Errata for "Nonlinear Dynamics and Chaos, Second Edition" by Steven. H. Strogatz

book front cover

p. 42, 2.5.6 c)
  Combining (a) and (b) ...

p. 43, 2.8.3 c)
  $\ln(E)$ vs. $\ln(t)$ should be $\ln(E)$ vs. $\ln(\Delta t)$.

p. 44, 2.8.6 d)
  Should say: Repeat part (c) ...

p. 51
  Just below equation (3): $a = \dfrac{\partial f}{\partial r}\Bigg|_{( x^{*}, r_{c})}$ (partial with respect to $r$, not $x$).

p. 80 and 81, 3.1.5
  The cases for $a = \dfrac{\partial f}{\partial r}\Bigg|_{( x^{*}, r_{c})} = 0$ and $a = \dfrac{\partial f}{\partial r}\Bigg|_{( x^{*}, r_{c})} \neq 0$ are backwards.

p. 82, 3.3.2
  Parameter $l$ should be $\lambda$ in the paragraph.

p. 87, 3.6.5 f)
  ...reduces to the approximate result in part (e).

p. 117, 4.4.2
  In line 2, change $\theta'$ to $\dot{\theta}$.

p. 129
  In the caption for Fig 5.1.5 (c): replace the equals sign with a minus sign, so that the caption reads:
(c) $-1 < a < 0$

p. 130
  Start of fifth paragraph: replace $\alpha$ with $a$ in the sentence: Finally when $a$ ...

p. 183, 6.3.9 c)
  Rennie Mirollo had a student who pointed out that part (c) is wrong. Many trajectories approach lines of the form $y = x + c$. Trajectories with large initial condition $x_0$, $y_0$ blow up to infinity in a finite time. Their value of $y-x$ at the blow-up time determines the value of $c$.

p. 184, 6.4.4
  First equation should have $b_{1}$, not $b$.

p. 185, 6.4.7
  The rate equations show capital $K$'s, but in the next sentence, when discussing the parameters, lower case $k$'s are given. Use lower case $k$'s in the rate equations.

p. 188, 6.5.6
  Use the straight "ell" in the displayed equation to replace the curly "ell" in the following line: "where $k$,$\ell$ ..."

p. 203
  V shouldn't be bold-faced in line 10: ...function ${\bf V}({\bf x})$ becomes ...function $V({\bf x})$

p. 232, 7.2.9 c)
  The dots are missing over $x$ and $y$ on the LHS of $\dot{x} = -2xe^{x^{2}+y^{2}}$ and $\dot{y} = -2ye^{x^{2}+y^{2}}$

p. 240, 7.6.16
  ${\bf v}$ should be bold in integral on the left side of the equation.

p. 242, 7.6.22, Line 3
  Insert minus sign before $\dfrac{1}{2}$. Equation should read $x \approx -\dfrac{1}{2}\epsilon a^{2}$.

p. 242, 7.6.25 c)
  Replace final $\sin(t+\theta)$ with $\sin(t+\phi)$.

p. 243, 7.6.26, Line 2
  There should be one dot over $x$, not two dots, in the equation. It should read $\dot{x} = -\epsilon x \sin^2 t$.

p. 264, Fig 8.4.1
  Fill in the open dot at $r = 0$ in top left and top middle panel; should be filled black dot.

p. 267
  "The generic scaling laws for bifurcation of cycles in two-dimensional systems are given in Table 7.4.1." -- should be Table 8.4.1, not 7.4.1.

p. 291
  Displayed equation for $n_B$ in Exercise 8.1.15: the final term should be $n_B$, not $n_{AB}$.

p. 302, 8.6.6
  In part (a), end. Write $\dot{\phi} = \omega$ instead of $\dot{\theta} = \omega$.

p. 318
  The number for (13) is missing, should be displayed to the right of the equation.

p. 344, Fig. 9.6.3
  Insert labels (a) and (b) on figure.

p. 349, 9.1.4
  Capitalize $p$ in $\dot{P} = \gamma_{1}(ED-p)$ to $\dot{P} = \gamma_{1}(ED-P)$.

p. 368
  Change $+r^2$ to $-r^2$ in the absolute value, in the line after the array of four displayed equations.

p. 370
  Saha and Strogatz (1994) - year should be 1995.

