Autonomous differential equation

In summary, the conversation discusses a pond formed by water collecting in a conical depression and its water flow and evaporation rates. The volume of water in the pond at a given time satisfies a differential equation, and the equilibrium depth and stability of the pond are also discussed. A condition for the pond not to overflow is also mentioned.
  • #1
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:


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.


[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.


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

Any help is highly appreciated.
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  • #2
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

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
  • #3
Thank you.

Great work so far! For part (b), we can determine the stability of the equilibrium by analyzing the sign of the derivative of the right-hand side of the differential equation at the equilibrium point. In this case, we have

\frac{d}{dV} \left( k - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3} \right) = -\frac{2}{3} \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{-1/3}

Substituting in the equilibrium point r = \sqrt{\frac{k}{\alpha \pi}}, we get

\frac{d}{dV} \left( k - \alpha \pi r^2 \right) = -\frac{2}{3} \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} \left( \frac{k}{\alpha \pi} \right) ^{-1/3} = -2\left( \frac{a}{h} \right) ^{2/3}

Since \left( \frac{a}{h} \right) ^{2/3} is always positive, we can see that the derivative is always negative at the equilibrium point. This means that the equilibrium point is stable, but we cannot determine if it is asymptotically stable without further information about the initial conditions.

For part (c), you are correct in stating that the condition for the pond to not overflow is

L = \frac{h}{a} \sqrt{\frac{k}{\alpha \pi}} \leq h

This means that the equilibrium depth of water must be less than or equal to the depth of the cone. If the equilibrium depth is greater than the cone's depth, then the pond will overflow.

What is an autonomous differential equation?

An autonomous differential equation is a type of differential equation where the independent variable does not appear explicitly in the equation. This means that the equation's solution is independent of the specific value of the independent variable.

What is the importance of autonomous differential equations?

Autonomous differential equations are important in a variety of scientific fields, such as physics, biology, and economics. They allow us to model and understand natural processes and systems, and make predictions about their behavior over time.

What are the key components of an autonomous differential equation?

The key components of an autonomous differential equation are the dependent variable, the derivative of the dependent variable, and any constants or parameters that may affect the behavior of the equation.

What is the difference between an autonomous and non-autonomous differential equation?

The main difference between autonomous and non-autonomous differential equations is that non-autonomous equations have an explicit dependence on the independent variable, while autonomous equations do not. This means that the solutions to non-autonomous equations are specific to a certain value of the independent variable, while solutions to autonomous equations are not.

What are some real-world applications of autonomous differential equations?

Autonomous differential equations are used to model and understand a wide range of phenomena, such as population growth, chemical reactions, and electrical circuits. They are also used in fields such as robotics and control systems to design and control autonomous systems.

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