12.8 Modeling: Membrane,
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1 SEC. 1.8 Modeling: Membrne, Two-Dimensionl Wve Eqution Modeling: Membrne, Two-Dimensionl Wve Eqution Since the modeling here will be similr to tht of Sec. 1., you my wnt to tke nother look t Sec. 1.. The vibrting string in Sec. 1. is bsic one-dimensionl vibrtionl problem. Eqully importnt is its two-dimensionl nlog, nmely, the motion of n elstic membrne, such s drumhed, tht is stretched nd then fixed long its edge. Indeed, setting up the model will proceed lmost s in Sec. 1.. Physicl Assumptions 1. The mss of the membrne per unit re is constnt ( homogeneous membrne ). The membrne is perfectly flexible nd offers no resistnce to bending.. The membrne is stretched nd then fixed long its entire boundry in the xy-plne. The tension per unit length T cused by stretching the membrne is the sme t ll points nd in ll directions nd does not chnge during the motion. 3. The deflection u (x, y, t) of the membrne during the motion is smll compred to the size of the membrne, nd ll ngles of inclintion re smll. Although these ssumptions cnnot be relized exctly, they hold reltively ccurtely for smll trnsverse vibrtions of thin elstic membrne, so tht we shll obtin good model, for instnce, of drumhed. Derivtion of the PDE of the Model ( Two-Dimensionl Wve Eqution ) from Forces. As in Sec. 1. the model will consist of PDE nd dditionl conditions. The PDE will be obtined by the sme method s in Sec. 1., nmely, by considering the forces cting on smll portion of the physicl system, the membrne in Fig. 31 on the next pge, s it is moving up nd down. Since the deflections of the membrne nd the ngles of inclintion re smll, the sides of the portion re pproximtely equl to x nd y. The tension T is the force per unit length. Hence the forces cting on the sides of the portion re pproximtely T x nd T y. Since the membrne is perfectly flexible, these forces re tngent to the moving membrne t every instnt. Horizontl Components of the Forces. We first consider the horizontl components of the forces. These components re obtined by multiplying the forces by the cosines of the ngles of inclintion. Since these ngles re smll, their cosines re close to 1. Hence the horizontl components of the forces t opposite sides re pproximtely equl. Therefore, the motion of the prticles of the membrne in horizontl direction will be negligibly smll. From this we conclude tht we my regrd the motion of the membrne s trnsversl; tht is, ech prticle moves verticlly. Verticl Components of the Forces. left side re (Fig. 31), respectively, These components long the right side nd the T y sin b nd T y sin. Here nd b re the vlues of the ngle of inclintion (which vries slightly long the edges) in the middle of the edges, nd the minus sign ppers becuse the force on the
2 576 CHAP. 1 Prtil Differentil Equtions (PDEs) Membrne y + Δy y x x + Δx TΔ x TΔy u TΔy α β TΔy α β y + Δy TΔ x TΔy y x x x + Δx x + Δx Fig. 31. Vibrting membrne left side is directed downwrd. Since the ngles re smll, we my replce their sines by their tngents. Hence the resultnt of those two verticl components is (1) T y (sin b sin ) T y (tn b tn ) T y [u x (x x, y 1 ) u x (x, y )] where subscripts x denote prtil derivtives nd nd re vlues between y nd y y. Similrly, the resultnt of the verticl components of the forces cting on the other two sides of the portion is y 1 y () T x [u y (x 1, y y) u y (x, y)] where nd re vlues between x nd x x. x 1 x Newton s Second Lw Gives the PDE of the Model. By Newton s second lw (see Sec..) the sum of the forces given by (1) nd () is equl to the mss r A of tht smll portion times the ccelertion u>t ; here r is the mss of the undeflected membrne per unit re, nd A x y is the re of tht portion when it is undeflected. Thus r x y u t T y [u x (x x, y 1) u x (x, y )] T x [u y (x 1, y y) u y (x, y)] where the derivtive on the left is evluted t some suitble point ( x, y) corresponding to tht portion. Division by r x y gives
