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A BRIEF OVERVIEW OF NONLINEAR ORDINARY
DIFFERENTIAL EQUATIONS
JOHN THOMAS
Abstract. This paper discusses the basic techniques of solving linear ordinary differential equations, as well as some tricks for solving nonlinear systems of ODE’s, most notably linearization of nonlinear systems. The paper proceeds to talk more thoroughly about the van der Pol system from Circuit Theory and the FitzHugh-Nagumo system from Neurodynamics, which can be seen as a generalization of the van der Pol system.
Contents
- General Solution to Autonomous Linear Systems of Differential Equations 1
- Sinks, Sources, Saddles, and Spirals: Equilibria in Linear Systems 4 2.1. Real Eigenvalues 5 2.2. Complex Eigenvalues 5
- Nonlinear Systems: Linearization 6
- When Linearization Fails 8
- The van der Pol Equation and Oscillating Systems 9
- Hopf Bifurcations 12
- Example: Neurodynamics 13 7.1. Ignoring I 13 7.2. Acknowledging I 14 7.3. Altering Parameters and Bifurcations 15 Acknowledgments 17 References 18
- General Solution to Autonomous Linear Systems of Differential Equations Let us begin our foray into systems of differential equations by considering the simple 1-dimensional case
(1.1) x′^ = ax
for some constant a. This equation can be solved by separating variables, yielding
(1.2) x = x 0 eat
Date: August 14, 2017. 1
2 JOHN THOMAS
where x 0 = x(0). Before proceeding to examine higher dimension linear, au- tonomous systems, it seems prudent to define ”linear” and ”autonomous” in this context. But first, a bit of notation.
Notation 1.3. Let x′ 1 = f 1 (t, x 1 , x 2 , ..., xn) x′ 2 = f 2 (t, x 1 , x 2 , .., xn) .. . x′ n = fn(t, x 1 , x 2 , ..., xn)
be a system of differential equations. I will write this as X′^ = F (t, X) where
X =
x 1 .. . xn
Unless otherwise specified, we will assume here that X ∈ Rn^ and F (t, X) : Rn+1^ → Rn.
Definition 1.4. An n-dimensional system of differential equations X′^ = F (t, X) is autonomous if F (t, X) in fact depends only on X.
Thus in discussion of autonomous systems, we write X′^ = F (X).
Definition 1.5. An n-dimensional system of differential equations X′^ = F (t, X) is linear if there exists A ∈ Rn×n^ such that X′^ = AX. That is, the system takes
the form
x′ 1 = a 11 x 1 + ... + a 1 nx 1 .. . x′ n = an 1 x 1 + ... + annxn
It is worth noting that any linear system of equations must also be autonomous. Let us now consider the very simple 2-dimensional system
(1.6) x′^ = ax y′^ = by
By repeating the 1-dimensional separation of variables and ”ignoring” either x or
y, we can see that X(t) = x 0 eat 0 and^ X(t) =^
y 0 ebt^ are solutions to (1.6). In fact,
we will show that X(t) =
x 0 eat y 0 ebt^ is also a solution to (1.6).
Theorem 1.7. Let X′^ = AX be a linear system of differential equations with solutions X(t) and Y (t). Then, (X + Y )(t) is also a solution to the system.
Proof. We know that (X + Y )′(t) = X′(t) + Y ′(t) and that X′(t) + Y ′(t) = AX + AY = A(X + Y ) by linearity. Therefore (X + Y )′(t) = A(X + Y ) as required.
Then, we have that x 0 eat y 0 ebt^ is indeed a solution to (1.6). This solution is more
interesting than it may at first appear. To clarify, let us rewrite (1.6) as
(1.8) X′^ =
a 0 0 b
X
4 JOHN THOMAS
We will not delve into the totality of the proof of this theorem in this paper. Suffice to say, the proof relies on the technique of Picard iteration. The basic idea of this technique is to construct a sequence of functions which converges to the solution of the differential equation. The sequence of functions pk(t) is defined by p 0 (t) = x 0 , our initial condition, and
pk+1 = x 0 +
∫ (^) t
0
pk(s)ds
This technique is useful not only for proving this theorem, but also for approximat- ing solutions to difficult or impossible to solve equations. Let us also consider an example which highlights the importance of the C^1 condition in theorem 1.14.
