### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# ELEM MATH PHYS PHYS 228

UW

GPA 3.96

### View Full Document

## 8

## 0

## Popular in Course

## Popular in Physics 2

This 46 page Class Notes was uploaded by Dr. Simeon Wiza on Wednesday September 9, 2015. The Class Notes belongs to PHYS 228 at University of Washington taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/192442/phys-228-university-of-washington in Physics 2 at University of Washington.

## Reviews for ELEM MATH PHYS

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/09/15

Lecture 24 Solving 2nd Order Partial Differential Equations I Homogeneous Equations with No Time Dependence Chapter 13 in Boas Our goal in the final Lectures is to review and synthesize what we have learned about solving the 2nd order partial differential equations characteristic of most of physics including satisfying the associated boundary conditions As outlined in previous lectures we start with the simple forms of Laplace s equation with no time dependence VZ P o 241 which applies to temperatures potentials gravitational and electrical etc in regions without sources The specific fOI H l of the solution depends on the number of relevant dimensions and the geometric structure of the problem s boundary conditions We proceed by separating variables in the appropriate number of dimensions and the required geometry In particular we want to choose the coordinate system typically rectangular cylindrical or spherical so that the boundaries corresponds to one of the coordinates being constant e g z 20 or r R0 Then we pick a representation of the solution to Laplace s equation that has completeorthogonal functions solutions of a SturmLiouville problem that vary along the boundary We then use these appropriate special functions to match the boundary conditions The essential feature of all such problems is that the solution to the differential equation that matches the fully specified boundary conditions is unique Once we find one such solution by any method we are done lD First consider the onedimensional problem the trivial case where the boundaries are points P gt yx or P gt xt 2 gyx0yxabx 242 The constants a and b are chosen to fit boundary conditions of the form e g y0 c y 0 0 which yields a c b 0 2D More interesting is the 2D problem where we can choose either rectangular coordinates x y or polar coordinates p Separation of variables in the former case yields LP gtX xY y and Laplace s equation can be written Physics 228 Lecture 24 1 Winter 2008 l d2 l d2 x Y k2 243 Mm We lt gt As usual this makes sense only if both sides of the above equation are the same constant written here as k2 From experience with the relevant special functions we know that the explicit forms of the solutions are I 0c eiikxei y COSkxCOSth k2 gt 0 sin kx sinhky or 244 I 0c eilklxe lkly COShkx COSky k2 lt 0 sinhkx sinky Note that the behavior in each case is sinusoidal completeorthogonal in one direction and hyperbolic not completeorthogonal in the other The specific choice depends on the boundary conditions to be matched As an example consider a 2D space defined by 0 S x S Lx 0 S y S Ly and assume that the boundary condition is P 0 except on the boundary at y Ly where we require P 2 Pl x which as noted may be a function of x This problem might describe the temperature distribution in a thin plate with 3 edges held at 0 and the 4th edge held at T 2 Pl This last boundary condition demands that we choose the case k2 gt 0 with the sine functions in the x coordinate It is in this coordinate ie the coordinate along the boundary with the nontrivial boundary condition that we find an eigenvalueeigenvector problem and a complete set of orthogonal functions This is just where we need them and this feature will remain true in all of our examples Overall the logic proceeds as follows As noted above we choose k2 gt 0the upper line in Eq 244 in order to obtain complete functions in x The boundary condition at y 0 LPx0 0 means select the sink ky form and not cosh ky The boundary condition at x 0 LP0 y 0 means select the sin kx form and not cos kx Finally the boundary condition at x Lx LP Lx y 0 provides the eigenvalue condition sin ka 0 gt kn mrLx Thus via linear superposition we have the general form ie this form satisfies Laplace s equation and the 3 vanishing boundary conditions atx0xanndy0 Physics 228 Lecture 24 2 Winter 2008 Pxy ia sinsinh 245 rpl x x The final nontrivial boundary condition says lI xLy 2 Pl x 2aquot sin sinhLmzLy 246 Note that as required on the physics side and automatically respected by the mathematics side the arguments of the sines and cosines both harmonic and hyperbolic ie the special functions of mathematics are dimensionless quantities even in a real physics problem where x y and LLy all have dimensions of length meters As usual the relevant eigenfunctions here the sin mmL functions are orthogonal Lx jdxsin m7 sin quot x 256 247 0 Lx Lx 2 and complete so that we can fit the boundary conditions on y Ly Thus we can solve for the coefficients at least implicitly from Eq 246 in the usual way by projecting onto the basis functions ie multiply through by the function on both sides of the equation integrate and use Eq 247 LX CO L Lx jdxsinm xrlx Zansinhmr ydxsinm x sin 0 Lx 711 L 0 L L 7rL 7rL Zansinh quot y an amsinh m y E 248 L 2 L 2 x Lx gt am 2 J dxsinm 7m l 1 SinhmLy L 0 L The resulting solution vanishes on 3 sides of the square and matches the nonzero boundary condition on the fourth side Physics 228 Lecture 24 3 Winter 2008 Note that if the boundary condition is zero on all 4 edges there is no nontrivial choice of functions we cannot have sines and cosines for both the x and y behavior which is the right answer ie P 0 everywhere If Vz l 0 is valid everywhere in some region no sources anywhere and P 0 on the boundary of the region then P 0 everywhere in the region Of course the discussion above corresponds to a special case with just one nonzero boundary condition In general the boundary conditions will be nonzero on all four sides e g 249 The essential point is that these conditions can be treated as 4 separate problems like the one above with zero boundary conditions on 3 sides and appropriate functions to fit the behavior on the fourth side Then we use linear superposition to obtain the solution that matches the full set of nonzero boundary conditions In a shorthand notation the individual functional forms for the 4 separate boundary condition problems are note the complete orthogonal functions