PartialDifferentialEquations

Transcription

1 C H A P T E R 13 PrtilDifferentilEqutions 13.1 Derivtion of the Het Eqution 75 We will now consider differentil equtions tht model chnge where there is more thn one independent vrible. For exmple, the temperture in n object chnges with time nd with the position within the object. The rtes of chnge led to prtil derivtives, nd the equtions relting them re clled prtil differentil equtions. The pplictions of the subject re mny, nd the types of equtions tht rise hve gret del of vriety. We will limit our study to the equtions tht rise most frequently in pplictions. These model het flow nd simple wves. The differentil eqution models for het flow nd the vibrting string will be derived in Sections 1 nd 3, where we will lso describe some of their properties. We will then systemticlly study ech of the equtions, solving them in some cses using the method of seprtion of vribles. Het is form of energy tht exists in ny mteril. ike ny other form of energy, het is mesured in joules (1 J 1 Nm). However, it is lso mesured in clories (1 cl J) or sometimes in British therml units (1 BTU 252 cl 1.54 kj). The mount of het within given volume is defined only up to n dditive constnt. We will ssume the convention of sying tht the mount of het is equl to when the temperture is equl to. Suppose tht ½V is smll volume in which the temperture u is lmost constnt. It hs been found experimentlly tht the mount of het ½Q in ½V is proportionl to the temperture u nd to the mss ½m ½V, where is the mss density of the mteril. Thus the mount of het in ½V is given by ½Q cu½v (1.1) The new constnt c is clled the specific het. It mesures the mount of het required

2 13.1 Derivtion of the Het Eqution 751 to rise 1 unit of mss of the mteril 1 degree of temperture. We will usully use the Celsius or Kelvin scles for temperture. et s consider thin rod tht is insulted long its length, s seen in Figure 1. If the length of the rod is, the position long the rod is given by x, where x Since the rod is insulted, there is no trnsfer of het from the rod except t its two ends. We my therefore ssume tht the temperture u depends only on x nd on the time t. x x Figure 1 The vrition of temperture in n insulted rod. Consider smll section of the rod between x nd x ½x. et S be the crosssectionl re of the rod. The volume of the section is S½x, so (1.1) becomes ½Q cus½x Therefore, the mount of het t time t in the portion U of the rod defined by x b is given by the integrl Qºt» S b cuºt x» dx (1.2) The specific het nd the density sometimes vry from point to point nd more rrely with time s well. However, we will usully be deling with homogeneous, time independent mterils for which both the specific het nd the density re constnts. The het eqution models the flow of het through the mteril. It is derived by computing the time rte of chnge of Q in two different wys. The first wy is to differentite (1.2). Differentiting under the integrl sign, we get d Q dt d b dt S cu dx S b ÌcuÍ dx t Of course, if the specific het nd the density do not vry with time, this becomes d Q dt S b c u t dx (1.3) The second wy to compute the time rte of chnge of Q is to notice tht, in the bsence of het sources within the rod, the quntity of het in U cn chnge only through the flow of het cross the boundries of U t x nd x b. The rte of het flow through section of the rod is clled the het flux through the section. Consider the section of the rod between x nd ½x. Experimentl study of het conduction revels tht the flow of het cross such section hs the following properties:

3 752 Chpter 13 Prtil Differentil Equtions Het flows from hot positions to cold positions t rte proportionl to the difference in the tempertures on the two sides of the section. Thus the het flux through the section is proportionl to uº ½x t» uº t». The het flux through the section is inversely proportionl to ½x, the width of the section. The het flux through the section is proportionl to the re of S of the boundry of the section. Putting these three points together, we see tht there is coefficient C such tht the het flux into U t x is given pproximtely by CS uº ½x t» uº t» (1.4) ½x The coefficient C is clled the therml conductivity. It is positive since, if uº ½x t» uº t», then the temperture is hotter inside U thn it is outside, nd the het flows out of U t x The therml conductivity is usully constnt, but it my depend on the temperture u nd the position x. If we let ½x go to in (1.4), the difference quotient pproches ux nd we see tht the het flux into U t x is CS u º t» (1.5) x The sme rgument t x b shows tht the het flux into U t x b is CS u ºb t» (1.6) x The totl time rte of chnge of Q is the sum of the rtes t the two ends. Using the fundmentl theorem of clculus, this is d Q dt S C u x ºb t» C u x º t» S b If the therml conductivity C is independent of x, this becomes d Q dt CS b x C u x dx (1.7) 2 u dx (1.8) x 2 In equtions (1.3) nd (1.8) we hve two formuls for the rte of het flow into U. Setting them equl, we see tht b u b c t dx C 2 u b x dx or u 2 c C 2 u dx t x 2 This is true for ll b, which cn be true only if the integrnd is equl to. Hence, c u t C 2 u x 2

4 Tble 1 Therml diffusivities of common mterils 13.1 Derivtion of the Het Eqution 753 Mteril k (cm 2 /sec) Mteril k (cm 2 /sec) Aluminum.84 Gold 1.18 Brick.57 Grnite.8.18 Cst iron.17 Ice.14 Copper 1.12 PVC.8 Concrete.4.8 Silver 1.7 Glss.43 Wter.14 throughout the mteril. If we divide by c, nd set k Cc, the eqution becomes u t k 2 u x 2 or u t k 2 u x 2 (1.9) The constnt k is clled the therml diffusivity of the mteril. The units of k re (length) 2 /time. The vlues of k for some common mterils re listed in Tble 1. Eqution (1.9) is clled the het eqution. As we hve shown, it models the flow of het through mteril nd is stisfied by the temperture. It should be noticed tht if we hve wll with height nd width tht re lrge in comprison to the thickness, then the temperture in the wll wy from its ends will depend only on the position within the wll. Consequently we hve one dimensionl problem, nd the vrition of the temperture is modeled by the het eqution in (1.9) A similr derivtion shows tht the diffusion of substnce through liquid or gs stisfies the sme eqution. In this cse it is the concentrtion u tht stisfies the eqution. For this reson eqution (1.9) is lso referred to s the diffusion eqution. Subscript nottion for derivtives We will find it useful to bbrevite prtil derivtives by using subscripts to indicte the vrible of differentition. For exmple, we will write u x u x u y u y u yx 2 u xy nd u x x 2 u x 2 Using this nottion, we cn write the het eqution in (1.9) quite succinctly s u t ku x x The inhomogeneous het eqution Eqution (1.9) ws derived under the ssumption tht there is no source of het within the mteril. If there re het sources, we cn modify the model to ccommodte them. If we look bck to eqution (1.7), which ccounts for the rte of flow of het into U, we see tht we must modify the right-hnd side to ccount for internl sources. We will ssume tht the het source is spred throughout the mteril nd tht het is being dded t the rte of pºu x t» therml units per unit volume per second. Notice tht we llow the rte of het inflow to depend on the temperture u, s well s on x nd t. An exmple would be rod tht is not completely insulted long its length. Then het would flow into or out of the rod long its length t rte

