# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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V(x,y) = (2Ax + By)(a11x + a12y) + (Bx + 2Cy)(a21x + a22y) = (2Aan + Ba21)x + [2(Aa12 + Ca21) + B(an + a22)]xy +

(2Ca22 + Ba12)y2. We choose A, B, and C so that

2Aan + Ba21 = -1, 2(Aa12 + Ca21) + B(an+a22) = 0, and 2Ca22 + Ba12 = -1. The first and third equations give us A and C in terms of B, respectively. We substitute in the second equation to find B and then calculate A and C. The

result is given in the text.

10c. Since a11a22 - a12a21 > 0 and a11 + a22 < 0, we see that

D < 0 and so A > 0. Using the expressions for A, B, and

C found in part (b) we obtain (4AC-b2)D 2 = [a21 + a22 + (a11a22-a12a21)][a11 + a12 + (a11a22-a12a21)]

- (a12a22+a11a21) 2 = (a11+a12+a21+a22)(a11a22-a12a21) + (a11+a12)(a21+a22)

+ (a11a22-a12a21) 2- (a12a22 + a11a21) 2 = (a11 + a12 + a21 + a22)(a11a22-a12a21) + 2(a11a22-a12a21) 2.

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Since a11a22 - a12a21 > 0 it follows that 4AC - B > 0.

11a. For V(x,y) = Ax2 + Bxy + Cy2 we have

V = (2Ax + By)(a11x+a12y + F1(x,y)) + (Bx+2Cy)(a21x+a22y + G1(x,y))

= (2Ax+By)(a11x+a12y) + (Bx+2Cy)(a21x+a22y)

+ (2Ax+By)F1(x,y) + (Bx+2Cy)G1(x,y)

= -x2-y2 + (2Ax+By)F1(x,y) + (Bx+2Cy)G1(x,y), if A,B and C are chosen as in Problem 10.

11b. Substituting x = rcos9, y = rsin9 we find that

V[x(r,0), y(r,9)] = -r2+r(2Acos9+Bsin9)F1[x(r, 9), y(r,9)]

+ r(Bcos9 + 2Csin9)G1[x(r,9), y(r,9)]. Now we make use of the facts that (1) there exists an M such that |2A| < M, |B| < M, and |2C| < M; and (2) given any e > 0 there exists a circle r = R such that |F1(x,y)| < er and |G1(x,y)| < er for 0 < r < R. We have |2Acos9 + Bsin9| < 2M and |Bcos9 + 2Csin9| < 2M. Hence

V[x(r,9), y(r,9)] < -r2 + 2Mr(er)+2Mr(er) = -r2(1 - 4Me).

Section 9.7

197

If we choose e = M/8 we obtain V[x(r,9), y(r,9)] < -r2/2

for 0 < r < R. Hence V is negative definite in 0 < r < R and from Problem 10c V is positive definite and thus V is a Liapunov function for the almost linear system.

Section 9.7, Page 530

1. Note that r = 1, 9 = t + t0 satisfy the two equations for all t and is thus a periodic solution. If r < 1, then dr/dt > 0, and the direction of motion on a trajectory is outward. If r > 1, then the direction of motion is inward. It follows that the periodic solution r = 1,

9 = t + t0 is a stable limit cycle.

2. r = 1, 9 = -t + t0 is a periodic solution. If r < 1,

then dr/dt > 0, and the direction of motion on a

trajectory is outward. If r > 1, the dr/dt > 0, and the

direction of motion is still outward. It follows that the solution r = 1, 9 = -t + t0 is a semistable limit cycle.

4. r = 1, 9 = -t + t0 and r = 2, 9 = -t + t0 are periodic

solutions. If r < 1, then dr/dt < 0, and the direction

of motion on a trajectory is inward. If 1 < r < 2, then dr/dt > 0, and the direction of motion is outward. Similarly, if r > 2, the direction of motion is inward.

It follows that the periodic solution r = 1,

9 = -t + t0 is unstable and the periodic solution r = 2,

9 = -t + t0 is a stable limit cycle.

7. Differentiating x and y with respect to t we find that

dx/dt = (dr/dt)cos9 - (rsin9)d9/dt and dy/dt = (dr/dt)sin9 + (rcos9)d9/dt. Hence ydx/dt - xdy/dt = (rsin9cos9)dr/dt - (r2sin29)d9/dt -

(rcos9sin9)dr/dt - (r2cos9)d9/dt = - r2d9/dt.

8a. Multiplying the first equation by x and the second by y

and adding yields xdx/dt + ydy/dt = (x2+y2)f(r)/r, or

rdr/dt = rf(r), as in the derivation of Eq.(8), and thus dr/dt = f(r). To obtain an equation for 9 multiply the first equation by y, the second by x and substract to obtain ydx/dt - xdy/dt = -x2-y2, or -r2d9/dt = -r2, using the results of Problem 7. Thus d9/dt = 1. It follows that periodic solutions are given by r = c, 9 = t + t0 where f(c) = 0. Since 9 = t + t0, the motion is counterclockwise.

198

Section 9.7

8b. First note that f(r) = r(r-2)2(r-3)(r-1). Thus r = 1,

0 = t + t0; r = 2, 0 = t + 10; and r = 3, 0 = t + t0 are

periodic solutions. If r < 1, then dr/dt > 0, and the direction of motion on a trajectory is outward. If

1 < r < 2, then dr/dt < 0 and the direction of motion is inward. Thus the periodic solution r = 1, 0 = t + t0 is a stable limit cycle. If 2 < r < 3, then dr/dt < 0, and the direction of motion is inward. Thus the periodic solution r = 2, 0 = t + t0 is a semistable limit cycle.

If r > 3, then dr/dt > 0, and the direction of motion is outward. Thus the periodic solution r = 3, 0 = t + t0 is unstable.

9. Setting x = rcos0, y = rsin0 and using the techniques of

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