3 SEC. 1.9 Rectngulr Membrne. Double Fourier Series 577 u t T r c u x(x x, y 1 ) u x (x, y ) x u y(x 1, y y) u y (x, y) d. y If we let x nd y pproch zero, we obtin the PDE of the model (3) u u u t c x y b c T r. This PDE is clled the two-dimensionl wve eqution. The expression in prentheses is the Lplcin u of u (Sec. 1.8). Hence (3) cn be written (3) u t c u. Solutions of the wve eqution (3) will be obtined nd discussed in the next section. 1.9 Rectngulr Membrne. Double Fourier Series Now we develop solution for the PDE obtined in Sec Detils re s follows. The model of the vibrting membrne for obtining the displcement u(x, y, t) of point (x, y) of the membrne from rest (u ) t time t is (1) u u c u t x y b () u on the boundry (3) (3b) u (x, y, ) f (x, y) u t (x, y, ) g (x, y). y b R Fig. 3. Rectngulr membrne x Here (1) is the two-dimensionl wve eqution with c T>r just derived, () is the boundry condition (membrne fixed long the boundry in the xy-plne for ll times t ), nd (3) re the initil conditions t t, consisting of the given initil displcement (initil shpe) f (x, y) nd the given initil velocity g(x, y), where u t u>t. We see tht these conditions re quite similr to those for the string in Sec. 1.. Let us consider the rectngulr membrne R in Fig. 3. This is our first importnt model. It is much simpler thn the circulr drumhed, which will follow lter. First we note tht the boundry in eqution () is the rectngle in Fig. 3. We shll solve this problem in three steps:
4 578 CHAP. 1 Prtil Differentil Equtions (PDEs) Step 1. By seprting vribles, first setting u(x, y, t) F(x, y)g(t) nd lter F(x, y) H(x)Q(y) we obtin from (1) n ODE () for G nd lter from PDE (5) for F two ODEs (6) nd (7) for H nd Q. Step. From the solutions of those ODEs we determine solutions (13) of (1) ( eigenfunctions ) tht stisfy the boundry condition (). u mn Step 3. We compose the u mn into double series (1) solving the whole model (1), (), (3). Step 1. Three ODEs From the Wve Eqution (1) To obtin ODEs from (1), we pply two successive seprtions of vribles. In the first seprtion we set u(x, y, t) F(x, y)g(t). Substitution into (1) gives FG ## c (F xx G F yy G) where subscripts denote prtil derivtives nd dots denote derivtives with respect to t. To seprte the vribles, we divide both sides by c FG: G ## c G 1 F (F xx F yy ). Since the left side depends only on t, wheres the right side is independent of t, both sides must equl constnt. By simple investigtion we see tht only negtive vlues of tht constnt will led to solutions tht stisfy () without being identiclly zero; this is similr to Sec Denoting tht negtive constnt by, we hve G ## c G 1 F (F xx F yy ). This gives two equtions: for the time function G(t) we hve the ODE () G ## l G where l c, nd for the mplitude function F (x, y) PDE, clled the two-dimensionl Helmholtz 3 eqution (5) F xx F yy F. 3 HERMANN VON HELMHOLTZ ( ), Germn physicist, known for his fundmentl work in thermodynmics, fluid flow, nd coustics.
5 SEC. 1.9 Rectngulr Membrne. Double Fourier Series 579 Seprtion of the Helmholtz eqution is chieved if we set F(x, y) H(x)Q( y). By substitution of this into (5) we obtin To seprte the vribles, we divide both sides by HQ, finding Both sides must equl constnt, by the usul rgument. This constnt must be negtive, sy, k, becuse only negtive vlues will led to solutions tht stisfy () without being identiclly zero. Thus This yields two ODEs for H nd Q, nmely, 1 H 1 H Q d H d Q H dx dy HQb. d H dx 1 Q d Q dy Qb. d H dx 1 Q Q d dy Qb k. (6) d H dx k H nd (7) d Q dy p Q where p k. Step. Stisfying the Boundry Condition Generl solutions of (6) nd (7) re H(x) A cos kx B sin kx nd Q(y) C cos py D sin py with constnt A, B,C, D. From u FG nd () it follows tht F HQ must be zero on the boundry, tht is, on the edges x, x, y, y b; see Fig. 3. This gives the conditions H(), H(), Q(), Q(b). Hence H() A nd then H() B sin k. Here we must tke B since otherwise H(x) nd F(x, y). Hence sin k or k mp, tht is, k mp (m integer).