along the boundary in each case mrx m PxLy l l x quot2116quot s1nL x Sll ll lL xy 00 L 1l x0 Pz x 2bquot sin sinhLyy quot1 Lx Lx 1 Lxy W3 y i0quot sinrlL sinh quot1 y y 2410 1l 0y 1l 4yncin sinnL sinh y We solve for the individual coefficients awbwcwdn by matching the 4 separate nonzero boundary conditions yielding expressions like Eq 248 Physics 228 Lecture 24 4 Winter 2008 Lx b 1 ijdxsin y x smh mrLy Lx 0 Lx LX 1 2L mry cnz L Idysin L 1 3y 2411 sinhmL y 0 y Ly L l 2 y mry d dysm 1 y L l l 4 Finally we simply sum the 4 series to obtain the full result On each edge we are summing 3 zeros plus the 1 correct nonzero behavior which is just what we want Since the solution is unique we are done If the boundary conditions are specified in polar form ie in terms of 0 and we should separate in those variables P gt R pqgt Here we can think of describing the time independent temperature distribution in a thin so no variation in 2 disk Laplace s equation now takes the separated form 2 13 a T6T0 Zap 5 MW d d 1 d2 2412 32 pRp zwgmz de dp d If we have a complete disk then the boundary condition is that LP should be periodic in This results in eigenfunctions that are oc ei39m or 0C sin m COS m with eigenvalues m integers We eXpect the nonzero boundary conditions to be specified at a fixed value of 0 LP p p0 2 PO gb Physics 228 Lecture 24 5 Winter 2008 Note that as promised the complete set of eigenfunctions em are functions of the coordinate along the boundary ie The radial equation has solutions of the form R p 0c p which yield note this is not Bessel s equation p paa2m2aim 2413 As we have now come to eXpect one solution is well behaved at 0 0 but poorly behaved asp gt so while the other solution has the opposite behavior In the special casem 0 where the obvious regular at the origin solution is R constant the irregular solution is R p 0c ln p the reader should verify that this is a solution of the above equation when m 0 For our consideration of the interior of a disk including the origin we choose the regular solutions We now have a typical Fourier series eXpansion T 27 lPP Zlamp cosm Zlmmsinm 2414 quot71 ml with the coefficients matched to the boundary condition at 0 p0 lP00 lPo Zampg cosm mepg sin m ml mil am Ci g g d cosm 1 o 2415 1 2 bm 7mg g d s1nm l 0 The factors of pg in the coefficients which arise automatically in the mathematics serve to remind us that a in a physics problem p and p0 will generally both have dimensions of length meters and b we cannot sum up temis with differing dimensions ie the real eXpansion parameter must be the dimensionless ratio Dp0 we must divide p by some dimensionful quantity and p0 is the only choice Physics 228 Lecture 24 6 Winter 2008 L110245 215 pi cosm 217 pi sinm LPPo To amp ia m cosm i5 sinm 2 W1 1 2416 12 am d cosm qj0 12 b 7 d smm Note that the temperature at the center of the disk can be nonzero a0 7 0 only if the boundary condition at p 2 p0 has a nonzero average value Johdgb l o 0 The reader is encouraged to consider how the form of the solution changes if the disk becomes an annulus not including the origin Now there are two boundaries in 0 and we must keep both solutions for R 0 039quot and pquotquot What if we consider a wedge a piece of the pie 0 S S b0 lt 27 instead of a full disk In this case there are three boundaries p p00 S S 0 0 S p S p0 0 and 0 S p S p0 0 As with the rectangle above we treat this situation as three separate problems with a nonzero boundary condition on only one boundary at a time and then use linear superposition The motivated student is encouraged to consider which sets of complete orthogonal functions are required for each configuration An example is exercise 13513 in Boas where 0 7r 4 and only the boundary p 2 po 0 S S 0 has a nonzero value In this case we need the complete set of angular function obeying f O f 7r 4 0 By inspection these are just the even sine functions fquot sin4n 3D Finally we can consider the case for structure in all 3 spatial dimensions If the boundary conditions are specified in rectangular coordinates we consider P gt X xY yZ z and Laplace s equation looks like 1 d2 1 d2 1 d2 X Y Z 0 2417 de2 xYdy2 y2d22 Z Physics 228 Lecture 24 7 Winter 2008 As before we solve this equation by assuming that each term is a constant say 163163163 and that the sum ofthe constants is equal to zero 16 16y2 1sz 2 0 Typically two of the constants are negative corresponding to sinusoidal completeorthogonal behavior in 2 dimensions while the third is positive corresponding to hyperbolic behavior As in the 2D case we think of the problem with boundary conditions specified on each of the 6 sides of the cube OSxSLxOSySLyOSzSLz I xyLz P1xy I xy0 P2xy Lxyz P3yz yZ L z I P0 P4yz x hE 2418 I y 2 5xz I x0z P6xz We treat each case as a separate problem with zero boundary conditions on 5 sides and the specified function on the sixth side For example for the first case we take 16 m7rLx 2 lt 0 k n7rLy2 lt 0 sinusoidal behavior in x and y and k k k m Lx 2 n Ly2 hyperbolic in the z direction Similarly to the 2D example above we take the sinh kzz function to match the vanishing boundary condition at z 0 We choose the sine functions in x and y to match the vanishing boundary conditions at x 0 and y 0 Finally the vanishing boundary conditions at x Lx and y Ly provide the eigenvalue constraints as above In the same shorthand as in Eq 2410 above the forms of the 6 solutions look like m x sin sinh z Lx Ly Physics 228 Lecture 24 8 Winter 2008 2 2 LP0yZLP4yZrZdensinmL ysmsmh Lx x LI xLyz I SxzZifmsinInijsinsinh y x z lsmlelmh w relief Expanding each of the boundary distributions 1 in the appropriate double Fourier oo LI x0z 2 1 6 xz Zigm sinmL7rx ml 711 expansion yields the 6 sets of coefficients For example we have 1 4 Z 2 2 LxLy sinh L quot 1 E z L L x y Lx Ly x Jde alysin n ijsinquotif y1 xy 0 0 Lx Ly After finding each of the 6 sets of coefficients we simple sum the resulting 6 series to find the overall solution A specific example where only one of the sides of the cube has a nonzero boundary condition is exercise 1359 in Boas Here Lx Ly Lz 10 2419 and P2 100 with all the other Pk s and corresponding coefficients vanishing We have Physics 228 Lecture 24 9 Winter 2008 10 10 LI 0yz LI lOyz LI x0z 2 11 x10z LI xy100 2420 of 4 22 LI xy0 100 Zme7ns1n 10 jsin 10 sinh7r m n ml quot1 1110042 2 iii Sinn 7 Csin jsinhM m ml quot1 Using the orthogonality of the sine functions 10 10 r r 2 J Idxdysin mzxjsinm xsin sinw 6W6quot 2421 0 10 10 10 10 2 0 we have b E 2sinh7nm2n2100 1J9gdxd sin 7 sin