Find the equilibrium solution for an autonomous equation.

[clik80clak]

Homework Statement

Consider a cylindrical water tank of constant cross section A. Water is pumped into the tank at a constant rate k and leaks out through a small hole of area a in the bottom of the tank. From Torcelli’s principle in hydrodynamics, it follows that the rate at which water flows through the hole is (alpha)(a)squareroot((2)(g)(h)) , where h is the current depth of water in the tank, g is the acceleration due to gravity, and alpha is a contraction coefficient that satisfies 0.5 < alpha < 1.0.

1. Show that the depth of water in the tank at any time satisfies the equation
dh/dt = ([k] - [(alpha)(a)squareroot{(2)(g)(h)}])/A
2. Determine the equilibrium depth h_e , of water, and show that it is asymptotically stable. Observe that h_e does not depend on A.

Homework Equations

We covered how to solve a DE using the integrating factor method after putting the eq. into standard form but now we have moved on to autonomous eq.'s and I'm a little unsure how to go about solving this.

The problem looks more complex than it is because of all the brackets I had to use but I think it is pretty straight forward for someone familiar with the subject.

The Attempt at a Solution

1.

dh/dt = rate in - rate out

= (volume in/min)(1/area) - (volume out/min)(1/area)
= [k(m³/min) / A(m²)] - [((alpha)(a)squareroot{(2)(g)(h)}(m³/min) / A(m²)]
= ([k] - [(alpha)(a)squareroot{(2)(g)(h)}])/A

(Seems straight forward have I done this wrong?)

2.

dh/dt = m(h) h

=> [(k/hA - (alpha)(a)squareroot{(2)(g)}/sqrt{h}A] h

(Skipped the intermediate steps. Is this the right form? What do I do now?)Thanks in advance :)

 

[mighty2000]

1. Your approach for part 1 looks correct. You have correctly applied the rate in - rate out principle to find the depth of water in the tank at any time.

2. To find the equilibrium depth, set dh/dt = 0 and solve for h. This will give you the value of h at which the rate in and rate out are equal, meaning the water level is not changing.

So, setting dh/dt = 0, we get:

0 = [k] - [(alpha)(a)squareroot{(2)(g)(h)}])/A

Solving for h, we get:

h = (k/(alpha^2)(a^2)g)

This is the equilibrium depth, which does not depend on A.

To show that this equilibrium is stable, we can take the derivative of dh/dt with respect to h and evaluate it at the equilibrium point h = he:

d(dh/dt)/dh = (-1/2)(alpha)(a)squareroot{(2)(g)}/(A(sqrt{h})^2)

At h = he, this becomes:

d(dh/dt)/dh = (-1/2)(alpha)(a)squareroot{(2)(g)}/(A(sqrt{he})^2)

Since alpha is between 0.5 and 1, this derivative is always negative, meaning that the equilibrium point is stable. As h increases or decreases from he, the rate of change of dh/dt also decreases, meaning that the water level will tend towards the equilibrium point.

I hope this helps!