6 58 CHAP. 1 Prtil Differentil Equtions (PDEs) In precisely the sme fshion we conclude tht C nd p must be restricted to the vlues p np>b where n is n integer. We thus obtin the solutions H H m, Q Q n, where H m (x) sin mpx nd Q n (y) sin npy b, m 1,, Á, n 1,, Á. As in the cse of the vibrting string, it is not necessry to consider m, n 1,, Á since the corresponding solutions re essentilly the sme s for positive m nd n, expect for fctor 1. Hence the functions (8) F mn (x, y) H m (x)q n ( y) sin mpx npy sin b, m 1,, Á, n 1,, Á, re solutions of the Helmholtz eqution (5) tht re zero on the boundry of our membrne. Eigenfunctions nd Eigenvlues. Hving tken cre of (5), we turn to (). Since p k in (7) nd l cv in (), we hve l ck p. Hence to k mp> nd p np>b there corresponds the vlue (9) m l l mn cp B n b, m 1,, Á, n 1,, Á, in the ODE (). A corresponding generl solution of () is G mn (t) B mn cos l mn t B* mn sin l mn t. It follows tht the functions u mn (x, y, t) F mn (x, y) G mn (t), written out (1) u mn (x, y, t) (B mn cos l mn t B* mn sin l mn t) sin mpx sin npy b l mn with ccording to (9), re solutions of the wve eqution (1) tht re zero on the boundry of the rectngulr membrne in Fig. 3. These functions re clled the eigenfunctions or chrcteristic functions, nd the numbers l mn re clled the eigenvlues or chrcteristic vlues of the vibrting membrne. The frequency of u mn is l mn >p. Discussion of Eigenfunctions. It is very interesting tht, depending on nd b, severl functions F mn my correspond to the sme eigenvlue. Physiclly this mens tht there my exists vibrtions hving the sme frequency but entirely different nodl lines (curves of points on the membrne tht do not move). Let us illustrte this with the following exmple.
7 SEC. 1.9 Rectngulr Membrne. Double Fourier Series 581 E X A M P L E 1 Eigenvlues nd Eigenfunctions of the Squre Membrne Consider the squre membrne with b 1. From (9) we obtin its eigenvlues (11) l mn cpm n. Hence l mn l nm, but for m n the corresponding functions F mn sin mpx sin npy nd F nm sin npx sin mpy re certinly different. For exmple, to l 1 l 1 cp15 there correspond the two functions F 1 sin px sin py nd F 1 sin px sin py. Hence the corresponding solutions u 1 (B 1 cos cp15t B * 1 sin cp15t)f 1 nd u 1 (B 1 cos cp15t B * 1 sin cp15t)f 1 hve the nodl lines y 1 nd x 1, respectively (see Fig. 33). Tking B 1 1 nd B* 1 B* 1, we obtin (1) u 1 u 1 cos cp15t (F 1 B 1 F 1 ) which represents nother vibrtion corresponding to the eigenvlue cp15. The nodl line of this function is the solution of the eqution F 1 B 1 F 1 sin px sin py B 1 sin px sin py or, since sin sin cos, (13) sin px sin py (cos py B 1 cos px). This solution depends on the vlue of B 1 (see Fig. 3). From (11) we see tht even more thn two functions my correspond to the sme numericl vlue of l mn. For exmple, the four functions F 18, F 81, F 7, nd correspond to the vlue F 7 l 18 l 81 l 7 l 7 cp165, becuse This hppens becuse 65 cn be expressed s the sum of two squres of positive integers in severl wys. According to theorem by Guss, this is the cse for every sum of two squres mong whose prime fctors there re t lest two different ones of the form n 1 where n is positive integer. In our cse we hve 65 5 # 13 ( 1)(1 1). B 1 = 1 u 11 u 1 u 1 u u 13 u 31 B 1 = 1 B 1 =.5 B 1 = B 1 =.5 B 1 = 1 Fig. 33. Nodl lines of the solutions Fig. 3. Nodl lines u 11, u 1, u 1, u, u 13, u 31 in the cse of of the solution (1) for the squre membrne some vlues of B 1
8 58 CHAP. 1 Prtil Differentil Equtions (PDEs) Step 3. Solution of the Model (1), (), (3). Double Fourier Series So fr we hve solutions (1) stisfying (1) nd () only. To obtin the solutions tht lso stisfies (3), we proceed s in Sec We consider the double series (1) u (x, y, t) m1 n1 m1 n1 u mn (x, y, t) (B mn cos l mn t B* mn sin l mn t) sin mpx sin npy b (without discussing convergence nd uniqueness). From (1) nd (3), setting t, we hve (15) u (x, y, ) m1 n1 B mn sin mpx sin npy b f (x, y). Suppose tht f (x, y) cn be represented by (15). (Sufficient for this is the continuity of f, f>x, f>y, f>x y in R.) Then (15) is clled the double Fourier series of f (x, y). Its coefficients cn be determined s follows. Setting (16) K m (y) B mn sin npy b n1 we cn write (15) in the form f (x, y) K m (y) sin mpx. m1 For fixed y this is the Fourier sine series of f (x, y), considered s function of x. From () in Sec we see tht the coefficients of this expnsion re (17) K m (y) f(x, y) sin mpx dx. Furthermore, (16) is the Fourier sine series of K m (y), nd from () in Sec it follows tht the coefficients re B mn b b K m (y) sin npy b dy. From this nd (17) we obtin the generlized Euler formul (18) B mn b b f (x, y) sin mpx sin npy dx dy b m 1,, Á n 1,, Á for the Fourier coefficients of f (x, y) in the double Fourier series (15).
