UMBC Dept of Math & Stat

Errata in Farlow's Differential Equations — Chapter 9

Page 510, Figure 9.4
The notation $S_1$, $S_2$, $S_3$ is not explicitly declared in the text but it is evident that these refer to the partial sums of one, two, and three terms in the infinite series (14).
Page 511, Equations (17) and (19):
(Spotted by Paul Rydeen) The two occurrences of $f(t)$ should be $f(x)$.
Page 512, Exercise 10:
(Spotted by Paul Rydeen) The periods of $\sin 2x$, $\sin 3x$, and $\sin 4x$ are $2\pi/2$, $2\pi/3$, and $2\pi/4$, respectively. Equivalently, those periods are $12\pi/12$, $8\pi/12$, and $6\pi/12$. The least common multiple of 6, 8, and 12 is 24, and therefore the period of $f(x)$ is $24\pi/12$, that is, $T=2\pi$. The book's answer of $T=\pi$ is incorrect.
Page 517, Figure 9.14:
(Spotted by Paul Rydeen) The notation $S_3$ refers to the partial sums of the first three terms of the infinite series (2a) and (2b).
Page 522, Exercise 24:
(Spotted by Paul Rydeen) In the book's answer, the leading term 1 should be $\frac{1}{2}$. Other than that, the answer is correct.

Remark: The immediate application of the equations (8) and (9) in the boxed prescription on page 520 leads to a solution in the form (8) where \[ a_0 = 1, \quad a_n=\frac{2}{n\pi} \sin \frac{n\pi}{2}, \quad n=1,2,\ldots. \] The form presented in the book's answer is obtained by noting that the values of $\sin \frac{n\pi}{2}$ cycle through $1, 0, -1, 0$, making every other term in the series zero.