Example 1.15. Consider the differential equation
x′^ = ln(x)
Since f ′^ = (^1) x is not continuous, this equation fails the condition of theorem 1.14. Next, consider the initial condition x(0) = −1. But then we have x′(0) = ln(−1), which is nonsense. Therefore, we have no solution with initial condition x(0) = −1.
One last important result from theorem 1.14 is that solution curves to a differ- ential equation which satisfies the conditions of theorem 1.14 do not intersect. Another important concept is the ”flow” of an n-dimensional differential equa- tion. The flow is a function
φ → R × Rn
such that φ(t, X 0 ) is the solution at time t with φ(0, X 0 ) = X 0. Then we have the theorem
Theorem 1.16. Consider the system X′^ = F (X) where F is C^1. Then φ(t, X) is C^1 , i.e. (^) ∂X∂φ and ∂φ∂t exist and are continuous.
Again, in the interest of time, we will not delve into the prof of this theorem. Worth noting is that we can calculate ∂φ∂t for any t provided we know the solution through X 0. We have ∂φ ∂t
(t, X 0 ) = F (φ(t, X 0 ))
We also have ∂φ ∂X
(t, X 0 ) = Dφt(X 0 )
where Dφt is the Jacobian of X → φt(X) and φt(X) is φ(t, X) with constant t. Note that (^) ∂X∂φ requires knowledge not only of the solution through X 0 , but also through all nearby initial conditions.
- Sinks, Sources, Saddles, and Spirals: Equilibria in Linear Systems
Definition 2.1. An equilibrium point of the n-dimensional autonomous system of differential equations X′^ = F (X) is a point Z ∈ Rn^ such that X′^ = 0 at X = Z.
In particular, 0 is always an equilibrium point of a linear system. Let us now restrict our discussion to 2-dimensional linear systems X′^ = AX. Specifically, let us look at the eigenvalues of A.
A BRIEF OVERVIEW OF NONLINEAR ORDINARY DIFFERENTIAL EQUATIONS 5
Theorem 2.2. Let X′^ = AX be a 2-dimensional linear system. If det(A) 6 = 0, then X′^ = AX has a unique equilibrium point (0,0).
Proof. An equilibrium point X = (x, y) of the system X′^ = AX is a point that satisfies AX = 0. We know from linear algebra that this system has a nontrivial solution if and only if det(A) = 0. Therefore if det(A) 6 = 0, the only solution to AX = 0 is (0, 0).
2.1. Real Eigenvalues. If we ignore for now the possibility that λi = 0 and that λ 1 = λ 2 , then we are left with three cases:
(1) 0 < λ 1 < λ 2 (2) λ 1 < λ 2 < 0 (3) λ 1 < 0 < λ 2
Let us first consider case (1):
Example 2.3. Let A have eigenvalues 0 < λ 1 < λ 2 and eigenvectors V 1 , V 2 which correspond to λ 1 and λ 2 respectively. Then the general solution is of the form
X(t) = α 1 eλ^1 tV 1 + α 2 eλ^2 tV 2
Then, any solution of the system tends to (0, 0) as t → −∞ and tends to (±∞, ±∞) as t → ∞. Thus, we call (0, 0) a ”source” in this case.
Next, we consider case (2):
Example 2.4. Let A have eigenvalues λ 1 < λ 2 < 0 and corresponding eigenvectors V 1 , V 2. Then the general solution tends to (0, 0) as t → ∞. In this case, we call the equilibrium point a ”sink”.
And finally, case (3):
Example 2.5. Let A have eigenvalues λ 1 < 0 < λ 2 and corresponding eigenvectors V 1 , V 2. Then, the general solution
X(t) = α 1 eλ^1 tV 1 + α 2 eλ^2 tV 2
tends to (0, 0) along V 1. That is, for the solution X(t) = α 1 eλ^1 tV 1 , (0, 0) is a sink. We call the line {X ∈ R^2 |X = βV 1 , β ∈ R} the ”stable line”. Similarly, the solution tends away from (0, 0) as t → ∞. Thus we call the line {X ∈ R^2 |X = βV 2 , β ∈ R} the ”unstable line”. Overall, we call the equilibrium point of this system a ”saddle”.