E m quot 2 0 0 y 10 10 20 2 andn odd 100 x mm 2422 7r 0otherwise 3b 100 42 im andn odd mzi x mm 51nh7 lm2 2 7r 0otherwise Thus Pxyz 100 2m 1 7m sin 2 101 sinh10 1 04 2m 12 2n if 00 00 sin 10 x2 0 2m 1 2n1 sinh7r2ml2 2nl2 If the boundary conditions are specified on cylindrical surfaces then we should separate variables in cylindrical coordinates p 2 We think in terms of a nite Physics 228 Lecture 24 10 Winter 2008 cylinder as composed of 3 surfaces the curved surface p 2 p0 O S lt 27 O S 2 S L2 and2 ends OSpSp0OS lt27rzO and OSpSp0OS lt27zzLZ As usual we split the problem up and treat only one nonzero boundary condition at a time On the curved surface we have the 2D orthogonal set of functions gm and cosm sin which can be used to match any behavior on that surface but vanishing on the ends of the cylinder 2 0 z L The corresponding radial dependence is given by J m n7rp Lz ie the separation constant k used earlier in our study of cylindrical coordinates has the value k mrLz arising from the eigenvalue problem of the z dependence vanishing at the ends The form of the solution used to match the 2D boundary conditions in pand on the ends of the cylinder was discussed in Lecture 23 In the shorthand notation used earlier to specify the form of the solution for each boundary condition the complete solution has 3 components that look like as in the earlier discussion x is the nth zero of Jm Jm xmyn 0 iam cosm sin n1 z sinm sin Ms L11p0 z 21111 Z H o m PP 0 Pzp 2 cosm smhMJm m0nl p0 p0 msmm 51 p0 m p0 LIJ0 Lz EPA02 i 00 fmvn COSm sinhJm M m M1 gmynsmm s1 p0 m p0 In each case the solution vanishes on 2 of the 3 surfaces and provides complete orthogonal functions in 2 dimensions on the third surface These orthogonal functions are in turn specified by two sets of eigenvalues This allows us to satisfy Physics 228 Lecture 24 11 Winter 2008 any set of boundary conditions on the 3 surfaces and provides the full solution Via linear superposition Finally consider the case of spherical boundary conditions as described in Lecture 22 If the boundary conditions are given on the surfaces of complete spheres then we know the solutions have the form again in our shorthand notation 00 1 Pr06 2 Pl 6 Z Z chmrl dmr 11m cos6 2423 0 quot771 where the spherical harmonics YLm cos 6 provide complete orthonormal functions on the surface of a sphere If the origin is included in the physical region then we know that all d 17quot 0 The reader is encouraged to consider how the solutions might look if we have only a fraction of a sphere ie when boundary conditions are specified on the section of a sphere In the next Lecture we will consider what changes when we include time dependence the Helmholtz equation and sources Poisson s equation Physics 228 Lecture 24 12 Winter 2008 Lecture 22 Legendre Polynomials H the General Eigenvalue Problem More of Chapter 12 in Boas In this lecture we want to finish our discussion of Legendre s polynomials and associated functions Of particular interest is the question of treating the polynomials as solutions of an eigenvalue problem or more generally a SturmLiouville Problem The underlying idea is to prove that when we solve an eigenvalue problem now in function space the corresponding eigenfunctions constitute a complete orthogonal function set on the specified coordinate space This is analogous to what we saw for the finite dimensional eigenvalue problem in terms of vectors and matrices In that case the problem was finite dimensional involving ncomponent vectors and n X n matrices Here we have a continuous free variable and the results are infinite dimensional We start by defining a general eigenvalue equation in terms of a second order differential operator Op via first the eigenvalue equation and then the definition of the operator Op 91quot x 1npxwn x d d 221 om x uxquot m q xm x where in is the eigenvalue 11 is the corresponding eigenfunction px is a general weight function and px and qx are general functions of x This equation is meant to be valid on an interval a S x S b which may be infinite For the Legendre case we have px l x2qx 0px 1 with a 1b 1 The general problem is further specified in terms of boundary conditions at x ab These may be expressed in terms of the behavior at the boundaries of wquot Dirichlet 14 Neumann or a combination For the Legendre case we required that the Illquot be finite at both a and b We proceed by considering the following integral for 2 of the eigenfunctions completely analogous to what we did in the finite dimensional matriX case We have b Icky209 vn010 w d idxwgpxgwnx WnPXEVmx Physics 228 Lecture 22 1 Winter 2008 dx dx pxwiw w if b 222 mdx quot quotdx m a b r 7 7 r d d b xp x 11quot lJm lml 7 x Vm VnVn Vm Id gt H gt where we have performed one integration by parts The final integrand in the next tolastline involves the expression 14 14 14 14 which vanishes because the order of the eigenfunctions does not matter ie they are ordinary functions not operators or matrices and commute For all cases of interest the boundary conditions are such as to guarantee that the final expression in this equation vanishes pxwiw V illl b 0 223 mdx n ndx m a 39 For example in the Legendre case the eigenfunctions are finite at a 1b 1 but px vanishes at both points Thus when we consider the corresponding combination of the righthandsides of the original equations we have 02 n jdxpxlI xlnx 224 From this equation just as in the matrix eigenvalue problem we are led to 2 conclusions First for nonidentical eigenvalues in i it the corresponding eigenfunctions must be orthogonal with respect to the weight function px b jdxpxwxwnx02n 722 225 By considering a single eigenfunction m n where jdxpxwxwnxldxpltx a M x2 i 0 226 Physics 228 Lecture 22 2 Winter 2008 we leam that the eigenvalues are real it it 2139 1min 0 227 For the case here of the real valued Legendre s polynomials we have m 2 n I dem xPn x 0 228 We can also see this result and the case m n by returning to Rodrigues formula Consider 1mquot 2 idem x 1 1 m 2 m dquot 2 quot 1gt lldxxx 1gt l 71 229 Let n be the larger integer m lt n and perform n integrations by parts noting that the factors of x2 1 ensure that the integrand always vanishes at the endpoints We find 1quot 1 dmn 2 m 2 n mnzzmgn nzlnljldxwx 1 1 J20 2210 where the last step follows from m n gt 2m ie m n derivatives on a polynomial of degree 2m yields zero This agrees with the general SturmLiouville analysis above On the other hand we have m 22 1m2j1dx6bxz1mlxz 1 2mm 2 Wildth 1 Physics 228 Lecture 22 3 Winter 