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Kavon Feest

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J. Metcalfe

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J. Metcalfe
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This 29 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 121B at University of California - Berkeley taught by J. Metcalfe in Fall. Since its upload, it has received 27 views. For similar materials see /class/226585/math-121b-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.




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Date Created: 10/22/15
Math 121b Mathematical Tools for the Physical Sciences Instructor J Metcalfe Spring 2007 March 13 2007 and March 15 2007 Lectures 1 PDE in unbounded domains 11 Review from the previous lecture Last week we began looking certain PDEs in unbounded domains Here we use the Fourier integral representation of a function 35 Ammo cosx B sinz dz where A i 35 cosx dz B asmm dz When approaching PDE in such domains we may separate variables once we have reduced to a homogeneous PDE with homogeneous boundary conditions When however we solve the eigen value problem we do not get a restriction to a discrete set of Thus when we form the linear combination we need to sum over a continuum ie integrate In order to then nd the coef cients we rely on the Fourier integral to provide our notion of orthogonality We rst looked at a heat equation in a semi in nite rod We also showed that the heat equation in an in nite rod satis es CO ux t 7 zt dz 00 where x is the initial temperature distribution and 1 19012 I t 7 77 7 147rkt eXplt 41 is called the fundamental solution to the heat equation 12 Another heat equation example As another example of the techniques let s show that um t erflt gt is the solution to 1 um7ut fooltmltoo tgt0 k 1 xgt0 71 mlt0 um 0 sgn From the previous example we have that lt 2 l m S M gt L132 d u m 7 z e z 47rkt 700 g Here we can show that 1 21 7 Z2 u zt 7 ex 7 dm Vanni p M Indeed it works best to graph the exponential function on the z plane Notice that it is symmetric about the vertical line at z Thus if you graph sgnzexp7z 7 z24kt and say eg that z gt 0 the portion to the right of 0 whose sign gets reversed by sgn cancels out the portion to the right of 2x due to the symmetry Using this latter formulation we do a change of variables z 7 m d 1 d 7 7 2 y x4kt y x4kt This yields 1 xjfkt 2 2 xjfkt 2 m uzti fl17 e yderf7 7 3 y o y 4195 Note that the error function provides solution to several other heat equations See eg as exercises Ummgut7 0ltltOO u0t 0 um0 1 and the complimentary error function solves Umm ut7 0ltltOO k u0t 1 um0 0 As we did in bounded domains you can use similar techniques to look at a variety of situations 7 like for the heat equation or Laplace s equation As on additional example we shall look at a potential equation in a slot in the next section 13 Potential in a slot As an additional example of a PDE for which the eigenvalue problem occurs in an unbounded domain we shall look at the following problem concerning potential in a slot umuyy0 Oltmlta ygt0 U70f90 1407 y 91207 W17 24 9224 As is usual we shall also assume that luz remains bounded as y a 00 This problem does not have the required homogeneity to separate variables We shall break it into two different problems each of which has homogeneity in one of the variables Let u 741 l 742 where Vzul 0 mm 0 1W u107y 0 ma y VZ LLZ 0 u2x 0 0 14207 y 91y7 ma y 92y and we assume that u1u2 are bounded You can get some intuition of what the solutions are going to look like by paying attention to the domains on which the functions are de ned For example x is de ned on 0 lt z lt a Thus we can expand f in terms of a Fourier series Thus for ul we expect a series solution On the other hand 91 and 92 are de ned on 0 lt y lt 00 A Fourier series is not possible without additional assumption and hence we must rely instead on the Fourier integral So here we expect an integral solution for U2 Let s solve the rst of the problems for 741 Setting 741 XzYy we can plug in and separate variables We obtain X7 7 7Y7 7 7A2 X Y 39 Here after separating we know that both sides must be constant We have chosen the negative constant and shall leave it to you to show that the positive constant yields only the trivial solution From the above we obtain two problems after bringing the homogeneous boundary conditions in as boundary conditions on X X AZX 0 X0 0 Xa Y 7 AZY 0 The solutions to the former are Xn sin with An For Y we have 717139 Yn expijygt39 We may ignore the seond solution to the Y equation due to the fact that it is unbounded at 00 Summing over all possible we have that 00 u1z y 2 an sin m explt7mrygt