2.2. Complex Eigenvalues. Of course, these examples cover only three of the four things I promised to discuss in this section. Next we turn to the possibility of complex eigenvalues and equilibria as spiral sources, spiral sinks, and centers (spiral saddles).
Example 2.6. Let X′^ = AX be a 2-dimensional liner system of differential equa- tions with
A =
0 β −β 0
with β 6 = 0. (This is the complex analogue of a saddle) Then, A has eigenvalues
±iβ. So, A has eigenvectors
i
and
i
to iβ and −iβ respectively. Then,
X′^ = AX has the solution
(2.7) X(t) = eiβt
i
i
A BRIEF OVERVIEW OF NONLINEAR ORDINARY DIFFERENTIAL EQUATIONS 7
This system has an equilibrium point at
−α −β
. Then, we can introduce a change
in coordinate ¯x 1 = x 1 + α, x¯ 2 = x 2 + β such that the new system
X¯′^ = X¯ +
α β
= X −
α β
α β
= X
Then, this new system X¯′^ = X¯ is linear with a unique equilibrium point at (0, 0).
Moreover, X¯′^ = X′^ for any X ∈ R^2 , since X¯′^ = (X −
α β
)′^ = X′.
Proposition 3.4. For any system of n-dimensional differential equations of the form X′^ = AX + V for some V ∈ Rn^ and with unique equilibrium point Xe, then the change of coordinates X¯ = X −Xe yields a linear system of differential equations X^ ¯′^ = A X¯ with unique equilibrium point 0.
Proof. Since Xe is an equilibrium point of the system, AXe + V = 0. Then, X¯′^ = A( X¯ + Xe) + V = A X¯ + (AXe + V ) = A X¯. Then, we have from 3.2 that X¯′^ = X′^ for any X ∈ Rn, as our proof in R^2 did not require any specific characteristics of R^2. Therefore 0 = X′ e = (Xe − Xe)′^ = ¯0, i.e. X¯′^ = A X¯ has equilibrium point 0. Since Xe is a unique equilibrium point, 0 is the only element of Rn^ with X¯ = 0.
More generally, we have the notion of conjugacy.
Definition 3.5. Let F : X → X and G : Y → Y. F and G are topologically conjugate if there exists a homeomorphism h : X → Y such that h ◦ F = G ◦ h
That is, two systems are conjugate if there exists a ”change of coordinates” from one system to the other. In studying nonlinear systems, we are notably interested in systems which are conjugate to linear systems. However, most nonlinear systems are not conjugate to linear equations. Most nonlinear systems, in fact, cannot be solved to arrive at a general equation.
Example 3.6. Consider the system
x′^ = x − 3 y + x^3 y′^ = −x + y − 2 y^4
We cannot solve this system explicitly, but we can at least discern the nature of the equilibrium point (0, 0). We know that as x, y → 0, x^3 and 2y^4 tend to 0 much faster than the linear terms. Therefore, sufficiently close to (0, 0), the system behaves similarly to
x′^ = x − 3 y y′^ = −x + y
The linear system has eigenvalues λ = 1 ± i
3, meaning the equilibrium point is a spiral source. Therefore, in the nonlinear system, we know at least that the equilibrium point is a source.
In fact, we have
8 JOHN THOMAS
Theorem 3.8. Suppose that the n-dimensional system X′^ = F (X) has an equi- librium point at X 0 and the eigenvalues of the Jacobian DF(X 0 ) have nonzero real part. Then the nonlinear flow is conjugate to the flow of the linearized system in a neighborhood of X 0.
The proof of this theorem is rather advanced and convoluted, so we will not discuss the details here. This style of linearizing nonlinear systems is a useful tool for determining the nature of equilibria and works well. Except when it doesn’t.
- When Linearization Fails
Example 4.1. Let us begin by considering a trivial example.
(4.2) x′^ = x^2 y′^ = −y^3
Then, the linearized system x′^ = 0, y′^ = 0 is rather boring to look at and, moreover, sheds no light on the behavior of the system even very close to the equilibrium point at the origin.
