Chapter 10 Introduction to Systems of 1st Order ODEs

10.1 Introduction to Systems of ODEs

Thus far, we have considered 1^st order ordinary differential equations as individual items. Just as in algebra we can think about solving systems of simultaneous equations, and doing so broadens the scope of problem we can solve, we can consider solving systems of differential equations. All of the unknown functions are still functions of a single independent variable. We will continue to use $t$ as the independent variable. Our investigation into systems of 1^st order ODE’s will generally have 2 (sometime 3) dependent functions. For example:

$$\begin{aligned} x'(t) &= f(x,y,t) \\ y'(t) &= g(x,y,t). \end{aligned}$$

where $x$ and $y$ are the unknown functions of $t$, and the right-hand sides, $f$ and $g$, give the rates of change for $x$ and $y$, respectively.

Given this basic form for a system of 1^st order ODE’s, we see that the solution is now a set of two functions, both $x(t)$ and $y(t)$. Let’s consider a simple example: $$\begin{aligned} x' &= y \\ y' &= x \end{aligned}$$ with $t$ as the independent variable. We seek the pair of functions, $x(t)$ and $y(t)$, that satisfy the system above. Certainly this problem is not as easy as $y' = y$, but it is closely related. We can immediately see one solution: $$\begin{aligned} x(t) &= C_{1}e^{t} \\ y(t) &= C_{1}e^{t} \end{aligned}$$

where $C_{1}$ is any constant. Notice the solution must contain both functions. We can verify this solution just as we did with a single 1^st order ODE: $$\begin{aligned} x' &= \frac{d}{dt}\left[x(t)\right]=\frac{d}{dt}\left[C_1e^t\right]=C_{1}e^{t} = y(t). \\ y' &= \frac{d}{dt}\left[y(t)\right]=\frac{d}{dt}\left[C_1e^t\right]=C_{1}e^{t} = x(t). \end{aligned}$$

Also, as we recall from our introduction to ODE’s, this solution represents an entire family of solutions. Any real valued constant, $C_{1}$, gives a correct solution.

There is another solution for our system. In this text we will not derive solutions to 1^st order ODE’s. However, it is important to realize that, in general, there are an many solutions for these systems as there are differential equations. In this example we have 2 ODE’s, and we can expect to find 2 solutions. Here is the second solution: $$\begin{aligned} x(t) &= C_{2}e^{- t} \\ y(t) &= - C_{2}e^{- t} \end{aligned}$$

where $C_{2}$ is any constant. Again, the solution must contain both functions, and we can verify the solution by applying the functions to the ODEs:

$$\begin{aligned} x' &= - C_{2}e^{- t} = y(t) \\ y' &= C_{2}e^{- t} = x(t) \end{aligned}$$

Because we have two solutions, we often number these solutions. A common form for the solutions to a system of 1^st order ODEs looks like this:

Solution #1	Solution #2
$$\begin{aligned} x_{1}(t) &= C_{1}e^{t} \\ y_{1}(t) &= C_{1}e^{t} \end{aligned}$$	$$\begin{aligned} x_{2}(t) &= C_{2}e^{- t} \\ y_{2}(t) &= - C_{2}e^{- t} \end{aligned}$$

These two solutions, each being a set of two functions $x(t)$ and $y(t)$, can be combined into a single, general solution. We create the general solution by taking a linear combination of the individual solutions. The specific procedure for creating the general solution is not important here. We simply state the general solutions as: $\begin{aligned} x(t) &= C_{1}e^{t} + C_{2}e^{- t}, \\ y(t) &= C_{1}e^{t} - C_{2}e^{- t}. \end{aligned}$

Again, we can verify that the general solution solves the differential equations. $$\begin{aligned} x' &= C_{1}e^{t} - C_{2}e^{- t} = y(t) \\ y' &= C_{1}e^{t} + C_{2}e^{- t} = x(t) \end{aligned}$$ Note that the linear combination of the individual solutions gives a complete solution to the problem. This general solution is still a family of solutions, but it has 2 constants instead of just 1 as we saw earlier.