a n1 a Using the remaining boundary condition 00 x u1z0 Zansinm a n1 which we recognize as a Fourier sine series Using the formulas for the coefficients of a sine series ie using orthogonality we have that an gOa z in dm We now turn to the U2 problem Letting U2 XmYy we have X 2 Y Y F 7 with Y0 0 and Yy bounded as y a 00 Notice that we have chosen the signs differently here As an exercise you should check that this is the only way to not have a triVial solution For X we have X 7 UZX 07 which has as a solution h h X Asp m Bsm w e z s1nhua s1nhua This is a convenient form of the solution which we saw when looking at potential in a rectangle You could equally well use exponentials or hyperbolic sine and cosine The algebra in order to nd the coe iicents becomes a bit more difficult though For Y we have Y qu 0 Thus Y 01 cosuy 02 sinuy Since Y0 O we must have 01 0 Thus YM sin My Taking the production solution and integrating over all possible u sinhua 7 sinhua w y MW Bo d sinhua lsm u y u We nd the coefficients by using the remaining boundary conditions 00 mm 14207 y 0 BM 811104 dw Recognizing the right side as the Fourier sine integral we have BM g A 91ysiH y dy Similarly 00 my may Am smog cm and Am 3 mggay smog dy As exercises you can try similar calcuations in other domains like eg a half plane a horizontal slot a quarter plane etc You can also try similar calcuations for unbounded regions for the wave equation 2 PDE using the Laplace transform 21 Introduction to the Laplace transform The Laplace transform is an integral transform meaning it takes in a function and yields function This new function is given by integrating the original function against another function which is called the kernal Speci cally the Laplace tranform L is given by unw mawmw We shall typically denote Fs Lfs That is we use the capital letter to denote the Laplace transform of the function given by the lower case of the same letter For fairly elementary functions f you can just compute the integral using eg integration by parts See the tables on p 469 471 of the text We then use properties of L to generate transforms of other functions based on those in the table Each of these can be derived using elementary methods like change of variable and integration by parts and I leave these derivations to you as important exercises Lel t t F8 7 5 L1 8 f 7 ND f lttgt 82 f 7 WW 7 H0 t L ft C175 8 0 L7705 fir8 1 00 Lft Fs ds S The second and equivalently the third of these is particularly important Using these proper ities if you apply the Laplace transform to an ODE you transform the ODE to an algebraic equation We shall do something similar using PDE For a one dimensional heat or wave equation we can apply the Laplace transform to the time variable to transform the PDE into an ODE which in theory is easier to solve Thus when solving a PDE directly is too hard we shall instead transform the solution to get a new equation an ODE for the examples mentioned above we shall solve the ODE and it remains to undo the Laplace transform ie apply the inverse Laplace transform L The last step is frequently the hardest Our main tools to this end are tables and algebraic methods like partial fractions and completing squares to t equations into the forms on the tables If you need to review partial fractions or completing the square please see me in office hours for assistance In many cases you can invert the Laplace transform using residue theory which you may have examined a bit in 121a We shall not use the latter here though We mentioned convolution brie y when we studied the heat equation in unbounded domains I want to say just a bit more here Recall that 9 0 z i 5 This is a multiplication that occurs frequently eg in hereditary systems where the current state of the system depends not only on current factors but on all factors that occurred up to that point Here are some properties of convolution and again I recommend the derivations as exercises 0 f 9 9 f 0 f9h f9h 0 f9hf9fh Not all properties of multiplication carry over though Eg o f f is not necessarily nonnegative o f 1 is not necessarily f indeed the identity for convolution is the Dirac delta function not 1 The key relation with regards to the Laplace transform is the following If Fs LU and 13 Lg then LU 9 FSG8 Thus L takes convolutions to multiplications 22 Examples Recall that Hf SHf 7 ND We ll transform in if only Thus Ux s Luzt 0 0 swam dt 814 7 0 istau 7 3 0 75 iaU L8mij e xtdt7J e umtdt78mms L Squaw 7 my 221 Example 1 Let s examine umut 0ltzlt1 tgt0 u0t 1 u1t 1 uz0 1 sin7rz We shall transform everything that has a 25 Thus Um Lut 3U 7 ux 0 3U 717 sin7rz U0 s L11s U1s Notice that we did not need to rst reduce to a homogeneous