Example 4.3. Next, consider the less trivial system
(4.4) x′^ = x^2 y′^ = y
Then, the system has a unique equilibrium point at the origin, and has all other solutions moving toward the x-axis and to the right. The linearized system x′^ = 0 y′^ = −y
has equilibrium points all along the x-axis, and all other solutions lie on verti- cal lines. Thus, however close to the origin we are, we have lost the horizontal movement of the system.
At this point, one might suspect that, logically, we cannot linearize a system with no linear dependence on a variable, but some nonlinear dependence. However, this suspicion is not quite correct.
Example 4.5. To see this, let us consider the system
(4.6) x′^ = −y + cx(x^2 + y^2 ) y′^ = x + cy(x^2 + y^2 )
for c 6 = 0. Then, the linearized system is x′^ = −y y′^ = x
Then, the linearized system has eigenvalues λ = ±i, yielding a center at the origin. The behavior of the nonlinear system, however, changes drastically. To see this clearly, we convert the system to polar coordinates. We have x = r cos θ y = r sin θ
10 JOHN THOMAS
Now, we must show that the second case is impossible. Let us suppose that it is possible. Then, let (x 0 , y 0 ) be a point in B on this solution and let us examine the function φt(x 0 , y 0 ) = (x(t), y(t)). Since x(t) can never be zero, the solution lies in the region R given by 0 < x ≤ x 0 and y ≤ y 0. We also have y(t) → −∞ as t → tl for some tl. We observe that
y(t) − y 0 =
∫ (^) t
0
y′(s)ds =
∫ (^) t
0
−x(s)ds
Since we know that 0 < x ≤ x 0 , it follows that y(t) → −∞ only when t → ∞. But then, we know that x′^ = y − x^3 + x and that −x^3 + x is bounded and y → −∞ as t → ∞. Then, x(t) − x 0 tends to −∞ as t → ∞, which contradicts our assumption that x > 0. Therefore, the solution must cross y−. Then, we can exploit the skew symmetry of the system about the origin. That is, if V (x, y) is the van der Pol vector field, then −V (x, y) = V (−x, −y). Thus, the solution must pass x−^ and return to y+. This proves the lemma.
Then, we can define a Poincar´e map P on y+. For (0, y 0 ) ∈ y+, we define P (y 0 ) as the first return of φt(0, y 0 ) to y+^ with t > 0. P is injective and surjective. Then, let P n(x, y) = P (P n−^1 (x, y)) represent the nth^ return of φt to y+. Now, we must prove
Theorem 5.4. The Poincar´e map has a unique fixed point in y+. The sequence P n(y 0 ) tends to this fixed point as n → ∞ for any y 0 ∈ y+.
Proof. We observe that any fixed point of P must lie on a periodic solution. Alter- natively, if P (y 0 ) 6 = y 0 , then y 0 must not lie on a periodic solution. We note, then, that if P (y 0 ) > y 0 , then P n(y 0 ) > P n−^1 (y 0 ) for any n. Thus, the solution spirals upward to infinity and, importantly, never returns to y 0. Similarly, if P (y 0 ) < y 0 , the solution cannot return to y 0. We then define a new map α : y+^ → y−^ by letting α(y) be the first intersection of φt with y−^ for t > 0. Then, define
δ(y) =
(α(y)^2 − y^2 )
Observe for future use that there exists a unique point (0, y∗) ∈ y+^ and time t∗ such that
(1) φt(0, y∗) ∈ A for 0 < t < t∗ (2) φt∗ (0, y∗) = (1, 0) ∈ x+
That is, there is a unique point such that the solution beginning at the point, after its first quarter rotation, hits (1, 0). The end draws nearer to sight, as the theorem will follow from the following proposition:
Proposition 5.5. δ(y) satisfies
(1) δ(y) > 0 if 0 < y < y∗ (2) δ(y) decreases monotonically to −∞ as y → ∞ In a bid to confound and irritate the reader, we will prove the proposition shortly. First, we show that Theorem 5.4 does indeed follow from the proposition.