p. 380, Line 6
  $\Delta_n = r_n - r_{n-1}$ (the final subscript is $n-1$, not $n+1$).

p. 393, Eqn. 10
  Insert $+$ between $-2$ and square root.

p. 396, 10.2.4
  Change $-r$ to $r$ in the map $x_{n+1} = x_n e^{-r(1-x_n)}$.

p. 403, 10.7.8, Line 2
  Delete subscript $n$ in the phrase "where $f(x_n, r)$ ..."

p. 434, 12.1.3
  The limit in calculating $d$ should be $n\rightarrow \infty$ instead of $n\rightarrow 0$.

p. 451, Last sentence
  "Figure 12.5.6 shows the Poincaré section for the system of Example 12.5.1." should be "Example 12.5.2."

p. 455, 12.1.7
  No shaded regions. Delete "shaded".

p. 463, 6.3.6
  Stable degenerate node

p. 465, 7.6.17, Line 2
  The superscript after $\Big(\gamma + \dfrac{1}{2} \cos 2\phi\Big)$ should be $-\dfrac{1}{2}$, not $-1$.

p. 465, 7.6.22
  The $+$ after $a \cos \omega t$ should be $-$.

p. 466, 8.5.4
  This should be numbered 8.5.5.

p. 468, second 11.2.4
  The third line from bottom should be numbered 11.2.5.

p. 468, 11.3.4
  Remove boldface on the first letter in $\ln 5$.

p. 474
  Harrison Biswas page number should be 394.

p. 484
  Kocak (1989) isn't explicitly referenced on page 34, but is relevant to the Practical Matters section.

p. 507
  In the index: under entry
    saddle-node bifurcation,
insert space between "of" and "cycles".

Errata for "Student Solutions Manual for Nonlinear Dynamics and Chaos, Second Edition" by Mitchal Dichter

book front cover

  The bifurcation diagram is incorrect. The correct bifurcation diagram is

3.2.1 bifurcation plot
  The last equation $= c_{1}-c_{2}x_{2}$ should be $= c_{1}x-c_{2}x^{2}$.

3.3.1 b)
  The "p" in the bottom of the fraction on the last line should not be there.

3.3.1 d)
  Extra parenthesis in $|GN-k| \ll |Gn+f|$ in the last paragraph.

3.5.7 c)
  The $\kappa = rt$ on the last line should be $\displaystyle \kappa = \frac{N_{0}}{K}$.

3.6.5 a)
  The second equation with $\displaystyle = -kx \left( 1 - \frac{L_{0}}{\sqrt{x^{2}+a^{2}}} \right)$ should be $\displaystyle = kx \left( 1 - \frac{L_{0}}{\sqrt{x^{2}+a^{2}}} \right)$ with no minus sign.

3.6.5 e) and 3.6.5 f)
  The fraction $\displaystyle \frac{u^{4}+3u^{2}}{2(1-u^{2})}$ near the end of parts e) and f) should be $\displaystyle \frac{u^{4} + 6u^{2}}{2(2-u^{2})}$.

4.5.1 b)
  The third equation $\displaystyle -\frac{\pi}{2}A \leq \frac{\Omega - \omega}{A} \leq \frac{\pi}{2}A$ should be $\displaystyle -\frac{\pi}{2}A \leq \Omega - \omega \leq \frac{\pi}{2}A$.

4.5.1 d)
  The bounds of integration were plugged in backwards in the last two equations. The second to last equation should be $\displaystyle = \frac{2}{A}\left(\ln\left(\Omega-\omega-A\frac{-\pi}{2}\right)-\ln\left(\Omega-\omega-A\frac{\pi}{2}\right)\right)$ and the last equation should be $\displaystyle = \frac{2}{A}\left(\ln\left(\Omega-\omega+A\frac{\pi}{2}\right)-\ln\left(\Omega-\omega-A\frac{\pi}{2}\right)\right)$.

5.1.11 a)
  The last line $\displaystyle \delta < \big|x_{0}\big| \Rightarrow \big|\big|\big(x(t),y(t)\big)\big|\big| < 2\big|x_{0}\big| = \epsilon$ should be $\displaystyle \big|\big|(x_{0},y_{0}\big)\big|\big| < \delta \Rightarrow \big|\big|\big(x(t),y(t)\big)\big|\big| < 2\delta = \epsilon$.