5 754 Chpter 13 Prtil Differentil Equtions tht is proportionl to the difference between the temperture in U nd the mbient temperture, so pºu x t» «Ìu T Í where T is the mbient temperture. Assuming there is source of het, eqution (1.7) becomes d Q dt CS u u ºb t» x x º t» S b pºu x t» dx The rest of the derivtion is unchnged, nd in the end we get c u t C 2 u x 2 p or u t k 2 u x 2 p c (1.1) Becuse of the term involving p, eqution (1.1) is clled the inhomogeneous het eqution, while eqution (1.9) is clled the homogeneous het eqution. Initil conditions We hve seen tht ordinry differentil equtions hve mny solutions, nd to determine prticulr solution we specify initil conditions. The sitution is more complicted for prtil differentil equtions. For exmple, specifying initil conditions for temperture requires giving the temperture t ech point in the mteril t the initil time. In the cse of the rod this mens tht we give function f ºx» defined for x nd we look for solution to the het eqution tht lso stisfies uºx» f ºx» for x (1.11) Types of boundry conditions In ddition to specifying the initil temperture, it will be necessry to specify conditions on the boundry of the mteril. For exmple, the temperture my be fixed t one endpoint of the rod s the result of the mteril being embedded in source of het kept t constnt temperture. The tempertures might well be different t the two ends of the rod. Thus if the temperture t x is T nd tht t x is T, then the temperture uºx t» stisfies uº t» T nd uº t» T for ll t. (1.12) Boundry conditions of the form in (1.12) specifying the vlue of the temperture t the boundry re clled Dirichlet conditions. In other circumstnces one or both ends of the rod might be insulted. This mens tht there is no flow of het into or out of the rod t these points. According to the discussion leding to eqution (1.5), this mens tht u x (1.13) t n insulted point. This type of boundry condition is clled the Neumnn condition. A rod could stisfy Dirichlet condition t one boundry point nd Neumnn condition t the other. There is third condition tht occurs, for exmple, when one end of the rod is poorly insulted from the exterior. According to Newton s lw of cooling, the flow

6 13.1 Derivtion of the Het Eqution 755 of het cross the insultion is proportionl to the difference in the tempertures on the two sides of the insultion. If this is true t the endpoint x, then rguing long the sme lines s we did in the derivtion of eqution (1.5), we see tht there is positive number «such tht u º t» «ºuº t» T» (1.14) x where T is the temperture outside the insultion nd u is the temperture t the endpoint x. Poor insultion t the endpoint x leds in the sme wy to boundry condition of the form u º t» ºuº t» T» (1.15) x where. Boundry conditions of the type in (1.14) nd (1.15) re clled Robin conditions. Robin boundry conditions lso rise when solid wll meets fluid or gs. In such cse thin boundry lyer is formed, which shields the rest of the fluid or gs from the temperture in the wll. The constnt is sometimes clled the het trnsfer coefficient. Initil/boundry vlue problems Putting everything together, we see tht the temperture uºx t» in n insulted rod with Dirichlet boundry conditions must stisfy the het eqution together with initil nd boundry conditions. The complete problem is to find function uºx t» such tht u t ºx t» ku x x for x nd t uº t» T nd uº t» T for t uºx» f ºx» for x. (1.16) The function f ºx» is the initil temperture distribution. The initil/boundry vlue problem is illustrted in Figure 2. As we hve indicted, the Dirichlet boundry condition t ech endpoint in (1.16) could be replced with Neumnn or Robin condition. The mximum principle. One of the mjor tenets of the theory of het flow is tht het flows from hot res to colder res. From this strting point, physicl resoning llows us to conclude tht the temperture uºt x» cnnot get too hot or too cold in the region where it stisfies the het eqution. To be precise, let m min x f ºx» nd M mx x f ºx» Then, if uºt x» is solution to the initil/boundry vlue problem in (1.16), minm T T uºt x» mxm T T for t nd x.

7 756 Chpter 13 Prtil Differentil Equtions t u(t,) = T u t = ku xx u(t,) = T u(x,) = f(x) x Figure 2 The initil/boundry vlue problem for the het eqution. This result is clled the mximum principle for the het eqution. In English it sys tht temperture uºt x» defined for t nd x must chieve its mximum vlue (nd its minimum vlue) on the boundry of the region where it is defined. Thus in Figure 2 The temperture uºt x» in the indicted hlf-strip must chieve its mximum nd minimum vlues on the three lines which forms its boundry. inerity If u nd Ú re functions nd «nd re constnts, then x º«u Ú» «u x Ú x (1.17) We will express this stndrd fct bout x by sying tht it is liner opertor. It is n opertor becuse it opertes on function u nd yields nother function ux. Tht it is liner simply mens tht (1.17) is stisfied. It follows esily tht more complicted differentil opertors, such s 2 x 2 nd 2 xy re lso liner. It then follows tht the het opertor t k 2 x 2 is liner. This implies the following theorem.

8 13.2 Seprtion of Vribles for the Het Eqution 757 THEOREM 1.18 The homogeneous het eqution is liner eqution, mening tht if u nd Ú stisfy u t ku x x nd Ú t kú x x nd «nd re constnts, then the liner combintion Û «u Ú stisfies Û t kû x x, so Û is lso solution to the homogeneous het eqution. We will mke frequent use of Theorem It will enble us to build up more complicted solutions s liner combintions of bsic solutions.... EXERCISES 1. Suppose tht the temperture t ech point of rod of length is originlly t 15 Æ. Suppose tht strting t time t the left end is kept t 5 Æ nd the right end t 25 Æ Write down the complete description of the initil/boundry vlue problem the temperture in the rod must obey. 2. Show tht the temperture in the rod in Exercise 1 must stisfy 5 uºx t» 25 for t nd x 3. Suppose the specific het, density, nd therml conductivity depend on x, nd re not constnt. Show tht the het eqution becomes t ÌcuÍ x C u x 4. If our rod is insulted t both ends, we would expect tht the totl mount of het in the rod does not chnge with time. Show tht this follows from eqution (1.7). 5. Prove Theorem 1.18 by showing tht Û t kû x x 6. Solutions to the Dirichlet problem in (1.16) re unique. This mens tht if both u nd Ú stisfy the conditions in (1.16), then uºx t» Úºx t» for t nd x Use the linerity of the het eqution nd the mximum principle to prove this fct. 7. Suppose we hve n insulted luminum rod of length. Suppose the rod is t constnt temperture of 15 Æ K, nd tht strting t time t, the left-hnd end point is kept t 2 Æ K nd the right-hnd endpoint is kept t 35 Æ K. Provide the initil/boundry vlue problem tht must be stisfied by the temperture uºt x». 8. Suppose we hve n insulted gold rod of length. Suppose the rod is t constnt temperture of 15 Æ K, nd tht strting t time t, the left-hnd end point is kept t 2 Æ K while the right-hnd endpoint is kept insulted. Provide the initil/boundry vlue problem tht must be stisfied by the temperture uºt x». 9. Suppose we hve n insulted silver rod of length. Suppose the rod is t constnt temperture of 15 Æ K, nd tht strting t time t, the righthnd end point is kept t 35 Æ K while the left-hnd endpoint is only prtilly insulted, so het is lost there t rte equl to 13 times the difference between the temperture of the rod t this point nd the mbient temperture T 15 Æ K. Provide the initil/boundry vlue problem tht must be stisfied by the temperture uºt x».

