Nonlinear Dynamics and Chaos by Steven H. Strogatz
Solutions by Aminur Rahman
Chapter 2: Flows on the Line
2.1
A Geometric Way of Thinking
2.1.1. The fixed points are: x∗ = nπ, n ∈ Z.
2.1.2. Points of greatest positive velocity are: x =2n+12 π, n ∈ Z. 2.1.3. (a) The flow’s acceleration is: ¨x = cos x.
(b) Points of greatest positive acceleration are: x = 2nπ, n ∈ Z. 2.1.4. (a) We plug in x0 = π/4 to get
t = ln csc x0+ cot x0 csc x + cot x x0=π/4 = ln 1 +√2 csc x + cot x ⇒ csc x + cot x = (1 +√2)e−t. ⇒ cotx 2 = (1 +√2)e−t. x 2 = cot −1h 1 +√2 i e−t . x = 2 cot−1 h 1 +√2 i e−t . = 2 tan−1 et 1 +√2 .
(b) The general solution is
x = 2 tan−1
et
csc x0+ cot x0
2.1.5. (a) The simple pendulum.
(b) At x∗ = 0 any perturbation throws the trajectory of the pendulum out of the ball
2.2.1. Unstable: x∗ = 4. Stable: x∗ = −4.
2.2.2. Unstable: x∗ = −1. Stable: x∗= 1.
2.2.3. Unstable: x∗ = 0. Stable: x∗ = ±1.
2.2.4. The fixed points are x∗ = nπ, n ∈ Z. To find the stability we take the derivative, f0(x) =
−e−xsin x + e−xcos x. Plugging in the fixed points shows that when n is even the fixed points are unstable and when n is odd the fixed points are stable.
2.2.5. This system has no fixed points.
2.2.6. Unstable x∗ = π4 + 2nπ, n ∈ Z. Stable: x∗ = −π4 + 2nπ, n ∈ Z.
2.2.7. We can’t explicitly find the fixed points, but one is encouraged to plot the function and analyze the qualitative behavior of the system.
2.2.8. Notice that a fourth order polynomial will satisfy the conditions. We need this function to be tangent to the x-axis at x = −1, and transverse to the x-axis at the other two fixed points. The function will be positive on x ∈ (−∞, −1) ∪ (−1, 0) ∪ (2, ∞) and negative on x ∈ (0, 2). So, consider the function,
f (x) = x(x + 1)(x − 2)(x + a) = −2ax − (2 + a)x2+ (a − 1)x3+ x4. Taking the derivative gives,
f0(x) = −2a − 2(a + a)x + 3(a − 1)x2+ 4x3.
To satisfy the tangency condition at x = −1 we set the derivative to zero at that point and solve for a which gives, a = 1. Therefore,
f (x) = x(x + 1)2(x − 2).
Testing the conditions we see that this does indeed satisfy them. We can also check this graphically by plotting the function.
2.2.9. It is perhaps easier to transform this plot into the flow line plot. By analyzing the flow on the real line we easily see that our ODE is: ˙x = x(x − 1).
2.2.10. (a) ˙x = 0. (b) ˙x = sin πx.
(c) Not possible because we need an unstable fixed point between any two unstable ones for smooth vector fields.
(e) ˙x =P100
n=1(x − n).
2.2.11. We consider the ODE
˙ Q + 1
RCQ = V0
R, Q(0) = 0.
The integrating factor is µ = exp(t/RC). By method of integrating factor, we have
µQ = Z µV0 Rdt = Z V 0 Re t/RCdt = V 0Cet/RC+ const. ⇒ Qet/RC = V0Cet/RC+ const. ⇒ Q = V0C + const.et/RC ⇒ Q = V0C 1 − e−t/RC .
2.2.12. Consider a nonlinear resistor for which I = g(V ) ⇒ V = g−1(V ). Then the voltage drop across the system becomes
−V0+ V + Q C = −V0+ g −1(I) +Q C = −V0+ g −1( ˙Q) + Q C = 0. Now, we solve for ˙Q,
g−1( ˙Q) = V0−
Q
C ⇒ ˙Q = g(V0− Q C.)
Then the fixed point is the same, Q∗ = CV0, because g is zero when its argument is zero.
Also, from the graph of g we see that g(V0− Q−∗/C) > 0 and g(V0− Q+∗/C) < 0, so the fixed
point is still stable. The nonlinearity doesn’t result in any qualitative changes. 2.2.13. (a) Dividing through by m gives,
˙v = g − k mv 2 ⇒ Z dv g −mkv2 = t + C ⇒ r m gktanh −1 k gmv = t + C. Since, v(0) = 0, C = 0, hence v = mg k tanh gk mt = rm k ert− e−rt ert+ e−rt , r = gk m. (b) Taking the limit we get, limt→∞v = mgk .
(c) The fixed point of the terminal velocity equation is v∗ = mgk , and it is stable. This
corresponds to the terminal velocity. Notice, we didn’t consider the negative value for v∗ since this would not be physical.
