Senior Seminar II Project

Analysis of the Trace Determinant Plane of

Systems of Linear Differential Equations

Joanna Sutton

Fall 2012

Abstract

This essay will analyze six different phase portraits of systems oflinear differential equations and how they are connected with the eigen-values of that system. The different cases of eigenvalues for each sys-tems and how these cases can be determined from the trace and de-terminant of the system will be demonstrated. Then, the link betweenthe trace, determinant, and phase plane for the linear system will beanalyzed.

1 Introduction

Trace-Determinant Planes are graphical representations of solutionsof planar systems. These systems demonstrate different behaviors inthe plane depending on the form of their eigenvalues. There are sixdifferent cases of eigenvalues that each result in different behavior.These cases are as follows: both eigenvalues are real and positive, bothare real and negative, both are real but one is positive and one negative,the eigenvalues are complex with a real part of zero, the eigenvalues arecomplex with a positive real part, and the eigenvalues are complex witha negative real part. The eigenvalue cases can be linked to six differentphase potraits. These portriats are nodal source, nodal sink, saddlepoint, center, spiral source, and spiral sink respectively. The trace anddeterminant of the system can be used to not only determine the caseof the eigenvalues, but the case of the phase potrait as well.

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2 Linear Systems

Trace-Determinant Planes are graphical representations of solutionsof planar systems. A planar system is a linear system of differentialequations of dimension 2. Differential Equations states that a systemof linear differential equations is any system of differential equationsthat has the following form:

x′1(t) = a11(t)x1(t) + · · ·+ a1n(t)xn(t) + f1(t)

x′2(t) = a21(t)x1(t) + · · ·+ a2n(t)xn(t) + f2(t)

x′n(t) = an1(t)x1(t) + · · ·+ ann(t)xn(t) + fn(t)

in which x1xn are unknown functions, and the coefficients aij(t) and fi(t)are known functions of an independent variable,t, all defined for tε(a, b),(a, b) is an interval in R. (Polking, Boggess, Arnold pg 425). In thispaper we will be analyzing systems of the form y = Ay,where

A =

(a11 a12a21 a22

)and y(t) =

(y1(t)y2(t)

)(Polking, Boggess, Arnold pg 452)

This system can be solved through first finding the characteristicpolynomial for the matrix A. This is done through taking the deter-minant of A− λI, where I is the identity matrix.

det(A− λI) = detA = det

(a11 − λ a12a21 a22 − λ

)= (a11 − λ)(a22 − λ)− a12a21

= λ2 − (a11 + a22)λ+ a11a22 − a12a21The quantity of (a11 +a22) is known as the trace of the matrix, abbre-viated as T . The quantity a11a22 − a12a21 is the determinant of thematrix A, abbreviated as D. This results in the equation

λ2 − Tλ+D (1)

To obtain the equation for the eigenvalues of A, we only need to setequation (1) to zero and solve. When applying the quadratic equation,we get

λ =T ±√T 2 − 4D

2. (2)

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This equation shows that if T 2 − 4D < 0, then the eigenvalues arecomplex. If T 2 − 4D = 0, then the eigenvalues are repeated, and ifT 2 − 4D > 0, then the eigenvalues are real and not repeated.

Now that there is an equation for the eigenvalues, the trace-deter-minant plane can be drawn and analyzed. The trace-determinant planeis key in analyzing the equilibrium points. The trace is set as thehorizontal axis, and the determinant is set as the vertical axis. Thecurve of T 2 − 4D = 0 is drawn on the graph separate the graph intotwo different regions (as shown below). The region above the curverepresents systems with complex eigenvalues, and the region belowthe curve represents systems with real eigenvalues. (Polking, Boggess,Arnold pg 452).

Figure 1: Trace-Determinant Plane (Polking, Boggess, Arnold p477)

3 The Six Different Phase Portraits

Before discussing how to solve for various cases of eigenvalues, it isimportant to know what kinds of behavior the different combinationswill show in the trace-determinant curve. If λ1 < 0 < λ2 (if there is

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one positive and one negative) the origin will be a saddle. If λ1 <λ2 < 0 (both negative eigenvalues) then the origin will be a sink. If0 < λ1 < λ2 (both positive eigenvalues) then the origin is a source.The following cases are for complex eigenvalues of the form λ = a± bi.If a < 0 and b 6= 0, then the origin is a spiral sink. If a > 0 and b 6= 0,then the origin is a spiral source. Lastly, is a = 0 and b 6= 0, then theorigin is the center. These behaviors are all shown on the table below.

Figure 2: Table of Phase Portraits in the Trace-Determinant Plane (Blan-chard, Devaney,Hall p341)

Now, to analyze the different equilibrium points, it is easier to firstlook at the eigenvectors of the system to find a solution.