In order to solve for the constants, $C_{1}$ and $C_{2}$, we apply initial conditions (IC’s) as we did with a single ODE. The IC’s must be applied simultaneously. Consider these initial conditions applied to our example: $x(0) = 3$ and $y(0) = 1$. Applying these conditions to the general solution yields: $$\begin{aligned} 3&=x(0) = C_{1}e^{0} + C_{2}e^{- 0} \\ 1&=y(0) = C_{1}e^{0} + C_{2}e^{- 0} \end{aligned}$$

Simplifying

$$\begin{aligned} 3 &= C_{1} + C_{2} \\ 1 &= C_{1} - C_{2} \end{aligned}$$ This system of algebraic equations can be solved by hand or using findZeros for systems of equations.

findZeros(c(C1+C2-3,C1-C2-1)~C1&C2)

##   C1 C2
## 1  2  1

The values for the constants has identified a single solution out of the family represented by the general solution. The particular solution to the initial value problem (IVP) is:

$$\begin{aligned} x(t) &= 2e^{t} + e^{- t} \\ y(t) &= 2e^{t} - e^{- t} \end{aligned}$$

The particular solution satisfies both the system of ODE’s and the set of initial conditions.

Finding solutions to systems of 1^st order ODEs is possible on occasion, but most systems of ODEs cannot be solved analytically. To learn more about analytical solutions to ODEs, we recommend the reader take a course in Ordinary Differential Equations.

10.1.1 Example #1

Consider the system of ODE $$\begin{aligned} x' &= - 3y \\ y' &= 2x - 5y. \end{aligned}$$ Does $$\begin{aligned}x &= 3e^{- 2t} \\ y &= 2e^{- 2t} \end{aligned}$$ solve this system of ODE?

Solution

For System A pair #1, we have: $x' = - 6e^{- 2t}$ and $y' = - 4e^{- 2t}$. Checking both sides of the system yields:

LHS:
$$\begin{aligned} x' &= - 6e^{- 2t} \\ y' &= - 4e^{- 2t} \end{aligned}$$

RHS: $$\begin{aligned} &- 3y = - 3 \cdot 2e^{- 2t} = - 6e^{- 2t} \\ &2x - 5y = 2 \cdot 3e^{- 2t} - 5 \cdot 2e^{- 2t} = - 4e^{- 2t} \end{aligned}$$

We can see that LHS=RHS, so the pair of functions $x = 3e^{- 2t}$, $y = 2e^{- 2t}$ is a solution to the system of ODEs.

10.1.2 Example #2

Consider the system of ODE $$\begin{aligned} x' &= - 3y \\ y' &= 2x - 5y. \end{aligned}$$ Does $$\begin{aligned} x &= e^{- 3t} \\ y &= 2e^{- 3t} \end{aligned}$$ solve this system?

Solution

Checking both sides of the system yields:

LHS:
$$\begin{aligned} x' &= - 3e^{- 3t} \\ y' &= - 6e^{- 3t} \end{aligned}$$ RHS: $$\begin{aligned} &- 3y = - 3 \cdot 2e^{- 3t} = - 6e^{- 3t} \\ &2x - 5y = 2 \cdot e^{- 3t} - 5 \cdot 2e^{- 3t} = - 8e^{- 3t}. \end{aligned}$$

Since LHS is not equal to RHS, the function pair $x = e^{- 3t}$, $y = 2e^{- 3t}$ does not solve the system of ODEs.

10.1.3 Example #3

Consider the system of ODE $$\begin{aligned} u' &= u - 5v \\ v' &= u - v. \end{aligned}$$ Does $$\begin{aligned} u &= 5\sin(2t) \\ v &= \sin(2t) - 2\cos(2t) \end{aligned}$$ solve this system?

Solution

Checking both sides of the system yields:

LHS: $$\begin{aligned} u' &= 10\cos(2t) \\ v' &= 2\cos(2t) + 4\sin(2t) \end{aligned}$$

RHS: $$\begin{aligned} u - 5v &= 5\sin(2t) - 5(\sin(2t) - 2\cos(2t)) = 10\cos(2t) \\ u - v &= 5\sin(2t) - (\sin(2t) - 2\cos(2t)) = 4\sin(2t) + 2\cos(2t) \end{aligned}$$

Since LHS=RHS, the function pair $u = 5\sin(2t)$, $v = \sin(2t) - 2\cos(2t)$ solves the system of ODE.