PDE and boundary conditions The method does not require this We have thus turned our PDE into an ODE boundary value problem UN 7 SU 717 sinrm since 3 is constant with respect to m which is the only variable for which we have derivatives To solve this nonhomogeneous ODE we have U where Up is a particular solution any solution that solves the ODE and U0 is the complimentary solution which is the solution to the homogeneous problem For U0 we must solve the homogeneous ODE Um7sU0 whose general solution is U cltfhg Cgem For Up we shall use the method of undetermined coe icients See me in office hours if review beyond these examples is needed Since the nonhomogeneous term is 717 sin 7m a zeroeth order polynomial and a sine we use Up A Bsinn39z Ccosn39m as an ansatz Differentiating we have U 7Bn392 sin 7m 7 C7r2 cos 7m Plugging into the ODE we have 7B7r2 sin 7m 7 C7r2 cos 7m 7 5B sin 7m 7 80 cos 7m 7 3A 717 sin 7m Thus we have a solution if CQ A Bwr Using these two solution we have sin 7m 1 U cleihg cgehg 7 2 3 7139 s We can now use the boundary conditions to resolve the constants 1 1 7U qw7 S S 1 1 7 U1 s 315quotg 3259g 7 s 3 That is 61 717 62 0 315quotg 3259g 0 Solving this system of linear equations we have 01 02 O Thus 1 sin 7m U 7 m7 3 S 712 S It remains to invert the Laplace tranform Using linearity we have 1 1 umt L 1U L 1 sinIrmL IJLm 1 577 sin7rm The last equality follows from using a table 222 Example 2 Let s look at the following wave equation umutt 0ltmlt1 tgt0 u0t 0 u1t 0 uz0 sin7rz utz 0 7 sin7rz We begin by applying L to the PDE and boundary conditions This yields Um Lutt SZU 7 ssin 7m 1 sin 7m U0 s 0 U1 s Rewriting the new differential equation we must solve UN 7 SZU 17 ssin7rz which is a nonhomogeneous ODE in m Writing U U0 1 Up where U0 is the complimentary solution and Up is a particular solution we rst see that U0 solves we 7 SZUC 0 Or 7 75m U0 015 625 For Up we shall use the method of undetermined coe icients Since the nonhomogeneous term is a sine we use as an ansatz Up Acos 7m Bsin 7m Thus U17 77r2A cos 7m 7 7r2B sin 7m Plugging into the equation we have 771392 32B sin 7m 7 71392 52Acos 7m 17 s sin 7m Thus A 0 771392 323 17 3 Hence 3 7 1 Up W s1n7rm Combining U0 and Up we have U 315751 025 1 sin 7m 32 71392 39 Using the boundary conditions we must have C1CgU0s0 61579 0259 U1 s 0 and thus 01 02 0 Thus Um s 371 ismwm 827r2 It remains to invert the Laplace transform Using tables we see that M35775 sin sz71 S J 2 sin7rm 71w8jr2 itiljf sinzrzcos mi 7 223 Partial fractions review Before moving on to more complicated examples let s pause to do a brief partial fractions review Let s start with a simple example of how to apply L 1 to s137539 We wish to apply partial fractions to break this into terms which are more likely to be in the tables To do so we apply partial fractions which is simply a way to undo nding a common denominator and adding fractions for proper rational functions ie a ratio of polynomials where the order of the numerator is less than that of the denominator For improper rational functions you may rst apply polynomial long division and then use partial fractions on the remainder For the given example we rst notice that it is a proper rational function Thus we write i B s1si5is1 3757 and we need to nd A and B Multiplying through by the denominator on the left we have 3Asi5Bs1 Plugging in s 5 we have 3 6B or B 12 Using 3 71 we have 3 76A or A 712 And thus 3 7 1 1 1 s1si572 375 317 Aim 13415 Vllsiill iea i What about the rational function 4s 1 7 32 28487139 Here the term involving the irreducible quadratic function is slightly more complicated In the numerator instead of taking a constant we must allow for a linear function Hence we write 4s1 A8B C 8228487178228487139 Multiplying through we have 431As Bsi 1 C322s4 We plug in s 1 to see that 5 7C or C 57 But there is no real number that we can plug in to isolate A and B We instead expand 431A32 37AsiBC322s4 Equating the coefficient of 52 on both sides we have 0AC or A757 Equating the constant term on both sides gives 17B4C or B4Cil137 Thus 4s1 71753135 1 822848717782284 737139 We can complete the square in the denominator to have 7175313 1 7175s11851 77s123 737177 s123 737139 Thus 5 1 g lll t Vll l t i 5 it 18 t i 5 t 77 3t 3t 7 7e COSf 7 e 8111f 75 In the case of the denominator haVing simple roots let s examine an alternative formula which extends nicely to when there are in nitely many roots So Us 1 where qp are polynomials with the degree of q less than that of p Assume also that 193 has simple roots r1 r2 rk Then we would wish to expand A A A M 18 1 2 k 193 sin 3772 377k Multiplying both sides by s 7 rm we have Sirmq8 14167767 A sirm Am 193 sin 3772 37m Now let 3 a