2008 2 39 ja nxqm 22 quot m 12 1 2ml 39 l 1 17x 74111711 22quot m 0 2211 This last expression defines a special function the Beta function of Chapter 117 in Boas this function was the basis of the original dual models circa 1970 that eventually led to string theory idxxpl 1 x 1 1 EBpq W W 2212 From the properties of the factorial Gamma functions we have mm 22m2 2071021 2 2 2 1 1 m m n 2213 2 I d P P 3 mn 1 xmx quotx mm The Legendre polynomials are also a complete set of functions on the interval 1 S x S 1 in the sense that 21 1 ZTBxPx5x x 1Sxx 1 2214 0 and we can eXpress any function on this interval as a sum of Legendre polynomials f x 2011310 0 1 2 2121 delxfx 2215 C We have not eXplicitly proved completeness here except by implication we have not Physics 228 Lecture 22 4 Winter 2008 found any nonzero function on the interval 1 S x 1 for which all of the cl vanish We can obtain the analogue of Parseval s Theorem which we interpreted as one way of stating completeness in the case of Fourier series expansions Consider the average square value of our generic function f x l 1 2 l 1 Ejdxf Ejdxl clcl x 7 2216 1 w 251139 w lcll2 2 26 6quot 211 Z21139 ll 0 10 The full length squared of the function f x is reproduced by the sum of the squares of the amplitudes along the basis vectors provided by the Legendre polynomials including the weight 1 21 1 which you will leam corresponds to one over the number of states with a given total angular momentum 1 Note that as a result of the fact that the Legendre polynomials are orthogonal and of degree I the expansion of an arbitrary polynomial of degree In will include Legendre polynomials only up to degree m m Poly xm ZCIP 2217 0 As a specific example consider the case discussed in Section 129 of Boas the step function at x 0 l0ltx l x 0 1sxlt039 2218 From Eq 2215 the coefficients in the Legendre polynomial expansion of this function are given by 21 1 1 21 11 T delxfx T delx 2219 Cl Physics 228 Lecture 22 5 Winter 2008 Perhaps surprisingly we can easily evaluate this nal integral using one of the identities we obtained in the preVious lecture Eq 2138 We have recall that Fn 1 nFn 11de 11dx1 00 51 000 51 7 21 1 1 1 1 clgt0 Tjdex EIdxltPM x Pl1 0 0 Z 131x P1x1 1110 4110 2 2 0 2 2220 Oleven gt 0 l l 11712 FEZJ 1112 FE2lj 1 Odd irwru rw re The last step follows from Eq 2131 We can simplify somewhat to obtain CHM 4 1 12 ll 12 211 212w rmrw 3 17 7 31111 11 clz gc3 c5 4 28 16 22212 32 These are the numbers obtained in Boas but here we have a closed expression for the coefficients in terms of the Gamma function Thus we have Physics 228 Lecture 22 6 Winter 2008 xz RxRx I x1 x m 1 my 22 23Fn1 4ltn1Pzn1x Next let us return again to Laplace s equation and include dependence on the azimuthal angle Recall that in spherical coordinates we have 2 V2T7i23r23 21 iKsinQ Tja ZT r 6r 6r r 1119 66 66 r s1n 95 2223 0 Again we separate variables 1 r 62 R r 6 and multiplying through by r2 and dividing by LP we have li rziRism i jd zz 0 2224 R dr dr s1n6 d6 d6 CDSll l 6 d In our discussion of the Legendre polynomials above we assumed that was a constant Here we want to remove that simplification and consider general but periodic behavior on the interval 0 S lt 27 Since the choice of the plane where O is arbitrary just like the choice of the orientation of the zaXis we eXpect that the functions describing physical systems will be smoothly periodic ie continuous through this plane We already know from our study of compleX numbers and Fourier transforms that such periodic functions can be faithfully represented by the complete set of basis functions em with integer m 00 lt m lt 00 So we should take one of these to represent the characteristic behavior for q and then again use linear superposition in the end to represent the most general behavior Thus with this Ansatz the last term in Laplace s equation becomes Physics 228 Lecture 22 7 Winter 2008 l d2 m2 CD CD sin2 9 d 2 sin2 9 2225 Hence with the same Ansatz for the radial terms as above 1 d 2 d r Rr lll Rr dr dr 2226 we obtain the Associated Legendre s equation 2 1 sinHiG 111 m2 0 s1n 9 d6 d6 sin 6 2 1 1x2i x 111 m2 x0 dx dx 1 x 2227 d2 2x d 1l1 m2 3g x WE x 1x2 W The fact that there are now double poles at x il in the coefficient of the zero derivatives term does not change the fact that x 0 is a regular point no singularities in the coefficient functions while x i1 are regular singular points only simple poles of order 1 in the first derivative term and order two in the zero derivative term Thus the analysis goes through much as in the ordinary Legendre case The boundary conditions are the same and the SturmLiouville analysis is identical except that now q x m2 1 x2 instead of zero It is straightforward to verify that the Associated Legendre s equation is solved by the Associated Legendre function defined by m2 dm leE1 x2 1 m dml 21 1x22Wx2 112mgl 2228 0mgtl Physics 228 Lecture 22 8 Winter 2008 Note that there is no associated function for gt1 and that the above formula actually works see problem 12108 in the HW for l S m lt 0 so that Pf x 1 l m P x 2229 As solutions of an equivalent SturmLiouville problem the Associated Legendre functions constitute complete orthogonal sets of functions on the interval 1 S x S1 for each value ofm 12 m 1 m m 2 lm J1 dxP xPn x glm5n 2230 We can now define an orthogonal and complete set of basis functions on the surface ofa sphere Q 0 S 6 S 7r0 S lt 27 in terms of the double sum 00 I Z Z chP cos 6equotm 2231 0 l l m7 The conventional normalization and symbol for these combinations called the spherical harmonics are YLm cos6 Ylm cos6 E Mpllml cosgeim 1m 1m 2 0 2232 4 1lml lmlt0 The spherical harmonics are the eigenfunctions with definite total angular momentum IZI 1 and z component of angular momentum L m L S L They have the following properties Physics 228 Lecture 22 9 Winter 2008 erm j dcosHqu Ylj m cos6 Yym cos6 J J dQYlj m Q 7 6 6 2233 11 mm39 gr opm Q 69 Q a 6cos6 cos6 6 39 F 9 231mm 9231 I I d916 QFQ In terms of the addition theorem of vectors the angle between two vectors r 9 cosy cos600s6 sin6sin6 cos gb we can also write 21 1 ZYMQYJ Q 259 9 22 4 Pcos 2234 lm 0 which just corresponds to choosing to measure the polar angle with respect to one of the defined vectors instead of the original z aXis so that only m 0 contributes The above analysis means that we can expand any function on the sphere as a sum of spherical harmonics Examples of the spherical harmonics for low eigenvalues are l 3 3 Y Y quot cos6Y quot Sll 166 0390 47 1390 47 U 87 