A BRIEF OVERVIEW OF NONLINEAR ORDINARY DIFFERENTIAL EQUATIONS 11
By using the skew symmetry of the vector field, we see that if (x(t), y(t)) is a solution curve, so too is (−x(t), −y(t)). By the intermediate value theorem and Proposition 5.5, there is a unique y 0 ∈ y+ such that δ(y 0 ) = 0. Then, α(y 0 ) = −y 0 and by symmetry, the solution through (0, y 0 ) is periodic. Since δ(y) 6 = 0 for y 6 = y 0 , φt(0, y 0 ) is the unique periodic solution. Next, we show that all other non-equilibrium solutions tend to this periodic solution. So, we define a map β : y−^ → y+, which sends a point in y−^ to its solution’s first intersection with y+. By symmetry, we know that β(y) = −α(−y). Also, P (y) = β(α(y)). We order y−^ ∪ y+^ such that y 1 > y 2 if y 1 is above y 2 on the y-axis. This ordering corresponds with our usual ordering of numbers. Note that α and β reverse ordering, while P preserves ordering. Now, we choose y ∈ y+^ with y > y 0. Then, we have α(y) < −y 0 = α(y 0 ). Also, δ(y) < 0, which implies that α(y) > −y. Therefore P (y) < y. We have already shown that y > y 0 implies y > P (y) > y 0. Then, P n(y) > P n+1(y) > y 0 for all n > 0. Then, P n(y) has a limit y 1 ≥ y 0 in y+. Note that, by continuity of P , y 1 is a fixed point of P. Then, since P has only one fixed point, we have y 1 = y 0. Thus, each solution that starts above y 0 spirals toward the periodic solution. We can see by similar methods that non-equilibrium solutions starting below y 0 spiral toward the periodic solution. This proves the theorem.
Now, all that remains is to prove Proposition 5.5.
Proof. Let γ : [a, b] → R^2 be a smooth curve in the plane and F : R^2 → R. We write γ(t) = (x(t), y(t)) and define ∫
γ
F (x, y) =
∫ (^) b
a
F (x(t), y(t))dt
If we have x′(t) 6 = 0 for all t ∈ [a, b], then along γ, y is a function of x. Then, we can write (^) ∫ b
a
F (x(t), y(t))dt =
∫ (^) x(b)
x(a)
F (x, y(x))
dt dx
dx
Then, ∫
γ
F (x, y) =
∫ (^) x(b)
x(a)
F (x, y(x)) dx/dt
dx
We have a similar expression if y′(t) 6 = 0. Now we introduce even another function
W (x, y) =
(x^2 + y^2 )
Let p ∈ y+. Suppose α(p) = φτ (p) for some τ > 0. Let γ(t) = (x(t), y(t)) for 0 ≤ t ≤ τ = τ (p) be the solution curve joining p to α(p). By definition,
δ(p) =
(y(τ )^2 − y(0)^2 ) = W (x(τ ), y(τ ))− = W (x(0), y(0))
Then,
δ(p) =
∫ (^) τ
0
d dt
W (x(t), y(t))dt
A BRIEF OVERVIEW OF NONLINEAR ORDINARY DIFFERENTIAL EQUATIONS 13
and μ ∈ [− 1 , 1]. Note that when μ = 1 we have the van der Pol system exactly. For any μ, as with the van der Pol system, the only equilibrium point is the origin and the linearized system is
X′^ =
μ 1 − 1 0
X
with eigenvalues
λ =
(μ ±
μ^2 − 4)
Then, for μ < 0, the origin is a spiral sink, and a spiral source for μ > 0. In fact, for μ < 0, the graph of fμ lies entirely in the first and third quadrant, meaning that the sign of x′^ is always the same as in the linearized system. Therefore, as in the linearized case, all solutions tend to the origin. In particular, no solutions tend to a periodic solution as they do in the van der Pol system. Note also that the system retains this spiral sink even when μ = 0 and the linearized system has a center at the origin. However, once μ becomes positive, the system oscillates as with the van der Pol system, since none of the proof of theorem 5.2 required that μ = 1. Just as with the van der Pol system, all non-equilibrium solutions tend to this new periodic solution. This vast change in behavior as we alter μ is called a Hopf Bifurcation.