rm This isolates a single term in the right leaVing A 1 i s 7 rmgtqltsgt m 7 lim lim q8 S TmMS qTm 39 W 298 Hm Ms Mm 10 For the second equality we used L Hopital s rule For the last step we use that the roots are simple to guarantee that p rm 7 0 You can refer back to the rst example to see that indeed works for that case What we shall do if we have a ratio of two powers series and thus there might be in nitely many roots which we shall assume to be simple in the denominator is to write q8 1 7 Aki 193 2 s 7 m where m are the roots This is an in nite partial fractions77 that is due to Heaviside Then using arguments similar to above it follows that 224 Example 3 Let s look at the heat equation umut 0ltzlt1 tgt0 u0t 1 u1t 1 ux 0 0 Taking the Laplace tranform of the equation and boundary conditions we have Um 3U 1 U1s U 0 7 lt sgt 8 Solving this homogeneous ODE we have U 01 cosh m 62 sinh x and using the boundary conditions 7 U 0 8 73 61 1 1 i 7 U1 s 7 cosh 62 s1nh S 8 Thus 1 17 cosh sinh m sinh 17 U 7 h h 3 cos s sinh sm m s sinh Here we have used a subtraction formula for sinh which we rst saw when studying the Laplacian in a rectangle Unfortunately many tables will not show you such a function Thus we need to decompose it into simpler parts using the in nte partial fractions We rst need to nd the zeros of the denominator ssinh O 11 The only real zero is s 0 But using that iA 57114 coshiA 5f cosA eiA sinhiA 2 isinA we have sinh in sinh coshin cosh sinhin sinh cosn icosh Esin 7 Thus for 0 sinh in sinhE cos 7 icosh 5 sin 7 we need sinhEcosn O and coshEsinn 0 Since coshE 7 0 for any 5 we must have sinn 0 or 7 imr Using then the former equation we have that cosn 7 O and thus sinhE 0 or E 0 Thus for sinh in O we must have 5 in iimr Setting E in it follows that s 7712712 Thus the roots are To O Tn 7n27r2 If we write 1 77 37 MM 2Ax n0 we shall argue as in the previous section For r0 O we multiply through by s 7 0 s to get 00 s sUsm A0 Letting s 7 O we have sinh m sinh 17 sinh m sinh 17 A0m l1 lim 970 s s1nh 970 s1nh Using L Hopital s rule this is i cosh m 17 m cosh 17 A0zlim2 1 M x17x1 970 Z cosh For Tn 7712712 we argue similarly by multiplying through by s 712712 and taking the limit Thus h zsinh 17z A 1 2 2 sin HEEWJS n W ssinh By L Hopital s this is sinh m sinh 17 z s 1 712712 cosh m cosh 17 lim SH WZWZ sinh 1 cosh 12 This evaluates to sinh iimrz sinh7rimr1 7 22 Him cosh imn39 since Tn is a root of sinh Using the identities we have 2sin mm sinmr17 mr cosmr Thus 1 00 2sinn7rzsinmr17z 1 U 7 2 AW 3 1 mrcosmr sn7r and applying L 1 to each term 7t1 i 2sin mr sin n7r1 7 ze2r2t39 1 717T COS 717T You can compare this solution to that which you could obtain from separation of variables and easily see that they coincide At this point we are still looking at simple examples which can be obtained both ways There will be problems in a bit for which the Laplace transform has distinct advant ages 225 Example 4 Let s look at one more example which can also be done using separation of variables before moving on to more complicated examples Let s solve the following wave equation umutt 0ltmlt1 tgt0 1402970 210070 x um00 m0m Taking L of the equation and boundary conditions we have Um SZU 7 m UOs 0 Uw1s 0 Letting U U0 Up we see that U0 01 cosh sz 62 sinh sz and using Up Am B as an ansatz it follows that Thus Uz s 01 cosh sz 02 sinh sz S Using the boundary conditions it follows that 0 U0 s cl and 1 0 Um1s cgscoshs 72 s or 1 c 77 2 s3 coshs Thus Uz s sinh sz S s3 cosh s We now need to apply L l Here we again need the in nite partial fractions Write U ismcoshs7sinhsmiiA 1 78 7 s3coshs in0 m 57 where Tn are the zeros of the denominator Setting 53 cosh s O we have that either 3 0 or cosh s cosh in 0 That is 0 cosh E cosh in sinh E sinh in cosh 5 cos 7 isinh 5 sin 7 Thus cosh Ecos 7 O sinh Esin 7 0 Since coshE never vanishes we must have 2 71 cosn0 or ni For such 7 sinn 7 O and thus sinhE 0 or E 0 Thus the roots are To 0 and m iiw Multiplying through and taking limits as above we have sz cosh s 7 sinh sz 1 sz cosh s 7 sinh sz 1m A0ltgt lim 970 s3 cosh 3 970 52 cosh 3 Using L Hopital s 1A zcoshs sz sinhs 7 mcosh sz zsinhs szcoshs 1 1m s70 25 cosh s 52 s1nh 3 970 2 cosh 2 43 s1nh s 52 cosh s For m ii Zn lw we ma use the formula 2 7 An MVPTn since it does not yield an indeterminate form Since ps 332 cosh s 53 sinh s 14 iiznz ln39s cosh m 7 sinh m i sin 7m Anx t 2 1 3 x 2 1 3 x 2 1 r cos r z 7 7r s1n r 7 7r sin 777139 3 n 2 h n j n2 h n n2 3 n2 Let s look at the inverse Laplace transform of these terms in pairs L71 isin Lilwm 1 isin Lilwm 1 2 1 3 2 1 siizn ln39 2 1 3 2 1 si2 17r 7quot sin 777139 2 s1n 777139 2 This is isinZLZ l39Irz i2n1rt isin7m 72n2717rt 3 V 1 3 2n71 2n71 