9 758 Chpter 13 Prtil Differentil Equtions 13.2 Seprtion of Vribles for the Het Eqution We will strt this section by solving the initil/boundry vlue problem tht we posed in (1.16). Stedy-stte tempertures u t ºx t» ku x x ºx t» for t nd x, (2.1) uº t» T nd uº t» T for t, (2.2) uºx» f ºx» for x (2.3) It is useful for both mthemticl nd physicl purposes to split the problem into two prts. We first find the stedy-stte temperture tht stisfies the boundry conditions in (2.2). A stedy-stte temperture is one tht does not depend on time. Then u t, so the het eqution (2.1) simplifies to u x x Hence we re looking for function u s ºx» defined for x such tht 2 u s ºx» 2 for x x u s º t» T nd u s º t» T for t (2.4) The solution to this boundry vlue problem is esily found, since the generl solution of the differentil eqution is u s ºx» Ax B where A nd B re rbitrry constnts. Then the boundry conditions reduce to u s º» B T nd u s º» A B T We conclude tht B T nd A ºT T» so the stedy-stte temperture is u s ºx» ºT T» x T It remins to find Ú u u s. It will be solution to the het eqution, since both u nd u s re, nd the het eqution is liner. The boundry nd initil conditions tht Ú stisfies cn be clculted from those for u nd u s in (2.2), (2.3), nd (2.4). Thus, Ú u u s must stisfy for x nd t Úº t» Úº t» for t Ú t ºx t» kú x x ºx t» Úºx» gºx» f ºx» u s ºx» for x. (2.5) The most importnt fct is tht the boundry conditions for Ú re Úº t» Úº t». When the right-hnd sides re equl to» we sy tht the boundry conditions re homogeneous. This will mke finding the solution lot esier. Hving found the stedy-stte temperture u s nd the temperture Ú, the solution to the originl problem is uºx t» u s ºx» Úºx t»

10 13.2 Seprtion of Vribles for the Het Eqution 759 Solution with homogeneous boundry conditions We will find the solution to the initil/boundry vlue problem with homogeneous boundry conditions in(2.5) using the technique of seprtion of vribles. It should be noted tht seprtion of vribles cn only be used to solve n initil/boundry vlue problem when the boundry conditions re homogeneous. Since this is the first time we re using the technique, nd since it is technique we will use throughout this chpter, we will go through the process slowly. The bsic ide of the method of seprtion of vribles is to hunt for solutions in the product form Úºx t» X ºx»T ºt» (2.6) where T ºt» is function of t nd X ºx» is function of x. We will insist tht the product solution Ú stisfies the homogeneous boundry conditions. Since Úº t» X º»T ºt» for ll t, we conclude tht X º». A similr rgument shows tht X º» This leds to two-point boundry vlue problem for X tht we will solve. In the end we will hve found enough solutions of the fctored form so tht we will be ble to solve the initil/boundry vlue problem in (2.5) using n infinite liner combintion of them. There re three steps to the method. Step 1: Seprte the PDE into two ODEs. When we insert Ú X ºx»T ºt» into the het eqution Ú t kú x x, we get X ºx»T ¼ ºt» kx ¼¼ ºx»T ºt» (2.7) The key step is to seprte the vribles by bringing everything depending on t to the left, nd everything depending on x to the right. Dividing (2.7) by kxºx»t ºt», we get T ¼ ºt» kt X ¼¼ ºx» ºt» X ºx» Since x nd t re independent vribles, the only wy tht the left-hnd side, function of t, cn equl the right-hnd side, function of x, is if both functions re constnt. Consequently, there is constnt tht we will write s such tht or T ¼ ºt» kt ºt» nd X ¼¼ ºx» X ºx» The first eqution hs the generl solution T ¼ kt nd X ¼¼ X (2.8) T ºt» Ce kt (2.9) We hve to work little hrder on the second eqution. Step 2: Set up nd solve the two-point boundry vlue problem. Since we insist tht the solution X stisfies the homogeneous boundry conditions, the complete problem to be solved in finding X is X ¼¼ X with X º» X º» (2.1)

11 76 Chpter 13 Prtil Differentil Equtions Notice tht the problem in (2.1) is not the stndrd initil vlue problem we hve been solving up to now. There re two conditions imposed, but insted of both being imposed t the initil point x, there is one condition imposed t ech endpoint of the intervl. Accordingly, this is clled two-point boundry vlue problem. It is lso clled Sturm iouville problem. 1 Another point to be mde is tht the constnt is still undetermined. Furthermore, s we will see, for most vlues of the only solution to (2.1) is the function tht is identiclly. Solving Sturm iouville problem mounts to finding the numbers for which there re nonzero solutions to (2.1). DEFINITION 2.11 A number is clled n eigenvlue for the Sturm iouville problem in (2.1) if there is nonzero function X tht solves (2.1). If is n eigenvlue, then ny function tht stisfies (2.1) is clled n eigenfunction. 2 The solution to Sturm iouville problem like (2.1) is the list of its eigenvlues nd eigenfunctions. Notice tht becuse of the linerity of the differentil eqution in (2.1), ny constnt multiple of n eigenfunction is lso n eigenfunction. We will usully choose the constnt tht leds to the lest complicted form for the eigenfunction. et s return to the exmple in (2.1). We will first show tht there re no negtive eigenvlues. To see this, set r 2 where r The eqution in (2.1) becomes X ¼¼ r 2 X which hs generl solution X ºx» C 1 e rx C 2 e rx The boundry conditions re X º» C 1 C 2 X º» C 1 e r C 2 e r From the first eqution, C 2 C 1. Inserting this into the second eqution, we get C 1 ºe r e r» Since r the fctor in prenthesis on the right is nonzero. Hence C 1 which in turn implies tht C 2 so the only solution is X ºx». This mens tht is not n eigenvlue. 3 This rgument cn be repeted if. In this cse the differentil eqution becomes X ¼¼, which hs the generl solution X ºx» x b where nd b re constnts. The boundry conditions become X º» b nd X º» b from which we esily conclude tht b. 1 This is our first exmple of Sturm iouville problem. We will study them in some detil in Sections 6 nd 7. 2 You will observe tht finding the eigenvlues nd eigenfunctions of Sturm iouville problem is similr in mny wys to finding the eigenvlues nd eigenvectors of mtrix. It might be useful to compre the sitution here with Section This grees with our physicl intuition bout het flow. If there were solution X with, then ccording to (2.6) nd (2.9), the product solution to the het eqution would be Úºx t» e kt X ºx» If this solution would grow exponentilly in mgnitude s t increses. In fct, we notice experimentlly tht tempertures tend to remin stble over time in the bsence of het sources.

