Partial Differential Equations (PDEs)

Transcription

1 c12.qxd 7/19/11 2:6 PM Page 238 CHAPTER 12 Partial Differential Equations (PDEs) SECTION 12.1 Basic Concets of PDEs, age 54 Purose. To familiarize the student with the following: Concet of solution, verification of solutions Suerosition rincile for homogeneous linear PDEs PDEs solvable by methods for ODEs SOLUTIONS TO PROBLEM SET 12.1, age c 1, u tt 4 u xx. Problems 2 13 should give the student a first imression of what kind of solutions to exect, and of the great variety of solutions comared with those of ODEs. It should be emhasized that although the wave and the heat equations look so similar, their solutions are basically different. It could be mentioned that the boundary and initial conditions are basically different, too. Of course, this will be seen in great detail in later sections, so one should erhas be cautious not to overload students with such details before they have seen a roblem being solved. 3. c Any c and k 5. c a>b. 6. c 2; u t 4u; u xx e 4t cos x 8. u t 9u c 2 v 2 u, c 3>v 14. Team Project. (a) Denoting derivatives with resect to the entire argument x ct and x ct, resectively, by a rime, we obtain by differentiating twice u xx vs ws, u tt vsc 2 wsc 2 and from this the desired result. (c) The student should realize that u 1> 2x 2 y 2 is not a solution of Lalace s equation in two variables, but satisfies the remarkable Poisson equation shown under (b). 16. Integrate twice with resect to y, u y c 1 (x), u c 1 (x)y c 2 (x) With the constants of integration c 1 (x) and c 2 (x) arbitrary. Problems will hel the student get used to the notations in this chater; in articular, y will now occur as an indeendent variable. Second-order PDEs in this set will also hel review the solution methods in Cha. 2, which will lay a role in searating variables. 18. u c 1 (x)e.4y c 2 (x)e.4y 2. The characteristic equation is 2l 2 9l 4 2(l 4)(l 1 2 ).

2 c12.qxd 7/19/11 2:6 PM Page 239 Instructor s Manual 239 Hence a general solution of the homogeneous PDE is u h (x, y) c 1 ( y)e 4x c 2 ( y)e.5x. A articular solution u of the nonhomogeneous PDE is obtained by the method of undetermined coefficients, u (x, y) 3 cos x sin x. 22. Set u to get v y v, v y >v 1, v c(x)e y x v, and u v dx c 1 (x)e y c 2 ( y). 24. By the given PDE and the chain rule, (A) yz x xz y y(z r r x z u u x ) x(z r r y z u u y ). Differentiate r 2 x 2 y 2 by arts and divide by 2r, (B) Now z r r x x>r, r y y>r. has in (A) the coefficients (use (B)) yr x xr y yx>r xy>r so that (A) reduces to z u. That is, z(r, u) deends only on r, not on the angle u, as for a shere, a circular cylinder, and so on. SECTION Solution by Searating Variables. Use of Fourier Series, age 545 Purose. This first section in which we solve a big roblem has several uroses: 1. To familiarize the student with the wave equation and with the tyical initial and boundary conditions that hysically meaningful solutions must satisfy. 2. To exlain and aly the imortant method of searation of variables, by which the PDE is reduced to two ODEs. 3. To show how Fourier series hel to get the final answer, thus seeing the reward of our great and long effort in Cha To discuss the eigenfunctions of the roblem, the basic building blocks of the solution, which lead to a deeer understanding of the whole roblem. Stes 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. SOLUTIONS TO PROBLEM SET 12.3, age If Assumtion 3 is violated, the string can move in various lanes. For large dislacements and> or angles the PDE would no longer be linear and have constant coefficients. Lack of elasticity entails loss of mechanical energy by conversion into heat (daming).