10.1.4 Example #4

$$\begin{aligned} u' &= u - 5v \\ v' &= u - v. \end{aligned}$$ Does $$\begin{aligned} u &= 2\sin(2t) + 4\cos(2t) \\ v &= 2\sin(2t) \end{aligned}$$ solve this system of ODE?

Solution

Checking both sides of the system yields:

LHS: $$\begin{aligned} u' &= 4\cos(2t) - 8\sin(2t) \\ v' &= 4\cos(2t) \end{aligned}$$

RHS:
$$\begin{aligned} u - 5v &= 2\sin(2t) + 4\cos(2t) - 5 \cdot 2\sin(2t) = - 8\sin(2t) + 4\cos(2t) \\ u - v &= 2\sin(2t) + 4\cos(2t) - 2\sin(2t) = 4\cos(2t) \end{aligned}$$

Since LHS=RHS, the function pair $u = 2\sin(2t) + 4\cos(2t)$, $v = 2\sin(2t)$ solves the system of ODE.

10.1.5 Example #5

Consider the system of ODEs $$\begin{aligned} u' &= u - 5v\\ v' &= u - v. \end{aligned}$$

This system has the general solution

$$\begin{aligned} u &= \left(C_{1} - 5C_{2}\right)\sin(2t) + 2C_{1}\cos(2t) \\ v &= \left(C_{1} - C_{2}\right)\sin(2t) + 2C_{2}\cos(2t). \end{aligned}$$

Solve for the constants $C_{1}$ and $C_{2}$ given the initial conditions: $u(0) = 2$ and $v(0) = - 4$.

We proceed by substituting the ICs into the general solution: $$\begin{aligned} &2=u(0) = \left( C_{1} - 5C_{2} \right)\sin(2 \cdot 0) + 2C_{1}\cos(2 \cdot 0) \\ &2 = 2C_{1}\\ &C_{1} = 1. \\\end{aligned}$$

$$\begin{aligned} &-4=v(0) = \left( C_{1} - C_{2} \right)\sin(2 \cdot 0) + 2C_{2}\cos(2 \cdot 0) \\ &- 4 = 2C_{2}\\ &C_{2} = - 2. \end{aligned}$$

Using the constants, we build the particular solution.

$$\begin{aligned} u &= \left( 1 - 5 \cdot (-2) \right)\sin(2t) + 2 \cdot 1 \cdot \cos(2t) \\ &= 11\sin(2t) + 2\cos(2t) \\ v &= \left( 1 - (-2) \right)\sin(2t) + 2 \cdot (-2) \cdot \cos(2t) \\ &= 3\sin(2t) - 4\cos(2t). \end{aligned}$$

The specific solution is: $u(t) = 11\sin(2t) + 2\cos(2t)$ ; $v(t) = 3\sin(2t) - 4\cos(2t)$

Note that we could also find the constants in the specific solution using findZeros, by manually inserting $t=0$. For example,

findZeros(c((C1-5*C2)*sin(2*0)+2*C1*cos(2*0)-2, 
            (C1-C2)*sin(2*0)  +2*C2*cos(2*0)-(-4))~C1&C2)

##   C1 C2
## 1  1 -2

10.2 Equilibrium Solutions

We focus on autonomous, systems of 1^st order differential equations:

$$\begin{aligned} x'(t) &= f(x,y) \\ y'(t) &= g(x,y) \end{aligned}$$

The definition for autonomous is the same as we saw earlier. Autonomous systems of ODE’s have rates of change that do not explicitly depend on the independent variable. Thus, the rate functions, $f(x,y)$ and $g(x,y)$, are never functions of $t$.

An equilibrium solution to a system of ODEs is a pair of functions, $x$ and $y$, that are constant and that satisfy the ODE.