Theorem 1

Suppose A is a 2x2 matrix with real eigenvalues λ1 6= λ2. Supposev1 and v2 are eigenvectors associated with these eigenvalues. Then,the general solution of the system y = Ay is

y(t) = C1eλ1tv1 + C2e

λ2tv2 (3)

(Polking,Boggess,Arnold p454)

With this theorem and equation 3, we can now start to explain whyeach system has the its specific type of equilibrium point.

3.1 Saddle Point

A system has a saddle point if it has one positive and one negativeeigenvalue. In other words, λ1 < 0 < λ2. To derive the saddle point,

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we must first let C2 = 0. This would cause y1(t) = C1eλ1tv1 to be the

exponential solution. Next, pick x to be any real number, and drawa solution for C1 = x and C1 = −x. The line must go through theorigin.When λ1 < 0 the line decreases from infinity to zero as t goesfrom negative infinity to infinity. Next, take C1 = 0. In this case weare now using y2(t) = C2e

λ2tv1 as our exponential solution. We solvefor it in a similar manner to how we solved for y1. We must draw theline for C2 = x, C2 = −x where x is a real number. The line must gothrough the origin. (Polking,Boggess,Arnold p469)

The saddle point is formed by taking the two half line solutionsand adding them. Curved lines are then draw along the ends of thesum vectors. The solutions for y1 and y2 are asymptotes of each other.As t → ∞, t → −∞ these solutions will never cross. These linesseparate the plane into four different regions. The lines are calledseparatrices. The two solutions that approach the origin as t → ∞are called the stable solutions. The two solutions which approach theorigin as t → −∞ are called the unstable solutions. The equilibriumpoint is called a saddle point because the topographic map looks likea horse’s saddle. (Polking,Boggess,Arnold p469)

Figure 3: Saddle Point1

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3.2 Nodal Sink

If a system has two negative eigenvalues, λ1 < λ2 < 0, then the planedisplays a nodal sink. Again, we must begin by analyzing equation(3). Both of the values for λ are negative, so y(t) is infinity largeat negative infinity. y(t) then converges to 0 on a half-line as t goesto infinity. Now, we must split this into two different cases, one casewhere t→∞ and the other where t→ −∞. In the first case, we mustfirst rewrite equation (3) as

y(t) = eλ2t(C1e(λ1−λ2)tv1 + C2v2)

In this case, as t → ∞, eλ2t → 0. Since we know λ1 < λ2 < 0, wealso know λ1 − λ2 < 0 which means(C1e

(λ1−λ2)t) → 0 . This resultsin y(t) → 0, but while converging the direction of y(t) gets closer toC2v2, so as y(t) → 0 the solution curve on the plane will be tangentto the half-line we get from C2v2. (Polking,Boggess,Arnold p471)

Now, for the second case we rewrite equation (3) as

y(t) = eλ1t(C1v1 + C2e(λ2−λ1)tv2)

As t → −∞,eλ1t → ∞ since λ1 < λ2 < 0 implies λ1 − λ2 > 0. Thisalso results in C2e

(λ2−λ1)t → 0. From this we know that (C1v1 +C2e

(λ2−λ1)tv2) → C1v1 and y(t) → 0, but as y(t) → 0, the directionof y(t) becomes that of C1v1.(Polking,Boggess,Arnold p471)

All solution curves in a system with two negative eigenvalues tendto the origin as t → ∞. If all solutions for an equilibrium point ofa system tend to the equilibrium point as t → ∞ and the eigenval-ues are real, then the equilibrium point is said to be a nodal sink.(Polking,Boggess,Arnold p471)

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Figure 4: Nodal Sink2

3.3 Nodal Source

A nodal source is the last phase portrait a system can have if it hasreal eigenvalues. In this case, 0 < λ1 < λ2. The properties for a nodalsource are similar to that of a nodal sink but reversed. As stated above,in a nodal sink the solution curves tend to the origin as t→∞, whilein a nodal source the solution curves tend to the origin as t → −∞.(Polking,Boggess,Arnold p472) This is explained below.

We must again split this into two different cases, one case wheret → ∞ and the other where t → −∞. In the first case, we must firstrewrite equation (3) as

y(t) = eλ2t(C1e(λ1−λ2)tv1 + C2v2)

In this case, as t→ −∞, eλ2t → 0. Since we know 0 < λ1 < λ2, wealso know λ1 − λ2 > 0 which means (C1e

(λ1−λ2)t) → 0 . This resultsin y(t) → 0, but while converging the direction of y(t) gets closer toC2v2, so as y(t) → 0 the solution curve on the plane will be tangentto the half-line we get from C2v2.