3 c12.qxd 7/19/11 2:7 PM Page Instructor s Manual If homogeneity were droed, it is hard to see what would haen; one would first have to be secific and state in what way homogeneity is changed and erhas suort theoretical results by hysical exeriments. 5. k cos 2t sin 2x 6. k(cos t sin x cos 2t sin 2x) k 8. (12 cos (t) sin (x) cos (2t) sin (2x) 4 cos (3t) sin (3x) 9 3 cos (4t) sin (4x) >3 12 cos(t) sin(x) cos(2t) sin(2x) 2> cos(3t) sin(3x) Á x a1 4 cos 2t sin 2x 1 1 cos 6t sin 6x 36 1 cos 1t sin 1x Á b There are more grahically osed roblems than in revious editions, so that CASusing students will have to make at least some additional effort in solving these roblems acos t sin x 1 9 cos 3t sin 3x 1 cos 5t sin 5x cos 7t sin 7x Á b (9>2 1>2 13) cos(t) sin(x) (9>2 13) cos(2t) sin(2x) 3 3>8 3 3 (9>2 2 13) cos(4t) sin(4x) 64 Á u a B n * sin nt sin nx, B n *.4 n 3 sin n 2 n1 16. F n sin (nx>l), G n a n cos (cn 2 2 t>l 2 ) 18. For the string the frequency of the nth mode is roortional to n, whereas for the beam it is roortional to n F() A C, C A, Fr() b(b D), D B. Then F(x) A(cos bx cosh bx) B(sin bx sinh bx) Fs(L) b 2 [A(cos bl cosh bl) B(sin bl sinh bl)] Ft(L) b 3 [A(sin bl sinh bl) B(cos bl cosh bl)]. The determinant (cos bl cosh bl) 2 sin 2 bl sinh 2 bl of this system in the unknowns A and B must be zero, and from this the result follows. 3

4 c12.qxd 7/19/11 2:7 PM Page 241 Instructor s Manual 241 From (23) we have cos bl 1 cosh bl because cosh bl is very large. This gives the aroximate solutions bl 1 (more exactly, 1.875, 4.694, 7.855, Á 2, 3 2, 5 2, Á ). SECTION D, Alembert, s Solution of the Wave Equation. Characteristics, age 553 Purose. To show a simler method of solving the wave equation, which, unfortunately, is not so universal as searation of variables. Comment on Order of Sections Section on the solution of the wave equation by the Lalace transform may be studied directly after this section. We have laced that material at the end of this chater because some students may not have studied Cha. 6 on the Lalace transform, which is not a rerequisite for Cha. 12. Comment on Footnote 1 D Alembert s Traité de dynamique aeared in 1743 and his solution of the vibrating string roblem in 1747; the latter makes him, together with Daniel Bernoulli ( ), 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, age u(, t) 1 2[ f (ct) f (ct)], f (ct) f (ct), so that f is odd. Also u(l, t) 1 2[ f (ct L) f (ct L)] hence f (ct L) f (ct L) f (ct L). This roves the eriodicity. 3. c 2 45>(1.5)>(3 9.8)) [m 2 >sec 2 ] n n 2 9. Ellitic, u f 1 (y 3ix) f 2 (y 3ix) 1. Hyerbolic, wave equation. Characteristic equation New variables are yr 2 16 ( yr 4)( yr 4). v y 4x, w y 4x.

5 c12.qxd 7/19/11 2:7 PM Page Instructor s Manual By the chain rule, u xx 16u vv 16u vw 16u wv 16u ww and 16u yy 16u vv 16u vw 16u wv 16u ww. Assuming u vw u wv, as usual, we have u vw, solvable by two integrations, as shown in the text. 11. Parabolic, u xf 1 (2x y) f 2 (2x y) 12. Parabolic. Characteristic equation New variables v x, w x y. By the chain rule, Substitution of this into the PDE gives the exected normal form 14. Hyerbolic. New variables x v and xy w. The latter is obtained from xyr y, u x 4u v 4u w yr 2 2yr 1 ( yr 1) 2. u x u v u w u xx u vv 2u vw u ww u xy u vw u ww u yy u ww. u vv. yr y 1 x, ln ƒ y ƒ ln ƒ x ƒ c. By the chain rule we obtain, in these new variables from the given PDE by cancellation of yu yy against a term in xu xy and division of the remaining PDE by x, the PDE u w xu vw. (The normal form is u vw u w >x u w >v.) We set u w z and obtain 1 c(w) z v v z, z v. By integration with resect to w we obtain the solution u 1 v f 1(w) f 2 (v) 1 x f 1(xy) f 2 (x). Note that the solution of the next roblem (Problem 15) is obtained by interchanging x and y in the resent roblem. 16. Ellitic. The characteristic equation is yr 2 2yr 1 3yr (1 3i)43yr (1 3i)4. Comlex solutions are y (1 3i) x const, y (1 3i) x const.