10.2.1 Example #6

Consider a park in which a population of squirrels, $S(t)$, and a population of hawks, $H(t)$, coexist. We assume that the the squirrels are preyed upon by hawks at a rate (squirrels/time) jointly proportional to both the number of squirrels and the number of hawks. We assume that the hawks reproduce at rate jointly proportional to the number of squirrels and hawks. In this case, a differential equation for the two populations is

$$\begin{aligned} S'&=0.3S\left(1-\frac{S}{225}\right) - 0.06SH\\ H'&=-0.5H+0.006SH. \end{aligned}$$ Find all the equilibrium solutions for this system of ODE.

Solution

We seek constant values for $S$ and $H$ that satisfy the differential equation. If both $S$ and $H$ are constant, then $S'=0$ and $H'=0$. Therefore we need to find values of $S$ and $H$ that solve $$\begin{aligned} 0&=0.3S\left(1-\frac{S}{225}\right) - 0.06SH\\ 0&=-0.5H+0.006SH. \end{aligned}$$ We could do this by hand using algebra, but it is far easier to use findZeros.

findZeros(c(0.3*S*(1-S/225)-0.06*S*H,
            -0.5*H+0.006*S*H)~S&H)

##        S     H
## 1  0.000 0.000
## 2 83.333 3.148

There are two equilibrium solutions. One equilibrium solution is $S=0$, $H=0$. We can interpret this to mean that our model predicts that if the population starts with $S=0$ and $H=0$, then the populations will never change. This makes sense.

The other equilibrium solution is more interesting. Our model predicts that if the populations of squirrels and haws is exactly $S=83.333$ and $H=3.148$ then the two populations will never change. They will balance each other perfectly. Of course, it isn’t possible to have this population of squirrels and hawks, since neither species offers fractional individuals. In the next section we will explore what happens if the two populations start close to, but not exactly at, this equilibrium solution.

10.2.2 Example #7

Consider a population of two species of tree, perhaps pine and aspen, growing in a forest. Pine and aspen trees compete for many of the same resources, sunlight, nutrients, and carbon dioxide.

Let $P(t)$ be the population of pine tree in the forest, in thousands, and $A(t)$ the population of Aspen, in thousands. One model for the two populations is $$\begin{aligned} A'&=0.004A(50-A-0.75P)\\ P'&=0.001P(100-P-3A) \end{aligned}$$ We can find the equilibrium solutions by solving $$\begin{aligned} 0&=0.004A(50-A-0.75P)\\ 0&=0.001P(90-P-3A) \end{aligned}$$ for $A$ and $P$.

findZeros(c(0.004*A*(50-A-0.75*P),
            0.001*P*(90-P-3*A))~A&P)

##    A  P
## 1  0  0
## 2  0 90
## 3 14 48
## 4 50  0

We can see that there are four equilibrium solutions. One, $A=0$, $P=0$, means that if there are no trees in the forest, neither population will change. This makes sense.

A second equilibrium solution, $A=0$, $P=90$, means that if the ever has a population of 0 Aspens and 90,000 Pines, then the pine trees will out compete the aspen to such an extent that the aspen never recover. Further, the pine trees will never increase their population either.

A third equilibrium solution, $A=50$, $P=0$, is the reverse situation. In this situation, the aspen trees have used all the resources in the environment, and there is no more room for either pine or more aspen trees.

The fourth equilibrium solution is interesting. In this case, $A=14$, $P=48$, the two species have both established populations and have used all of the available resources in the environment.

10.2.3 Example #8

Consider the motion of a frictionless pendulum with length 2 meters. Let $x(t)$ be the angle the pendulum makes with a vertical line through the pivot point, in radians, and let $y(t)$ be the angular velocity of the pendulum, in radians per second. A model that describes this motion is $$\begin{aligned} x'&=y\\ y'&=-4.9\sin(x). \end{aligned}$$ The equilibrium solutions to this system of ODEs can be found by solving $$\begin{aligned} 0&=y\\ 0&=-4.9\sin(x). \end{aligned}$$

findZeros(c(y,-4.9*sin(x))~x&y)

There are many equilibrium solutions to this system of equations. What position/velocity pairs do they correspond to? Do those position velocity pairs make sense?

