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by: Vivien Bradtke V


Vivien Bradtke V
Texas A&M
GPA 3.6


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This 58 page Class Notes was uploaded by Vivien Bradtke V on Wednesday October 21, 2015. The Class Notes belongs to MATH 412 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 38 views. For similar materials see /class/226039/math-412-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
Math 412 501 Theory of Partial Differential Equations Lecture 33 Bessel functions Spectral properties of the Laplacian in a circle Eigenvalue problem V2 A 0 in DXy X2y2 3 R2 uaD 0 In polar coordinates r76 2 2 iZ f 71237 w 0 0ltrlt R 7rlt6lt7r R0 0 7T lt 0 lt 7T Additional boundary conditions 06 lt oo 7T lt 0 lt 7T r77r r77r1 r77rr77r 0lt rlt Separation of variables 076 frh6 Substitute this into the equation f rh0 r lfrh0 r 2frh 6 Afrh6 0 Divide by frh0 and multiply by r2 r2f r rfr Ar2fr h 6 7 f0 h0 i It follows that r2f r rfr Ar2fr i h 6 f0 7 We u const The variables have been separated r2f H M2 uf 0 h uh Boundary conditions R0 0 and 06 lt 00 hold if fR 0 and f0 lt oo Boundary conditions r 7T r 7T and r 7r r7r hold if 174 hm and h 7r hTF Eigenvalue problem h uh h 7T h7T h 7T hTF Eigenvalues um m2 m 0172 uo 0 is simple the others are of multiplicity 2 Eigenfunctions ho 1 hm6 cos m0 and hm6 sin m6 for m 2 1 Dependence on r r21 H Ar2 uf 0 fR 0 f0 lt oo We may assume that u m2 m 071727 Also we know that A gt 0 Rayleigh quotientl New variable 2 r removes dependence on A df 7 df d2f i d2f dr 7 dz dr2 7 dz2 d2f df 2 2 2 z dz2zdzz mf 0 This is Bessel39s differential equation of order m Solutions are called Bessel functions of order m d2f df 2 2 2 7 z E2Ez mf70 Solutions are well behaved in the interval 000 Let f1 and f2 be linearly independent solutions Then the general solution is f C1f1 C2f2 where C1 C2 are constants We need to determine the behavior of solutions as z gt0and asz gtoo In a neighborhood of 0 Bessel s equation is a small perturbation of the equidimensional equation Equidimensional equation d2f df 2 2 z z m f 0 dz2 dz For m gt 0 the general solution is fz C12quot7 ng m where c1 c2 are constants For m 0 the general solution is fz c1 c2 log 2 where c1 c2 are constants We hope that Bessel functions are close to solutions of the equidimensional equation as z gt 0 Theorem For any m gt 0 there exist Bessel functions f1 and f2 of order m such that f1z N z quot and 132 N 2 quot as z gt 0 Also there exist Bessel functions f1 and f2 of order 0 such that f1z N 1 and 132 N logz as z gt 0 Remarks f1 and f2 are linearly independent ii fl is determined uniquely while f2 is not Jmz Bessel function of the first kind Ymz Bessel function of the second kind Jmz and Ymz are certain linearly independent Bessel functions of order m Jmz is regular while Ymz has singularity at 0 Jmz and Ymz are special functions Asz gt0 we havefor mgt0 2 quotm 1 7 Z 1 m JmZ N mz YmZ N W m Also J0z N 1 Y0z N glogz Jmz is uniquely determined by its asymptotics as z gt 0 Original definition by Bessel 1 7T Jmz cosz sin 739 m7 d7 7T 0 Behavior of the Bessel functions as z gt 00 does not depend on the order m Any Bessel function f satisfy fz Az 12cosz B 02 1 as z gt 00 where A B are constants The function f is uniquely determined by A B and its order m As 2 gt 00 we have Jmz gcos lt2 g 0271 Ymz gsin lt2 g 0271 Let 0 ltjm71 ltjm72 lt be zeros of Jmz and 0 lt ym71 lt ym72 lt be zeros of Ymz Then the zeros are interlaced m lt ym71ltjm71lt ym2 ltjm2 lt Asymptotics of the nth zeros as n gt 00 1mm n m hr ymm n m W Eigenvalues of the Laplacian in a circle Intermediate eigenvalue problem r2f rf M2 m2f 0 fR 0 f0 lt oo New variable 2 r reduced the equation to