6 c12.qxd 7/19/11 2:7 PM Page 243 Instructor s Manual 243 This gives the solutions of the PDE: u f 1 ( y (1 3i)x) f 2 ( y (1 3i)x). Since the PDE is linear and homogeneous, real solutions are the real and the imaginary arts of u. 18. Parabolic. Characteristic equation yr 2 6yr 9 ( yr 3) 2. New variables v x, w y 3x. By the chain rule, u x u v 3u w u xx u vv 6u vw 9u ww u xy u vw 3u ww u yy u ww. Substitution into the PDE gives the exected normal form u vv 6u vw 9u ww Solution 6u vw 18u ww 9u ww u vv. u f 1 (v) f 2 (w) f 1 (x) f 2 ( y 3x) f 1 f 2 where and are any twice differentiable functions of the resective variables. 2. The Tricomi equation is ellitic in the uer half-lane and hyerbolic in the lower, because of the coefficient y. u F(x)G( y) gives yfsg FGs, and k 1 gives Airy s equation. Fs F Gs yg k SECTION Heat Equation: Solution by Fourier Series. Steady Two-Dimensional Heat Problems. Dirichlet Problem, age 558 Purose. This section has two uroses: 1. To solve a tyical heat roblem by stes similar to those for the wave equation, ointing to the two main differences: only one initial condition (instead of two) and u t (instead of u tt ), resulting in exonential functions in t (instead of cosine and sine in the wave equation). 2. Solution of Lalace s equation (which can be interreted as a time-indeendent heat equation in two dimensions). Comments on Content Additional oints to emhasize are More raid decay with increasing n, Difference in time evolution in Figs. 295 and 291,

7 c12.qxd 7/19/11 2:7 PM Page Instructor s Manual Zero can be an eigenvalue (see Examle 4), Three standard tyes of boundary value roblems, Analogy of electrostatic and (steady-state) heat roblems. Problem Set 12.6 includes additional heat roblems and tyes of boundary conditions. SOLUTIONS TO PROBLEM SET 12.6, age (c 2 2 >L 2 )2 ln 2, c 2.35L 2 u 32 2 asin.1xe.17522t 1 sin 2 t.3xe.1752(3) sin.5xe.1752(5)2t Á b 8. u I U 1 (U 2 U 1 )x>l. This is the solution of (1) with u>t satisfying the boundary conditions. 1. u(x, ) f (x) 1, U 1 1, U 2, u I 1 1x. Hence B n [1 (1 1x)] sin nx 1 dx x sin nx 1 dx 2 n cos n (1)n1 n # This gives the solution u(x, t) 1 1x a For x 5 this becomes u(5, t) e 1.729t 1 3 e1.556t 1 5 e4.323t Á 4. Obviously, the sum of the first few terms is a good aroximation of the true value at any t. We find: t u(5, t) 1 99 n (1) n1 n sin nx 2t 1 e1.752(n>1) u 1>2 2 4 cos (x) e t 4>9 cos (3x) e 9t 4 25 cos (5x) e25t Á 13. u u(x, t) cos 4x e 16t (Due to orthogonality all the terms excet for n 4 vanish. When n 4, the integral evaluates to 1).

8 c12.qxd 7/19/11 2:8 PM Page 245 Instructor s Manual ( 2) 8 cos (x) e t 8 9 cos (3x) e9t 8 25 cos (5x) e25t Á 16. c 2 v xx v t, v(, t), v(, t), v(x, ) f (x) Hx(x )>(2c 2 ), so that, as in (9) and (1), 18. u 44 a n1 Hx(x ) u(x, t) 2c 2 B n 2 af (x) a B n sin nx e c2 n 2 t n1 Hx(x ) 2c 2 b sin nx dx. where 1 (2n 1)x (2n 1)y sin sinh (2n 1) sinh 2(2n 1) u 5 asin 1 4 x sinh 1 4 yb>sinh 2. CAS Project. (a) u 8 (sin x sinh y)>sinh 2 (b) u y (x,, t), u y (x, 2, t), u 21. u 128 a 1 (2n 1)x (2n 1)y n1 sin sinh (2n 1) sinh(2n 1) u u I u II, where u I 4U 1 u II 4U a n1 a n1 1 (2n 1)x sin 2n (2n 1)x sin 2n 1 24 sinh 3(2n 1)y>244 sinh (2n 1) sinh 3(2n 1)(1 y>24)4. sinh (2n 1) 24. u F(x)G( y), F A cos x B sin x, u x (, y) Fr()G( y), B, G C cosh y D sinh y, u y (x, b) F(x)Gr(b), C cosh b, D sinh b, G cosh ( b y). For u cos x cosh (b y) we get u x (a, y) hu(a, y) ( sin a h cosh a) cosh (b y). Hence must satisfy, which has infinitely many ositive real solutions g 1, g 2, Á tan a h>, as you can illustrate by a simle sketch. Answer: u n cos g n x cosh g n (b y), where g g n satisfies g tan ga h. To determine coefficients of series of u n s from a boundary condition at the lower side is difficult because that would not be a Fourier series, the g n s being only aroximately regularly saced. See [C3], , 167. SECTION Heat Equation: Modeling Very Long Bars. Solution by Fourier Integrals and Transforms, age 568 Purose. Whereas we solved the roblem of a finite bar in the last section by using Fourier series, we show that for an infinite bar (ractically, a long insulated wire) we

