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Mathematics 3F03 Advanced Differential Equations, Fall 2012

INSTRUCTORS SOLUTIONS TO ASSIGNMENT 2

October 17, 2012 @ 23:04

1 Pages 5758, Exercise 2(i), (iii) and (v).

For each of the following systems of the form X= AX,

(a) Find the eigenvalues and eigenvectors ofA.

(b) Find the matrix Tthat putsAin canonical form.

(c) Find the general solution of both X= AXand Y = (T1AT)Y.

(d) Sketch the phase portraits of both systems.

(i) A=

0 11 0

(iii) A =

1 11 0

(v)A=

1 11 3

Solution: We obtain the eigenvalues ofA by solving for in the characterisitic equation,det(A I) = 0. For each eigenvalue , we find associated eigenvectors by solving the linearequation (AI)v= 0 forv R2. If the eigenvalues are real and distinct, as in parts (i) and(v), then there are two linearly independent real eigenvectors, the Jordan canonical form ofAis J= diag(1, 2), and the similarity transformationTthat convertsA to Jhas columnsequal to linearly independent eigenvectors ofA. If the eigenvalues are complex (= i,, R, = 0), as in part (iii), then the real Jordan canonical form ofA is

J=

and the similarity transformation Thas columns equal to the real and imaginary parts ofeither of the complex eigenvectors.Carrying out the computations for the matrices in question yields the results summarized

in the table below. The type of phase portrait is determined by the eigenvalues in these cases:with two real eigenvalues of opposite sign we have a saddle; complex eigenvalues with positivereal part yield a spiral source.

A Eigenvalues J T Type

(i)

0 11 0

1

1 00 1

1 11 1

saddle

(iii)

1 11 0

1

2(1 i3)

12

32

32

12

1

2

32

1 0

spiral source

(v)

1 11 3

1 3

13 00 1 + 3

2 + 3 2 31 1

saddle

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Before going any further, it is important to check that the computations above are actuallycorrect. Recalling the inverse of a 2 2 invertible matrix T is

T1 = 1

det T

d bc a

,

we can easily check that the table above is correct by verifying in every case that

J=T1AT .

To obtain the general solutions, we use the important result (proved in class, but whichyou should be able to prove yourself) that etV0 is a solution of X

= AX if V0 is aneigenvector ofA associated with eigenvalue. This led us to the conclusion that in the caseof distinct real eigenvalues 1 and 2, the general solution to X

= AX is

X(t) =c1e1tV1+c2e

2tV2,

where c1, c2

R and V1, V2 are (linearly independent) eigenvectors ofA. The table above

therefore tells us everything we need in order to write down the solutions in the case of thesaddles (since the columns ofTare linearly independent real eigenvectors ofA). Thus, wehave

(i) X(t) =c1et1

1

+c2e

t

11

Y(t) =c1et

10

+c2e

t

01

.

Note here that the only difference between the form of the solution in original and canonicalcoordinates is that in the latter the eigenvectors are the coordinate axes. Similarly, we have

(v) X(t) =c1e(1+

3)t

2 + 31

+c2e

(13)t

2 31

Y(t) =c1e(1+

3)t

10

+c2e

(13)t

01

.

In the case with complex eigenvalues, we have = 1232

and the general solution isthe complex function

Z(t) =c1e+tV++c2e

tV , ()whereV

are complex eigenvectors associated with

,

V=

1

2 i

32

1

=

12

1

i

32

1

.

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Inserting this in (*), and using the standard notation[z] for the real part ofz, we have

(iii) X(t) = [Z(t)]

=

c1e( 12+i

3

2 )t

12

1

+i

32

1

+c2e

( 12i

3

2 )t

12

1

i

32

1

=et/2

c1ei3

2

t1

21

+i3

21

+c2ei3

2

t1

21 i

3

21

=et/2

c1

cos

3

2 t+i sin

3

2 t1

2

1

+i

32

1

+ c2

cos

3

2 t+i sin

3

2 t 1

2

1

i

32

1

=et/2

(c1+c2)cos

3

2 t

12

1

+ (c1+c2)sin

3

2 t

32

1

=et/2

C1cos

32 t121 +C2sin

32 t3

21

where we have renamed the arbitrary constants in the final step. In canonical coordinateswe have

(iii) Y(t) =et/2

C1cos

3

2 t

10

+C2sin

3

2 t

01

.

Phase portraits in original and canonical coordinates are shown below.

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x

yOriginal Coordinates

(i) x

yCanonical Coordinates

x

y

(iii) x

y

x

y

(v) x

y

2 Page 58, Exercise 7.

Consider the system X= AX, where

A=

10

,

with = 0. Show that all solutions tend to (respectively, away from) the origin tangentiallyto the eigenvector (1, 0) when 0).

Solution: The matrix A is already in Jordan canonical form. The unique eigenvalue is and the 1 above the main diagonal tells us that we are in the situation in which the repeatedeigenvalue has geometric multiplicity 1 (there is only one eigendirection).