12 13.2 Seprtion of Vribles for the Het Eqution 761 Next suppose tht nd set 2 Then the differentil eqution in (2.1) is X ¼¼ 2 X which hs the generl solution X ºx» cos x b sin x For this solution the boundry condition X º» becomes. Then the boundry condition X º» becomes b sin We re only interested in nonzero solutions, so we must hve sin This occurs if n for some positive integer n. When this is true we hve the eigenvlue 2 n For ny nonzero constnt b, X ºx» b sinºn x» is n eigenfunction. The simplest thing to do is to set b 1 In summry, the eigenvlues nd eigenfunctions for the Sturm iouville problem in (2.1) re n x n n2 2 2 nd X n ºx» sin for n (2.12) Finlly, by incorporting (2.9) nd (2.12), we get the product solutions, Ú n ºx t» e n2 2 kt 2 n x sin for n (2.13) to the het eqution, tht lso stisfy the boundry condition Ú n º t» Ú n º t». Step 3: Stisfying the initil condition. Hving found infinitely mny product solutions in (2.13), we cn use the linerity of the het eqution (see Theorem 1.18) to conclude tht ny finite liner combintion of them is lso solution. Hence, if b n is constnt for ech n, then for ny N the function Úºx t» N n1 b n Ú n ºx t» N n1 b n e n2 2 kt 2 n x sin is solution to the het eqution tht stisfies the homogeneous boundry conditions. We re nturlly led to consider the infinite series ½ ½ Úºx t» b n Ú n ºx t» b n e n2 2 kt 2 n x sin (2.14) n1 n1 We will ssume tht the coefficients b n re such tht this series converges, nd tht the resulting function Ú stisfies the het eqution nd the homogeneous boundry conditions. These fcts re true formlly. 4 They re lso true in the cses tht we will consider, but we will not verify this. To do so requires some lengthy mthemticl rguments tht would not significntly dd to our understnding of the issue. 4 Formlly mens tht we ignore the mthemticl niceties of verifying tht we cn differentite the function Ú by differentiting the terms in the infinite series.

13 762 Chpter 13 Prtil Differentil Equtions Referring bck to our originl initil/boundry vlue problem in (2.5), we see tht the function Ú defined in (2.14) stisfies everything except the initil condition Úºx» gºx» f ºx» u s ºx» However, we hve yet to determine the coefficients b n. Using the series definition for Ú in (2.14), the initil condition becomes ½ n x gºx» Úºx» b n sin for x (2.15) n1 Eqution (2.15) will be recognized s the Fourier sine expnsion for the initil temperture g. According to Section 12.3, nd in prticulr eqution (3.7), the vlues of b n re given by b n 2 n x gºx» sin dx (2.16) Substituting these vlues into (2.14) gives complete solution to the homogeneous initil/boundry vlue problem in (2.5). As indicted previously, the function uºx t» u s ºx» Úºx t» stisfies the originl initil/boundry vlue problem in equtions (2.1), (2.2), nd (2.3). EXAMPE 2.17 Suppose rod of length 1 meter (1cm) is originlly t Æ C. Strting t time t, one end is kept t the constnt temperture of 1 Æ C, while the other is kept t Æ C. Find the temperture distribution in the rod s function of time nd position. Assume tht the therml diffusivity of the rod is k 1 cm 2 sec. If we use the meter s the unit of length, then k 1 m 2 /sec. The temperture in the rod, uºx t», must solve the initil/boundry vlue problem u t ºx t» 1 u x x ºx t» for t nd x 1, uº t» nd uº1 t» 1 for t, uºx» for x 1. (2.18) Following the discussion t the beginning of this section, we write the temperture distribution s u u s Ú, where u s ºx» is the stedy-stte temperture with the sme boundry conditions s u, nd Ú is temperture with homogeneous boundry conditions, nd the sme initil condition s u u s. The stedy-stte temperture u s must stisfy u ¼¼ s with u s º» nd u s º1» 1 We esily see tht u s ºx» 1 x Then the temperture Ú u u s must stisfy Ú t ºx t» 1Ú x x ºx t» for t nd x 1, Úº t» nd Úº1 t» for t, Úºx» 1 x for x 1. (2.19) The boundry vlues re homogeneous, so we cn use the formul for the solution in (2.14), with k 1 nd 1, to get Úºx t» ½ n1 b n e 1n2 2t sin n x (2.2)

14 13.2 Seprtion of Vribles for the Het Eqution 763 The coefficients re determined by the initil condition. Setting t in (2.2) nd using Úºx» 1 x, we obtin 1 x ½ n1 b n sin n x Therefore, the b n re the Fourier sine coefficients of 1 x on the intervl º 1», which by (2.16) re Thus, b n 2 1 º 1 x» sin n x dx 2 Úºx t» 2 ½ n1 Finlly, the temperture in the rod is º 1» n uºx t» u s ºx» Úºx t» 1 x 2 n 1 x sin n x dx º 1» n 2 n e 1n2 2t sin n x ½ n1 º 1» n n e 1 n2 2t sin n x (2.21) u 1 5 x.5 1 Figure 1 The temperture in the rod in Exmple The temperture is plotted in Figure 1. The initil temperture is uºx» Æ C. The stedy-stte temperture is plotted in blue. The blck curves represent the temperture distribution fter 2 second intervls. Notice how the temperture increses with time throughout the rod to the stedy-stte temperture. Het flows from hot to cold, so to mintin the new temperture of 1 Æ C t the right endpoint, het must flow into the rod t this point. It then flows through the rod, rising the temperture in the process. Some het hs to flow out of the rod t the left endpoint to mintin the temperture there. Eventully the rod reches stedy stte, t which point s much het flows out of the rod t x s flows in t x 1.