Y i3cos26 l Y 1 cos6sin66 22 35 2390 V167 M V87 39 Y2 2 2 Vii sin2 66H quot 327 To close this discussion let us retum again to Laplace s equation To find a solution to the radial part of the equation we try the Ansatz R r N r Substituting we find Physics 228 Lecture 22 10 Winter 2008 REVr2Rrgtr5r2r l ll1 1 C V 2236 gt l 1 1 r dlrll As suggested earlier we find two solutions one that is well behaved for r gt 0 and one well behaved for r gt so Depending on the specific boundary conditions the solution to Laplace s equation will have the general form see Section 137 in Boas 00 1 yr6 Z Z clumrl dhmr I IJYM COS 11 2237 0 m7l with the constants Chm and d1 chosen to match the boundary conditions For example if the function wis desired between 2 concentric spheres of radii V1 and r2 on which wis speci ed we will need both the clquot and the d1quot to match the boundary conditions on the spheres If the function wis desired inside a sphere of radius r1 on which wis specified and wis finite everywhere inside of the sphere another form of boundary condition the dim all vanish and we will need only the clquot to match the boundary conditions on the sphere If the function wis desired outside of a sphere of radius r1 on which wis specified and wis finite everywhere outside of the sphere another form of boundary condition the clquot all vanish except perhaps 00 0 and we will need only the dim to match the boundary conditions on the sphere Physics 228 Lecture 22 11 Winter 2008 Lecture 15 Inhomogeneous 2nd order linear ordinary differential equations with periodic driving functions Fourier Series Chapter 7 in Boas Our goal is to be able to solve differential equations with the general form 156 bic 0x F t which appear so often in physics In this lecture as in Lecture 5 we will focus on the special case where the righthandside is periodic F t F t T and typically satisfies the smoothness constraints of Dirichlet see below and Section 76 in Boas Our approach is to find a basis set of periodic functions the analogue in function space of basis vectors in the usual vector space such that we can expand any periodic function in terms of this basis set as a linear combination with constant coefficients If we know the particular solutions corresponding to these basis functions we can use linear superposition to write down the particular solution to the general problem as the corresponding linear combination of the individual basis function particular solutions this feature is a major reason why linear equations are so much easier to analyze than nonlinear equations While we will use the language of periodic in time there is nothing special about time as the independent periodic variable We could as well think about x as the independent periodic variable as in Boas We require two general properties of these periodic basis functions orthonormality and completeness in complete analogy with what we required of unit basis vectors These properties are expressed in terms of integrals which replace the more familiar scalar products of vectors as a sum over products of components If we label the basis functions as fquot t n an integer the orthonormal property is typically expressed as note the complex conjugate sz 0k l 71710 I6kl1kzla 151 where the symbol on righthandside is the Kronecker delta function The property of completeness is often expressed in terms of the Dirac delta function see Chapter 811 in Boas Zzz39Ti 50 14 152 Completeness can also be expressed in temis of the average modulus squared of a Physics 228 Lecture 15 1 Winter 2009 function being the sum of the squares of the individual components Parseval s theorem see Section 711 in Boas and below ASIDE The Dirac delta function see Section 811 in Boas is an especially important function and concept that we will touch on several times in this course It is a singular highly peaked function de ned by the integral relations 0 otherwise defx5x x0fxo2xl ltx0 ltng fdx 5x x01x1 ltx0 ltx2 X1 0 otherw1se The integral returns the function f x evaluated at the argument of the delta function as long as the range of integration includes the point x0 x1 lt x0 lt x2 and the function f x is defined at x0 If the point x0 is not in the range of integration the integral vanishes We assume that the periodic functions of interest are well behaved as defined by Dirichlet Section 76 a F bF c F t is single valued O S t lt T 39 t has a finite number of maxima and minima O S t lt T 39 A AAA t has a finite number of discontinuities where it exhibits jumps 0 S t lt T 39 T d F t is absolutely integrable IO dtlF lt00 If these properties hold the expansion of F t in terms of the basis functions ie the Fourier series will converge to F t at all points where the function is continuous and converge to the midpoint at points where F t displays a discontinuous jump Note that on either side of such a jump there is an unavoidable amount of overshoot the series always overestimates the magnitude of the jump by about 9 on both sides This result is called the Gibbs phenomenon and we will study it in detail in an Extra Credit HW exercise Physics 228 Lecture 15 2 Winter 2009 Returning to the question of the periodic basis functions we note that they can be expressed in two ways real sines and cosines or complex eXponentials We will discuss these cases in order Sine and Cosine Series The basis functions sin ngtjpos zgt usually for real functions of t are clearly periodic Using the combined angle formulae sina ib sinacosb icosasinb 153 cosaibcosacosbsinasinb and sina 17 sina b 2 b b cosa 2 cosa 154 cosa b cosa 17 2 sinacosb cosacosb sinasinb we can easily verify periodicity sin n2 t TU Si n2 t2n 2511101243 T T T 155 cos n2 tT cos n2 t2n7r cos nz t T T T We also need the integral properties of these periodic functions integer m Physics 228 Lecture 15 Winter 2009 1T 2 T dtsin m t T l T 1T wsmgzt i 0 ndm 0 T27rm T 0 27rm 056 T 1m0 1 27 T dtcos m T l 7 lZgnn33t 2 1 F0Qm 0 T 27rm T 0 27rm With Eqs 154 and 156 we can easily obtain the other desired properties 0 The sinusoidal basis functions are orthonormal for any integers m and n note the factor of 2 T gIdlsin m2 tsinnZ 7rt T 0 T T ljdtcos m nZt cosmn t26mquot m n T0 T T 0m n T 2Idicos m2 tcosnZ 7rt T0 T T r 1 158 ljdtcosUmn tcosm n t25mquot m n T0 T T 0m n 2T 2n 2 Jdts1nm tcosn tj T0 T T 159 1T 2n 2n Jdts1n mn t s1n m n t 0 T0 T T Note as we have discussed earlier that these results tell us that the average value of S111227Z39I llT is equal to the average value of cos2 27mtT is equal to 1 2 averaged over a