9 c12.qxd 7/19/11 2:8 PM Page Instructor s Manual can use the Fourier integral for the same urose. Figure 299 shows the time evolution for a rectangular initial temerature ( 1 C between x 1and 1, zero elsewhere), giving bell-shaed curves as for the density of the normal distribution. We also show tyical alications of the Fourier transform and the Fourier sine transform to the heat equation. Short Courses. This section can be omitted. SOLUTIONS TO PROBLEM SET 12.7, age From (8) and (6) we obtain 3. and thus 4. e v >2 cos v dv 4 (1 4 2 ), B. Hence 5. A 2 u e c2 2t cos x d A 2 u 4 cos x e c2 2 t d (1 4 2 ) A 2 1 A 1 a cos v dv 2 cos v 4 v e2, sin(a) u 2 2 t cos(x)ee 2 d (1 v) cos v dv 2 (1 cos ), etc. sin () cos() 6. A, B 1 1 (v) sin v dv 2, etc Simle straightforward integration. 1. See (36) in A. A u(x, t) U, where t 2 c erf 1 x erf 1 x 2c1t 2c1t d a 2 sin a, B SECTION Modeling: Membrane, Two-Dimensional Wave Equation, age 575 Purose. A careful derivation of the two-dimensional wave equation governing the motions of a drumhead, from hysical assumtions (the analog of the modeling in Sec. 12.2).

10 c12.qxd 7/19/11 2:8 PM Page 247 Instructor s Manual 247 SECTION Rectangular Membrane. Double Fourier Series, age 577 Purose. To solve the two-dimensional wave equation in a rectangle x a, y b ( rectangular membrane ) by searation of variables and double Fourier series. Comment on Content New features as comared with the one-dimensional case (Sec. 12.3) are as follows: 1. We have to searate twice, first by u F(x, y)g(t), then the Helmholtz equation for F by F H(x)Q( y). 2. We get a double sequence of infinitely many eigenvalues l mn and eigenfunctions u mn ; see (9), (1). 3. We need double Fourier series (easily obtainable from the usual Fourier series) to get a solution that also satisfies the initial conditions. SOLUTIONS TO PROBLEM SET 12.9, age Modeling is the art of recognizing and neglecting minor factors and circumstances, and formulating major factors so that they become mathematically accessible, leading to a model that can be solved. No assumtion in any model can be satisfied exactly; in articular, in Assumtion 2 the tension will change during the motion. 4. B mn 8>(mn 2 ) if m, n odd, otherwise 5. B mn 8 m odd, n even mn 2, B mn 24 when both m and n are odd. B mn otherwise. mn 2, 6. B mn 8 when m is odd and n even mn 2, B mn 8 mn 2, when m is even and n is odd; otherwise. 7. For general B mn 4 a(1)m a(1) mn b(1) n b(1) mn m, n, n 2 m B when m is even and n is odd; 8 mn 8a when m is even mn 2, mn 2, and n is odd; otherwise. 8. B mn 64a 2 b 2 >(m 3 n 3 6 ) if m and n are odd, B mn otherwise 1. The rogram will give you etc # # # # # #

