Systems of Differential Equations (Part 13. Modeling)
Posted by Suhaivi Hamdan on 5:58 PM with No comments
Modeling
In this section we’re going to go back and revisit the idea
of modeling only this time we’re going to look at it in light of the fact that
we now know how to solve systems of differential equations.
We’re not actually going to be solving any differential
equations in this section. Instead we’ll
just be setting up a couple of problems that are extensions of some of the work
that we’ve done in earlier modeling sections whether it is the first order modeling or the vibrations
work we did in the second order chapter.
Almost all of the systems that we’ll be setting up here will be nonhomogeneous systems (which we only
briefly looked at), will be nonlinear (which we didn’t look at) and/or will
involve systems with more than two differential equations (which we didn’t look
at, although most of what we do know
will still be true).
Mixing Problems
Let’s start things by looking at a mixing problem. The last time we saw these was back in the first
order chapter. In those problems we had
a tank of liquid with some type of contaminate dissolved in it. Liquid, possibly with more contaminate
dissolved in it, entered the tank and liquid left the tank. In this situation we want to extend things
out to the following situation.
We’ll now have two tanks that are interconnected with liquid
potentially entering both and with an exit for some of the liquid if we need it
(as illustrated by the lower connection).
For this situation we’re going to make the following assumptions.
 The inflow and outflow
from each tank are equal, or in other words the volume in each tank is
constant. When we worked with a
single tank we didn’t need to worry about this, but here if we don’t well
end up with a system with nonconstant coefficients and those can be quite
difficult to solve.
 The concentration of the
contaminate in each tank is the same at each point in the tank. In reality we know that this won’t be
true but without this assumption we’d need to deal with partial
differential equations.
 The concentration of
contaminate in the outflow from tank 1 (the lower connection in the figure
above) is the same as the concentration in tank 1. Likewise, the concentration of
contaminate in the outflow from tank 2 (the upper connection) is the same
as the concentration in tank 2.
 The outflow from tank 1 is split and only some of the liquid exiting tank 1 actually reaches tank 2. The remainder exits the system completely. Note that if we don’t want any liquid to completely exit the system we can think of the exit as having a value that is turned off. Also note that we could just as easily done the same thing for the outflow from tank 2 if we’d wanted to.
Let’s take a look at a quick example.
Example 1 Two
1000 liter tanks are with salt water.
Tank 1 contains 800 liters of water initially containing 20 grams of
salt dissolved in it and tank 2 contains 1000 liters of water and initially
has 80 grams of salt dissolved in it.
Salt water with a concentration of
Solution
Okay, let
Recall that the basic differential equation is the rate of
change of salt (
Here is the differential equation for tank 1.
In this differential equation the first pair of numbers is
the salt entering from the external inflow.
The second set of numbers is the salt that entering into the tank from
the water flowing in from tank 2. The
third set is the salt leaving tank as water flows out.
Here’s the second differential equation.
Note that because the external inflow into tank 2 is fresh
water the concentration of salt in this is zero.
In summary here is the system we’d need to solve,
This is a
nonhomogeneous system because of the first term in the first
differential equation. If we had fresh water flowing into both of
these we would in fact have a homogeneous system.

Population
The next type of problem to look at is the population
problem. Back in the first order modeling section we looked at some population
problems. In those problems we looked at
a single population and often included some form of predation. The problem in that section was we assumed
that the amount of predation would be constant.
This however clearly won’t be the case in most situations. The amount of predation will depend upon the
population of the predators and the population of the predators will depend, as
least partially, upon the population of the prey.
So, in order to more accurately (well at least more accurate
than what we originally did) we really need to set up a model that will cover
both populations, both the predator and the prey. These types of problems are usually called predatorprey problems. Here are the assumptions that we’ll make when
we build up this model.
 The prey will grow at a
rate that is proportional to its current population if there are no
predators.
 The population of
predators will decrease at a rate proportional to its current population
if there is no prey.
 The number of encounters
between predator and prey will be proportional to the product of the
populations.
 Each encounter between the predator and prey will increase the population of the predator and decrease the population of the prey.
So, given these assumptions let’s write down the system for
this case.
Example 2 Write
down the system of differential equations for the population of predators and
prey using the assumptions above.
Solution
We’ll start off by letting x represent the population of the predators and y represent the population of the
prey.
Now, the first assumption tells us that, in the absence of
predators, the prey will grow at a rate of
Next, the third and fourth assumptions tell us how the
population is affected by encounters between predators and prey. So, with each encounter the population of
the predators will increase at a rate of
Putting all of this together we arrive at the following
system.
Note that this
is a nonlinear system and we’ve not (nor will we here) discuss how to solve
this kind of system. We simply wanted
to give a “better” model for some population problems and to point out that
not all systems will be nice and simple linear systems.

