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Partial Diff Equations

by: Otilia Murray I

Partial Diff Equations MAT 118A

Otilia Murray I
GPA 3.88


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This 10 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 118A at University of California - Davis taught by Staff in Fall. Since its upload, it has received 70 views. For similar materials see /class/187393/mat-118a-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15
5 Fourier series Just like Taylor expansions7 Fourier series is a way of expressing a function as an in nite sum of other functions In the case of the Taylor expansion7 we write n m E f mow which means that we expand f with respect to the basis 1717127 I I In the case of Fourier cosine series we write a0 m7 fzgtltlamcoslt b m1 which means that we expand f with respect to the basis 17 cos 7 cos 7I I I in the case of Fourier sine series we write m7 Z ams1nlt b gt m1 which means that we expand f with respect to the basis sin 7 sin 7 I I I and in case of the full Fourier series we write a m7 m7 7 0 fz7 Zamc0slt b gt mes1nlt b m1 m1 which means that we expand f with respect to the basis 17 cos 7cos 7I I I 7 sin 7sin 7 I I Let us now nd out how to compute the coef cients a07 a17 I II and 12171227 I I II The following is more or less from m 51 The coe icients The reason it is useful to expand in terms of cosines sines or cosines and sines is that they possess the following property DEFINITION 51 Let f andg be functions Then we say that f andg are orthogonal in L2Q f7gL2Q 07 where 7 39L2Q is the L2Qinnerproduct Let us now see that any two distinct elements in the collection l7cos 7cos 7 I I and any two distinct elements in the collection sin 7sin 7I I are orthogonal on 07bI LEMMA 52 For distinct m and n we have Ob COS COS and we also have for m f 0739 and for m 0 we have Also7 for distinct m and n we have 7 sin dz 0 1 sin2 dz and we also have PROOF See homework I Also7 any two distinct elements in the collection 1 cos 7cos thogonal on 7 Hi7sin7isin are or LEMMA 53 For all m and n we have for distinct m and n we have and m7rz mrz 7bbsinlt b gtsinltTgt dz0 and m7rz m7rz 7bbcoslt b gtdz7 s1nlt b gtdz0 Also and for m 0 PROOF See homework I Let us now see how lemma 52 will help us expand a function f de ned on 0712 Suppose that we have already found an expansion in terms of cosines for That means that we can write 51 g Z amcos on 0712 n7rz Fix n and multiply 51 by cos T and integrate over 07 12 Then we have 52 0b cos dz 0b cos dz 5 3 Abamcos COS dz The value on the right hand side of 52 depends on what n is If n 07 then by lemma 52 we have 2 2 a0 54 7 zdz7 idza lt gt bM gt b M 2 0 which determines what do must be If n gt 0 we have 55 gmb cos dz gig0 11 am cos cos dz an which determines what an must be Thus if we can expand f in the form 51 Then the coef cents a0 a1 a2 i i must be determined by 54 and 55 Similarly if we can write 56 f2bmanlt fzmqm then the coef cients must be given by 2 mrz 57 In Emb s1n dzi And if 58 70 2 am cos 2 12m sin on 71212 m1 m1 then the coef cients must be given by 1 7mm i n 7 7 gt 5 9 a bibbfzcoslt b gtdz forn70 1 5 10 12 Eiw sin dz for n 2 1 Now we know what the coef cients must looks like we can compute some examples EXAMPLE Let us nd the sine Fourier series expansion of the function f I on 012 The Fourier sine series coef cients are given by 57 and therefore 2 mrz an 34M mkflm 2 712 7mm 2 12 7mm 5 12 a ll Tll 0 t W Am Tl dz 2b 2 b 2 mrz 5 13 7 cos mr g s1n0bdx 5 14 mr Therefore if the Fourier series converges then we have 2b7l 1 7mm Z i lt7 n1 mr on 0 12 EXAMPLE Let us nd the cosine Fourier series expansion of the function f I on 0 12 The Fourier cosine series coef cients are given by 54 and 55 Therefore 2 2122 7 zdz77b b 0 5 b2 00 and 515 an my 1 cos dz am who 51 772 Therefore if the Fourier series converges then we have 12 4b nnz 1272n27r2008lt 12gt n1 on 0 by But just like the case of the Taylor expansion there are issues with convergence 7 for the Taylor expansion there is a socalled radius of convergence as you may recallI Let us rst see what we mean by convergenceI DEFINITION 54 We say that a sequence of function f1 f2III converges to a function f in LNG if 