Part 24 Partial Differential Equations-Advanced Engineering Mathematics-Solution Manual, Exercises of Engineering Mathematics

Download Part 24 Partial Differential Equations-Advanced Engineering Mathematics-Solution Manual and more Exercises Engineering Mathematics in PDF only on Docsity! Instructor’s Manual 225 Note that convolution ƒ
g differs from that in Chapter 6, and so does the formula (12) in the convolution theorem (we now have a factor 2
). Short Courses. Sections 11.4–11.9 can be omitted. SOLUTIONS TO PROBLEM SET 11.9, page 528 2. By integration of the defining integral we obtain
0
 e(k
iw)x dx  • j0
  • . 4. Integration of the defining integral gives
1
1 e(2
w)ix dx  (e(2
w)i  e
(2
w)i)  2i sin (2  w)   . 6. By integration by parts,
1
1 xe
iwx dx  ( j1
1 
1
1 e
iwx dx)  (   e
iwxj1
1 )  (  (e
iw  eiw))   (  )   • (w cos w  sin w). 8. By integration by parts we obtain
0
1 xe
x
iwx dx  ( j0
1 
0
1 e
(1iw)x dx)  ((1)  j0
1 )  (  )  (1  e1iw(1  i(w  i)))  (1  iwe1iw). 1  2
 (w  i)2 1  2
 (w  i)2 1  e1iw  (1  iw)2 e1iw  1  iw 1  2
 e
(1iw)x  (1  iw)(1  iw) e1iw  (1  iw) 1  2
 1  1  iw xe
(1iw)x  (1  iw) 1  2
 1  2
 i  w2 2 
i sin w  w2 i cos w  w 2 
1  w2 2 cos w  iw 1  2
 1  (iw)2 eiw  iw e
iw  iw 1  2
 1  iw xe
iwx  iw 1  2
 1  2
 sin (w  2)  w  2 2 
i  2
 (2  w) i  2
 (2  w) 1  2
 1  k  iw 1  2
 e(k
iw)x  k  iw 1  2
 1  2
 im11.qxd 9/21/05 12:33 PM Page 225 docsity.com 226 Instructor’s Manual Problems 1 to 9 should help the student get a feel for integrating complex exponential functions and for their transformation into cosine and sine, as needed in this context. Here, it is taken for granted that complex exponential functions can be handled in the same fashion as real ones, which will be justified in Part D on complex analysis. The problems show that the technicalities are rather formidable for someone who faces these exponential functions for the first time. This is so for relatively simple ƒ(x), and since a CAS will give all the results without difficulty, it would make little sense to deal with more complicated ƒ(x), which would involve increased technical difficulties but no new ideas. 10. ƒ(x)  xe
x (x  0), g(x)  e
x (x  0). Then ƒ  e
x  xe
x  g  ƒ, and by (9), iw
(ƒ) 
(ƒ ) 
(g) 
(ƒ) hence (iw  1)
(ƒ) 
(g)  ,
(ƒ)  . 12. We obtain •  •   • . 14. Team Project. (a) Use t  x  a as a new variable of integration. (b) Use c  3b. Then (a) gives e2ibw
(ƒ(x))     • . (c) Replace w with w  a. This gives a new factor eiax. (d) We see that ƒ̂(w) in formula 7 is obtained from ƒ̂(w) in formula 1 by replacing w with w  a. Hence by (c), ƒ(x) in formula 1 times eiax should give ƒ(x) in formula 7, which is true. Similarly for formulas 2 and 8. SOLUTIONS TO CHAP. 11 REVIEW QUESTIONS AND PROBLEMS, page 532 12.  (cos x  cos 3x  cos 5x   • • •) 14. 1  (cos  cos  cos  • • •) 16. (sin
x  sin 2
x  sin 3
x  • • •) 18. (1  cos x  sin x  cos 2x  sin 2x  • • •) 20.
 2(sin x  1_2 sin 2x  1_ 3 sin 3x  1_ 4 sin 4x  • • •) 22. 1/2 by Prob. 17. Convergence is slow. 4  5 2  5 sinh