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Now, for the second case we rewrite equation (3) as

y(t) = eλ1t(C1v1 + C2e(λ2−λ1)tv2)

As t → ∞, eλ1t → ∞ since 0 < λ1 < λ2 implies λ1 − λ2 <0. This also results in C2e

(λ2−λ1)t → 0. From this we know that(C1v1 + C2e

(λ2−λ1)tv2) → C1v1 and y(t) → 0, but as y(t) → 0, thedirection of y(t) becomes that of C1v1.

Figure 5: Nodal Source3

3.4 Center

There are now three cases of phase portraits a system can have if thesystem has complex eigenvalues of the form λ = a+ ib. For all of thesecases, we rewrite equation 3 to be

y(t) = C1eat(cos(bt)v1 − sin(bt)v2) + C2e

at(sin(bt)v1 + cos(bt)v2) (4)

(Polking,Boggess,Arnold p473) Where b 6= 0 The first case for com-plex eigenvalues we let a = 0. This cases equation (4) to become

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y(t) = C1(cos(bt)v1 − sin(bt)v2) + C2(sin(bt)v1 + cos(bt)v2)

(Polking,Boggess,Arnold p473) We already know that cos(bt) andsin(bt) are periodic functions that have a period of 2π

|b| and a frequency

of |b|. We also know that y(t) must have the same period and fre-quency. This means that the solution is a closed curve around the ori-gin. When the equilibrium point of a system is surrounded by closedcurve solutions it is referred to as a center. A linear system withpurely imaginary eigenvalues has a center around the origin. Thesecurves to not need to be circles though, they can be ellipses as well.(Polking,Boggess,Arnold p474)

Figure 6: Center4

3.5 Spiral Sink

We must now consider the cases where a 6= 0. In the first case, we willsuppose that a < 0. We rewrite equation (4) to be.

y(t) = eat(C1(cos(bt)v1 − sin(bt)v2) + C2(sin(bt)v1 + cos(bt)v2)) (5)

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(Polking,Boggess,Arnold p474). Here, y(t) is just the equation fora center multiplied by eat. We already know from above that this partof y(t) generates a phase portrait of ellipses centered around the ori-gin. Since a < 0, we know that eat → 0 as t → ∞. This means thatas the solution curves are circling the origin, they are being drawn into the origin as well. This results in a spiral phase portrait. (Polk-ing,Boggess,Arnold p474) The solutions are spiraling around the originwhile also approaching the origin. This is referred to as a spiral sink.A spiral sink is present anytime a system has a complex eigenvaluewith a negative real part.

Figure 7: Spiral Sink5

3.6 Spiral Source

The last case we have is if a > 0. The general solution formula is againgiven by equation (5) above. This time however, since a > 0, eat →∞as t→∞. Thus, the solutions again circle the origin as with the spiralsink, but grow away from the origin as opposed to approaching it. Thisis called a spiral source. Any system that has a complex eigenvaluewith a positive real part has a spiral source as a phase portrait.

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We must remember that the trace-determinant plane is a parameterplane. This means that changing the parameter of the system changesthe values of a11, a12, a21, and a22 of the matrix A, and thereforechanges the point (T,D). Small changes to the parameter don’t makemuch of a difference. A small parameter change will usually not changethe form of the equilibrium point of the system. A spiral sink will staya spiral sink, a center will stay a center, etc. (Blanchard,Devaney,Hallp345)

Figure 8: Spiral Source6

4 The Trace-Determinant Plane

4.1 Real Eigenvalues

If T 2 − 4D > 0 this signifies that the eigenvalues are real, so whenanalyzing these systems we will always use T 2−4D > 0. There are sixdifferent combinations of real eigenvalues for a system. The eigenvaluescan be both positive, one positive and one zero, both negative, onenegative and one positive, one negative and one zero, and both zero.All of these cases can be solved for using equation 2.

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First we will analyze the cases with a positive eigenvalue. If T > 0in equation 2, then there must be one positive eigenvalue as the equa-tion would be the sum of two positive numbers (Blanchard,Devaney,Hallp344), but we still need to solve for the other eigenvalue of the system.If D = 0, the equation becomes

λ =T ±√T 2

2= 0

The other eigenvalue in the case is zero. Next consider the case ofD > 0. In this case,

T 2 − 4D < T 2

We are still only evaluating the system for when T > 0, so this equationcan be solved further as follows.√

T 2 − 4D < T

If this is combined with equation 2, we see that

T −√T 2 − 4D

2> 0

This shows that both eigenvalues will be positive. (Blanchard,Devaney,Hallp344)

Now that we have shown what happens when D > 0 and D = 0, itis time to show what happens when D < 0. When D < 0,

T 2 − 4D > T 2

Again, we are only evaluating the system for when T > 0, so thisequation can be solved further as follows:√

T 2 − 4D > T

T −√T 2 − 4D

2< 0 (6)

Since the system comes out as less than 0, there is one positive eigen-value and one negative eigenvalue. (Blanchard,Devaney,Hall p344)

The above covers all the cases of positive eigenvalues, so we mustnow evaluate those with negative eigenvalues. This time, we will beanalyzing the equation 2 with T < 0. If T < 0, we know that Equation(6) is the sum of two negative numbers, and therefore negative itself.Now, we only need to consider the same when

T +√T 2 − 4D

2

Similar as above, if D = 0, then there is one zero eigenvalue. IfT < 0 and D = 0, the system has one negative and one zero eigenvalue.