9 SEC. 1.9 Rectngulr Membrne. Double Fourier Series 583 The B mn in (1) re now determined in terms of f (x, y). To determine the B* mn, we differentite (1) termwise with respect to t; using (3b), we obtin u t ` t m1 n1 B* mn l mn sin mpx sin npy b g (x, y). Suppose tht g (x, y) cn be developed in this double Fourier series. Then, proceeding s before, we find tht the coefficients re (19) B* mn bl mn b g (x, y) sin mpx npy sin dx dy m 1,, Á n 1,, Á. Result. If f nd g in (3) re such tht u cn be represented by (1), then (1) with coefficients (18) nd (19) is the solution of the model (1), (), (3). E X A M P L E Vibrtion of Rectngulr Membrne Find the vibrtions of rectngulr membrne of sides ft nd b ft (Fig. 35) if the tension is 1.5 lb>ft, the density is.5 slugs>ft (s for light rubber), the initil velocity is, nd the initil displcement is () f (x, y).1 (x x )(y y ) ft. y R y u x x Membrne Fig. 35. Exmple Initil displcement Solution. c T>r 1.5>.5 5 [ft >sec ]. Also B* mn from (19). From (18) nd (), B mn # 1 (x x ) sin mpx dx (y y ) sin npy dy..1(x x )(y y ) sin mpx sin npy dx dy Two integrtions by prts give for the first integrl on the right 18 m 3 p 3 [1 (1)m ] 56 m 3 p 3 (m odd) nd for the second integrl 16 n 3 p [1 3 (1)n ] 3 n 3 p 3 (n odd).
10 58 CHAP. 1 Prtil Differentil Equtions (PDEs) For even m or n we get. Together with the fctor 1> we thus hve B mn if m or n is even nd B mn 56 # 3.65 m 3 n 3 6 p m 3 n 3 (m nd n both odd). From this, (9), nd (1) we obtin the nswer u (x, y, t).65 m,n odd 1 5p cos m n b t sin mpx m 3 3 n sin npy (1).65 cos 15p cos 15p113 t sin px py sin 1 7 t sin 3px py sin 1 79 cos 15p137 cos 15p15 t sin px 3py sin t sin 3px 3py sin Á b. To discuss this solution, we note tht the first term is very similr to the initil shpe of the membrne, hs no nodl lines, nd is by fr the dominting term becuse the coefficients of the next terms re much smller. The second term hs two horizontl nodl lines ( y 3, 3), the third term two verticl ones (x 3, 8 3), the fourth term two horizontl nd two verticl ones, nd so on. P R O B L E M S E T Frequency. How does the frequency of the eigenfunctions of the rectngulr membrne chnge () If we double the tension? (b) If we tke membrne of hlf the density of the originl one? (c) If we double the sides of the membrne? Give resons.. Assumptions. Which prt of Assumption cnnot be stisfied exctly? Why did we lso ssume tht the ngles of inclintion re smll? 3. Determine nd sketch the nodl lines of the squre membrne for m 1,, 3, nd n 1,, 3,. 8 DOUBLE FOURIER SERIES Represent f (x, y) by series (15), where. f (x, y) 1, b 1 5. f (x, y) y, b 1 6. f (x, y) x, b 1 7. f (x, y) xy, nd b rbitrry 8. f (x, y) xy ( x) (b y), nd b rbitrry 9. CAS PROJECT. Double Fourier Series. () Write progrm tht gives nd grphs prtil sums of (15). Apply it to Probs. 5 nd 6. Do the grphs show tht those prtil sums stisfy the boundry condition (3)? Explin why. Why is the convergence rpid? (b) Do the tsks in () for Prob.. Grph portion, sy, x 1, y 1, of severl prtil sums on common xes, so tht you cn see how they differ. (See Fig. 36.) (c) Do the tsks in (b) for functions of your choice. 1. CAS EXPERIMENT. Qudruples of F mn. Write progrm tht gives you four numericlly equl l mn in Exmple 1, so tht four different F mn correspond to it. Sketch the nodl lines of F 18, F 81, F 7, F 7 in Exmple 1 nd similrly for further tht you will find SQUARE MEMBRANE Find the deflection u (x, y, t) of the squre membrne of side p nd c 1 for initil velocity nd initil deflection y.3..1 sin x sin y.1 sin x sin y xy (p x) (p y).3..1 x Fig. 36. Prtil sums S, nd S 1,1 in CAS Project 9b F mn