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Writing out the system X= AX in coordinates, we have

x= x+y (1a)

y= y (1b)

The second equation is independent of the first and can be solved immediately:

y(t) =y0et , (2)

wherey0 = y(0). Given this expression for y(t), the first equation can now be written as asimple non-autonomous linear first order ODE in one variable,

x= x+y0et , (1a)

which is a type of equation studied ad nauseam in previous courses. One solution methodis to guess a solution of the form

x(t) =c1et +c2te

t , (3)

and then solve for c1 and c2 by plugging this guess into the differential equation. First notethat Equation (3) impliesx(0) =c1, so we can write

x(t) =x0et +c2te

t . (2)

The derivative of this function is

x(t) =x0et +c2te

t +c2et (4a)

=[x0et +c2te

t] +c2et (4b)

=x(t) +c2et . (4c)

Plugging this expression forx into Equation (1a) and simplifying, we findy(t) =c2e

t ,

and hence Equation (2) impliesc2= y0.

Thus, the general solution ofX = AX is

X(t) =

x(t)y(t)

=

x0e

t +y0tet

y0et

(5a)

=x0et1

0 +y0ett

1 (5b)

Ify0 = 0 then y(t) = 0t, so the solution starts and remains on the x-axis (the eigendirec-tion); if 0 then x(t)sign(x0) (and thus travelsaway from origin along the eigendirection regardless of the sign ofx0). Ifx0 = 0 then wecan write the solution

X(t) =y0tet

11/t

,

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i.e.,X(t) is proportional to the vector (1, 1/t), which approaches the eigenvector (1, 0) ast . Moreover, the proportionality factor y0tet 0 ast if 0. If x= 0 and y0= 0 then the solution is the sum of the two special cases justconsidered, hence retains the properties to be established.

Although not required, the graph below shows the phase portrait of the system. Thedirection of flow is not marked, but will be inwards if < 0 and outwards if > 0. This

is an example of special sink or special source in which there is only one invariant line(thex-axis in this case).

x

y

3 Page 59, Exercise 9.

Determine a computable condition that guarantees that, if a matrix A has complex eigen-values with nonzero imaginary parts, then solutions ofX= AXtravel around the origin inthe counterclockwise direction. Note: Although not stated in the question, assume thatA isa2 2 real matrix.

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Solution: Recall the meaning of the RHS of a system of ODEs X = F(X). F(X) is avector field that points in the direction of motion at any pointX. In our present situation,F(X) =AX, with A a 2 2 matrix. Consider X= (1, 0)T and note that AX is then equalto the first column ofA. So establishing the direction of rotation in the phase plane reducesto the question of whether the first column ofA is a vector that points down (clockwise) orup (counter-clockwise), which is determined by the sign of the bottom left entry ofA:

a21 < 0 = clockwise ,a21 > 0 = counter-clockwise .

Note that this implies that multiplying A by1 (which does not change the eigenvalues)will change the sense of rotation of the phase portrait. Consequently, the sense of rotationin canonical coordinates is not necessarily the same as the sense of rotation in the originalcoordinates.

4 Page 59, Exercise 10.

Consider the system X= AX, where

A=

a bc d

,

where a+d= 0 but ad bc = 0. Find the general solution of this system and sketch thephase portrait.

Solution: The characteristic polynomial of the matrixA is

p() = det(A

I)

=2 (a+d)+ (ad bc)=2 (a+d)= [ (a+d)] ,

so the eigenvalues of A are 1 = 0 and 2 = a+ d = 0. Since there are distinct realeigenvalues, there is a basis of real eigenvectors; writing V for an eigenvector associatedwith eigenvalue , one eigenbasis is{V0, Va+d} where

V0= ba

, Va+d= ac

. ()

Indeed, using the fact that ad bc= 0, we have

AV0=

0

ad bc

=

00

= 0 V0,

AVa+d=

a2 +bc

c(a+d)

=

a2 +adc(a+d)

= (a+d)

ac

= (a+d)Va+d.

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All seems well, but we have made an implicit assumption in (*) that V0 and Va+d are notzero vectors. So a bit more care is actually required. Ifa = b = 0 then V0 = (1, 0) is aneigenvector of = 0. Ifa = c = 0 then Va+d= (b, d) is an eigenvector of = a + d(and notethat d = 0 here because we cannot have both a = 0 and d = 0 given that a+d = 0).

With the understanding that V0 andVa+d are defined as above in the special cases wheresome entries ofAare zero, the general solution ofX= AX is

X(t) =c1V0+c2e(a+d)tVa+d.

Ifa+d 0 we have an attracting line. Seethe example phase portraits below.

Note that this question occurs in the textbook before the discussion of the trace-determinantplane, but it is really asking us to determine the nature of solutions on the trace axis of thetrace-determinant plane (ignoring the origin).

x

y

A = 1 2

3 6

x

y

A = 1 2

3 6

END OF ASSIGNMENT

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