15 764 Chpter 13 Prtil Differentil Equtions The rte of convergence The generl term in the infinite series in eqution (2.14) is b n e n2 2 kt 2 n x sin (2.22) Since the sine function is bounded in bsolute vlue by 1, this term is bounded by b n e n2 2 kt 2 By the Riemnn-ebesgue lemm (see Theorem 2.1 in Section 12.2), the Fourier coefficient b n s n ½. On the other hnd, the exponentil term e n2 2 kt 2 extremely rpidly s n ½, t lest if the product kt is reltively lrge. As result the series in eqution (2.14) converges rpidly for lrge vlues of the time t. The result is tht the sum of the series in (2.14) cn be ccurtely pproximted by using reltively few of the terms of the infinite series. Sometimes one term is enough. EXAMPE 2.23 For the rod in Exmple 2.17 how mny terms of series in (2.21) re needed to pproximte the solution within one degree for t 1 1 nd 1. Estimte how long will it tke before the het in the rod is everywhere within 5 Æ of the stedystte temperture? The generl term in the series in (2.21) is bounded by 2e 1n2 2 t n We will estimte the error by computing the first omitted term. 5 Thus we wnt to find the smllest integer n for which 2e 1ºn1»2 2 t ̺n 1»Í 1 Since we cnnot solve this inequlity for n, we compute the left-hnd side for vlues of n nd t until we get the correct vlues. For t 1 we discover tht we need 12 terms, while for t 1 we need 5, nd for t 1 one term will suffice. For the temperture of the rod to be within 5 Æ of the stedy-stte temperture, we will certinly need the first term in the infinite series in (2.21) to be less thn 5. If we solve 2e 12 t 5 we obtin t sec. We compute tht for t 2 578, the second term in the series is bout 12, so t sec is good estimte. However, in view of the fct tht we re ignoring terms, nd n estimte is not expected to hve four plce ccurcy, 2 6 sec might be preferble, nd since 2 58 sec is 43 minutes, tht might be even better. Insulted boundry points As mentioned in Section 1, if the boundry points of the rod re insulted, there is no flow of het through the endpoints of the rod, nd the correct boundry conditions re the Neumnn conditions u x º t» u x º t» The initil/boundry vlue problem to be solved is now u t ºx t» ku x x ºx t» for t nd x, u x º t» nd u x º t» for t, uºx» f ºx» for x. (2.24) 5 This is rough estimte nd is not usully good ide. It is justified in this cse becuse the terms re decresing so rpidly.

16 13.2 Seprtion of Vribles for the Het Eqution 765 We will use the method of seprtion of vribles gin, strting by looking for product solutions uºx t» X ºx»T ºt» Notice tht since the Neumnn boundry conditions re homogeneous, it is not necessry to find the stedy-stte solution first. Step 1: Seprte the PDE into two ODEs. This first step is unchnged. The product uºx t» X ºx»T ºt» is solution only if the fctors stisfy the differentil equtions T ¼ kt nd X ¼¼ X (2.25) where is constnt. The first eqution hs the generl solution T ºt» Ce kt (2.26) Step 2: Set up nd solve the two-point boundry vlue problem. We will gin insist tht the product solution stisfy the boundry conditions. Since u t º t» X ¼ º»T ºt» for ll t, we must hve X ¼ º». A similr rgument shows tht X ¼ º», so we wnt to solve X ¼¼ X with X ¼ º» X ¼ º» (2.27) This is the two-point or Sturm iouville boundry vlue problem for the Neumnn problem. As before, we find tht there re no negtive eigenvlues. If the differentil eqution in (2.27) becomes X ¼¼, which hs the generl solution X ºx» x b. The first boundry condition is X ¼ º», leving us with the constnt function X ºx» b. This function lso stisfies the second boundry X ¼ º», so is n eigenvlue. We will choose the simplest nonzero constnt b 1 nd set X ºx» 1 The corresponding function in (2.26) is T C, which is lso constnt. Once more we choose C 1 so the resulting product solution to the het eqution is the constnt function u ºx t» X ºx»T ºt» 1 For, we set 2 where Then the differentil eqution in (2.27) is X ¼¼ 2 X which hs the generl solution X ºx» cos x b sin x The boundry condition X ¼ º» becomes b. Since, we hve b Then the boundry condition X ¼ º» becomes sin Since we re only interested in nonzero solutions, we must hve sin Therefore, n for some positive integer n. When this is true we hve 2 n nd X ºx» cosºn x» Agin cn be ny nonzero constnt, nd the simplest choice is 1 In summry, the eigenvlues nd eigenfunctions for the Sturm iouville problem in (2.27) re n x n n2 2 2 nd X n ºx» cos for n (2.28) Notice tht in the cse n, X ºx» 1, s we found erlier. For every nonnegtive integer n we get the product solution u n ºx t» e n2 2 kt 2 n x cos (2.29)

17 766 Chpter 13 Prtil Differentil Equtions to the het eqution by using (2.26). Observe tht this solution lso stisfies the boundry conditions u n x º t» u n º t» x Step 3: Stisfying the initil conditions. Hving found infinitely mny product solutions in (2.29), we cn use the linerity of the het eqution (see Theorem 1.18) to conclude tht ny liner combintion of the product solutions is lso solution. Hence if n is constnt for ech n, the function uºx t» 2 ½ n1 2 ½ n1 n u n ºx t» (2.3) n e kn2 2 t 2 n x cos is formlly solution. Setting uºx» f ºx», we obtin the eqution f ºx» 2 ½ n n x n cos (2.31) This is the Fourier cosine expnsion of f on the intervl x. From Section 3 of Chpter 12, nd especilly eqution (3.2) in tht section, we see tht the coefficients n re given by n 2 n x f ºx» cos dx for n (2.32) Substituting these vlues into (2.3) gives complete solution to the het eqution with Neumnn boundry conditions. Notice tht ech term in the infinite sum in (2.3) tends to s t ½ Using this nd the definition of the coefficient we see tht lim t½ uºx t» 2 1 f ºx» dx Thus s t increses in n insulted rod, the temperture tends to constnt equl to the verge of the initil temperture. EXAMPE 2.33 Suppose rod of length 1 meter mde from mteril with therml diffusivity k 1 cm 2 sec is originlly t stedy stte with its temperture mintined t Æ C t x nd t 1 Æ C t x 1. (See Exmple 2.17.) Strting t time t, both ends re insulted. Find the temperture distribution in the rod s function of time nd position. Find the constnt temperture which is pproched s t ½ Estimte how long it will tke for ll portions of the rod to get to within 5 Æ C of the finl temperture?