11 SEC. 1.1 Lplcin in Polr Coordintes. Circulr Membrne. Fourier Bessel Series RECTANGULAR MEMBRANE 1. Verify the discussion of (1) in Exmple. 15. Do Prob. 3 for the membrne with nd b. 16. Verify B mn in Exmple by integrtion by prts. 17. Find eigenvlues of the rectngulr membrne of sides nd b 1 to which there correspond two or more different (independent) eigenfunctions. 18. Minimum property. Show tht mong ll rectngulr membrnes of the sme re A b nd the sme c the squre membrne is tht for which u 11 [see (1)] hs the lowest frequency. 19. Deflection. Find the deflection of the membrne of sides nd b with c 1 for the initil deflection f (x, y) sin 6px py sin nd initil velocity. b. Forced vibrtions. Show tht forced vibrtions of membrne re modeled by the PDE u tt c u P>r, where P (x, y, t) is the externl force per unit re cting perpendiculr to the xy-plne. 1.1 Lplcin in Polr Coordintes. Circulr Membrne. Fourier Bessel Series It is generl principle in boundry vlue problems for PDEs to choose coordintes tht mke the formul for the boundry s simple s possible. Here polr coordintes re used for this purpose s follows. Since we wnt to discuss circulr membrnes (drumheds), we first trnsform the Lplcin in the wve eqution (1), Sec. 1.9, (1) u tt c u c (u xx u yy ) (subscripts denoting prtil derivtives) into polr coordintes r, u defined by x r cos u, y r sin u; thus, r x y, tn u y x. By the chin rule (Sec. 9.6) we obtin u x u r r x u u u x. Differentiting once more with respect to x nd using the product rule nd then gin the chin rule gives () u xx (u r r x ) x (u u u x ) x (u r ) x r x u r r xx (u u ) x u x u u u xx (u rr r x u ru u x )r x u r r xx (u ur r x u uu u x )u x u u u xx. Also, by differentition of r nd u we find r x x x y x r, u x 1 1 ( y>x) y y b x r.
12 586 CHAP. 1 Prtil Differentil Equtions (PDEs) Differentiting these two formuls gin, we obtin r xx r xr x r 1 r x y 3 r r 3, u xx y r 3b r x xy r. We substitute ll these expressions into (). Assuming continuity of the first nd second prtil derivtives, we hve u ru u ur, nd by simplifying, (3) u xx x r u xy rr r u 3 ru y r u uu y r u xy 3 r r u u. In similr fshion it follows tht () u yy y r u xy rr r u 3 ru x r u uu x r u xy 3 r r u u. By dding (3) nd () we see tht the Lplcin of u in polr coordintes is (5) u u r 1 r u r 1 u r u. Circulr Membrne Circulr membrnes re importnt prts of drums, pumps, microphones, telephones, nd other devices. This ccounts for their gret importnce in engineering. Whenever circulr membrne is plne nd its mteril is elstic, but offers no resistnce to bending (this excludes thin metllic membrnes!), its vibrtions re modeled by the two-dimensionl wve eqution in polr coordintes obtined from (1) with u given by (5), tht is, y (6) u t c u r 1 r u r 1 u r u b c T r. R x Fig. 37. Circulr membrne We shll consider membrne of rdius R (Fig. 37) nd determine solutions u(r, t) tht re rdilly symmetric. (Solutions lso depending on the ngle u will be discussed in the problem set.) Then u uu in (6) nd the model of the problem (the nlog of (1), (), (3) in Sec. 1.9) is (7) (8) u (R, t) for ll t (9) (9b) u u t c r 1 r u (r, ) f (r) u t (r, ) g (r). u r b Here (8) mens tht the membrne is fixed long the boundry circle r R. The initil deflection f (r) nd the initil velocity g(r) depend only on r, not on u, so tht we cn expect rdilly symmetric solutions u(r, t).