full period or many full periods Physics 228 Lecture 15 4 Winter 2009 00 We can expand any sufficiently smooth Dirichlet function F t as Z 27239 j 27239 ancos n t bnsm n t nl T T T an 2 2J Ftcosnz 7Itjdt T0 T T 2J Ftsin nz tjdt T 0 T Ft 1510 bquot The integrals for the coefficients follow directly from substituting the series in the integral and applying the orthonormality conditions of Eq 157 159 Note the conventional factor of 12 in the first term of the expansion so that do is defined like the other coefficients Completeness can be expressed most easily in terms of Parseval s Theorem assuming F t is real 1 5 1 1 gig 173 n which is the analogue of the familiar statement that the square of the length of a vector is the sum of the squares of the individual components defined by a complete set of orthonormal basis vectors Thus there is no extra length along any other not included direction Eq 1511 like its vector analogue follows simply from substituting the series in the integral writing out the terms in F 2 t and using the orthonormal properties of the sines and cosines It is useful for lazy and smart people to recognize some special cases Physics 228 Lecture 15 o Fodd t is an odd function Fodd t F0dd t since the sine function is odd and the cosine is even only the sine series contributes and a 0 for all n Winter 2009 o F t is an even function F t F t now only the cosine series even even even contributes and b 0 for all n 0 Also the integrals for the nonzero coefficients can usually be simplified T2 bquot 2 I Fodd tsinn2 t 0 m 1512 an 2 2 Fmquot tcosn2 t Now consider the alternative expansion more natural for complex functions but recall we can always take real or imaginary parts at the end Complex Exponential Series Now the basis functions are emlMm Due to the simple properties of the exponential function the analogue of Eqs 153 and 154 is trivial eimn27rTt eim27rTtein27rTt 1513 We also have lld em m i T TL T 0 em2 Tt0 21102n 0 1514 Thus we can readily derive the following desired properties be sure to note the complex conjugates o The exponential basis functions are orthonormal where in this case the relevant integral involves a function times the complex conjugate function 1 T magT manT 1 T im7n27rTt Lm 17 71656 e 7 dfe 5m 0mg 1515 Physics 228 Lecture 15 6 Winter 2009 for any integers m and n 0 Again we can expand any sufficiently smooth Dirichlet function F t as F0 2 i Cnein27rTt IT 1516 717127th 91than H where the expression for the coefficient follows from the definition of the expansion and Eq 1515 0 In this case the completeness can be expressed in terms of the Dirac delta function i e quot2 T equotquot2 Tquot i 61 1 nT 1517 where the second sum guarantees the periodic structure If we know the form of a function F t on the interval 0 S t lt T then we can always construct a periodic version from T 00 oo FinnOdie Z 6t t39nT Z cneiquot2 T39 0 11700 11700 The student is encouraged to think about what this result means 0 Completeness can also be expressed in terms of Parseval s Theorem 2 cquot T 00 1Ff2 df 2 1519 0 n7oo 0 Special case If F t is a real valued function then cquot 0 Physics 228 Lecture 15 7 Winter 2009 As an example consider the function in Exercise 1 7 S x lt 0 f x 00ltxlt7r 7 1 n M 1520 In this case the period is T 27 and we use x as the independent variable We have coefficients a 2i i fxcosnxdx i j cosnxdx n 72771 72 77 Ln 0 1521 2 0 M 0n 0 m 7 and 1 1 bquot Eifxsmnxdx sm nxdx 2 1522 cosnx 0 n Odd 717239 n 0neven Thus the function can be expanded as 1 2 sin2n1x x 1523 f 2 7r 2n1 Note that we have expressed the sum over all odd integers as a sum over 2n 1 all n This is a useful way to proceed and has an analogue in using 2n all n to sum over the even integers Physics 228 Lecture 15 Winter 2009 For fun we can use Mathematica see the H notebook for this lecture to plot this series M first for 0 S n S 4 which looks like the figure M to the right And next for 0 S n S 49 note that the oscillations have largely damped out except at the discontinuities where the Gibbs phenomenon remains It is illustrative to also try expanding this function in terms of the complex exponential series We have 1 Sim L 0 finx cquot Eidxf 2 rdxe ln0 2 1524 39 On even 0 l 0 l l n imx 1em7r 1 1 I 27m 7 27m 27m L Odd 7m Hence we have the following expansion 1 l 00 ei2nlx x f 2 7rquot 2n1 1 1 1 ei2quot1x l 700 ei2nlx 2 72M 2nl an12n1 1525 1Limlz W 2 71M 2nl 2 71M 2nl 39 Physics 228 Lecture 15 9 Winter 2009 The last 2 steps make clear that we have obtained the same result as above ASIDE It is informative to consider why the sine and cosine or the complex exponential have these properties of orthonormality and completeness ie why are they so useful As we will study in more detail the answer is that they are solutions of an eigenvalueeigenvector problem These functions are the eigenfunctions of the eigenvalue problem defined by the equation d2 2 were m with eigenvalue m2 and the boundary condition specified by periodicity ftTftwnn7 The eigenfunction solutions have the eXpected orthonormality and completeness properties just as in the finite dimensional case with matrices This will be a common theme in our search for useful functions Now we can return to the starting point of differential equations and write a56bxcxaD alDa2xFt icnemanTy quot700 a i b 2 1526 1392 2all4a2 a where the 05k are either both real or compleX conjugates of each other As usual we try xpyn t 0c 6 m27zT r and find by linear superposition ein27rTt 00 Cquot 3510 2 Z 1527 quot a Inf 051 In 0 2 T Physics 228 Lecture 15 10 Winter 2009 If the original F t is real valued so that cquot c then it follows that particular solution is real and of the form C ein27rTt C 00 t 2 Re quot xpreal C quot1 2 2 1528 a In O1 In O2 ie the term for n mm gt O is the complex conjugate of the n m term In the above example we implicitly assumed that the parameter b in the original equation the damping is nonzero and that the 0le have a nonzero real part Thus the factors in the denominator of Eqns 1527 and 1528 cannot vanish If this is not the case it is possible that a term in the Fourier expansion of the righthandside matches the natural frequency of the original equation ie of the complementary solution We should brie y recall from the previous lecture how to proceed in this case As an example consider sample Exercise 8633 which is included in Lecture 14 Appendix B rewritten to match the notation here The homogeneous equation is 56 x 0 with complementary solution x t A sint B cost If we drive this system