18 13.2 Seprtion of Vribles for the Het Eqution 767 According to our nlysis in Exmple 2.17, the stedy-stte temperture t t is f ºx» 1 x, with x mesured in meters. This will be the initil temperture. With length mesured in meters, k 1 m 2 /sec. Our new initil/boundry vlue problem is u t ºx t» 1 u x x ºx t» for t nd x 1, u x º t» u x º1 t» for t, uºx» f ºx» 1 x for x 1. (2.34) The solution s given in (2.3) with k 1 nd 1 is uºx t» The initil condition becomes 2 ½ n1 uºx» 1 x n e 1 n2 2t cos n x (2.35) 2 ½ n1 n cos n x The n re the Fourier cosine coefficients of 1 x on the intervl Ì 1Í, so 1, nd 1 for n even, n 2 1 x cos n x dx 4 n 2 2 for n odd. Substituting into (2.35), using n 2p 1 we get the solution uºx t» ½ p 1 1 º2p1» 2 e 2t cosº2p 1» x (2.36) º2p 1» 2 Notice tht ech of the terms in the series, with the exception of the constnt first term, include n exponentil fctor tht pproches s t ½ Thus the temperture in the rod pproches the constnt, stedy-stte temperture of 5 Æ C s t ½ Notice lso tht 5 Æ C is the verge of the initil temperture over the rod. This reflects the fct tht the ends re insulted, nd no het flows into or out of the rod. We suspect tht one of the exponentil terms (with p ) in eqution (2.36) will suffice to find how long it tkes for the temperture to be within 5 Æ of the constnt stedy-stte temperture. We solve 4 e 12 t 2 5 to get t 2 12 sec. We check tht the contribution to the temperture of the p 1 term is less thn 3 1 8, so 2 12 sec is good estimte. The temperture is shown in Figure 2. The initil temperture f ºx» 1 x nd the constnt stedy-stte temperture of 5 Æ C re shown plotted in blue. The blck curves re the temperture profiles plotted t time intervls of 3s.... EXERCISES

19 768 Chpter 13 Prtil Differentil Equtions u 1 5 x.5 1 Figure 2 The temperture for the rod in Exmple Consider rod 5 cm long with therml diffusivity k cm 2 sec. Originlly the rod is t constnt temperture of 1 Æ C. Strting t time t the ends of the rod re immersed in n ice bth t temperture Æ C. Show tht the temperture uºx t» in the rod for t is given by uºx t» ½ p 4 º2p 1» e kº2p1» 2 2t25 n x sin 5 (2.37) If the rod is mde of gold, find the therml diffusivity in Tble 1 on pge 753, nd estimte how long it tkes the temperture in the rod to decrese everywhere to less thn 1 Æ C. How mny terms in the series for u re needed to pproximte the temperture within one degree t t 1 sec. On one figure, plot the temperture versus x for t Estimte how long it tkes the temperture in the rod in Exercise 1 to decrese everywhere to less thn 1 Æ C if it is mde of luminum, silver, or PVC. For luminum nd silver, how mny terms of the series in (2.37) re needed to pproximte the temperture throughout the rod within 1 Æ when t 1 sec. For PVC, how mny terms re needed to pproximte the temperture throughout the rod within 1 Æ when t 1 dy. 3. Consider wll mde of brick 1 cm thick, which seprtes room in house from the outside. The room is kept t 2 Æ. () Originlly the outside temperture is 1 Æ C nd the temperture in the wll hs reched stedy-stte. Wht is the temperture in the wll t this point? (b) There is sudden cold snp nd the outside temperture drops to 1 Æ C. Find the temperture in the wll s function of position nd time. 4. The wll of furnce is 1 cm thick, nd built from refrctory mteril with therml diffusivity k cm 2 sec Originlly there is no fire in the furnce nd the temperture of the furnce nd the outside re both 2 Æ C. At

20 13.2 Seprtion of Vribles for the Het Eqution 769 t, fire is lit nd the inside of the furnce is quickly rised to 42 Æ C. Find the temperture in the wll for t. In Exercises 5 8, find the temperture uºt x» in rod modeled by the initil/boundry vlue problem u t ºx t» ku x x ºx t» for t nd x, uº t» T nd uº t» T for t, uºx» f ºx» for x with the indicted vlues of the prmeters. 5. k 4, 1 T T nd f ºx» xº1 x» 6. k 2, T T nd f ºx» sin 2x sin 4x 7. k 1, T T nd f ºx» sin 2 x 8. k 1, 1 T T 2 nd f ºx» x In Exercises 9 12 use the temperture computed in the given exercise. Plot the initil temperture versus x nd dd the plots of the tempertue versus x for number of time vlues like those in the text tht show the significnt portion of the chnge of the temperture. (Approximte the solution with n pproprite prtil sum.) In ddition, plot y u x º t» nd y u x º t» s functions of t. Recll from (1.5) nd (1.6) tht these terms re proportionl to the het flux through the endpoints of the rod. Give physicl description of wht is hppening to the tempertue s time increses. Include the informtion from the grphs of the flux nd the grphs of the solution. 9. Exercise 5 1. Exercise Exercise Exercise 8 In Exercises 13 18, find the temperture uºt x» in rod modeled by the initil/boundry vlue problem u t ºx t» ku x x ºx t» for t nd x, u x º t» u x º t» for t, uºx» f ºx» for x with the indicted vlues of the prmeters. Plot the solution for number of time vlues like those in the text tht show the significnt portion of the chnge of the temperture. Give physicl explntion of wht is hppening to the solution s time progresses. 13. k 1 1, nd f ºx» x x 12 º1 x» 12 x 1

21 77 Chpter 13 Prtil Differentil Equtions 14. k 1 2, nd f ºx» 15. k 1 1, nd f ºx» sinº x» 1 x 1 1 x k 1 1, nd f ºx» cosº x» x k 1 2, nd f ºx» ºx 1» 1 x k 13 2, nd f ºx» xº2 x» In Exercises 19 21, we will consider het flow in rod of length, where n internl het source, given by pºx», is present. As indicted in eqution (1.1), this leds to the initil/boundry vlue problem u t k 2 u x pºx» 2 c uº t» A uºx» f ºx» x t uº t» B x (2.38) for the inhomogeneous het eqution, where f nd p re given (known) functions of x nd A nd B re constnts. 19. The corresponding stedy-stte solution is the function Úºx» tht stisfies the prtil differentil eqution nd the boundry conditions. Show tht Úºx» stisfies Ú ¼¼ ºx» pºx» with Úº» A nd Úº» B C (Remember tht k Cc, where C is the therml conductivity.) Suppose tht u h ºx t» is the solution to the initil/boundry vlue problem u h t k 2 u h x 2 x t u h º t» u h º t» t u h ºx» f ºx» Úºx» x for the homogeneous het eqution. Show tht the function uºx t» u h ºx t» Úºx» is solution to the initil/boundry vlue problem in (2.38). 2. Use Exercise 19 to find the solution to the initil/boundry vlue problem in (2.38) with k 1, 1 pºx»c 6x, A, B 1, nd f ºx» sin x. 21. Use Exercise 19 to find the solution to the initil/boundry vlue problem in (2.38) with k 1, 1 pºx»c e x, A 1, B 1e, nd f ºx» sin 2 x.