11 c12.qxd 7/19/11 2:9 PM Page Instructor s Manual cos 5t sin 4x sin 3y cos 15 t sin 2x sin y 18. A ab, b A>a, so that from (9) with m n 1 by differentiating with resect to a and equating the derivative to zero, we obtain a l 2 11 ; c 2 br a 1 2 a 1 br a 1 2 b 2 a a 2 br 2 2a 2 A 2 a 3 A 2 hence a 4 A 2, a 2 A, b A>a a. SECTION Lalacian in Polar Coordinates. Circular Membrane. Fourier Bessel Series, age 585 Purose. Detailed derivation of the transformation of the Lalacian into olar coordinates. Derivation of the function that models vibrations of a circular membrane. Comment on Content The transformation is a tyical case of a task often required in alications. It is done by two alications of the chain rule. In solving the wave equation we concentrate on the simler case of radially symmetric vibrations, that is, vibrations indeendent of the angle. (For eigenfunctions deending on the angle, see Probs ) We do three stes: 1. u W(r)G(t) gives for W Bessel s equation with, hence solutions W(r) J (kr). 2. We satisfy the boundary condition W(R) by choosing suitable values of k. 3. A Fourier Bessel series (18) hels to get the solution (17) of the entire roblem. Short Courses. This section can be omitted. SOLUTIONS TO PROBLEMS SET 12.1, age If u u(r) and we set ur v, then Hence ln v ln r ~ 2 u us ur>r vr v>r. vr>v 1>r, c ln (c 1 >r), v c 1 >r. By integration, u c 1 ln r c Team Project. (a) r 2 cos 2u r 2 (cos 2 u sin 2 u) x 2 y 2, r 2 sin 2u 2xy, etc. (c) u 4 ar sin u 1 3 r 3 sin 3u 1 5 r 5 sin 5u Á b (d) The form of the series results as in (b), and the formulas for the coefficients follow from u r (R, u) a n1 nr n1 (A n cos nu B n sin nu) f (u). (f) u (r 4>r)(sin u)>3 by searating variables sin(x) 22 3 sin(3x) sin(5x) 44 Á

12 c12.qxd 7/19/11 2:9 PM Page 249 Instructor s Manual r 2 cos 2u r cos(x) 2 r sin(u) 2 9 r 3 cos(3x) 1>2 r 2 sin (2u) 2>9 r 3 sin(3u) 8 r 5 cos(5x) 1>4 r 4 sin (4u) Á Excet for the resence of the variable r, this is just another imortant alication of Fourier series, and we concentrate on a few simle ractically imortant tyes of boundary values. Of course, earlier roblems on Fourier series can now be modified by introducing the owers of r and considered from the resent oint of view. 1. To get u on the x-axis, the idea is to extend the given otential from u skew-symmetrically to the whole boundary circle r 1; that is, 11u( u) if u (given) u(1, u) b 11u( u) if u. Then you obtain (valid in the whole disk and thus in the semidisk) u(r, u) 88 ar sin u r 3 sin 3u r 5 sin 5u Á b. 12. CAS Project. (b) Error.4863 (m 1),.2229,.1435,.156,.691,.589,.513,.454,.48 (m 1) (c) The aroximation of the artial sums is oorest for r. (d) The radii of the nodal circles are u 2 : a 1 >a Comarison.435>.5.87 u 3 : a 1 >a > a 2 >a > u 4 : a 1 >a > a 2 >a >.5.94 a 3 >a > We see that the larger radii are better aroximations of the values of the nodes of the string than the smaller ones. The smallest quotient does not seem to imrove (to get closer to 1); on the contrary, e.g., for u 6 it is.8. The other ratios seem to aroach 1 and so does the sum of all of them divided by m l m >(2) ca m >(2R) increases with decreasing R. 16. The reason is that f () 1. The artial sums equal Á the last value having 3-digit accuracy. Musically the values indicate substantial contributions of overtones to the overall sound. 18. Differentiation brings in a factor 1>l m R>(ca m ). 22. On Notation. n is standard for Legendre olynomials and for Bessel functions of integer order. Hence we needed another letter for numbering the zeros of J 1, J 2, Á, and we took m. Hence, for examle, the ositive zeros of are numbered a 21, a 22, J 2.835,