2n71 2n71 77139 SIHTW 77139 SIHTW 2n71 sini39n39z 2 71 2 71 32 751 2 7fteil n2 7rtgt 2n71 2n71 77139 SIHTW sinm 2 71 3 lt2s1n 2 mi 2n71 2n71 lt 2 7r sin 2 7139 And thus 2sinm 27171 uzt E 3 s1n 2 mi n1 sin hing 23 More complicated examples In the above situations we could have also separated variables and it is a good exercise to do so and compare the results When separation of variables is available the Laplace transform methods appear to be unnecessarily difficult There are situations however where they present a distinct advantage One such situation where they are particularly convenient is when you have time dependent boundary conditions or inhomogeneities With such time dependence one is not able to nd a steady state and reducing to a homogeneous equation so that separation of variables is possible might be tough or impossible 231 Example 1 Let s look at the following heat equation umut 0ltzlt1 0ltt um 07 t Vut 07 t u1t 1 ux 0 0 This corresponds physically to an insulated rod with the left end attached to an insulated container of uid where the uid is circulated in a way that its temperature uniformly matches that of the end of the rod Notice the t derivative in the boundary condition This is a timedependent boundary condition If you attempted separation of variables here you would nd that the eigenfunctions were not orthogonal So we instead use Laplace transform methods Applying L we have Um 3U m0 s 7 v8U07 s 1 U1 s 7 S The solution to the homogeneous ODE is U 01 cosh m 02 sinh m Then 62 Um0 s 39ysU0 s 39yscl or 02 W561 And 1 7 U1 s 01 cosh 1 62 sinh xE 01 cosh 1 yxgcl sinh 8 Thus 7 1 Cl 7 scosh 1 39y sinh 7 397 C cosh y mEsinh and cosh m 39y sinh m qs scosh 1 mE sinh 193 39 Clearly s 0 is the only real root of p Letting E in we see that 0 298 E in2cosh5 271 WE in sinh 17 Thus we must solve when the second term vanishes That is when 0 cosh Ecosh in sinhE sinh in 39y insinh E cosh in cosh E sinh in cosh Ecos 7 i sinh Esin 7 39y insinh 5 cos 7 i cosh 5 sin 7 cosh Ecos 7 i sinh Esin 7 1 vi sinh 5 cos 7 1 Mn sinh Ecos 7 1 WE cosh 5 sin 7 7 yr cosh 5 sin 7 Equating real and imaginary parts we have 0 cosh 5 cos 7 1 vi sinh 5 cos 7 7 yr coshE sinn 0 sinEsinn 39ynsinhEcosn 39yEcoshEsinn cosh E 1 vi sinh E 7w cosh 5 cos 7 7 0 yr sinh E sinh E 1 vi cosh 5 sin 7 T 0 39 For this to have a solution7 we need the determinate of the matrix to vanish That is7 0 cosh 5 V5 sinh 5sinh 5 V5 cosh 5 772 cosh 5 sinh 5 1 52 772 sinh 5 cosh 5 39y5cosh2 5 sinh2 5 If E gt 07 then both terms above are positive If E lt 07 then both terms above are negative Thus7 the only way to get 0 is to have 5 0 With E 07 the equation from the real parts reads 0cosn739ynsinn 1 tann 7 7W Let n1 n2 be the solutions to this equation Then7 r0 07 m ink2 771g and we set 618 00 1 7 A p 3 0 k 3 7 m For Ao7 we have A0z 11 M 1 we use that 1 5 39y i Y 7 7 7 193 cosh wgsmhypr 5lt2 sinh 2 sinh 2 cosh gt Using the de nition of mmie that cosh 1 y sinh W 07 we have 1 MW g1 v 77 WSW Thus7 cosnkz 7 mm sin 77km Ak 2 V 1 v 771 cosnk Using that 1 U A 7 2 WE S 7 m we have 00 1 008mm 7 nwsmwkm 2 25172 7 t Wm v 1Vnwcosnk exp m after inverting L term by term 232 Example 2 Here we shall only look for the part of the solution to the following which exists for a long time This is like steady state but it might not be steady with timedependent boundary conditions The initial condition does not affect the long time behavior and thus we may take it to vanish Thus we look to solve umut 0ltx tgt0 1407 t 1 05 ux 0 0 Taking L here we have Um 3U U0 s Thus as a general solution we have Uz s 01 exp7 z 62 exp m At this point we make two further assumptions about the solution 0 That um t is bounded as x a 00 o and that denotes the square root with nonnegative real part This forces us to take 02 0 Thus it must be that Us z Fs exp7 z Suppose now that ft17 57m 1 asinwt and thus 1 1 0w F 7 7 7 7 S s s B 32 LUZ and 1 1 0w Ux s 7 g 7 m m exp7 Notice that this becomes 00 and thus you are dividing by zero77 and effectively have a root in the denominator at s O 73 iiw Since 73 corresponds to a term whose inverse Laplace t 91 transform is 5quot which decays exponentially and is thus not a signi cant part of the long time behavior we may drop it Thus we need to write 1A 27 sizw s2w UA01A1 3 Here A0 lim0 sFs exp7 z 1 97 A1 3sz 7 iwFs exp7 x exp7 m A2 shijS iwFs exp m 7 8Xp77iwx 18 Since VEMZ em4 121i Si and similarly 1 72 7 in 72gt um t 1 exp72Zmem 7 exp77iwze m 1expiwt7 w21im7 gexp7iwt7 mi2n 7 m 2 2 1 aexp7mm sinwt 7 233 Example 3 brie y In this nal example we look at something with repeated roots brie y um utt 7 sinwt 0 lt z lt1 tgt 0 u0 t 0 u1t um0 0 utm0 This corresponds to the movement of a steel wire in a sinusoidal magnetic