with a periodic function F t period 27 whose Fourier expansion contains ann 1 term eg Ftancosmza1 0 1529 we must be careful with this specific term If we naively try a particular solution of the form xp1 2 d1 cost we find 56171 xpl 0 75 all cost Instead as suggested in the previous lecture we must include an extra factor of the independent variable t Further in this real notation we need to use xp1 dlt sint for it to match It is simpler ie smarter and lazier to employ complex notation and take the real part at the end We find Physics 228 Lecture 15 11 Winter 2009 it quot it it 2P1 the gtZp1 2131 Z1 f21fe a1e a a a 1530 gtZ12 1gtxplRejte jts1nt z 1 Thus a complete particular solution of the equation x F t based on Eqn 1529 is given by 2 112 n a a an xpt3031tsmtzl cosnt 1531 The reader is encouraged to verify that this result satisfies the original inhomogeneous equation Physics 228 Lecture 15 12 Winter 2009 Lecture 15 Inhomogeneous 2nd order linear ordinary differential equations with periodic driving functions Fourier Series Chapter 7 in Boas Our goal is to be able to solve differential equations with the general form 156 bic 0x F t which appear so often in physics In this lecture as in Lecture 5 we will focus on the special case where the righthandside is periodic F t F t T and typically satisfies the smoothness constraints of Dirichlet see below and Section 76 in Boas Our approach is to find a basis set of periodic functions the analogue in function space of basis vectors in the usual vector space such that we can expand any periodic function in terms of this basis set as a linear combination with constant coefficients If we know the particular solutions corresponding to these basis functions we can use linear superposition to write down the particular solution to the general problem as the corresponding linear combination of the individual basis function particular solutions this feature is a major reason why linear equations are so much easier to analyze than nonlinear equations While we will use the language of periodic in time there is nothing special about time as the independent periodic variable We could as well think about x as the independent periodic variable as in Boas We require two general properties of these periodic basis functions orthonormality and completeness in complete analogy with what we required of unit basis vectors These properties are expressed in terms of integrals which replace the more familiar scalar products of vectors as a sum over products of components If we label the basis functions as fquot t n an integer the orthonormal property is typically expressed as note the complex conjugate sz 0k l 71710 I6kl1kzla 151 where the symbol on righthandside is the Kronecker delta function Completeness is often expressed in terms of the Dirac delta function see Chapter 811 in Boas Zurmrwiso Mr 152 or in terms of the average modulus squared of a function being the sum of the squares of the individual components Parseval s theorem see Section 711 in Boas Physics 228 Lecture 15 1 Winter 2008 ASIDE The Dirac delta function see Section 811 in Boas is an especially important function and concept that we will touch on several times in this course It is a singular highly peaked function de ned by the integral relations 0 x x lt x lt x jdxfx xx0 fl 0 1 0 2 xi 0 otherw1se 01 lzx lt x lt x Idx6x x0 1 0 2 XI 0 otherw1se The integral returns the function f x evaluated at the argument of the delta function as long as the range of integration includes the point x0 x1 lt x0 lt x2 and the function f x is defined at x0 If the point x0 is not in the range of integration the integral vanishes We assume that the periodic functions of interest are well behaved as defined by Dirichlet Section 76 a F b F c F d F t is single valued O S t lt T 39 t has a finite number of maxima and minima O S t lt T 39 A AAA t has a finite number of discontinuities where it exhibits jumps 0 S t lt T 39 T t is absolutely integrable J1 dtlF lt 00 If these properties hold the expansion of F t in terms of the basis functions ie the Fourier series will converge to F t at all points where the function is continuous and converge to the midpoint at points where F t displays a discontinuous jump Note that on either side of such a jump there is an unavoidable amount of overshoot the series always overestimates the magnitude of the jump by about 9 on both sides This result is called the Gibbs phenomenon and we will study it in detail in an Extra Credit HW exercise Returning to the question of the periodic basis functions we note that they can be expressed in two ways real sines and cosines or complex exponentials We will discuss these cases in order Physics 228 Lecture 15 2 Winter 2008 Sine and Cosine Series The basis functions sin ngtjpos zgt usually for real functions of t are clearly periodic Using the combined angle formulae sina ib sinacosb icosasinb i 153 cosa b cosacosb s1nas1nb and Sinacosbsinabsina b b b cosa 2c0sa 154 cosa b cosab 2 cosacosbz sinasinbz we can easily verify periodicity sin n2 t TU sin n2 t2n7r sin n2 t T T T 155 2 2 2 cos n tT cos n t2n7r cos n t T T T We also need the integral properties of these periodic functions integer m 0m0 27 T dtsin m t l i T 12T cosm2T t 21 10m 0 7239 7rm 0 156 lm0 l T dtcos m t l i i l T smm2 t 0 00m 0 T27rm 0 27rm Physics 228 Lecture 15 Winter 2008 With Eqs 154 and 156 we can easily obtain the other desired properties 0 The sinusoidal basis functions are orthonormal for any integers m and n note the factor of 2 T Jdtsinm2tsinn2t 0 ljdtcosUm nZt cosmn t26mquot m n T0 T T 0m n T Jdtcosm2T 7rtcosn2T 7rt 0 r 1 158 ljdtcosUmn tcosm n t25mquot m n To T T 0m n T Idtsinm2 tjcosn2 tj 0 T ljdtsinmn tjsinm n tj0 T0 T T 159 Note as we have discussed earlier that these results tell us that the average value of Si11227mtT is equal to the average value of cos2 27mtT is equal to 1 2 averaged over a full period or many full periods 0 We can expand any sufficiently smooth Dirichlet function F t as Ft an cos n27 tj bquot sinEngtj Physics 228 Lecture 15 4 Winter 2008 T an 2 2 Ftcosnz 7Itjdt T 0 T 2 T 2 1510 bquot IFtsin n tjdt T 0 T The integrals for the coefficients follow directly from substituting the series in the integral and applying the orthonormality conditions of Eq 157 159 Note the conventional factor of 12 in the first term of the expansion so that do is defined like the other coefficients 0 Completeness can be expressed most easily in terms of Parseval s Theorem assuming F t is real T 2 oo JF2tdtCjT Zan2b 1511 0 n1 