13 SEC. 1.1 Lplcin in Polr Coordintes. Circulr Membrne. Fourier Bessel Series 587 Step 1. Two ODEs From the Wve Eqution (7). Bessel s Eqution Using the method of seprtion of vribles, we first determine solutions u(r, t) W (r)g(t). (We write W, not F becuse W depends on r, wheres F, used before, depended on x.) Substituting u WG nd its derivtives into (7) nd dividing the result by c WG, we get ## G c G 1 W Ws 1 r Wrb where dots denote derivtives with respect to t nd primes denote derivtives with respect to r. The expressions on both sides must equl constnt. This constnt must be negtive, sy, k, in order to obtin solutions tht stisfy the boundry condition without being identiclly zero. Thus, This gives the two liner ODEs G ## c G 1 W Ws 1 r Wrb k. (1) nd G ## l G where l ck (11) Ws 1 r Wr k W. We cn reduce (11) to Bessel s eqution (Sec. 5.) if we set s kr. Then 1>r k>s nd, retining the nottion W for simplicity, we obtin by the chin rule Wr dw dr dw ds ds dr dw ds k nd Ws d W ds k. By substituting this into (11) nd omitting the common fctor k we hve (1) d W ds 1 s dw ds W. This is Bessel s eqution (1), Sec. 5., with prmeter. Step. Stisfying the Boundry Condition (8) J Solutions of (1) re the Bessel functions nd Y of the first nd second kind (see Secs. 5., 5.5). But Y becomes infinite t, so tht we cnnot use it becuse the deflection of the membrne must lwys remin finite. This leves us with (13) W (r) J (s) J (kr) (s kr).
14 588 CHAP. 1 Prtil Differentil Equtions (PDEs) On the boundry r R we get W (R) J (kr) from (8) (becuse G would imply u ). We cn stisfy this condition becuse hs (infinitely mny) positive zeros, s 1,, Á J (see Fig. 38), with numericl vlues 1.8, 5.51, , , nd so on. (For further vlues, consult your CAS or Ref. [GenRef1] in App. 1.) These zeros re slightly irregulrly spced, s we see. Eqution (13) now implies (1) Hence the functions kr m thus k k m m R, m 1,, Á. (15) W m (r) J (k m r) J m R rb, m 1,, Á re solutions of (11) tht re zero on the boundry circle r R. W m Eigenfunctions nd Eigenvlues. For in (15), corresponding generl solution of (1) with l l m ck m c m >R is G m (t) A m cos l m t B m sin l m t. Hence the functions (16) u m (r, t) W m (r)g m (t) (A m cos l m t B m sin l m t)j (k m r) with m 1,, Á re solutions of the wve eqution (7) stisfying the boundry condition (8). These re the eigenfunctions of our problem. The corresponding eigenvlues re l m. The vibrtion of the membrne corresponding to u m is clled the mth norml mode; it hs the frequency l m >p cycles per unit time. Since the zeros of the Bessel function J re not regulrly spced on the xis (in contrst to the zeros of the sine functions ppering in the cse of the vibrting string), the sound of drum is entirely different from tht of violin. The forms of the norml modes cn esily be obtined from Fig. 38 nd re shown in Fig. 39. For m 1, ll the points of the membrne move up (or down) t the sme time. For m, the sitution is s follows. The function W (r) J ( r>r) is zero for r>r 1, thus r 1 R>. The circle r 1 R> is, therefore, nodl line, nd when t some instnt the centrl prt of the membrne moves up, the outer prt (r 1 R> ) moves down, nd conversely. The solution u m (r, t) hs m 1 nodl lines, which re circles (Fig. 39). J (s) α α 3 α α 1 Fig α 1 α α 3 α Bessel function J (s) s
15 SEC. 1.1 Lplcin in Polr Coordintes. Circulr Membrne. Fourier Bessel Series 589 Fig. 39. m = 1 m = m = 3 Norml modes of the circulr membrne in the cse of vibrtions independent of the ngle Step 3. Solution of the Entire Problem To obtin solution u (r, t) tht lso stisfies the initil conditions (9), we my proceed s in the cse of the string. Tht is, we consider the series (17) u (r, t) W m (r)g m (t) (A m cos l m t B m sin l m t)j m R rb m1 m1 (leving side the problems of convergence nd uniqueness). Setting t nd using (9), we obtin (18) u (r, ) m1 A m J m rb f (r). R Thus for the series (17) to stisfy the condition (9), the constnts must be the coefficients of the Fourier Bessel series (18) tht represents f (r) in terms of J ( m r>r); tht is [see (9) in Sec with n,, m m, nd x r], A m (19) A m R rf (r)j R J m rb dr 1 ( m ) R (m 1,, Á ). Differentibility of f (r) in the intervl r R is sufficient for the existence of the development (18); see Ref. [A13]. The coefficients B m in (17) cn be determined from (9b) in similr fshion. Numeric vlues of A m nd B m my be obtined from CAS or by numeric integrtion method, using tbles of J nd J 1. However, numeric integrtion cn sometimes be voided, s the following exmple shows.