(d) The average velocity is vavg= ∆s∆t = 29300116 = 253 ft/sec.
(e) Solving for the terminal velocity numerically gives V ≈ 265 ft/sec. Plugging this into the terminal velocity formula gives k = mgV2 = .120 kg/ft.
2.3.1. (a) We solve the ODE ˙N = rN (1 − N/K) via separation of variables. Z dN rN (1 − N/K) = t + C ⇒ 1 rln N N − K = t + C. Plugging in the initial conditions gives, c = 1rln N0
N0−k, then 1 rln N N − K = t + 1 r ln N0 N0− K ⇒ N N − K = N0 N0− K ert⇒ N = N0 N0− K ertN − K N0 N0− K ert ⇒ N = −K N0 N0−Ke rt 1 − N0 N0−Ke rt = −KN0ert N0− K − N0ert .
(b) Let x = 1/N , then the ODE becomes
˙ x = −N2N = −˙ 1 N2N 1 −N K = −1 N 1 −N K = −x 1 − 1 Kx = −x + 1 K; x(0) = 1/N0 ⇒ Z dx −x + 1/K = t + C ⇒ − ln −x + 1 K = t + C ⇒ −x + 1 K = αe −t ⇒ x = βe−t+ 1 K Plugging in the initial condition gives, β = N1
0 − 1 K, then x = 1 N0 − 1 K e−t+ 1 K.
2.3.2. (a) We find the fixed points, ˙x = k1ax − k−1x2 = 0 ⇒ x∗ = 0,kk−11a. For stability we
differentiate, f0(x) = k1a − 2k−1x, which shows the fixed points are unstable and stable
respectively. (b) Sketching.
2.3.3. (a) In the Gompertz equation, a is the growth constant (the ”ability” of cells to grow) and b−1 is the carrying capacity (1/b is as large as the cell can feasibly get).
(b) The fixed points are N∗ = 0, 1/b. Differentiating gives, f0(N ) = −a ln(bn) − a, so the
fixed points are unstable and stable respectively. Now we can sketch the vector field. 2.3.4. We set the derivative to zero to find the critical points, f0(N ) = −2a(N − b) = 0 at N = b,
which corresponds to the local maxima because f00(N ) = −2a.
2.3.5. We find the fixed points,NN˙ = r−a(N −b)2 = 0 ⇒ (N −b)2 = ar ⇒ N −b = ±ra ⇒ N∗ = ±ra+b.
N∗ = ra + b is stable. N∗ = −ar + b is unstable if b > ra, otherwise it’s unphysical, since
2.4
Linear Stability Analysis
2.4.1. The derivative is f0(x) = 1 − 2x, so the fixed points, x∗ = 0, 1 are unstable and stable
respectively.
2.4.2. The derivative is f0(x) = 3x2− 6x + 2, so the fixed points, x∗ = 0, 1, 2 are unstable, stable,
and unstable respectively.
2.4.3. The derivative is f0(x) = sec2x, so the fixed point, x∗= 0, is unstable.
2.4.4. The derivative is f0(x) = −3x2+ 12x, so x∗ = 6 is unstable, however the derivative test is
inconclusive for x∗ = 0. Graphically, we see that x∗= 0 is half-stable.
2.4.5. The derivative is f0(x) = 2x exp(−x2), so the derivative test is inconclusive for the unique fixed point x∗= 0. Graphically, we see that it is half-stable.
2.4.6. The derivative is f0(x) = 1/x, so the unique fixed point x∗= 1 is unstable.
2.4.7. For a > 0, the derivative is f0(x) = a − 3x2, so x∗ = ±
√
a is stable, and x∗ = 0 is unstable.
If a < 0, x∗ = 0 is stable, but the other two fixed points vanish since they are now imaginary.
For a = 0, the dynamical system becomes ˙x = −x3, so the only fixed point is x∗ = 0 and it
is stable. Note: this is a supercritical pitchfork bifurcation. 2.4.8. This is already done. See 2.3.3 b in the previous section. 2.4.9. (a) We solve the ODE via separation of variables,
˙ x = −x3 ⇒ Z x−3dx = −t + C1⇒ − 1 2x −2 = −t + C1 ⇒ 2x2 = 1 t + C2 ⇒ x2 = 1 2t + C3 ⇒ x = ± r 1 2t + C3 → 0 as t → ∞ (b) Numerics.
2.5.1. (a) The fixed point x∗ = 0 is stable for any c > 0.
(b) We solve the ODE, ˙ x = x−c ⇒ Z xcdx = t + k ⇒ x c+1 c + 1 = t + k.
For x = 0, t = −k, and for x = 1, t = c+11 − k, so the time it takes the flow to go from x = 1 to x = 0 is T = 1/(c + 1).