- Example: Neurodynamics Let us now take an in-depth look at the FitzHugh-Nagumo system used to model neurodynamical behavior. The system is given by
(7.1) x
′ (^) = y + x − x^3 3 +^ I y′^ = −x + a − by
where I is a parameter and
0 <
(1 − a) < b < 1
This system is a simplification of neurodynamical systems and is reminiscent of differential equations used to model circuits. Specifically, the system shares many similarities with the van der Pol system. Here, x is analogous to voltage in circuits and y represents an amalgam of forces which restore the system to rest. I represents the ”excitability” of the system, for reasons we will see later; a and b are constants. We can then immediately see that 13 < a < 1. Let us begin by examining the system when I = 0.
7.1. Ignoring I. First, let us find all equilibria points of the system (7.1) when I = 0. We know that any equilibria will occur at the intersection of the x- and y-nullclines. The x-nullcline is given by
(7.2) yx =
x^3 3 − x
and the y-nullcline by
(7.3) yy =
a b
x b First, we observe that, because yx → ∞ as x → ∞ and → −∞ as x → −∞, and yy → −∞ as x → ∞ and vice versa, the nullclines must intersect at some point.
14 JOHN THOMAS
Then, we observe that the x-nullcline is monotonically increasing except on the region [− 1 , 1], on which it is monotonically decreasing. The y-nullcline is monotoni- cally decreasing. Then, if the nullclines have multiple intersections, one intersection must occur on the region [− 1 , 1] with respect to x. In fact, we must also have that the y-nullcline passes through both vertical edges of the rectangle [− 1 , 1] × [− 23 , 23 ], since the x-nullcline has yx(1) = − 23 and yx(−1) = 23. But, we see that, since (^1) b > 1, if yy (−1) = 23 , then yy (1) < 23 − 2 = − 43. Therefore, the y-nullcline is too steep to intersect the x-nullcline twice. So we have
Proposition 7.4. For I = 0, the system has a unique equilibrium point (x 0 , y 0 ).
Next, we examine the nature of this equilibrium point. Now, we consider the vec- tor field at points slightly displaced from equilibrium along the x- and y-nullclines. Along the x-nullcline, we have
(7.5) y′^ = −x + a − b(
x^3 3
− x) = a −
bx^3 3
Then, we see that for small, y′(x 0 + ) < y′(x 0 ) and y′(x 0 − ) > y′(x 0 ). Along the y-nullcline, we have
(7.6) x′^ =
a b
x b
x^3 3
a b
b )x −
x^3 3 Then, we have that x′(x 0 + ) < x′(x 0 ) and x′(x 0 − ) > x′(x 0 ). So, minor displacement along the nullclines pushses solutions back toward the equilibrium. Then we have
Proposition 7.7. The equilibrium point (x 0 , y 0 ) is always a sink.
7.2. Acknowledging I. Our first order of business now that we allow I 6 = 0 is to discern whether or not this changes the number of equilibrium points. We now have the x-nullcline
(7.8) yx = x^3 3
− x − I
Thus, the addition of I shifts the graph of the x-nullcline up or down. However, because the end behavior of the nullclines is unchanged, we must still have at least one equilibrium point. Then, if we are to have multiple equilibria, we must have the y-nullcline pass through the vertical edges of [− 1 , 1] × [− 23 − I, 23 − I]. However, our argument from the previous subsection still applies here, so we still have a unique equilibrium point xI , yI. In fact, we also see that, if I > 0, the graph of the x-nullcline shifts down, shifting the equilibrium point down. Then, since the equilibrium must lie along the y-nullcline, the equilibrium point is shifted right. Similarly, if I < 0, the equilibrium point is shifted left. Therefore, we have
Proposition 7.9. The system has a unique equilibrium point (xI , yI ) and xI varies monotonically with I.