10.2.4 Example #9

Find the equilibrium solutions for this system of 1^st order ODEs $$\begin{aligned} x' &= 3y^{2} + 6y - xy \\ y' &= 3x - xy \end{aligned}$$ To find the equilibrium solutions, we set the rates of change equal to zero and solve for $x$ and $y$. $$\begin{aligned} 0 &= 3y^{2} + 6y - xy\\ 0 &= 3x - xy \\ \end{aligned}$$ Factoring, $$\begin{aligned} 0 &= y(3y + 6 - x)\\ 0 &= x(3 - y) \end{aligned}$$ We see from the second equation, $0 = x(3 - y)$, that equilibria can only occur when $x = 0$ or $y = 3$. We will check the first equation for each situation. With $x = 0$, the first equation becomes $$\begin{aligned} 0 &= y(3y + 6 - 0) \\ \end{aligned}$$ This has solutions $y=0$, $y=-2$. This gives two equilibrium solutions: $x = 0\ ;y(t) = 0$ and $x = 0\ ;y = - 2$.

Now we check for equilibrium solutions featuring $y = 3$. $$\begin{aligned} 0 &= 3(3 \cdot 3 + 6 - x) \\ x &= 15 \end{aligned}$$ This provides an additional equilibrium solution: $x = 15\ ;y = 3$.

The three equilibrium solutions are: $x = 0\ ;y = 0$ and $x = 0\ ;y = - 2$ and $x = 15\ ;y = 3$.

We could also find the equilibrium solutions using findZeros.

findZeros(c(3*y^2 + 6*y - x*y,
            3*x - x*y)~x&y)

##    x  y
## 1  0 -2
## 2  0  0
## 3 15  3

10.3 Exercises

Problems 1) – 3). Complete two steps per problem: a) Verify that the given solution satisfies the system of 1^st order ODEs, and b) find the particular solution by solving for the constants, $C_{1}$ and $C_{2}$.

	System	Solution to the system	ICs
1)	$$\begin{aligned} u' &= 3u - 2v \\ v' &= 4u - 3v\ \end{aligned}$$	$$\begin{aligned} u &= C_{1}e^{- t} + C_{2}e^{t} \\ v &= 2C_{1}e^{- t} + C_{2}e^{t} \end{aligned}$$	$$\begin{aligned} u(0) &= - 2 \\ v(0) &= 1 \end{aligned}$$
2)	$$\begin{aligned} p' &= - 3p + 3q \\ q' &= 2p - 2q\ \end{aligned}$$	$$\begin{aligned} p &= 3C_{1}e^{- 5t} + C_{2} \\ q &= -2C_{1}e^{- 5t} + C_{2} \end{aligned}$$	$$\begin{aligned} p(0) &= - 1 \\ q(0) &= 4 \end{aligned}$$
3)	$$\begin{aligned} x' &= - x - 2y \\ y' &= x + y\ \end{aligned}$$	$$\begin{aligned} x &= C_{1}\cos(t) - (C_{1} + 2C_{2})\sin(t) \\ y &= C_{2}\cos(t) + (C_{1} + C_{2})\sin(t) \end{aligned}$$	$$\begin{aligned} x(0) &= - 1 \\ y(0) &= 2 \end{aligned}$$

Problems 4) – 9). Find the equilibrium solution(s) for each system of 1^st order ODEs

4. $$\begin{aligned} y' &= - 3y + z \\ z' &= 2y - 5z\ \end{aligned}$$
5. $$\begin{aligned} x' &= x - 1 \\ y' &= 2x + xy \end{aligned}$$
6. $$\begin{aligned} u' &= - 3v + uv \\ v' &= 4u - 2uv \end{aligned}$$
   
7. $$\begin{aligned} p' &= 6p + 0.1pq \\ q' &= - 12q + 0.4pq \end{aligned}$$
8. $$\begin{aligned} r' &= r - rs \\ s' &= 2s(3 - s) - rs \end{aligned}$$
   
9. $$\begin{aligned} x' &= y^{2} - y + 3xy \\ y' &= x^{2} + x\ \end{aligned}$$