1. Qualitative Analysis of Systems of Ordinary Differential Equations

2. Example
Equilibrium Analysis in One Dimension: Example
A predator-prey model says that a prey population 𝑥 and a predator population 𝑦, as functions of 𝑡 satisfy the
differential equations
𝑥 ′ =𝛼𝑥−𝛽𝑥𝑦
𝑦 ′ =𝛿𝑥𝑦−𝛾𝑦
Visualize the solutions to these equations with a phase plane and find all equilibrium solutions, given
𝛼=2, 𝛽=6, 𝛿=0.4, 𝛾=1.3. (Assume 𝑥(𝑡) and 𝑦(𝑡) measure the populations in thousands of individuals.)
Solution
To find equilibrium solutions, we set the derivatives equal to zero.
𝛼𝑥−𝛽𝑥𝑦=0
𝛿𝑥𝑦−𝛾𝑦=0
Solving, we obtain:
𝑥,𝑦 = 0,0 & (3.250,0.333)
Thus, we are at equilibrium if both species are extinct (0,0) or if the populations are at 3250 prey and 333 predators. (At equilibrium, it is expected that the predator population will be much smaller than the prey population.)

3. Example
A predator-prey model says that a prey population 𝑥 and a predator population 𝑦, as functions of 𝑡 satisfy the
differential equations
𝑥 ′ =𝛼𝑥−𝛽𝑥𝑦
𝑦 ′ =𝛿𝑥𝑦−𝛾𝑦
Visualize the solutions to these equations with a phase plane and find all equilibrium solutions, given
𝛼=2, 𝛽=6, 𝛿=0.4, 𝛾=1.3. (Assume 𝑥(𝑡) and 𝑦(𝑡) measure the populations in thousands of individuals.)
𝑥 0 ,𝑦 0 = 3.3 , 0.3
Solution
A solution can be visualized by a parametric plot 𝑥 𝑡 ,𝑦 𝑡 . Let’s assume, for example,

4. The plot indicates that the two populations are intertwined in a cyclic pattern. The two populations oscillate about the equilibrium (3.250,0.333). The initial condition (3.3,0.3) is indicated with a black point.

5. Instead of copy/pasting the code above and changing the constants, we can make it into a Module.

6. Example
(a) Visualize the differential equations in 𝑥(𝑡) and 𝑦(𝑡):
𝑥 ′ = 1 2 𝑥−𝑦
𝑦 ′ =− 5 4 𝑥− 3 2 𝑦
(b) Verify that 𝐱 1 𝑡 = 𝑒 −2𝑡 is a solution; verify that 𝐱 2 𝑡 = 𝑒 𝑡 −2 1 is a solution as well; verify that any linear combination of these is again a solution.
(c) Visualize several solutions from part (b).

7. Example
(a) Visualize the differential equations in 𝑥(𝑡) and 𝑦(𝑡):
𝑥 ′ = 1 2 𝑥−𝑦
𝑦 ′ =− 5 4 𝑥− 3 2 𝑦
Solution
If we find a solution, we can call it 𝐱 𝑡 = 𝑥 𝑡 ,𝑦 𝑡 .
This is a parametric curve.
Its velocity at time 𝑡 will be 𝑥 ′ 𝑡 , 𝑦 ′ 𝑡 .
On the other hand, to be a solution, it must satisfy the differential equations, which means:
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦= 𝑥 ′ 𝑡 , 𝑦 ′ 𝑡 must be equal to 𝑥−𝑦,− 5 4 𝑥− 3 2 𝑦 at the point (𝑥,𝑦)
This is a bit like a slope-field, but now it’s a full-fledged velocity field!

8. Solution parametric curves will travel through
the vector field with the indicated velocity.

9. Example
(a) Visualize the differential equations in 𝑥(𝑡) and 𝑦(𝑡):
𝑥 ′ = 1 2 𝑥−𝑦
𝑦 ′ =− 5 4 𝑥− 3 2 𝑦
(b) Verify that 𝐱 1 𝑡 = 𝑒 −2𝑡 is a solution; verify that 𝐱 2 𝑡 = 𝑒 𝑡 −2 1 is a solution as well; verify that any linear combination of these is again a solution.
The solution 𝐱 1 𝑡 = 𝑒 −2𝑡 has 𝑥-component 𝑥 𝑡 =2 𝑒 −2𝑡 and 𝑦-component 5 𝑒 −2𝑡 . Plug in for 𝑥 and 𝑦.
Solution
−4 𝑒 −2𝑡 = 𝑒 −2𝑡 −5 𝑒 −2𝑡
−10 𝑒 −2𝑡 =− 𝑒 −2𝑡 − 𝑒 −2𝑡
A bit of arithmetic verifies that these equations are indeed true. So we do have a solution to the differential equation (!)
The details of verifying 𝑒 𝑡 −2 1 and the linear combinations of these two solutions is left as an exercise.