6.3.9 c)
Due to a very interesting and subtle assumption in the exercise, there is something wrong with the reasoning in part (c). Finding $u(t)$ this way assumes that $x(t)$ and $y(t)$ are defined for all time $t$, which isn't true. Turns out the $y^{3}$ terms in the $\dot{x}$ and $\dot{y}$ equations cause $x(t)$ and $y(t)$ to go to infinity in finite time for most initial conditions. (Solve $\dot{y}=y^{3}$ and you'll see why solutions blow up in finite time.) This is also apparent from the phase portrait in part (e).

So in fact what happens is the trajectories that escape to infinity asymptote to lines of the form $y = x + c$, since $y(t) - x(t)$ will decay exponentially to some constant $c$ as $t$ approaches the finite blow up time.

6.4.11 b)
  The eigenvalues for matrix $\displaystyle A_{(0,0,z)}$ should be changed from $\displaystyle \lambda_{1} = 0 \quad \lambda_{2} = 0 \quad \lambda_{3} = rz$ to $\displaystyle \lambda_{1} = rz \quad \lambda_{2} = rz \quad \lambda_{3} = 0$.

7.2.7 a)
  The partial derivatives should be reversed to $\displaystyle \frac{\partial}{\partial y}\big(y+2xy\big) = 1+2x = \frac{\partial}{\partial x}\big(x+x^{2}-y^{2}\big)$.

7.3.1 e)
  The bottom half should be
$\displaystyle \frac{1}{1+\sin^{2}(2\theta)} = \frac{-1}{4\cos(\theta)\sin^{3}(\theta)} \Rightarrow 4\cos(\theta)\sin^{3}(\theta) = -1-\sin^{2}(2\theta)$

Determining the solutions of this equation is difficult analytically, but graphically it's pretty clear that there are no real solutions to the equation.

7.3.7 a)
  The last line $= 2ab\sin(2\theta)-1$ should be $= ab\sin(2\theta)-1$.

7.3.9 a)
  The last line $\displaystyle r_{1} = \frac{1}{5}\sin(t)+\frac{2}{5}\cos(\theta) + Ce^{-2t}$ should be $\displaystyle r_{1} = \frac{1}{5}\sin(\theta)+\frac{2}{5}\cos(\theta) + Ce^{-2\theta}$.

  The $\displaystyle 2k \ln\big(8 − \ln(27)\big)$ should be $\displaystyle 2k \big(8 − \ln(27)\big)$ without the first $\displaystyle \ln$.

7.6.3 b)
  The initial conditions were applied incorrectly. The correct equations are,
\begin{align} &\ddot{x}_{0}+x_{0} = 0 \Rightarrow x_{0} = \cos(t) \\ &\ddot{x}_{2}+x_{2} = 0 \Rightarrow x_{2} = 0 \\ &\ddot{x}_{1}+x_{1} - 1 = 0 \Rightarrow x_{1} = 1-\cos(t) \end{align}   and \begin{align} x(t,\epsilon) &= x_{0}(t,\epsilon) + \epsilon x_{1}(t,\epsilon) + \epsilon^{2}x_{2}(t,\epsilon) + O\big(\epsilon^{3}\big) \\ &\approx \cos(t) + \epsilon\big(1-\cos(t)\big) = \epsilon + (1-\epsilon)\cos(t) \end{align}
7.6.15 a)
  There were many errors. Here is the corrected segment. \begin{align} \frac{\text{d}r}{\text{d}T} &= \langle h\sin(\theta) \rangle = \frac{1}{2\pi} \int_{0}^{2\pi} h\big(r\cos(\theta),-r\sin(\theta)\big)\sin(\theta) \text{d}\theta \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} -r^{3}\cos^{3}(\theta)\sin(\theta) \text{d}\theta = 0 \Rightarrow r(T) = r_{0} \\ r\frac{\text{d}\phi}{\text{d}T} &= \langle h\cos(\theta) \rangle = \frac{1}{2\pi} \int_{0}^{2\pi} h\big(r\cos(\theta),-r\sin(\theta)\big)\cos(\theta) \text{d}\theta \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} -r^{3}\cos^{4}(\theta) \text{d}\theta = -\frac{3}{8}r^{3} \\ \frac{\text{d}\phi}{\text{d}T} &= -\frac{3}{8}r^{2} \end{align} \begin{equation} r(0) = \sqrt{x(0)^{2}+\dot{x}(0)^{2}} = a \qquad r(T) = a \end{equation} \begin{equation*} \omega = 1 + \epsilon\phi' = 1 - \epsilon \frac{3}{8}r_{0}^{2} + O\big(\epsilon^{2}\big) \approx 1 - \frac{1}{16}a^{2} \end{equation*}
8.1.15 c)
  One of the matrices at the bottom of part (c) was transposed. There was also an unnecessary matrix at the bottom. Here is the corrected version. \begin{equation*} A = \begin{pmatrix} \frac{\text{d}}{\text{d}n_{A}}\dot{n}_{A} & \frac{\text{d}}{\text{d}n_{B}}\dot{n}_{A} \\ \frac{\text{d}}{\text{d}n_{A}}\dot{n}_{B} & \frac{\text{d}}{\text{d}n_{B}}\dot{n}_{B} \end{pmatrix} = \begin{pmatrix} -2n_{A}-2n_{B}-2p+1 & -2n_{A}-2p \\ -2n_{B} & -2n_{A}-2n_{B}-p+1 \end{pmatrix} \end{equation*} \begin{equation*} A_{(1-p,0)} = \begin{pmatrix} -1 & -1 \\ 0 & p-1 \end{pmatrix} \qquad \lambda_{1} = -1 \quad \lambda_{2} = p-1 \end{equation*}
8.7.3 c)
  The explanation for the limits as $T \rightarrow 0$ and $T \rightarrow \infty$ were poorly written, wrong, and ambiguous. Here is the rewritten part (c).