eld Taking L we have w 32 LUZ U0 s 0 U1 s Um SZU 7 Then you can check that w U cosh sinh cl 8m 2 8m 8282 wz Moreover using the boundary conditions you can check that w wsinhs 5232 Adz 232 Adz cosh s w coshs2 7coshs127m 3232 Adz coshs2 Uz s 7 cosh sz sinh sz S S w 282 w2 Let s take an 7139 which turns out to be one of the natural frequencies of the wire Then we have 7 7r coshs2 7 coshs12 7 7 5232 7r coshs2 39 This is unde ned at s 0 iin39 ii2n 7 17r The only one of these that differs from the preVious techniques is iiIr and thus let s restrict our attention to this one These turn out to be double rootsl Instead of expecting a partial fraction decomposition with A0 A1 si7r 372471397 we instead need to allow for 1408 i n39 Bo 1418 7i7r B1 si7r2 s7iw2 7 which after applying L l yields Aoeim Boteim Alem Blteiwt39 To nd the coel39licients7 we argue as in section 223 to see that B1 lim 3 7 2471392Uz7 s s7vl7r B1 7 872W A1 lim 3 7 247139Um7 s 7 and similarly for A0 and B0 Eg7 for Bl7 we get 7 i 7 2 7r coshs2 7 coshs12 7 Bl 51 M 3232 M2 coshs2 7r coshs2 7coshs127m 323 i n39 WEEKS2 7 7r coshi7r2 7 coshi7r12 7 7 iw22i7r sinhi7r2 lim 1 i 77 s1n7rm 712 The corresponding term in u then has the form 1 39 t fitem 712 and you can see a resonance happening due to the t in front Math 121b Mathematical Tools for the Physical Sciences Instructor J Metcalfe Spring 2007 January 30 2007 Lecture 11 Review from the previous lecture Last week we looked at several examples of initial value boundary value problems for the heat equation In order to determine the in nite number of constants that appear at the end we use orthogonality At times this is incorporated into a Fourier series but we saw examples were the resulting series was not a Fourier series Nevertheless we saw that in these examples orthogonality still held and thus we were able to nd the constants appropriately To show orthogonality more general we began a study of Sturm Liouville problems Here we look at the eigenvalue problem 8m gt ml 7 qm gtm VPWWW 07 l lt m lt r 041 l DIMl 0 BMW BMW 0 Recall that we can put any linear second order homogeneous equation into the form above In addition to the problems that we have seen involving gtH Z gt 0 we also have two very important examples which we will later discuss in detail in Bessel s equation 2 77L 2 mu lt 0 and Legendre s equation 17 xz m A2ux 0 For eigenfunctions lt1 gtm solutions to the above equation and boundary condition for some An and Am respectively we were able to show a weighted orthogonality 7 nltzgt mltzgtpltzgt dz o m y n 12 Series expansions Recall that Fourier series are roughly just a change of basis in an in nite dimensional vector space Here we simply change to a different basis which is based on the eigenvalue problem above If we have a1 gtl 7 ampl 8 gt WEWQE AZPlt gt gtltgt 07 1lt 90 lt T 0 BMW BMW 07 we have weighted orthogonality T dm 0 m 7 n I Let f be a given function on 1 lt z lt r We want to know if we can expand 1 into a combination of eigenfunctions That is can we write M 2mm n1 In particular we ask the same questions that we asked for Fourier series 1 What must the Gas be 2 To what does the series converge The rst answer follows from orthogonality Here instead of just multiplying through by another eigenfunction and integrating we need to also account for the weight Thus we multiply both sides by gtmzpz and integrate Thus we see that fltzgt mltzgtpltzgtdzCn nltzgt mltzgtpltzgtdzcW fnltzgtpltzgtdm Thus 7 ff f gtmp 195 T f np 195 39 Here we are again ignoring a technical point concerning commuting the integral with the in nite sum Cm The convergence question can also be answered similar to that for Fourier series While we shall not go into detail let s just mention that if things eg smpz qm are sufficiently nice then the series converges to ak 2 7 where fm resp fm7 is the right handed limit resp left handed limit In particular the series converges to the value of the function at any point where the function is continuous This is enough to allow you to do separation of variables for some variable coefficient problems such as 3 814 814 ung mung Such problmes occur eg when the material properties are not uniform 13 Schrodinger equation Having just looked at several examples for the heat equation we shall brie y pause to do a similar example for the Schrodinger equation 814 712 hi 7V2 V 0 2 at 2m u where V2 is the Laplacian This is one of the fundamental equations in quantum mechanics It is an equation whose solutions share certain properties with both those of the heat equation and the wave equation though it technically is not in the class of either of these equations The Vm is called the potential When the potential vanishes the equation