotically stable.

Thanks anyway
 

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