10. Example
(a) Visualize the differential equations in 𝑥(𝑡) and 𝑦(𝑡):
𝑥 ′ = 1 2 𝑥−𝑦
𝑦 ′ =− 5 4 𝑥− 3 2 𝑦
(c) Visualize several solutions from part (b).
Some solutions are plotted by picking various linear combinations.

11. How do we find them? Discussion
In the previous problem, we found that solutions to the linear system
𝑥 ′ = 1 2 𝑥−𝑦
𝑦 ′ =− 5 4 𝑥− 3 2 𝑦
How do we find them?
could be found of the form 𝐱= 𝑒 𝜆𝑡 𝐮 for certain values of 𝜆 and certain vectors 𝐮.
First write the differential equation in matrix form.
where 𝐱= 𝑥 𝑦 and 𝐀= 1/2 −1 −5/4 −3/2
𝐱 ′ =𝐀𝐱
Now suppose we “guess” 𝐱= 𝑒 𝜆𝑡 𝐮 and plug this into the differential equation.
𝜆 𝑒 𝜆𝑡 𝐮= 𝑒 𝜆𝑡 𝐀𝐮
Divide both sides by 𝑒 𝜆𝑡 .
𝐀𝐮=𝜆𝐮
This is called the characteristic equation for the matrix 𝐀
See the Crash Course on Linear Algebra presentation for a complete description of how to solve this.

12. Example
Solve the linear system of differential equations. Draw a phase portrait.
𝑥 ′ = 𝑥− 1 80 𝑦
𝑦 ′ = 𝑥 𝑦
Rewrite the system as 𝐱 ′ =𝐀𝐱 where 𝐀= 11/200 −1/80 1/100 1/40 .
Solution
Guess 𝐱= 𝑒 𝜆𝑡 𝐮 and solve the eigen-equation:
𝐀𝐮=𝜆𝐮
First find the eigenvalues by solving det 𝐀−𝜆𝐈 =0:
𝜆 1 =3/100, 𝜆 2 =1/20.
The eigenvectors corresponding to these are:
𝐮 1 = , 𝐮 2 =
Conclude that the general solution is:
𝐱= 𝐶 1 𝐱 1 + 𝐶 2 𝐱 2
𝐱= 𝐶 1 𝑒 𝑡 2 𝑒 𝑡 + 𝐶 𝑒 𝑡 2 𝑒 𝑡

13. Example
Solve the linear system of differential equations. Draw a phase portrait.
𝑥 ′ = 𝑥− 1 80 𝑦
𝑦 ′ = 𝑥 𝑦
EQUILIBRIUM TYPE: UNSTABLE NODE
Solution
𝐱= 𝐶 1 𝑒 𝑡 2 𝑒 𝑡 + 𝐶 𝑒 𝑡 2 𝑒 𝑡

14. Example (Complex Eigenvalues)
Consider the system of differential equations (derivatives are with respect to 𝑡)
𝑥 ′ =3𝑥−4𝑦
𝑦 ′ =4𝑥+𝑦
Find the general solution; visualize it.
Solution
Again, the equation is 𝐱 ′ =𝐀𝐱.
We guess 𝐱= 𝑒 𝜆𝑡 𝐮 and get the eigen-equation 𝐀𝐮=𝜆𝐮.
Find eigenvalues by solving det 𝐀−𝜆𝐈 =0:
𝜆 1,2 =2± 15 𝑖

15. FACTS FROM COMPLEX EIGENVALUE THEORY
Fact 1: Complex eigenvalues will come in complex conjugate pairs 𝜆 +,− =𝑎±𝑏𝑖
Fact 2: Corresponding eigenvectors will come in complex conjugate pairs 𝐮, 𝐮
Fact 3: If we use the + eigenvalue 𝜆 + =𝑎+𝑏𝑖 and the corresponding eigenvector 𝐮, then the complex solution 𝑒 𝜆 + 𝑡 𝐮 is correct if interpreted through Euler’s Identity: the real part will be a solution and the imaginary part will be a solution.
Fact 4: No new solutions are found by using 𝜆 − .