For points sufficiently close to the equilibrium point (the origin under the new coordinates), the system behaves similarly to
(7.10) x′^ = y + (1 − x^2 I )x + I y′^ = −x + a − by
16 JOHN THOMAS
Thus, we must in this case be cautious about trusting the result of the linearized system. The full system in this case is
x′^ = y + x − x
3 3 y′^ = −x − y
Then, along the x-nullcline of the system, we have
y′^ = −
x^3 3
Then, displacing slightly from the equilibrium to the right along the x-nullcline yields y′^ < 0. Since the x-nullcline is decreasing around the equilibrium point, this displacement pushes solutions farther away from the equilibrium point. This is also true of displacement to the left. Along the y-nullcline, we have
x′^ = −
x^3 3
Then, we see that slight displacement from the origin along the y-nullcline restores solutions to equilibrium. Therefore, we have a real saddle at the equilibrium for b = 1. To summarize, we have
Proposition 7.17. The system has a saddle for b ≥ 1 and b < − 3 , a spiral source for − 3 < b < 1 , and a sink for b = − 3.
Next, we allow the same restriction on b as above and also allow I to vary again. Then we have the x-nullcline
y = x^3 3
− x − I
and the y-nullcline
y = −
x b Along the x-nullcline, we have
y′^ = −
bx^3 3
and along the y-nullcline
x′^ = (1 −
b )x −
x^3 3
+ I
First, we consider the case that I < 0. In this case, the equilibrium point has xe > 0 and ye < 0. Then, for b < 1, we have a (spiral) sink. For b > 1, we have a saddle if xe ∈ (− 1 , 0) and a sink otherwise. For b = 1, we have a real sink. Next, if I > 0, we have an equilibrium point with xe > 0 and ye < 0, but the system’s behavior as b changes mimics the previous case. Now, we extend this inquiry to the case where b ≤ 0. Firstly, we observe that for b = 0, we have
x′^ = y + x − x
3 3 +^ I y′^ = −x
which is the van der Pol system with the addition of I. This system has an equi- librium point at (0, −I) and the equilibrium is a sink for any I.
A BRIEF OVERVIEW OF NONLINEAR ORDINARY DIFFERENTIAL EQUATIONS 17
Then, we examine the system when b < 0. To clarify the sign of terms, let d = −b. Then the system is
(7.19) x
′ (^) = y + x − x^3 3 +^ I y′^ = −x + dy
For I = 0, the system has an equilibrium at the origin and linearizes with eigenval- ues
λ =
(−g ±
g^2 − 4 g)
where g = d + 1. Then, the equilibrium is always a sink, and is a spiral sink for b > −3. However, in this case the origin is not a unique equilibrium. Examining the intersection of nullclines, we find that the system has three equilibrium points, the other two at (± 3
1 + h, ± 3 h
1 + h), where h = (^1) d. Examining small displacement along nullclines from these equilibria, we see that both points are saddles, with solutions tending toward the equilibrium along the y-nullcline and away from the equilibrium along the x-nullcline. Next, we consider I 6 = 0. Then, depending on |I|, we may have 1 or 3 equilibria. Similarly to our argument concerning equilibrium uniqueness in § 7 .2, we have mul- tiple equilibria when the y-nullcline passes through the region [− 1 , 1]×[− 23 −I, 23 −I] though we no longer require passage through the vertical edges of the region. Then, we have a single equilibrium as long as
1 b
− I)
Then we have
Proposition 7.20. The system has one equilibrium point, a saddle, when (^1) b < −( 23 − I), and three equilibria, two saddles and a sink, otherwise.
Lastly, we turn our attention to the case that b = 0 and I, a 6 = 0. Then, the system is
(7.21) x
′ (^) = y + x − x^3 3 +^ I y′^ = −x + a
Firstly, we observe that now, since the y-nullcline is a vertical line, the system will always have a single equilibrium (xe, ye). Then, for points sufficiently close to the equilibrium, the system behaves similarly to
x′^ = y + (1 − x^2 e)x + I y′^ = −x + a
Then, by examining displacement along the nullclines, we see that the equilibrium is a spiral sink for |x| < 1, a center for |x| = 1, and a spiral source otherwise. This concludes our analysis of this system.
Acknowledgments. It is with utmost gratitude that I extend my thanks to my mentor Brian Chung for his eagerness to help and the clarity of his explanations, without which I would be lost in this endeavor.