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If D < 0,T 2 − 4D > T 2

This can be simplified as follows :√T 2 − 4D > T and

√T 2 − 4D > −T .

Since we know√T 2 − 4D > 0, we only use the case

√T 2 − 4D > −T .

So, T +√T 2 − 4D > 0 which results in

T +√T 2 − 4D

2> 0

This results in one positive and one negative eigenvalue. We saw abovethat the system has a positive and negative eigenvalue if T > 0 andD < 0. This shows that a system having one positive and one negativeeigenvalue is solely dependent on D. As long as D < 0 this will be thecase.

In the last case, take D > 0. This means that

T 2 − 4D < T 2

So, √T 2 − 4D < T and

√T 2 − 4D < −T

We know that√T 2 − 4D must be greater than 0, so we only con-

sider the case0 <√T 2 − 4D < −T

This results inT +√T 2 − 4D

2< 0

This system would have two negative eigenvalues.

The last thing we have not considered is if T = 0. If T = 0, D mustalso be zero to keep

√T 2 − 4D > 0, and both eigenvalues are zero.

4.2 Complex Eigenvalues

If T 2− 4D < 0, then the eigenvalues are both complex numbers. Evenfurther, both eigenvalues are complex conjugates of each other. Onlythe real part of the number has an impact on the equilibrium pointwhen the eigenvalues are complex. All complex numbers are of theform a + bi. If we look again at equation 2, (Polking,Boggess,Arnoldp476)

λ =T ±√T 2 − 4D

2.

The real part of the eigenvalue, a, is equal to T/2. If T < 0, thenwe know a must also be less than zero. This results in a complex

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eigenvalue with a negative real part. If T > 0, then a > 0 as well, andthis gives us a complex eigenvalue with a positive real part. Lastly,if T = 0, then the eigenvalue is purely imaginary. (Polking, Boggess,Arnold p477)

All of the above cases of eigenvalues are summarized in the graphbelow

Figure 9: Eigenvalues on the Trace-Determinant Plane

5 Phase Portraits and Eigenvalues

If we look at both the table in Figure 2 and Figure 9, it is very clearto see how they are related. This relation between eigenvalues andphase potraits results in a relation between phase potraits, trace anddeterminant. Real eigenvalues are present when T 2 − 4D > 0. WhenT > 0 there must be at least one positive eigenvalue. We know thatwhen T > 0 and D > 0 there must be two positive eigenvalues. Wealso know that two positive eigenvalues result in a nodal source. Wecan now relate the trace and determinant with this phase portrait.When the trace and determinant are both greater than zero, we knowwe will have a nodal source. Then same follows for the rest of thephase portraits. If T < 0 and D > 0, then both eigenvalues arenegative which results in a nodal sink. When D < 0, then there isone positive eigenvalue and one negative eigenvalue and this resultsin a saddle point. This is independent of the value of T . Complexeigenvalues and phase portraits follow similarily. When T = 0, a = 0,since a = T/2 and the system would have purely imaginary eigenvalues.

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Purely imaginary eigenvalues result in the system displaying a center.If T < 0, then then a < 0 as well and the system’s phase portrait is aspiral sink. Similarly, if T > 0, a is greater than zero as well, and wehave showed this results in a spiral source. This above relationship isshown through the graph below, where the complex region is shadedin blue and the real region is shaded in tan.

Figure 10: Phase Portraits and Eigenvalues on the Trace-Determinant Plane

6 Conclusion

The Trace-Determinant Plane is a graphical representation of solutionsof planar systems. There are six important different types of behaviorthat a system can demonstrate in the trace-determinant plane. Thereare other types of behavior, but these are the most important. Thesesix plane portraits are as follows: saddle, nodal sink, nodal source,center, spiral sink, and spiral source. The portrait a system has isdecided by the form of the eigenvalues for that system. If a system hastwo positive, real eigenvalues its portrait is a nodal source, two real,negative results is a nodal source, both real with one positive and onenegative results in a saddle, complex eigenvalues with a = 0 resultsin a center, complex with a < 0 results in a spiral sink, and finally

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complex with a > 0 results in a spiral source.

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