Mechanical Vibrations
When we first looked at mechanical
vibrations we looked at a single mass hanging on a spring with the
possibility of both a damper and/or an external force acting on the mass. Here we want to look at the following
situation.
In the figure above we are assuming that the system is at
rest. In other words all three springs
are currently at their natural lengths and are not exerting any forces on
either of the two masses and that there are no currently any external forces
acting on either mass.
We will use the following assumptions about this situation
once we start the system in motion.
will measure the displacement of mass from its equilibrium (i.e. resting) position and will measure the displacement of mass from its equilibrium position.
 As noted in the figure
above all displacement will be assumed to be positive if it is to the
right of equilibrium position and negative if to the left of the
equilibrium position.
 All forces acting to the
right are positive forces and all forces acting to the left are negative
forces.
 The spring constants,
, , and , are all positive and may or may not be the same value.
 The surface that the system is sitting on is frictionless and so the mass of each of the objects will not affect the system in any way.
Before writing down the system for this case recall that the
force exerted by the spring on the each mass is the spring constant times the
amount that the spring has been compressed or stretched and we’ll need to be
careful with signs to make sure that the force is acting in the correct
direction.
Example 3 Write
down the system of differential equations for the spring and mass system
above.
Solution
To help us out let’s first take a quick look at a
situation in which both of the masses have been moved. This is shown below.
Before
proceeding let’s note that this is only a representation of a typical case,
but most definitely not all possible cases.
In this case
we’re assuming that both
Also, we’ve
shown the external forces,
Before
proceeding we need to talk a little bit about how the middle spring will
behave as the masses move. Here are
all the possibilities that we can have and the affect each will have on
1. If both mass move the same amount
in the same direction then the middle spring will not have changed length and
we’ll have
2. If both masses move in the
positive direction then the sign of
3. If both masses move in the
negative direction we’ll have pretty much the opposite behavior as #2. If
4. If
5. Finally, if
Now, we’ll use
the figure above to help us develop the differential equations (the figure
corresponds to case 2 above…) and then make sure that they will also hold for
the other cases as well.
Let’s start off
by getting the differential equation for the forces acting on
In this case
Next, because
we’re assuming that
The
differential equation for
Note that this
will also hold for all the other cases.
If
Let’s now write
down the differential equation for all the forces that are acting on
In this case
The
differential equation for
We’ll leave it
to you to verify that this differential equation does in fact hold for all
the other cases.
Putting all of
this together and doing a little rewriting will then give the following
system of differential equations for this situation.
This is a
system to two linear second order differential equations that may or may not
be nonhomogeneous depending whether there are any external forces,

We have not talked about how to solve systems of second
order differential equations. However,
it can be converted to a system of first order differential equations as the
next example shows and in many cases we could solve that.
Example 4 Convert
the system from the previous example to a system of 1^{st} order
differential equations.
Solution
This isn’t too
hard to do. Recall that we did this for
single higher order differential equations earlier in the chapter when we
first started to look at systems. To
convert this to a system of first order differential equations we can make
the following definitions.
We can then
convert each of the differential equations as we did earlier in the chapter.
Eliminating the
“middle” step we get the following system of first order differential
equations.
The matrix form
of this system would be,
While we never
discussed how to solve systems of more than two linear first order
differential equations we know most of what we need to solve this.
In an earlier section we discussed briefly solving
nonhomogeneous systems and all of that information is still valid here.
For the
homogenous system, that we’d still need to solve for the general solution to
the nonhomogeneous system, we know most of what we need to know in order to
solve this. The only issues that we
haven’t dealt with are what to do with repeated complex eigenvalues (which
are now a possibility) and what to do with eigenvalues of multiplicity
greater than 2 (which are again now a possibility).

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