13 c12.qxd 7/19/11 2:9 PM Page Instructor s Manual a 23, Á. (In the 9th Edition we used the robably less advantageous oosite order a 12, a 22, a 32, Á.) For consistency, we should have numbered the ositive zeros of by a but this would make formulas unnecessarily clumsy, and we wrote a 1, a 2, Á 3, Á J a 1, a 2,,, in articular since the roblem occurred only at the very end, in the last roblems of Sec SECTION Lalace s Equation in Cylindrical and Sherical Coordinates. Potential, age 593 Purose. 1. Transformation of the Lalacian into cylindrical coordinates (which is trivial because of Sec. 12.9) and sherical coordinates; some remarks on areas in which Lalace s equation is basic. 2. Searation of the Lalace equation in sherical coordinates and alication to a tyical boundary value roblem. For simlicity we consider a boundary value roblem for a shere with boundary values deending only on. We do three stes: 1. u G(r)H() and searation gives for H Legendre s equation. 2. Continuity requirements restrict H to Legendre olynomials. 3. A Fourier Legendre series (18) hels to get the solution (17) of the interior roblem. Similarly for the exterior roblem, whose solution is (2). Short Courses. Omit the derivation of the Lalacian in cylindrical and sherical coordinates. SOLUTIONS TO PROBLEM SET 12.11, age By (11r) in Sec. 5.3 we have (cf. Fig. 312) u 1 A 1 r cos if 1 2. This is the xy-lane. Similarly, u 2 A 2 r 2 2 (3 cos2 1) if cos 1 13 and u 3 A 3 r 3 2 (5 cos3 3 cos ) if cos and B By (5), 2 u us ur>r. Searation and integration gives us>ur 1>r, ln ƒ ur ƒ ln ƒ r ƒ c 1. Taking exonents and integrating again gives ur c>r and u c ln r k. 12. u 8 ln r>(ln 2) v F(r)G(t), Fs # k 2 F, G c 2 k 2 G, F n sin (nr>r), G n B n ex (c 2 n 2 2 t>r 2 ), B n 2 R R rf (r) sin nr R dr

14 c12.qxd 7/19/11 2:1 PM Page 251 Instructor s Manual f ~ 2n 1 (w) w, A n Since w P 1 (w) and the P n (w) are orthogonal 2 1 wp n (w) dw. 1 on the interval 1 w 1, we obtain A 1 1, A n (n, 2, 3, Á ). Answer: u r cos. Of course, this is at once seen by integration. 18. By definition, P 2 (cos ) 3 2 cos Hence and 1 cos P 2 (cos ) 2 3 u 2 3 r 2 P 2 (cos ) 2 3 r 2 (cos 2 ) 1 3) u 4r 3 P 3 (cos ) 2r 2 P 2 (cos ) rp 1 (cos ) In Prob. 16, f () cos ; hence u int r cos, u ext r 2 cos. In Prob. 19, f () cos 2; hence f () 2 cos 2 1, so that and thus and 2 cos P 2(cos ) 1 3 u int 4 3 r 2 P 2 (cos ) 1 3 u ext 4 3r 3 P 2 (cos ) 1 3r. 24. Team Project. (a) The two dros over a ortion of the cable of length x are Ri x and L(i>t) x, resectively. Their sum equals the difference u x x u x. Divide by x and let x :. (c) To get the first PDE, differentiate the first transmission line equation with resect to x and use the second equation to relace and u xx Ri x Li xt R(Gu Cu t ) L(Gu t Cu tt ). Now collect terms. Similarly for the second PDE. 1 (d) Set c2. Then u t c 2 u xx, the heat equation. By (9), (1), Sec. 12.6, RC i x i xt: (e) u 4U asin x l e l 1 2t 1 3 u U cos (t>(l1lc sin (x>l) sin 3x l e l 3 2t Á b, l n 2 n2 2 l n RC. SECTION Solution of PDEs by Lalace Transforms, age 6 Purose. For students familiar with Cha. 6 we show that the Lalace transform also alies to certain PDEs. In such an alication the subsidiary equation will generally be an ODE. Short Courses. This section can be omitted.