Using the equation derived from part (a) \begin{align*} &\lim_{T\rightarrow 0} x(T) = \lim_{T\rightarrow 0} e^{-T}x_{0} - A\big(1-e^{-\frac{T}{2}}\big)^{2} = x_{0} - A\big(1-1\big)^{2} = x_{0} \\ &\lim_{T\rightarrow \infty} x(T) = \lim_{T\rightarrow \infty} e^{-T}x_{0} - A\big(1-e^{-\frac{T}{2}}\big)^{2} = (0)x_{0} - A\big(1-0\big)^{2} = -A \end{align*} The first result is quite plausible. As the period goes to zero then the forcing function $F(t)$ has almost no time to do anything and $x(t)$ has no time to move anywhere within one $T$-length period. If $x(0) = x_{0}$ then $x(T) \approx x(0) = x_{0}$ as $T \rightarrow 0$. The solution has virtually no time to change.

The second result is intuitive by letting $T$ be large, which in this context means $T \gg 1$. Why is that the correct condition? Because for the equation in question, the rate constant for exponential decay, and hence its inverse, the time scale, are both equal to 1. We can see that by looking at the coefficient of $x$ in the equation \begin{equation*} \dot{x} + x = \pm A. \end{equation*} So, how does $x(t)$ behave for large $T$? Well, at first the value of $x(t)$ relaxes exponentially fast, on a time scale of 1, to a value of $x \approx A$. It stays there as the time dawdles its way up to $t = \frac{T}{2}$. Then, in the second half of the forcing cycle, we switch the value of $F(t)$ to $-A$. By the same reasoning as before, $x$ then moves exponentially fast to $x \approx -A$. It essentially stays at $x = -A$ until the end of the cycle at $t =T$.

What makes this reasoning valid (asymptotically, for $T \gg 1$) is that the relaxation to the equilibrium happens on a much faster time scale than the switching of $F$.

This all becomes very clear visually by plotting $x(t)$ for a moderately large forcing period. For instance, $T = 20$, $A = 3$, and $x(0)=0$.