16. Example (Complex Eigenvalues)
Consider the system of differential equations (derivatives are with respect to 𝑡)
𝑥 ′ =3𝑥−4𝑦
𝑦 ′ =4𝑥+𝑦
Find the general solution; visualize it.
Solution
Again, the equation is 𝐱 ′ =𝐀𝐱.
We guess 𝐱= 𝑒 𝜆𝑡 𝐮 and get the eigen-equation 𝐀𝐮=𝜆𝐮.
Find eigenvalues by solving det 𝐀−𝜆𝐈 =0:
𝜆 1,2 =2± 15 𝑖
From the theory, we know we only have to use 𝜆 1 = 𝑖.
𝐮 1 = 𝑖 4
Find the eigenvector:
𝐱 = 𝑒 𝜆 1 𝑡 𝐮 1 ; 𝐱 1 =Re 𝐱 ; 𝐱 2 =Im 𝐱
𝐱 1 = 𝑒 2𝑡 cos 𝑡 + 𝑒 2𝑡 sin 𝑡 4 𝑒 2𝑡 cos 𝑡
𝐱 2 = 𝑒 2𝑡 cos 𝑡 + 𝑒 2𝑡 sin 𝑡 4 𝑒 2𝑡 sin 𝑡
𝐱= 𝐶 1 𝐱 1 + 𝐶 2 𝐱 2

17. Example (Complex Eigenvalues)
Consider the system of differential equations (derivatives are with respect to 𝑡)
𝑥 ′ =3𝑥−4𝑦
𝑦 ′ =4𝑥+𝑦
Find the general solution; visualize it.
Solution
𝐱 1 = 𝑒 2𝑡 cos 𝑡 + 𝑒 2𝑡 sin 𝑡 4 𝑒 2𝑡 cos 𝑡
𝐱 2 = 𝑒 2𝑡 cos 𝑡 + 𝑒 2𝑡 sin 𝑡 4 𝑒 2𝑡 sin 𝑡
𝐱= 𝐶 1 𝐱 1 + 𝐶 2 𝐱 2
EQUILIBRIUM TYPE: UNSTABLE SPIRAL

18. Example
Solve, analyze, visualize:
𝑥 ′ =3𝑥−6𝑦
𝑦 ′ =3𝑥−3𝑦
Solution
Write the equation in matrix-vector notation 𝐱 ′ =𝐀𝐱; guess 𝐱= 𝑒 𝜆𝑡 𝐮 and solve the eigen-equation 𝐀𝐮=𝜆𝐮.
Values of 𝜆 must satisfy det 𝐀−𝜆𝐈 =0, yielding 𝜆=±3𝑖.
We may focus on 𝜆 1 =3𝑖.
The corresponding eigenvector 𝐮 is 𝐮 1 = 1+𝑖 1 .
Two solutions are found from real and imaginary parts:
𝐱 1 =Re 𝑒 𝜆 1 𝑡 𝐮 1 , 𝐱 2 =Im 𝑒 𝜆 1 𝑡 𝐮 1
𝐱 1 = cos 3𝑡 − sin 3𝑡 cos 3𝑡 , 𝐱 2 = cos 3𝑡 + sin 3𝑡 sin 3𝑡
𝐱= 𝐶 1 𝐱 1 + 𝐶 2 𝐱 2
Analysis
It can be shown through trig identities that
the ellipses have major axis at angle 𝑜 and that the major radii are times the length of the minor radii.
Solutions have period 𝜏/3=2.094 s
EQUILIBRIUM TYPE: STABLE CENTER

19. Example
Solve, analyze, visualize:
𝑥 ′ =−2𝑦
𝑦 ′ =−𝑥+𝑦
Solution
Write the equation in matrix-vector notation 𝐱 ′ =𝐀𝐱; guess 𝐱= 𝑒 𝜆𝑡 𝐮 and solve the eigen-equation 𝐀𝐮=𝜆𝐮.
Values of 𝜆 must satisfy det 𝐀−𝜆𝐈 =0, yielding real eigenvalues 𝜆 1 =2, 𝜆 2 =−1.
The eigenvectors, found by finding the nullspaces of 𝐀−𝜆𝐈 are 𝐮 1 = 1 −1 and 𝐮 2 =
𝐱 1 = 𝑒 𝜆 1 𝑡 𝐮 1 = 𝑒 2𝑡 − 𝑒 2𝑡 , 𝐱 2 = 𝑒 𝜆 2 𝑡 𝐮 2 = 2 𝑒 −𝑡 𝑒 −𝑡
𝐱= 𝐶 1 𝐱 1 + 𝐶 2 𝐱 2
EQULIBRIUM TYPE: (UNSTABLE) SADDLE NODE

20. Phase Portraits
We are going to study the following non-linear system in a fair amount of detail:
𝑥 ′ = 𝑥 2 −3𝑥𝑦−4
𝑦 ′ =𝑥𝑦+ 𝑦 2
Our study will go through the following steps:
1. Identify the equilibrium points
2. Draw a phase portrait with software and label the equilibrium points on it
3. Linearize the system about each equilibrium point
4. Solve the linearizations to understand local and global behavior of the original non-linear system

21. Phase Portraits
We are going to study the following non-linear system in a fair amount of detail:
𝑥 ′ = 𝑥 2 −3𝑥𝑦−4
𝑦 ′ =𝑥𝑦+ 𝑦 2
1. Identify the equilibrium points
2. Draw a phase portrait with software and label the equilibrium points on it
Solution
We set 𝑥 ′ and 𝑦 ′ to zero…
𝑥 2 −3𝑥𝑦−4=0
𝑥𝑦+ 𝑦 2 =0
𝑥,𝑦 : −2,0 , −1,1 , 1,−1 , 2,0
3. Linearize the system about each equilibrium point
Let’s study the point (−1,1) first.