2.5.2. Notice 1 + x10≥ 1 + x2, so we solve the inequality,
t − C ≤ Z
dx
1 + x2 = tan
−1x ⇒ t − C ≤ tan−1x.
So, as x → ∞, t − C ≤ π/2 ⇒ t ≤ π/2 + C. Since this holds for any valid constant, C, t is finite.
2.5.3. We solve the ODE, ˙ x = rx + x3⇒ t − C = Z dx x(r + x2) = Z dx rx − Z xdx r(r + x2) = 1 rln x − 1 2rln(r + x 2) = 1 r[ln x − ln p r + x2] = 1 r ln x r + x2 → 0 as x → ∞
So, t = C, and hence is a constant. Note: this is a subcritical pitchfork bifurcation. 2.5.4. We solve the ODE,
˙ x = x1/3 ⇒ Z x−1/3dx = t + C1⇒ 3 2x 2/3 = t + C ⇒ x2/3= 2 3t + C2⇒ x = 2 3t + C2 3/2 . Since x(0) = 0, x = 32t3/2
, but x = 0 is also a solution. So, we have the following solutions,
x = ( 0 for t < t0, 2 3(t − t0) 3/2 for t ≥ t0; ∀ t0 ≥ 0.
2.5.5. We solve the ODE, ˙ x = |x|p/q ⇒ t + C = Z |x|−p/qdx = Z x−p/qdx = 1 −p/q + 1x −p/q+1 .
However, x = 0 is also a solution. Notice, we need only consider x ≥ 0 because x(0) = 0 and ˙
x ≥ 0 for any x.
(a) For p < q, we have the following solutions,
x = (
0 for t < t0,
[(−p/q + 1)(t − t0)]1/(−p/q+1) for t ≥ t0;
∀ t0 ≥ 0.
Since this holds for any t0, there are infinitely many solutions.
(b) For p > q, −p/q + 1 < 0, so for x(0) = 0, the only solution is x = 0.
2.5.6. (a) By the conservation of mass, the volume is also conserved, and the equations given are just the time derivatives of the volume.
(b) The potential energy of the system is ∆mgh and the kinetic energy is 12(∆m)v2. If all
the potential energy is converted to kinetic energy, v2 = 2gh.
(c) From (a) and (b) we have, av = A ˙h and v = √2gh. Then, −a√2gh = A ˙h ⇒ ˙h =
−a A
√ 2gh.
(d) We solve the ODE, ˙h = −Ch1/2⇒ Z h−1/2= −Ct + C1 ⇒ 2h1/2 = −Ct + C1 ⇒ h = −C 4 t 2+ C 1t + C2.
Since h(0) = 0, C2 = 0, hence h = −C4 t2+ C1t. Here, C is known from (c), however C1
2.6.1. Any second order ODE can be broken down into two first order ODEs, and hence is a two dimensional dynamical system.
2.6.2. Proof. Suppose the system ˙x = f (x) has a nontrivial periodic solution x(t), i.e. x(t) = x(t+T ) for some T > 0, and x(t) 6= x(t + s) for all 0 < x < T . ConsiderRt+T
t f (x) dx dtdt. By the chain rule we have, Z t+T t f (x)dx dtdt = Z x(t+T ) x(t) f (x)dx = 0 because x(t) = x(t + T ). However, Z t+T t f (x)dx dtdt = Z t+T t f (x)2dt > 0
because T > 0 and f is not identically zero. This forms a contradiction. Therefore, there can be no periodic solutions to one dimensional dynamical systems.
2.7
Potentials
2.7.1. The potential is V (x) = 12x2− 1 3x
3. The fixed points are x
∗ = 0, 1 and unstable and stable
respectively.
2.7.2. The potential is V (x) = 3x, and the system has no fixed points.
2.7.3. The potential is V (x) = − cos x. The fixed points are x∗ = ±nπ, n ∈ Z. The fixed points are
stable when n is odd and unstable when n is stable.
2.7.4. The potential is V (x) = 2x − cos x, and the system has no fixed points. 2.7.5. The potential is V (x) = − cosh x. The fixed point x∗= 0 is stable.
2.7.6. The potential is V (x) = rx +12x2−14x4. This must be done graphically and/or numerically. 2.7.7. Proof. Suppose the system ˙x = f (x) has a periodic solution x(t) such that x(t) = x(t + T ) for some T > 0 and x(t) 6= x(t + s) for all 0 < s < T . Consider the integral Rtt+Tf (x)dxdtdt. Assume there is a nontrivial potential function V (x) such that dV (x)/dx = f (x). Then
Z t+T t f (x)dx dtdt = Z x(t+T ) x(t) dV dxdx = V (x) x(t+T ) x(t) = 0, because x(t) = x(t + T ), but Z t+T t f (x)dx dtdt = Z t+T t dV dx dx dtdt = Z t+T t dV dt dt = Z t+T t − dV dx 2 dt ≤ 0. Therefore, V ≡ 0, hence the system can not have a periodic solution.