22 13.3 The Wve Eqution The Wve Eqution We will strt with the derivtion of the wve eqution in one spce dimension. We will be modeling the vibrtions of wire or string tht is stretched between two points. A violin string is very good exmple. We will lso look t two techniques for solving the wve eqution. Derivtion of the wve eqution in one spce vrible We ssume the string is stretched from x to x. We re looking for the function uºx t» tht describes the verticl displcement of the wire t position x nd t time t. We ssume the string is fixed t both endpoints, so uº t» uº t» for ll t. We will ignore the force of grvity, so t equilibrium we hve uºx t» for ll x nd t, which mens tht the string is in stright line between the two fixed endpoints. To derive the differentil eqution tht models vibrting string, we hve to mke some simplifying ssumptions. In mthemticl terms the ssumptions mount to ssuming tht both uºx t», the displcement of the string, nd ux, the slope of the string, re smll in comprison to, the length of the string. T u(x) T x x + x Figure 1 The forces cting on portion of vibrting string. T u T T x θ Consider the portion of the string bove the smll intervl between x nd x ½x, s illustrted in blue in Figure 1. The forces cting on this portion come from the tension T in the string. The tension is force tht the rest of the string exerts on this prticulr prt. For the portion in Figure 1, tension cts t the endpoints. We ssume tht the tension is so lrge tht the string cts s if it were perfectly flexible nd cn bend without the requirement of bending force. With tht ssumption, the tension cts tngentilly to the string. The tension t the point x is resolved into its horizontl nd verticl components in Figure 2. We re ssuming tht the positive direction is upwrd. The verticl component is T u T sin, nd the horizontl component is T x T cos. The slope of the grph of u t the point x is u x tn We re ssuming tht the slope is very smll, so is smll. Therefore, cos 1, Figure 2 The resolution of the tension t the point x.

23 772 Chpter 13 Prtil Differentil Equtions nd tn sin As result, we hve T u T u x ºx t» nd T x T In similr mnner, we find tht horizontl component of the force t x ½x is pproximtely T, which cncels the horizontl component t x. More interesting is the fct tht the verticl component of the force t x ½x is pproximtely T u ºx ½x t» x so the totl force cting in the verticl direction on the smll portion of the string is F T u u ºx ½x t» x x ºx t» The length of the segment of string is close to ½x. If the string is uniform nd hs liner mss density, then the mss of the segment is m ½x. The ccelertion of the segment in the verticl direction is 2 ut 2 By Newton s second lw, we hve F m, which trnsltes into ½x 2 u t 2 u T u ºx ½x t» x x ºx t» Dividing by ½x nd tking the limit s ½x goes to, we hve 2 u t 2 T lim ½x 1 ½x If we set c 2 T, the eqution becomes u u ºx ½x t» x x ºx t» T 2 u x 2 u tt c 2 u x x (3.1) This is the wve eqution in one spce vrible. The constnt c hs dimensions length/time, so it is velocity. Notice tht the homogeneous wve eqution in (3.1) is liner. Once gin we cn build complicted solutions out of simpler ones. Solution to the wve eqution by seprtion of vribles et s turn to the solution of the eqution for the vibrting string. Since the wve eqution is of order 2 in t, we re required to specify the initil velocity of the string s well s the initil displcement. Thus we re led to the initil/boundry vlue problem u tt ºx t» c 2 u x x ºx t» for x nd t, uº t» nd uº t» for t uºx» f ºx» nd u t ºx» gºx» for x (3.2) We will find the solution using seprtion of vribles. Since the process is similr to tht used in previous exmples, we will omit some of the detils. Notice tht the boundry conditions in (3.2) re homogeneous, so we cn proceed directly with the seprtion of vribles. The strting point is to look for product solutions of the form uºx t» X ºx»T ºt»

24 13.3 The Wve Eqution 773 Step 1: Seprte the PDE into two ODEs. Inserting uºx t» X ºx»T ºt» into the wve eqution nd seprting vribles gives X ¼¼ ºx» X ºx» T ¼¼ ºt» c 2 T ºt» Since x nd t re independent vribles, ech side of this eqution must equl constnt, which we will denote by. Thus the fctors must stisfy the differentil equtions X ¼¼ X nd T ¼¼ c 2 T (3.3) Step 2: Set up nd solve the two-point boundry vlue problem. The first eqution in (3.3) together with the boundry condition uº t» uº t» implies tht X must solve the two-point boundry vlue problem X ¼¼ ºx» X ºx» with X º» X º» (3.4) We hve seen this Sturm iouville problem before in (2.1). The solutions, given in (2.12), re n x n n2 2 2 nd X n ºx» sin for n Step 3: Stisfying the initil conditions. With n n 2 2 2, the second eqution in (3.3) is T ¼¼ cn 2 T The functions cosºcnt» nd sinºcnt» form fundmentl set of solutions. Consequently, we hve found the product solutions n x cnt n x cnt u n ºx t» sin cos nd Ú n ºx t» sin for n Since the wve eqution is liner, the function uºx t» ½ n1 ½ n1 [ n u n ºx t» b n Ú n ºx t»] n x sin n cos cnt sin cnt b n sin (3.5) is solution to the wve eqution for ny choice of the coefficients n nd b n tht ensures tht the series will converge. Further, uºx t» lso stisfies the homogeneous boundry conditions. The first initil condition is f ºx» uºx» ½ n1 n sin n x

