Differential Equations in Our Life

10/17/08 Revised

Differential equations - easy to hear but often impossible to solve- are around there in our life. If you are an econ learner, you have much time to express the economy by using differential equations. Especially in economic growth theory (if you want the examples, see Barro & Sala-i-Martin, Economic Growth, 2th edition, MIT Press), we use many differential equations.

However, we can also use them to express the things around our life other than tedious economic growth theory. The differential equations that are used to express them could usually be relatively easy to solve. Let me show you some examples:

［1］ Memory

In 1885, Hermann Ebbinghaus discovered the exponential nature of forgetting. The following formula can roughly describe the forgetting:

R(t)=R(0)exp(-λt)

where R(0) is an initial condition.

As time goes by, we lose some of our memory. The fraction of what we forget is called a forgetting rate λ. Our memory level at period t is R(t). The amount we forget can be expressed by our memory times λ, that is, λR(t). So we can show how we forget by using the following formula:

dR(t)/dt = -λR(t) .....(1)

The formula (1) means that we lose the fraction of our memory λR(t) as time goes by. How we get the memory function R(t) is as follows:

Step1, dR(t)/dt/R(t) =-λ .....(2)

Step2, (1/R(t)) dR(t) = -λdt .....(3)

Step3, lnR(t)=-λt+C .....(4)

where ln is a natural logarithm and C is a constant. Note that dlnR(t)/dR(t) = 1/R(t).

Finally R(t)=R(o)exp(-λt) where R(0)=exp(C). .....(5)

Note that R(t) = exp(-λt+C). Our memory can be expressed by the first-order autonomous differential equation. Other examples are:

［2］Bathtub and running water

Assume we have a bathtub full of water with the area of the base, S. Once we put off the stopper, the water begins to run out. We can see the change of the water level of bathtub by using differential equations.

Let a the water level of bathtub at the first period, y(t) the water level at period t, and V(t) the amount of the water at period t. Then we can express the speed of running water as dV(t)/dt. This is proportional to the depth of bathtub, y(t).

dV(t)/dt =-ky(t) .....(6)

where k is a constant.

At this time we have to notice that V(t)=S×y(t).

dV(t)/dt = S×dy(t)/dt .....(7)

Substituting the formula (6) for (7),

-ky(t)= S×dy(t)/dt .....(8)

Manipulating the formula (8),

(1/y(t))dy(t) = -k/s dt .....(9)

And integrating the formula (9),

ln(y(t)/a) = -(k/S)t, and

y(t)=a×exp［(-k/S)t］.....(10)

The water level of our bathtub can be expressed by the function of t.

［3］ Population growth

N(t) is the population level at period t, n is the population growth rate (Here n is assumed constant. ) and N(0) is an initial condition.

The rate of change of population is the rate of the derivative of N(t) with respect to N(t) to the level, N(t):

dN(t)/dt = n×N(t) .....(11)

And then, (1/N(t)) dN(t) = ndt.

Thus, ln N(t) = nt +C .....(12)

where C is a constant.

N(t) = N(0)exp(nt) .....(13)

where N(0) is an initial condition and N(0) = expC. The population growth is expressed by the formula (13).

By the way, Robert Malthus was an English political economist best known for his highly influential views on population growth. His Principle of Population was based on the above idea: if population grows at a geometric rate whereas the food supply grows at an arithmetic rate the population growth will make us very poor. This is a very famous prediction.

［4］ Hot coffee
A hot coffee left in the restroom is getting colder as time goes by. Newton's law of cooling states that the speed of cooling of a hot coffee at period t is proportional to the difference in temperatures between the coffee and its surrounding.

According to the law, dF(t)/dt = -λF(t) .....(14)

where F(t) is the t-period difference in temperatures between the coffee and its surrounding and λ is a constant.

At period 0, the difference is F(0). And then,

F(t)=F(0)exp(-λt) .....(15)

The temperature in your room is 26℃.And the temperature of your coffee is 63℃. In 20 minutes your coffee is at 45℃. If you left your coffee in your room and t minutes passed, at what temperature would it be?

Using the formula F(0) = 63℃-26℃=37℃,

F(t) = 37×exp(-kt) .....(16)

and using F(20) =19,

37×exp(-kt) = 19 .....(17),

and exp(-k) = (19/37)^(1/20) = 0.967.....

Finally,

F(t) = 37×(0.967)^t .....(18)

According to the formula (18), in t minutes your coffee would be at about (26+37×0.967^t)℃. The change of the temperature of your coffee can be expressed by the first-order differential equation.

The above equations are all similar with each other. We can see that many things in our life can be expressed by differential equations. If you know another example of what seems to be expressed by differential equations, please let me know! (The photos are all from Wikipedia.)