8.7.3c) plot

8.7.9 a)
  The expression for $r(t)$ is incorrect. The $2\pi$'s should be $t$'s. \begin{equation*} r(t) = \frac{e^{t}r_{0}}{(e^{t}-1)r_{0}+1} \end{equation*}
9.5.5 a)
  The was a mistake in the derivative of the $z$ term. A 1 should have turned into a 0 which propagated through the problem. Here are the corrected pieces. \begin{equation*} \dot{z} = \frac{1}{\sigma\epsilon^{3}}\frac{\text{d}Z}{\text{d}\tau} = xy-bz = \frac{X}{\epsilon}\frac{Y}{\epsilon^{2}\sigma}-b\frac{1}{\epsilon^{2}}\left(\frac{Z}{\sigma}+1\right) \end{equation*} \begin{equation*} \frac{1}{\sigma\epsilon^{3}}\frac{\text{d}Z}{\text{d}\tau} = \frac{X}{\epsilon}\frac{Y}{\epsilon^{2}\sigma}-b\frac{1}{\epsilon^{2}}\left(\frac{Z}{\sigma}+1\right) \Rightarrow \frac{\text{d}Z}{\text{d}\tau} = XY-b\epsilon\left(Z+\sigma\right) \qquad \lim_{\epsilon \rightarrow 0} \frac{\text{d}Z}{\text{d}\tau} = XY \end{equation*} 10.1.9 \begin{equation} f'(x) = \frac{2(1+x)-2x(1)}{(1+x)^{2}} = \frac{2}{(1+x)^{2}} \end{equation} 10.1.11 d) \begin{align*} |x_{1}| &= |3x_{0}-x_{0}^{3}| = |x_{0}||3-x_{0}^{2}| = |x_{0}||3-(2 + \epsilon)^{2}| \\ &= |x_{0}||3-(4 + 4\epsilon + \epsilon^{2})| = |x_{0}||-1 - 4\epsilon - \epsilon^{2})| \\ &= |x_{0}|(1 + 4\epsilon + \epsilon^{2}) \end{align*} 10.2.1 a) \begin{equation*} x_{n+1} - x_{n} = -r\epsilon(1 + \epsilon) - (1 + \epsilon) = -(r \epsilon+1)(1 + \epsilon) < -r \epsilon \end{equation*} So the sequence decreases by at least $r \epsilon$ every iteration.

  I incorrectly wrote the square roots as $\pm\sqrt{1-r}$ when it should be $\pm\sqrt{r-1}$ which changed several parts of the problem. \begin{equation*} x = 0,\pm\sqrt{r-1} \end{equation*} So there are three fixed points for $1< r$ and one fixed point for $r \le 1$. \begin{equation*} f'(\pm\sqrt{r-1}) = \frac{2}{r} - 1 \qquad r < 1 \Rightarrow x^{*} = 0 \text{ is stable} \end{equation*} \begin{equation*} 1 < r \Rightarrow x^{*} = 0 \text{ is unstable, and } x^{*} = \pm\sqrt{r-1} \text{ are stable} \end{equation*} 10.3.11 a) \begin{equation*} -1 < -(1+r) < 1 \Rightarrow -2 < r < 0 \end{equation*} \begin{equation*} -2 < r < 0 \Rightarrow \text{Stable} \qquad r < -2 \text{ or } 0 < r \Rightarrow \text{Unstable} \end{equation*} 10.3.11 b) \begin{align*} &r < 0 \Rightarrow f'(0) = -(1+r) > -1 \Rightarrow \text{Stable} \\ &0 < r \Rightarrow f'(0) = -(1+r) < -1 \Rightarrow \text{Unstable} \end{align*} 11.4.1
  The diagrams of the boxes overlaid on iterations of the Koch snowflake should be overlaid on the complete Koch snowflake. The box counting is computed for the fractal, not each iteration of constructing the fractal.

  Missing power of $n$ added to $N(\epsilon) = (p^{2}-m^{2})^{n}$.

  Change $x_{n}$ to $x_{0}$ in $x_{1} = y_{0}+1-ax_{0}^{2}$.

Much thanks is owed to Alon Ray as he is almost the sole discoverer of errata for the solutions manual.

As of June 2017, Mitchal Dichter is no longer an instructor of math at Campus Learning Assistance Services at the University of California, Santa Barbara.

External links (This is a special special purpose webpage so these links allow Google to associate this page with related pages, thereby making this webpage possible for people to find.)
Student Solutions Manual for Nonlinear Dynamics and Chaos, 2nd edition on Amazon
Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering (Studies in Nonlinearity) on Amazon
Nonlinear Dynamics and Chaos, 2nd ed. SET with Student Solutions Manual (Studies in Nonlinearity) on Amazon
Nonlinear Dynamics and Chaos, Second Edition on Taylor & Francis
Student Solutions Manual for Nonlinear Dynamics and Chaos, Second Edition on Taylor & Francis
Steven Strogatz personal webpage