7 meme A 0 EXAMPLE Let us see if the partial sums of 2211 7 zz 1 convergeI We have N 518 217 zz 1 17 z17 zz 17 zz2 I I I 1 7 zzN 1 n1 519 17111III1N1 520 17 zNI For lt 1 we have IN 7gt 0 as N 7gt 00 for z 71 the partial sum oscillates for z 1 we have IN 1 and for gt 1 the partial sum divergesI Let us therefore only consider I such that 71 S I S L For 0 S p lt 00 we have 17 N 521 17 Zn 7 any which converges to 0 as N 7gt Go For p 00 we have N 17 217 zz 1 n1 This means that the sum converges in Lp711 for p lt 00 but not in L 711I 2 lszldz 7 71 711 NP1 5I22 maxlle Iz 7111I L 711 In our case the approximating functions f1 f2 I I I are going to be the partial sums a0 a0 7m a0 7m 27m 77a1coslt7gt3a1coslt7gtagcos 7 III 2 7 2 b for example and we are hoping that the partial sums will converge in L17 for p 2 or 00 to the function we are expanding or Fourier series we will see two different theorems regarding convergence one for convergence in the L2 norm and one for convergence in the L normI Let us begin with the convergence result for the L2normI 21 52 LZconvergence In this section7 we will prove the following theorem for the sine series on 07 b THEOREM 55 Let f be in L207b Then the Fourier sine series converges to f in L207 b The proofs of the analogous results of the Fourier cosine series and the full Fourier series are very similar We begin with a lemma which shows that the Fourier sine series of f is the closest thing to f which is in the form of a Fouries series LEMMA 56 The least squares approximation Let f be a function in L207b Fix an integer N From the collection of all possible c17 7 cN7 the Fourier coe cients b17 7 bN minimize N ENltClyuchgt Hf nglcmsm L2ob PROOF We have N 2 2 7 7 mrrz 523 EN 7 E cm s1nlt b m1 L20b 524 0b lfzl2d172mI1cmOb fzsinltmrzgtdz 525 i i677an Sin Sin d1 01 Because of lemma 527 the last term in 523 is equal to N 2 2 mrrz d 7b N 2 72cmObs1nlt b gt r7 Zcm m1 Thus we have N 2 uplg cm 7 gOM sin dz OM sin dz by completing the square The right hand side of 526 is minimized when 2 mnz cm Emb fzs1nlt b gtdz that is when c17 7 cN are the Fourier coef cients I 526 Ei lfrl2dr 1 01 b N 527 7 5 Z m1 This means that the Fourier expansion is the closest thing to f in the subspace spanned by vectors sin 7sin 7 Now we prove that the Fourier sine series converges somewhere we do not yet know what it converges to7 but we hope that it is LEMMA 57 Let f be a function in L207b and let 21 bm sin Lg be the Fourier sine series of Then 21 bm sin Lg is in L207b PROOF Let7s measure the size of 21 12m sin Lg in the L207 blnorm N 2 N N mwz 7mm mrz 5 28 7an s1n Lamb Egbmbn Ob s1nlt b gt s1n dz b N 529 E W bf b 00 531 E 2 b3 S llflliqoiy Thus from 528 and 531 we have S llflquoiy L2ob 0 mwz 2 12m s1n lt b gt m1 Since f is in L207 b HfHsz lt 00 which means that Do mwz 2 12m sin lt b l converges to something in L207 by I A second essential property of the bases we are considering the rst was the orthogonality is that in some sense they cover all possible functions in L DEFINITION 58 A collection offunctions fl7 f2 I in L2Q is said to be complete in L2Q f7 fmL2Q 0 for all m implies that f 0 g 27 its We a are complete in L20b and Lcos 7cos IIWsin 7sin LEMMA 59 is complete in L27b7 b PROOF This is a complicated result7 and we will not prove it here For a proof7 see I LEMMA 510 Let f be a function in L20b and let 21 12m sin Lg be the Fouiier sine series of Then bmsin in L207 b PROOF Fix it Then 532 AM 161 7 quot12me sin dz 5 33 My sin dz 7 i hm 0b sin Sin d1 m1 bn bnb 7 534 7 7 7 7 7 0i This means that for all n7 535 in 7 min my gt a b 7 or Since sin 7 sin i l is complete7 we therefore know that bmsin l Finally7 from the orthogonality of the basis elements7 we can deduce Parsevalls identityi PROPOSITION 511 Let f be in L20b and let 21 12m sin Lg be the Fouiier sine series of Then b 00 llfllf oi i Z I 1 m Let us now work on some examples EXAMPLE Consider the function l on 0712 It is clear that f is in L207 by We can therefore nd the Fourier sine expansion of f and expect it to be wellbehaved The coef cients given by 57 5 12m gO b sin dz b 537 7 cos 0 538 l 7 cosm7r 5 39 17 71mi This means that if m is even we have 12m 4m7r and if m is odd7 12m 0i 53 Lmconvergence In this