only differs from the heat equation by the presence of the i and by a sign From the perspective of separation of variables neither of these cause particular di iculties Let s look at a one dimensional example Take V E V0 where V0 is a positive constant Then let s study the lVBVP hZ ihutiumVou0 0ltmlta 2m u0t 0 uat T0 M9070 1 This is not homogeneous so we need to nd a steady state solution though the physical meaning is less relevant in this context A better term is a stationary solution Making the usual substitutions u a 12m at a O um a 1 um a 12 yields hz 7U V01 0 2m 110 0 1101 T0 The general solution to the ODE using the characteristic is 2mV0 2mV0 vz 7 01 cos 2 sin We now use the boundary conditions to solve for the constants 0 110 cl 2 V To va 62 sinltagt Thus gt 5 m sin fail0 a sin lt2mV0 gt 71 h We now need to nd the transient solution w u 7 1 Taking derivatives of u w v and plugging into the equation we see that w satis es hZ ihwt iwzm Vow 0 2m w0t 0 wa t 109070 1 W90 3 WE We now make our ansatz wz t gtmTt Plugging this into the equation we have hZ mm n V0 gtT 0 Toewmwmm ma m Since the variables are separated it must be that the two sides are equatl to a constant let s call the said constant E Then we have T h7 E z T Tt exp7iEth and we also have 2 h if V 7 E 0 2m 0 gt gt0 0 Ma Here it must be that E gt V0 Otherwise you only get the trivial solution I ll leave the veri cation of this statement to you as an exercise The general solution to the eigenvalue problem is 1 Ci COSltWgt 02 sinltzgt Using the boundary conditions 0 gt0 01 E 7 V02m lt h gt Since it cannot be that 62 0 otherwise the solution is trivial we must have iEi V02ma h 0 gta 2 sin 717139 h2n7r2 En Qmaz V0 Set k Then our solution has the form 00 WE t 1 expwEntwmmnx n1 It remains to nd the coefficients To do so we shall use the initial condition and orthogonality Thus we see that i foagz sinknz dz 7 f0 sin2knz dz 39 To get the solution to our initial problem we need only piece things together u w u bn 14 Wave equation In arbitrary dimension the wave equation is 1 721m VZ LL C where c is the wave speed This equation is used to describe waves of many sorts light sound electromagnetic waves elasticity etc In the one dimensional setting we have 1 gutt Umm which describes for example the position of a vibrating string lmagining a guitar string we have the wave equation boundary conditions describing having xed ends and we specify an initial position and an initial velocity notice that there are two time derivatives thus we need to pieces of if data We shall look at the vibrating string problem 1 umm7utt Oltmlta tgt0 C u0t 0 ua t 140570 1 8tuz70 9W This is a linear equation and the PDE and boundary conditions are homogeneous lf eg the boundary conditions were not homogeneous you would rst try to nd a stationary solution ie a steady state solution if this were the heat equation it just doesn t have the physical meaning In fact as friction is not taken into account the equation never steadies Since the equation is homogeneous we may proceed immediately to separation of variables Set ux t Plugging this into the equation we have 1 gt ltxgtTlttgt gar5W0 or 1 TN 7 27 339 As the variables are separated it must be that both sides are constant We shall set both sides to the negative constant 7A2 and I ll leave it to you to check that other constants 2 0 yield only the trivial solution We now have two equations T AZCZT 0 gt A2 gt0 gtlt0gt gtagt0 The latter is an eigenvalue problem we ve seen frequently lt s solutions are 1 sinnz An E a The general solution to the former equation is plugging in An Tn an cosnct 1 bn sinnct Notice that as t a 00 these do not decay which is quite in contrast to the behavior for the heat equation Using the above solutions we have that un sinnzan cosnct 1 sinnct are solutions for each n and thus by the principle of superposition we have u Z sinnman cosnct 1 sinnct n1 It remains to nd the coef cients an and I for which we shall use the initial conditions Using the initial position we have ux 0 2 an sinn7rxa You should recognize this as the Fourier sine series for f or else you can use orthogonality to nd the coef cients Thus 2 1 an 7 sinmrma dz a 0 Using the other initial condition we have 00 9m utz 0 Z bnkncsinnm n1 There again is a technicality here which you should be aware of but we are not going to concern ourselves with too much That is we are assuming that you can pull the derivative inside of the in nite series which generally requires a stronger than usual understanding of the convergence If we write B bnA c we have 00 9z Z aninnx n1 Thus you recognize Bn as the Fourier sine coef cient for 9 Thus bnknc B 2 agm sinmrza dm 0 a 2 a bn mO gzsinmrma dm Let