25 774 Chpter 13 Prtil Differentil Equtions To stisfy this condition, we choose the coefficients n to be n 2 f ºx» sin n x dx (3.6) the Fourier sine coefficients for f. The second initil condition involves the derivtive u t ºx t». Differentiting (3.5) term by term, we see tht ½ cnt cnt u t ºx t» n1 cn sin n x The second initil condition now becomes gºx» u t ºx» n sin ½ n1 b n cn b n cos n x sin Therefore, b n cn should be the Fourier sine coefficients for g, or b n 2 cn gºx» sin n x dx (3.7) Inserting the vlues of n nd b n into (3.5) gives the complete solution to the wve eqution. Notice tht every solution is n infinite liner combintion of the product solutions cnt n x cnt n x sin sin nd cos sin These solutions re periodic in time with frequency n nc All of these frequencies re integer multiples of the fundmentl frequency 1 c. In music the contributions for n 1 re referred to s higher hrmonics. It is the fundmentl frequency tht our ers focus on, but the higher hrmonics dd body to the sound. This coupling of fundmentl frequency with the higher hrmonics is thought to be ccountble for the plesing sound of vibrting string. We will see lter tht the sitution is different for the vibrtions of drum. EXAMPE 3.8 Suppose tht string is stretched nd fixed t x nd x. The string is plucked in the middle, which mens tht its shpe is described by 6 f ºx» x x if x 2 if 2 x At t the string is relesed with initil velocity gºx» Find the displcement of the string s function of x nd t. Assume tht for this string we hve c 2 6 The wve eqution ws derived under the ssumption tht the displcement nd the slopes were smll. While this is not true of this nd the other exmples tht we will exmine, it is true for n initil displcement of, sy, 1 f ºx». Since the wve eqution is liner, the solution with this initil condition is 1 the solution we find in Exmple 3.8.

26 13.3 The Wve Eqution 775 The solution is given by (3.5). We hve only to find the coefficients n nd b n. Since gºx», we hve b n. The coefficients n re the Fourier sine coefficients of f on the intervl º», nd they re given by n 2 f ºx» sin nx dx Inserting the definition of f, nd evluting the integrl, we find tht n if n is even, nd if n 2k 1 is odd we hve Substituting into (3.5), we see tht uºx t» is the solution. ½ k 2k1 º 1» k 4 º2k 1» 2 º 1» k 4 sinº2k 1»x cos 2º2k 1»t (3.9) º2k 1» 2 The rte of convergence The generl term in the series in eqution (3.5) is n x cnt cnt sin n cos b n sin (3.1) The first fctor, sinºn x» is bounded in bsolute vlue by 1. We cn express the second fctor in terms of its mplitude nd phse, cnt cnt cnt n cos b n sin A n cos n (3.11) Ô where the mplitude A n n 2 b2 n Thus, the generl term in (3.1) is bounded by A n for ll t. We cn judge È the convergence of the solution in eqution È (3.5) by ½ the rte of convergence of n1 A ½ n. Notice tht the rte of convergence of n1 A n does not chnge s t increses. EXAMPE 3.12 The displcement of the string in Exmple 3.8 is given by the series in (3.9). How mny terms must be included if we pproximte the solution by the sum including ll terms stisfying A 2k1 1? How mny if we include ll terms stisfying A 2k1 1? We see tht A 2k1 4̺2k 1» 2 Í For ny cceptble error e, we hve A 2k1 e if k Ô 1 e 12 Thus for n cceptble error of e 1 we must keep ll terms with k 5, nd for e 1 terms with k 17 re needed. Compring Exmples 2.23 nd 3.12, we see tht mny more terms re needed to get the required ccurcy for solutions to the wve eqution thn re needed for solutions to the het eqution. The exponentil decy of the terms in the solution to the het eqution mkes the series converge much fster.

27 776 Chpter 13 Prtil Differentil Equtions D Alembert s solution et s exmine nother pproch to solving the wve eqution in one spce vrible. We strt by finding ll solutions to the wve eqution u tt ºx t» c 2 u x x ºx t» (3.13) without worrying bout initil or boundry conditions. We do this by introducing new vribles x ct nd x ct By the chin rule, u x u x u x u u Similrly, we hve u t c u u Differentiting once more using the chin rule, we see tht u x x Ìu u Í x Ìu u Í Ìu u Í u 2u u Similrly, u tt c 2 Ìu 2u u Í. Therefore, u tt c 2 u x x 4c 2 u Consequently, in the new vribles the wve eqution hs the form u If we red this eqution s u we cn integrte to find tht u º» H º» where H º» is n rbitrry 7 function of. We cn now integrte once more to find tht uº» H º» d Gº» where Gº» is n rbitrry function of. If we set Fº» Ê H º» d, we find tht uº» Fº» Gº» where F nd G re rbitrry functions. In terms of the originl vribles, we see tht every solution to the wve eqution (3.13) hs the form uºx t» Fºx ct» Gºx ct» (3.14) where F nd G re rbitrry functions. It is esily verified tht ny function of the form in (3.14) is solution to the wve eqution. The generl solution to the wve eqution in (3.14) is clled the d Alembert solution. 7 Although the rgument used requires tht H is differentible function, it is relly true tht H cn be n rbitrry function. The sme is true for the functions F nd G tht follow.

28 13.3 The Wve Eqution 777 Trveling wves If we choose F in (3.14), we see tht uºx t» Gºx ct» is solution to the wve eqution. et s get n ide of wht this solution looks like. Figure 3 shows the grph of function Gºx» tht is nonzero bump centered t x. Figure 4 shows the grph of Gºx ct», where t is fixed. Notice tht the grph of Gºx ct» is now centered t x ct. From this we see tht s t increses, the solution uºx t» Gºx ct» to the wve eqution hs grph versus x tht is bump moving to the right s t increses. Furthermore, since the wve hs moved distnce ct in time t, it is moving to the right with speed c. y y y = G(x) y = G(x ct) x ct x Figure 3 The grph of G(x). Figure 4 The grph of G(x ct) for t. Similrly, the solution Fºx ct» represents wve moving to the left with speed c s t increses. We will cll solutions of the form Gºx ct» nd Fºx ct» trveling wves. As result, we see tht the d Alembert solution in (3.14) represents the generl solution to the wve eqution (3.13) s the sum of two trveling wves, one moving to the right with speed c nd the other moving to the left with speed c. Solving the initil/boundry vlue problem The d Alembert solution in (3.14) cn be used to find the solution to the initil/boundry vlue problem tht we encountered in (3.2). To mke the rgument somewht esier to follow, we will mke the ssumption tht the initil velocity is, so the initil/boundry vlue problem we will solve is u tt ºx t» c 2 u x x ºx t» for x nd t, uº t» nd uº t» for t uºx» f ºx» nd u t ºx» for x. (3.15) In the process we will gin dditionl informtion bout the solution. We strt with d Alembert solution uºx t» Fºx ct»gºx ct» from (3.14). We will use the initil nd boundry conditions in (3.15) to find out wht F nd G hve to be. We will ssume tht F nd G re defined for ll vlues of x. Observe tht u t ºx t» cìf ¼ ºx ct» G ¼ ºx ct»í. Therefore, the initil conditions imply tht f ºx» uºx» Fºx» Gºx» nd u t ºx» cìf ¼ ºx» G ¼ ºx»Í