section we state a theorem concerned with L convergence THEOREM 512 Let f f and f be continuous on 0712 The Fourier sine series converges in L 0b to PROOF This is a complicated theoremi We will omit the proof I 54 Existence for Laplace7s equation Now we have the technology to consider the existence proof in one special setting In this section we will consider two problems 1 Find go such that Ago 0 on Q with boundary condition go g on 89 2 Find go such that Ago 0 on Q with boundary condition Vgo N g on 89 The following discussion is from 2 and can also be found in 7 and We will consider these two on the very special domain 9 E R2 z 0 S I S a and 0 S y S b We will consider a very special form of solution gozy That is7 the solution will split into two parts one part which only depends on I and another which only depends on y This technique is known as variables separation Since we know that solutions are unique7 we know that if we were to nd a solution of this form7 it is actually the only solution to this problemi Let us apply the two dimensional Laplacean to gozy The equation we wish to solve then is X IYy XzYy 0 on 9 This means that we want to solve X I 7 Y y X 1 Wu 540 Since the left hand side of 540 depends only on I and the right hand side of 540 depends only on y both sides must be equal to a constant or Thus we have 541 X z cXz and Y y foly The ODE 541 is a differential eigenproblem and depending on the sign of the eigenvalue 5 eigenfunction X and Y changes 1 If c lt 07 then we can write 5 7A2 with A gt 0 and from 541 we have 542 Xz A sinz B cosz and Yy Csinhy D coshyi 2 If c 07 then 543 Xz A Br and Yy C Dy 3 If c gt 07 then we can write 5 A2 where A gt 07 and we have 544 Xz A sinhz B coshz and Yy Csiny D cosyi To see which solution we want7 we have to be more speci c about the boundary values we want 25 541 A Dirichlet boundary condition Suppose that we are considering the Dirichlet problem problem 1 from above where the boundary data 9 takes the following form 545 9170 0 546 gzb 0 547 NM 0 5 48 90 My where h f 0 If h 0 then the obvious solution would be so 0i Let7s begin by checking if option 1 can satisfy the above boundary conditions From 545 we see that 549 0 90170 XIY0A Suppose that Xz 0 Then gammy XIYy 0 This means that the boundary condition 548 will be violated because h f 0 So we must have Y0 0 This means that 550 0 Y0 Csinh0 Dcosh0 Di From 546 we see that 551 0 IJI XIYbl Suppose that Xz 0 Then gammy XzYy 0i Again7 this means that the boundary condition 548 will be violated because h f 0 So we must have Yb 0 This means that 552 0 W Csinhbl Since sinh is only zero in one place and that place is not Ab gt 0 we must have C 0 This means that Yy 07 which means that gammy XIYy 07 which is a violation of the boundary condition 548 This means that option 1 will not satisfy the boundary conditions 545 548 If we were to try option 2 we would nd the same thing We therefore hope that the third option can provide a solution which satis es the boundary conditions Let us therefore see what happens if we let Yy C siny D cosyi From 545 and 546 we can ascertain what A and D must be First we have 0 Y0 Di We also have 0 Yb Csinbl Here we have a choice Since we do not want C to be zero 7 since this would mean that the solution is identically zero7 we must have 0 sinb which means that Abn7r where n a positive integer Let us now consider X From 547 we have 0 X0 Bi Thus for each positive n7 we have a harmonic function 553 gonzy an sinh sin i which satis es 5457 5 46 and 547 Since the Laplacean is a linear operator7 Do mrz mry 5 54 7 an sinh lt b gt sm lt b gt 26 is also harmonici To make this function t the condition 548 we use what we know about Fourier series We suppose that h is wellbehaved enough that it has a Fourier sine expansion see theorem 5 5 and theorem 512 To satisfy 548 we need to nd ahag i H such that 555 hy i amsinh Sin By letting 12m am sinh Lg we see that what we are looking for are Fourier coef cients 542 A Neumann boundary condition The method of separation of variables is also amenable to the Neumann problem problem 2 from above7 and you will be asked to obtain such a solution in a homewor i


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