s look at a common example 1 umm7utt Oltmlta tgt0 c u0t 0 ua t 2 hf 0 lt m lt 12 uz0f90 hlt 7 a2ltxlta 8tum00 Here clearly we have 1 E 0 You can compute 2 a i an gO x s1nmrza dz i 7 sinmn39 a 7 7T2 7742 39 2aja2 I12m sjnn7ma dz i hlt2 7 37m sinn7rma dm Thus 00 i um t 7 Z W sinmrza cosmrcta n1 This is a perfectly legitimate solution but it is awfully dif cult to picture what the solution looks like It also requires in nite summation which we shall show shortly isn t strictly required to understand solutions to the wave equation We shall use the trig identity which follows from the sine addition formula 1 sinA cosB sinA 7 B sinA This coincidentally is the same identity used to prove the orthogonality used in Fourier series If we use this we see that 1 8h 2 1 8h 2 m t m 2 sinn7rax 7 025 W 2 sinn7rax 025 Recall that 8h 2 E w sinmrz a is the Fourier sine series of 1 that is where the an s came from after all Thus the series represents the odd periodic extension f0 of 1 And since 1 is continuous and vanishes at 0 and a f0 is continuous and thus the series converges to 1 0 at every point Using this observation in the above equation we have ux t 7 ct 1 ct Thus u is the superposition of two waves One travelling to the left with speed c and one travelling to the right with speed 0 See the gures This also allows you to evaluate u without summing You just need to be careful to use the odd periodic extension of 1 when x 7 ct and z 1 ct land outside of 001 15 d7Alembert7s solution Here we attempt to generalize what we did above and allow us to write the solution of the wave equation without using the in nite series representation Again we look to write the solution in terms of two waves travelling in opposite directions We rst note that 0782 282 7lt8 8gtlt8 13gt at C 8352 at 68x 8 Cam 7 quot Thus our goal seems plausible as we can factor this into factors depending on derivatives in 257 oz and t 1 oz We rst do a change of variables w z ct z z 7 ct We let um t vw By the chain rule practice this it is a must to do any study in PDE we have iviwamp 3m 7 3w 3m 323z 7 3w 3239 314 3U 31 3t 7 3w 3t 32 3t 3w 3239 82 3 gt 3x2 7 3m 3w 32 7 3 3w 3 32 3 3w 321132 802 t azawa t 7810825 wa 2va U227 using that 12W Um provided that v the second derivatives of v are continuous Similarly exercise we have um 02vww 2va U22 Thus 102utt um if and only if U ww 7 2112112 l Uzz U ww l 2112112 l Uzz or equivalently Uwz 0 This equation implies an independence of w and 2 Indeed we can just integrate it Since 3 31 i 7 0 32 310 we have 8 U 7 0 W W a constant with respect to Thus v 0ltwgt dw M wltwgt W The gtz is the constan 7 with respect to w that results from the integration Thus we have ux t ct 7 ct This is what we will refer to as d Alembert s solution but we rst need to nd what 1 and g5 are in terms of the initial data Using the initial data 1 149070 We WE and 92 7149570 Gil90 7 ewe We rewrite the latter equation as 1 12 was 7 ltzgt g gltygt dy A Gltzgt A 0 8 Solving the system we have on 0 lt z lt a 1W 50 C95 A mm m7mm7m We shall be plugging in z i ct into these equations These values can land outside of 0 a when t is large yet 1 and j are only de ned on Q lt z lt a Thus we need to nd a proper extension to use By an extension we mean a function 1 which is de ned everywhere but 0 lt m lt a To determine the extension we shall use the boundary conditions We have ux t ct 7 ct Thus 1 1 0 140 10375 Mimi Emmi CCt A 5fct Cict A Hence 0 Nit fct CCt GCt As 1 and G are independent and this sums identically to 0 for all t it follows that m77 wm mm76em This says that for f we need the odd extension f0 and for G we need the even extension G5 At z a we have 0 uat 1a ct gta7 ct a ct fa7 ct a ct 7 3a7 ct Thus fa ct 7fa7 ct f7a ct Ga ct Ca7 ct Cct7 a Here we have used that we have an odd extension for f and even extension for G The above two equations say that the values remain same at a ct and ct 7 a for any 25 That is if you move 2a down the line the values repeat ie we need to extend to periodic functions with periods 2a Hence for f we use the odd periodic extension f0 and for G the even periodic extension Ge Hence 11z ct 2 ct l 15x ct A 1 5f0 Ct G5x 7 ct 7 A Hence the d Alembert solution is m t 7 f0x 7 ct f0z 7 225 11435 7 ct 7 1525 7 ct If one of f or G vanishes this is quite easy to plot Notice that um t the position at point z and time if only depends on f at z ct and z 7 ct and on g in between z 7 ct and z ct since G is the integral of g In particular 1 and g at some point z only affect the solutoin between z 7 ct and z ct at time t later This is nite propagation speed which is quite different from what happens for eg the heat equation


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