16 59 CHAP. 1 Prtil Differentil Equtions (PDEs) E X A M P L E 1 Vibrtions of Circulr Membrne Find the vibrtions of circulr drumhed of rdius 1 ft nd density slugs>ft if the tension is 8 lb>ft, the initil velocity is, nd the initil displcement is f (r) 1 r [ft]. Solution. c T>r 8 [ft >sec ]. Also B m, since the initil velocity is. From (1) in Sec. 11.6, since R 1, we obtin A m 1 r (1 r )J ( m r) dr J 1 ( m ) J ( m) m J 1 ( m ) 8 3 m J 1 ( m ) where the lst equlity follows from (1c), Sec. 5., with 1, tht is, J ( m ) m J 1 ( m ) J ( m ) m J 1 ( m ). Tble 9.5 on p. 9 of [GenRef1] gives m nd J r ( m ). From this we get J 1 ( m ) J r( m ) by (1b), Sec. 5., with, nd compute the coefficients A m : m m J 1 ( m ) J ( m ) A m Thus f (r) 1.18J (.8r).1J (5.51r).5J (8.6537r) Á. We see tht the coefficients decrese reltively slowly. The sum of the explicitly given coefficients in the tble is The sum of ll the coefficients should be 1. (Why?) Hence by the Leibniz test in App. A3.3 the prtil sum of those terms gives bout three correct decimls of the mplitude f(r). Since l m ck m c m >R m, from (17) we thus obtin the solution (with r mesured in feet nd t in seconds) u (r, t) 1.18J (.8r) cos.897t.1j (5.51r) cos 11.t.5J (8.6537r) cos t Á. In Fig. 39, m 1 gives n ide of the motion of the first term of our series, m of the second term, nd m 3 of the third term, so tht we cn see our result bout s well s for violin string in Sec. 1.3.
17 SEC. 1.1 Lplcin in Polr Coordintes. Circulr Membrne. Fourier Bessel Series 591 P R O B L E M S E T RADIAL SYMMETRY 1. Why did we introduce polr coordintes in this section?. Rdil symmetry reduces (5) to u u rr u r >r. Derive this directly from u u xx u yy. Show tht the only solution of u depending only on r x y is u ln r b with rbitrry constnts nd b. 3. Alterntive form of (5). Show tht (5) cn be written u (ru r ) r >r u uu >r, form tht is often prcticl. BOUNDARY VALUE PROBLEMS. SERIES. TEAM PROJECT. Series for Dirichlet nd Neumnn Problems () Show tht 1, Á u n r n cos nu, u n r n sin nu, n,, re solutions of Lplce s eqution u with u given by (5). (Wht would u n be in Crtesin coordintes? Experiment with smll n.) (b) Dirichlet problem (See Sec. 1.6) Assuming tht termwise differentition is permissible, show tht solution of the Lplce eqution in the disk r R stisfying the boundry condition u(r, u) f (u) (R nd f given) is () u(r, u) b n r R b nsin nu d where n, b n re the Fourier coefficients of f (see Sec. 11.1). (c) Dirichlet problem. Solve the Dirichlet problem using () if R 1 nd the boundry vlues re u (u) 1 volts if p u, u (u) 1 volts if u p. (Sketch this disk, indicte the boundry vlues.) (d) Neumnn problem. Show tht the solution of the Neumnn problem u if r R, u N (R, u) f (u) (where u N u>n is the directionl derivtive in the direction of the outer norml) is u(r, u) A r n (A n cos nu B n sin nu) n1 n1 c n r R b ncos nu with rbitrry A n B n A nd (e) Comptibility condition. Show tht (9), Sec. 1., imposes on f (u) in (d) the comptibility condition (f) Neumnn problem. Solve u in the nnulus 1 r if u r (1, u) sin u, u r (, u). 5 8 ELECTROSTATIC POTENTIAL. STEADY-STATE HEAT PROBLEMS The electrosttic potentil stisfies Lplce s eqution u in ny region free of chrges. Also the het eqution u t c u (Sec. 1.5) reduces to Lplce s eqution if the temperture u is time-independent ( stedy-stte cse ). Using (), find the potentil (equivlently: the stedy-stte temperture) in the disk r 1 if the boundry vlues re (sketch them, to see wht is going on). 