22. Phase Portraits
𝑥 ′ = 𝑥 2 −3𝑥𝑦−4
𝑦 ′ =𝑥𝑦+ 𝑦 2 [
𝑢 ′ =−5𝑢+3𝑣
𝑣 ′ =𝑢+𝑣

23. Phase Portraits
𝑢 ′ =−5𝑢+3𝑣
𝑣 ′ =𝑢+𝑣

24. Phase Portraits
𝑢 ′ =−5𝑢+3𝑣
𝑣 ′ =𝑢+𝑣
“Guess” 𝐮= 𝑒 𝜆𝑡 𝐰
𝐮 ′ = − 𝐮
𝜆 1 =−2−2 3 , 𝜆 2 =−2+2 3
𝐰 𝟏 = −3− , 𝐰 2 = −
𝐮= 𝐮 1 + 𝐮 2
𝐮 𝑡 = 𝑒 −2−2 3 𝑡 −3− 𝑒 − 𝑡 −
𝐮 𝑡 = 𝑒 −5.46𝑡 − 𝑒 1.46𝑡

25. Before we move on to study the equilibrium point (2,0)
𝑥 ′ = 𝑥 2 −3𝑥𝑦−4
𝑦 ′ =𝑥𝑦+ 𝑦 2
=: 𝐹 1 𝑥,𝑦
we need a formula for the linearization.
=: 𝐹 2 (𝑥,𝑦)
Here it is:
𝐷 𝐅 𝐱 0 = 𝜕 𝐅 1 /𝜕𝑥 𝜕 𝐅 1 /𝜕𝑦 𝜕 𝐅 2 /𝜕𝑥 𝜕 𝐅 2 /𝜕𝑦 2,0
𝐮 ′ =𝐷 𝐅 𝐱 0 𝐮
Local Linearization Formula for Equilibria
If 𝐱 0 = 𝑥 0 , 𝑦 0 is an equilibrium point of the system of differential equations
𝐱 ′ =𝐅 𝐱
then, in local coordinates 𝑢=𝑥− 𝑥 0 , 𝑣=𝑦− 𝑦 0 , the linearization is
𝐮 ′ =𝐷 𝐅 𝐱 0 𝐮
where 𝐷 𝐅 𝐱 0 is the Jacobian matrix
𝐷 𝐅 𝐱 0 = 𝜕 𝐅 1 /𝜕𝑥 𝜕 𝐅 1 /𝜕𝑦 𝜕 𝐅 2 /𝜕𝑥 𝜕 𝐅 2 /𝜕𝑦
𝐷 𝐅 𝐱 0 = 𝜕 𝜕𝑥 𝑥 2 −3𝑥𝑦−4 𝜕 𝜕𝑦 𝑥 2 −3𝑥𝑦−4 𝜕 𝜕𝑥 𝑥𝑦+ 𝑦 2 𝜕 𝜕𝑦 𝑥𝑦+ 𝑦 ,0
𝐹 1 𝑥,𝑦 𝐹 2 𝑥,𝑦
𝐱 ′
𝐱 ′ =𝐅 𝑥,𝑦
𝐱 ′ =𝐅 𝐱
𝐱 ′ = 𝑥 ′ 𝑦 ′
𝐱= 𝑥 𝑦
= 2𝑥−3𝑦 −3𝑥 𝑦 𝑥+2𝑦 2,0
𝐹 1 𝑥,𝑦 = 𝑥 2 −3𝑥𝑦−4
𝐹 1 𝑥,𝑦
𝐱 ′ 𝐲 ′
𝑥 ′ = 𝑥 2 −3𝑥𝑦−4
𝑦 ′ =𝑥𝑦+ 𝑦 2
𝐹 2 𝑥,𝑦 =𝑥𝑦+ 𝑦 2
= 4 −6 0 2
𝐹 2 𝑥,𝑦
𝐮 ′ = 4 − 𝐮

26. 𝐮 ′ = 4 − 𝐮
𝜆 1 =2, 𝜆 2 =4
𝐮= 𝑒 𝜆𝑡 𝐰
𝐰 1 = , 𝐰 2 = 1 0
𝐮= 𝐶 1 𝐮 1 + 𝐶 2 𝐮 2
𝐮= 𝐶 1 𝑒 2𝑡 𝐶 2 𝑒 4𝑡 1 0

27. Just two more equilibrium points to study.
Try them yourself!