which is the analogue of the familiar statement that the square of the length of a vector is the sum of the squares of the individual components along a complete set of orthonormal basis vectors Thus there is no extra length along any other not included direction Eq 1511 like its vector analogue follows simply from substituting the series in the integral writing out the terms in F 2 t and using the orthonormal properties of the sines and cosines It is useful for lazy and smart people to recognize some special cases 0 Fodd t is an odd function Fodd t F0dd t since the sine function is odd and the cosine is even only the sine series contributes and a 0 for all n t F t now only the cosine series even 0 F t is an even function F even even contributes and b 0 for all n Also the integrals for the nonzero coefficients can usually be simplified Physics 228 Lecture 15 5 Winter 2008 T2 bquot 2 I Fodd tsinn2 t 0 4 m 1512 an 2 2 Even tcosn t Now consider the alternative expansion more natural for complex functions but recall we can always take real or imaginary parts at the end Complex Exponential Series in27rTt Now the basis functions are e Due to the simple properties of the exponential function the analogue of Eqs 153 and 154 is trivial eimn27rTt eim27rTtein2 Tt 1 513 We also have lm0 T il l0m 039 1514 0 27m in271Tt 1T J39dtem2 Tt l T T 0 T 27m Thus we can readily derive the following desired properties be sure to note the complex conjugates The exponential basis functions are orthonormal where in this case the relevant integral involves a function times the complex conjugate function l m 9 11 nm 1 T mmT manT 1 T im7n27rTt ngte e dte 6 0m i n for any integers m and n 0 Again we can expand any sufficiently smooth Dirichlet function F t as Physics 228 Lecture 15 6 Winter 2008 F0 2 i Cnein27rTt 1 Two 7 WT 1516 cnEJthte where the expression for the coefficient follows from the definition of the expansion and Eq 1515 In this case the completeness can be expressed in terms of the Dirac delta function i eiquot2 T equotquot2 Tquot i 61 1 nT 1517 14700 where the second sum guarantees the periodic structure If we know the form of a function F t on the interval 0 S t lt T then we can always construct a periodic version from T 00 oo genOdie f Jdf FU39 Z 6t t39nT Z cneiquot2 T39 1518 0 11700 11700 The student is encouraged to think about what this result means 0 Completeness can also be expressed in terms of Parseval s Theorem 1 T F 2 d 0 2 9 t t n0cn 1519 0 Special case If F t is a real valued function then cquot 0 As an example consider the function in Exercise 751 Physics 228 Lecture 15 7 Winter 2008 fx l 7 S x lt 0 1520 00ltxlt7r In this case the period is T 27r and we use x as the independent variable We have coefficients an 2fxcosnxdx icosnxdx ln0 1521 0 smnx Zoynio n7 7 and b i if fxsinnxdx i ii sinnxdx n 7277 7277 cosnx 0 in odd 1522 n72quot n 7 014 even Thus the function can be expanded as 1 2 sin 2nl x fx Z 1523 2 7r quot0 211 1 Note that we have expressed the sum over all odd integers as a sum over 2n 1 all n This is a useful way to proceed and has an analogue in using 2n all n to sum over the even integers It is illustrative to also try expanding this function in terms of the complex exponential series We have Physics 228 Lecture 15 8 Winter 2008 1 2 1524 On even i O z e 27m 0 lm I imr n 421 6 27m1 1 Ln odd 7m Hence we have the following expansion 1 l 00 ei2nlx x f 2 n2n1 1 l39 1 ei2quot1x l 700 ei2nlx 2 7r 02nl 72quot12n1 1525 2 720 2nl 2 72M 2nl The last 2 steps make clear that we have obtained the same result as above ASIDE It is informative to consider why the sine and cosine or the complex exponential have these properties of orthonormality and completeness ie why are they so useful As we will study in more detail the answer is that they are solutions of an eigenvalueeigenvector problem These functions are the eigenfunctions of the eigenvalue problem defined by the equation d2 Wftw2ft with eigenvalue m2 and the boundary condition specified by periodicity ftTftann7 The eigenfunction solutions have the expected orthonormality and completeness Physics 228 Lecture 15 9 Winter 2008 properties just as in the finite dimensional case with matrices This will be a common theme in our search for useful functions Now we can return to the starting point of differential equations and write ai biccxaD alDa2xFIZ 0 Cnein27rTt quot49 b b2 C 1526 0 12 i 2 2a 4a a where the 05k are either both real or complex conjugates of each other As usual we try xpyn t 0c emlhT and find by linear superposition in27rTt 3510 2 i 1527 quot inf 061 inf 062 T T If the original F t is real valued so that cquot c then it follows that particular solution is real and of the form C ein27rTt C 00 t 2 Re quot XPreal C quot1 272 272 a In O1 Inf 062 ie the term for n mm gt O is the complex conjugate of the n m term In the above example we implicitly assumed that the parameter b in the original equation the damping is nonzero and that the 01km have a nonzero real part Thus the factors in the denominator of Eqns 1527 and 1528 cannot vanish If this is not the case it is possible that a term in the Fourier expansion of the righthandside matches the natural frequency of the original equation ie of the complementary solution We should brie y recall from the previous lecture how to proceed in this case As an example consider sample Exercise 8633 which is included in Lecture Physics 228 Lecture 15 10 Winter 2008 14 Appendix B rewritten to match the notation here The homogeneous equation is 56 x 0 with complementary solution x6 t A sint B cost If we drive this system with a periodic function F t period 27 whose Fourier expansion contains ann 1 term eg Ft ian cosntza1 70 1529 nl we must be careful with this specific term If we naively try a particular solution of the form xp1 2 d1 cost we find 56131 xp1 0 7 all cost Instead as suggested in the previous lecture we must include an extra factor of the independent variable t Further in this real notation we need to use xp1 dlt sint for it to match It is simpler ie smarter and lazier to employ compleX notation and take the real part at the end We find 2P1 the gt 39Z39p1 2p1 Z1 I 2139 te ale a a a 1530 gtZ12 1gtx11Rejte jtsmt z 1 Thus a complete particular solution of the equation x F t based on Eqn 1529 is given by a a w a x t 01tsin n COSI lt 1531 I 2 2 gl nz I The reader is encouraged to verify that this result satisfies the original inhomogeneous equation Physics 228 Lecture 15 11 Winter 2008

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.