Page 522, Exercise 26:
(Spotted by Paul Rydeen) The book's answer is incorrect. The correct solution is \[ f(x) = \sum_{ \stackrel{ \text{only odd values} }{n=1} }^\infty b_n \sin\frac{n x}{4}, \] where $b_n$ is evaluated according to equation (17) as \[ b_n = - \frac{8}{n} \cos \frac{n\pi}{2} + \frac{16}{n^2\pi} \sin \frac{n\pi}{2} \quad \text{(odd $n$)}. \] That's good enough for an answer to the problem, however it's possible to simplify the result. For one thing, $\cos \frac{n\pi}{2}$ is zero for odd $n$, and therefore the expression for $b_n$ reduces to \[ b_n = \frac{16}{n^2\pi} \sin \frac{n\pi}{2} \quad \text{(odd $n$)}. \] Furthermore, we get rid of the “odd $n$” qualifier by replacing $n$ by $2n-1$: \[ b_{2n-1} = \frac{16}{(2n-1)^2 \pi} \sin \frac{(2n-1)\pi}{2}, \quad n=1,2,3\ldots. \] We also observe that \[ \sin \frac{(2n-1)\pi}{2} = \sin\Bigl( n\pi - \frac{\pi}{2} \Bigr) = (-1)^{n+1}, \] whence \[ b_{2n-1} = \frac{16(-1)^{n+1}}{(2n-1)^2 \pi} , \] and therefore \[ f(x) = \sum_{n=1}^\infty \frac{16(-1)^{n+1}}{(2n-1)^2 \pi} \sin\frac{(2n-1) x}{4}. \]
Page 522, Exercise 27:
(Spotted by Paul Rydeen) In the problem's statement, all occurrences of $t$ are supposed to be $x$. Furthermore, the book's answer is incorrect. The correct solution is \[ f(x) = \sum_{ \stackrel{ \text{only odd values} }{n=1} }^\infty b_n \sin\frac{n \pi x}{4}, \quad\text{ where } b_n = \frac{4}{n\pi} \Bigl[ \cos \frac{n\pi}{4} - \cos \frac{n\pi}{2} \Bigr]. \] This simplifies, in the manner described in the previous comment, to \[ f(x) = \sum_{n=1}^\infty \frac{4}{(2n-1)\pi} \sin\Bigl( \frac{n\pi}{2} + \frac{\pi}{4} \Bigr) \sin\frac{(2n-1) \pi x}{4}. \]
Page 522, Exercise 28:
(Spotted by Paul Rydeen) In equation (18), the initial condition $\ddot{x}(0)=0$ is meant to be $\dot{x}(0)=0$.
Page 528, Exercise 24(d):
(Spotted by Paul Rydeen) The $=0$ at the end shouldn't be there. The equation is meant to be $u_t = u_{xx} + u$.
Page 538, Exercise 13:
(Spotted by Paul Rydeen) We are given $u(x,t) = \sin(\pi x)\,(\cdots)$. That should be $u(x,t) = \sin(n \pi x)\,(\cdots)$.
Page 539, Exercise 18:
(Spotted by Paul Rydeen) Despite the exercise's intent, the equation given there is separable. The following slightly modified equation, however, is not separable: \[ u_{xy} + u_{xx} + u_{yy} = 0. \]
Page 543, Example 3:
(Spotted by Paul Rydeen) The calculations are correct but the final answer has a few typos. The correct answer is \[ u(x,t) = \frac{4\epsilon}{\pi^2} \biggl\{ \cos(\alpha \pi t) \sin(\pi x) - \frac{1}{3^2} \cos(3 \alpha \pi t) \sin(3 \pi x) + \cdots \biggr\}. \] Alternatively, the solution may be expressed compactly as \[ u(x,t) = \frac{4\epsilon}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \sin \Bigl( \frac{n\pi}{2} \Bigr) \cos(\alpha n \pi t) \sin(n \pi x). \]
Page 545, Exercise 17:
(Spotted by Paul Rydeen) The answer given in the book is what comes from $u(x,0)$. To that we need to add the contribution from $u_t(x,0)$, which is $ \frac{2}{3\pi} \sin(3 \pi x) \sin(3 \pi t). $
Page 545, Exercise 22:
The book's answer says $n=2,3,\cdots$. It should say $n=1,2,\cdots$.
Page 551, Equation 21:
(Spotted by Paul Rydeen) The exponential terms are missing $\alpha$ in the exponents. They should be $e^{-(\pi\alpha)^2 t}$, $e^{-(3\pi\alpha)^2 t}$, $e^{-(5\pi\alpha)^2 t}$.
Page 553, Exercises 10–14:
(Spotted by Paul Rydeen) The exercises ask for the steady-state temperatures, but they are not given in the answers. The steady-state temperatures $u_\text{ss}$, may be determined by letting $t\to\infty$ in equation (23), whereby we obtain $u_\text{ss} = \lim_{t\to\infty} u(x,t) = \frac{a_0}{2}$. Referring to the equation that gives the values of the coefficients $a_n$ for all $n$, we see that $a_0 = 2\int_0^1 f(x)\,dx$, where $f(x)$ is the initial condition. We conclude that \[ u_\text{ss} = \int_0^1 f(x)\,dx, \] that is, when the ends are insulaterd, the steady-state temperature is a constant, and equals the average of the initial temperature. Consequently, the steady-state temperatures for problems 10 through 14 are 1, 0, 1, 1/2, 1/2, respectively.

Furthermore, in the book's answers to exercises 11–14, the $\alpha$ is missing in the exponents.

Page 553, Exercise 15:
(Spotted by Paul Rydeen) The book's answer has a few typos. The correct answer is \[ u(x,t) = e^{-(\pi\alpha/2)^2 t} \sin\Bigl(\frac{\pi x}{2} \Bigr) + \frac{1}{3} e^{-(3\pi\alpha/2)^2 t} \sin\Bigl(\frac{3\pi x}{2} \Bigr). \]



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Author: Rouben Rostamian