15 c12.qxd 7/19/11 2:1 PM Page Instructor s Manual SOLUTIONS TO PROBLEM SET 12.12, age w w(x, t), W l{w(x, t)} W(x, s). The subsidiary equation is W and w(x, ) 1. x xl{w t(x, t)} W x x(sw w(x, )) xl(1) x s By simlification, W. x xsw x x s By integration of this first-order ODE with resect to x we obtain For x we have w(, t) 1 and Hence c(s) 1>s 2, so that W c(s)e sx2 >2 1 s 2 1 s. W(, s) l {w(, t)} l {1} 1 s c(s) 1 s 2 1 s. 1 W >2 1. s 2 esx2 s 1 2 s The inverse Lalace transform of this solution of the subsidiary equation is w(x, t) (t 1 2 x 2 ) u(t 1 2 x 2 ) t 1 c t 1 if t 1 2 x 2 1 2x 2 1 if t 1 2 x W C(s)x s 2x (s 1)s 2, W(, s), C(s), w(x, t) 2x(t 1 et ) w(x, t) 1>2 1>2t 1>2 u (t 2 x 2 ) (1 t 2x 2 ) as obtained from W (x, s) s 1 2s 2 e 2x 2 s c(s) with c(s) (s 1)>(2s 2 as obtained from 7. w f(x)g(t), xf r g # ) w(, t) 1, W (, s) 1>s. 2xt, take f(x) x to get g C 1 e t 2t 2 and C 1 2 using the initial condition w(x, ), i.e, g(). 8. W l{w}, W xx (1s 2 1s 25)W (1s 5) 2 W. The solution of this ODE is W c 1 (s)e (1s5)x c 2 (s)e (1s5)x with c 2 (s), so that the solution is bounded. c 1 (s) follows from W(, s) l{w(, t)} l{sin t} 1 s 2 1 c 1(s).

16 c12.qxd 7/19/11 2:11 PM Page 253 Instructor s Manual 253 Hence W 1. s 2 1 e(1s5)x The inverse Lalace transform (the solution of our roblem) is w l 1 {W} e 5x u(t 1x) sin (t 1x), a traveling wave decaying with x. Here u is the unit ste function (the Heaviside function). 1. From W F(s)e (x>c)1s and the convolution theorem we have From this and formula 39 in Sec. 6.9 we get, as asserted, 12. W (x, s) s 1 e 1sx>c, l{u(t)} 1>s, and since w(x, ), Now aly the convolution theorem. w f * l 1 {e k1s }, k x c. t w k f (t t) 2 22t 3 ek >(4t) dt. W(x, s) F(s)sW (x, s) F(s)[sW (x, s) w(x, )] F(s) l e w f. t SOLUTIONS TO CHAPTER 12 REVIEW QUESTIONS AND PROBLEMS, age u A( y) sin 3x B( y) cos 3x 17. u C 1 (x)e 3y C 2 (x)e y u g( y)(1 e x ) f ( y) 19. Hyerbolic, f 1 (x) f 2 (y 2x) 2. Parabolic, yr 2 4yr 4 ( yr 2) 2, v x, z y 2x, u xf 1 ( y 2x) f 2 ( y 2x) 21. Hyerbolic, f 1 (y 3x) f 2 (y 3x) 22. cos 4t sin 2x 23. 5>8 sin (x) cos (2t) 5 sin (3x) cos (6t) 1>16 sin (5x) cos (1t) u 4 acos 2t sin x 1 9 cos 6t sin 3x 1 25 cos 1t sin 5x Á b 26. u 4 x asin 2 1 e.1143t 1 3x sin 9 1 e.129t Á b 28. u 2 12(cos x e t 1 4 cos 2x e 4t 1 9 cos 3x e 9t Á )

17 c12.qxd 7/19/11 2:11 PM Page Instructor s Manual u 32 a1 4 cos 2x e4t 1 36 cos 6x e36t Á b u 64 2 a 34. l 11 >(2) c(11 1)>(2) 1> Area R 2 >2 1, R 12>, and 38. u (u u 1 )r r 1 u 1r 1 u r where r is the distance from the center of the (r 1 r )r r 1 r, sheres 4. f () 4 cos 3. Now, by (11r), Sec. 5.3, Answer: a m1 n1 m,n odd 1 sin mx sin ny 2 (m 2 n 2 )t m 3 ec 3 n ck 11 >(2) k 11 >(2) a 11 >(2R) 3.832>(212>) 3.832> 18. cos P 3(cos ) 3 5 P 1(cos ). u 8 5 r 3 P 3 (cos ) 12 5 rp 1(cos ).

18 c12.qxd 7/19/11 2:11 PM Page 255 Instructor s Manual 255 Author Query AQ1 Please rovide solution