1  3 1  2 2 
5
x  2 1  25 3
x  2 1  9
x  2 8 
2 1  5 1  3 2 
1  2 sin bw  w 2 
2i sin bw  iw2
 eibw  e
ibw  iw2
 sin b(w  a)  w  a 2 
2i sin b(a  w)  a  w i  2
 e
ib(a
w)  eib(a
w)  a  w i  2
 1  2
 (1  iw)2 1  2
 (1  iw) im11.qxd 9/21/05 12:33 PM Page 226 docsity.com (c) The student should realize that u
1/x2  y2 is not a solution of Laplace’s equation in two variables. It satisfies the Poisson equation with ƒ
(x2  y2)
3/2, which seems remarkable. 28. A function whose first partial derivatives are zero is a constant, u(x, y)
c
const. Integrate the first PDE and then use the second. 30. Integrating the first PDE and the second PDE gives u
c1(y)x  c2(y) and u
c3(x)y  c4(x), respectively. Equating these two functions gives u
axy  bx  cy  k. Alternatively, uxx
0 gives u
c1(y)x  c2(y). Then from uyy
0 we get uyy
c1x  c2
0; hence c1
0, c2
0, and by integration c1
y  , c2
y  and by substitution in the previous expression u
c1x  c2
yx  x  y  . SECTION 12.2. Modeling: Vibrating String, Wave Equation, page 538 Purpose. A careful derivation of the one-dimensional wave equation (more careful than in most other texts, where some of the essential physical assumptions are usually missing). Short Courses. One may perhaps omit the derivation and just state the wave equation and mention of what c2 is composed. SECTION 12.3. Solution by Separating Variables. Use of Fourier Series, page 540 Purpose. This first section in which we solve a “big” problem has several purposes: 1. To familiarize the student with the wave equation and with the typical initial and boundary conditions that physically meaningful solutions must satisfy. 2. To explain and apply the important method of separation of variables, by which the PDE is reduced to two ODEs. 3. To show how Fourier series help to get the final answer, thus seeing the reward of our great and long effort in Chap. 11. 4. To discuss the eigenfunctions of the problem, the basic building blocks of the solution, which lead to a deeper understanding of the whole problem. Steps of Solution 1. Setting u
F(x)G(t) gives two ODEs for F and G. 2. The boundary conditions lead to sine and cosine solutions of the ODEs. 3. A series of those solutions with coefficients determined from the Fourier series of the initial conditions gives the final answer. Instructor’s Manual 229 im12.qxd 9/21/05 5:16 PM Page 229 docsity.com SOLUTIONS TO PROBLEM SET 12.3, page 546 2. k(cos t sin x  1_3 cos 3 t sin 3 x) 4. (cos t sin x  cos 2 t sin 2 x  cos 3 t sin 3 x   • • •) 6. (cos t sin x  cos 3 t sin 3 x  cos 5 t sin 5 x   • • •) 8. ( cos 2 t sin 2 x  cos 6 t sin 6 x  cos 10 t sin 10 x  • • •) There are more graphically posed problems (Probs. 5–10) than in previous editions, so that CAS-using students will have to make at least some additional effort in solving these problems. 10. (sin cos t sin x  sin cos 2 t sin 2 x  sin cos 3 t sin 3 x  sin cos 4 t sin 4 x   • • •) 12. u
 n
1 Bn* sin nt sin nx, Bn*
sin 14. Team Project. (c) From the given initial conditions we obtain Gn(0)
Bn

L 0 ƒ(x) sin dx, G . n(0)
nBn* 
0. (e) u(0, t)
0, w(0, t)
0, u(L, t)
h(t), w(L, t)
h(t). The simplest w satisfying these conditions is w(x, t)
xh(t) /L. Then v(x, 0)
ƒ(x)  xh(0) /L
ƒ1(x) vt(x, 0)
g(x)  xh (0) /L
g1(x) vtt  c2vxx
xh/L. 16. Fn
sin (n x/L), Gn
an cos (cn 2 2t/L2) 18. For the string the frequency of the nth mode is proportional to n, whereas for the beam it is proportional to n2. 20. F(0)
A  C
0, C
A, F (0)
(B  D)
0, D
B. Then F(x)
A(cos x  cosh x)  B(sin x  sinh x) F(L)
2[A(cos L  cosh L)  B(sin L  sinh L)]
0 F(L)
3[A(sin L  sinh L)  B(cos L  cosh L)]
0. The determinant (cos L  cosh L)2  sin2 L  sinh2 L of this system in the unknowns A and B must be zero, and from this the result follows. From (23) we have cos L
 0 1  cosh L 2A(1  cos  )  n ( n 2  2) n x  L 2  L n  2 0.04  n3  5 1  16 2  5 1  9 2  5 1  4  5 25  2 2 1  100 1  36 1  4 8  2 1  25 1  9 8  2 1  27 1  8 12k  3 230 Instructor’s Manual im12.qxd 9/21/05 5:16 PM Page 230 docsity.com because cosh L is very large. This gives the approximate solutions L  1_2 , 3_ 2 , 5_ 2 , • • • (more exactly, 1.875, 4.694, 7.855, • • •). SECTION 12.4. D’Alembert’s Solution of the Wave Equation. Characteristics, page 548 Purpose. To show a simpler method of solving the wave equation, which, unfortunately, is not so universal as separation of variables. Comment on Order of Sections Section 12.11 on the solution of the wave equation by the Laplace transform may be studied directly after this section. We have placed that material at the end of this chapter because some students may not have studied Chap. 6 on the Laplace transform, which is not a prerequisite for Chap. 12. Comment on Footnote 1 D’Alembert’s Traité de dynamique appeared in 1743 and his solution of the vibrating string problem in 1747; the latter makes him, together with Daniel Bernoulli (1700–1782), the founder of the theory of PDEs. In 1754 d’Alembert became Secretary of the French Academy of Science and as such the most influential man of science in France. SOLUTIONS TO PROBLEM SET 12.4, page 552 2. u(0, t)
1_2 [ƒ(ct)  ƒ(ct)]
0, ƒ(ct)
ƒ(ct), so that ƒ is odd. Also u(L, t)
1_2 [ƒ(ct  L)  ƒ(ct  L)]
0 hence ƒ(ct  L)
ƒ(ct  L)
ƒ(ct  L). This proves the periodicity. 4. (1/2 )(n /2) • 80.83
20.21n 10. The Tricomi equation is elliptic in the upper half-plane and hyperbolic in the lower, because of the coefficient y. u
F(x)G(y) gives yFG
FG,

k and k
1 gives Airy’s equation. 12. Parabolic. Characteristic equation y 2  2y  1
(y  1)2
0. New variables v

x, w

x  y. By the chain rule, ux
uv  uw uxx
uvv  2uvw  uww uxy
uvw  uww uyy
uww. Substitution of this into the PDE gives the expected normal form uvv
0. G  yG F  F Instructor’s Manual 231 im12.qxd 9/21/05 5:16 PM Page 231 docsity.com