5. u (1, u) if 1 p u 1 p nd otherwise 6. u (1, u) cos 3 u 7. u (1, u) 11 ƒ u ƒ if p u p 8. u (1, u) u if 1 p u 1 p nd otherwise 9. CAS EXPERIMENT. Equipotentil Lines. Guess wht the equipotentil lines u (r, u) const in Probs. 5 nd 7 my look like. Then grph some of them, using prtil sums of the series. 1. Semidisk. Find the electrosttic potentil in the semidisk r 1, u p which equls 11u (p u) on the semicircle r 1 nd on the segment 1 x Semidisk. Find the stedy-stte temperture in semicirculr thin plte r, u p with the semicircle r kept t constnt temperture u nd the segment x t. CIRCULAR MEMBRANE 1 pnr n1 1 pnr n1 p f (u) cos nu du, p p f (u) sin nu du. p p f (u) du. p 1. CAS PROJECT. Norml Modes. () Grph the norml modes s in Fig. 36. u, u 5, u 6
18 59 CHAP. 1 Prtil Differentil Equtions (PDEs) (b) Write progrm for clculting the A m s in Exmple 1 nd extend the tble to m 15. Verify () F rr 1 r F r 1 r F uu k F. numericlly tht m (m 1 )p nd compute the error for m 1, Á, 1. (c) Grph the initil deflection f (r) in Exmple 1 s Show tht the PDE cn now be seprted by substituting F W (r)q(u), giving well s the first three prtil sums of the series. Comment on ccurcy. (d) Compute the rdii of the nodl lines of u, u 3, u when R 1. How do these vlues compre to those of (5) (6) Qs n Q, r Ws rwr (k r n )W. the nodes of the vibrting string of length 1? Cn you estblish ny empiricl lws by experimenttion with Periodicity. Show tht Q (u) must be periodic with further u? period p nd, therefore, n, 1,, Á in (5) nd m (6). Show tht this yields the solutions 13. Frequency. Wht hppens to the frequency of n Q eigenfunction of drum if you double the tension? n * sin nu, W n J n (kr), n, 1, Á Q n cos nu,. 1. Size of drum. A smll drum should hve higher 1. Boundry condition. Show tht the boundry condition fundmentl frequency thn lrge one, tension nd density being the sme. How does this follow from our (7) u (R, u, t) formuls? 15. Tension. Find formul for the tension required leds to k k mn mn >R, where s nm is the mth positive zero of J n (s). to produce desired fundmentl frequency f 1 of. Solutions depending on both r nd U. Show tht drum. solutions of () stisfying (7) re (see Fig. 31) 16. Why is A 1 A Á 1 in Exmple 1? Compute the first few prtil sums until you get 3-digit ccurcy. Wht does this problem men in the field u nm (A nm cos ck nm t B nm sin ck nm t) J n (k nm r) cos nu of music? (8) u* 17. Nodl lines. Is it possible tht for fixed c nd R two nm (A* nm cos ck nm t B* nm sin ck nm t) or more u m [see (16)] with different nodl lines J n (k nm r) sin nu correspond to the sme eigenvlue? (Give reson.) 18. Nonzero initil velocity is more of theoreticl interest becuse it is difficult to obtin experimentlly. Show tht for (17) to stisfy (9b) we must hve (1) B m K m R rg (r)j ( m r>r) dr u 11 u 1 u 3 where K m >(c m R)J 1 ( m ). Fig. 31. Nodl lines of some of the solutions (8) VIBRATIONS OF A CIRCULAR MEMBRANE DEPENDING ON BOTH r AND U 19. (Seprtions) Show tht substitution of into the wve eqution (6), tht is, () u tt c u rr 1 r u r 1 r u uu b, gives n ODE nd PDE (3) G # # l G, where l ck, u F (r, u)g (t) 3. Initil condition. Show tht u t (r, u, ) gives B nm, B* nm in (8).. Show tht u* m nd u m is identicl with (16) in this section. 5. Semicirculr membrne. Show tht u 11 represents the fundmentl mode of semicirculr membrne nd find the corresponding frequency when c 1 nd R 1.
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