Bessel39s equation of order m Hence the general solution is fr c1JmX r c2 YmX r where c1 c2 are constants Singular condition f0 lt 00 holds if c2 0 Nonzero solution exists if JmX R 0 Thus there are infinitely many eigenvalues Am717Am727 where MAW R jmm ie AW jmmR Summary Eigenvalue problem v2 l 0 in D Xiy x2 y2 3 R2 UlaD 0 Eigenvalues Am jm7nR2 where m012 n 1727 andjmm isthe nth zero of the Bessel function Jm Eigenfunctions O7nr76 JOUOW rR Eor m 2 139 m7nr76 JmUmm rR COS m0 and m7 7r76 Jm0m7n Sin Math 412 501 Theory of Partial Differential Equations Lecture 42 More on the Dirac delta function Green39s functions for ODEs Dirac delta function 6X is a function on R such that o 6X 0 for all X y 0 0 60 00 o ff 6X dX 1 For any continuous function f and any X0 6 R fX6X X0 dX fX0 6X is a generalized function or distribution That is 6 is a linear functional on a space of test functions f such that 6f f0 Distributions Class of test functions 8 consists of infinitely smooth rapidly decaying functions on R To be precise f E S if sup kamx lt 00 for any integers k m 2 0 Convergence in S we say that f7 gt f in S as n gt 00 if sup Xkfmx flmx gt 0 as n gt 00 for any integers k m 2 0 Class of distributions 8 consists of continuous linear functionals on 8 That is a linear map E S gt R belongs to S if fn gt f whenever f7 gt f in 8 Convergence in 8 we say that 6 gt E in 8 if 61 gt f for any f E S Examples Delta function 6f f0 ii Shifted 6 function 6X0X 6X X0 6X0f fX0 iii Let g be a bounded locally integrable function on R Then f gt fXgX dX is a distribution which is identified with g Delta sequence is a sequence of functions g17g27 such that g7 gt 6 in S as n gt 00 That is for any f E S quotIii 0 fxgx dx f0 Delta family is a family of functions he 0 lt e S 50 such that linghE 6 in 8 8H How to differentiate a distribution If g is a piecewise differentiable bounded function on R then fXgX dX fXgx dX for any test function f E 8 Let 7 be a distribution Then S 3 f n gt 7f is also a distribution which is denoted 7 and called the derivative of 7 in 8 In the case when 7 is a differentiable function the derivative in S coincides with the usual derivative Heaviside step function 0 if Xlt0 HX 1 If X20 The Heaviside function is a regular distribution For any test function f E 8 HM 1fXHXdx fX dx fx 0 a0 Thus the derivative of the Heaviside function is the delta function H 6 Green39s functions for ODEs Boundary value problem d2u W Definition 1 Green s function of the problem is a function GXX0 X7X0 E 0 L such that for any f L uXA fX0GXXOdXO fX u0 uL 0 Definition 2 Green s function GXX0 of the problem is its solution for fX 6X X0 82GXX0 8X2 6x X0 C07X0 CLXO 039 Definition 1 shows how to use Green s function Definition 2 shows how to find Green s function Both definitions are equivalent Definition 2 means that 82G o 0forxltxo andXgtX0 8X2 0 GXX0 is continuous at X X0 BGXX0 BGXX0 1 o 8X XXg 8X XX07 axb if XltX7 GXX0 O Cxd If XgtX07 where a b C d may depend on X0 BGXXO i a if X lt X07 19X i C if X gtX0 Besides G0Xo b and GLXo CL d Therefore C a1 aX0 LL aX0bCX0d b0 b0 CX0L CLd0 d X0 0 x0 L i i I l Gxxo XIV 23 Lxo L X0 if X lt X07 L GX7X0 X0 TL X if X gt X0 GXX0 GX0X Maxwell s reciprocity Hilbert space L20 L h fOL hX2 dX lt oo Dense subspace 39H h E C20 L h0 hL 0 Linear operator L 39H gt L20 L h h L is selfadjoint hggt h ggt for all hg E H L lt h1ggt LILxde L L Hoo 0 1700de 0 1700de L hxgtgxgt L0 magxx dx lth ggt Inverse operator L 1 L20 L gt L20 L If 1f u then u f u0 uL 0 L 71m O GXX0fxodxo Since the operator L is self adjoint so is 4 L L lt 1f17ggt Gx7xogtfxogt dxodx L L f 1ggt fxcxxomdxodx L 1 is self adjoint if and only if GXX0 GX0X Nonhomogeneous boundary value problem u X fX u0 oz uL B We have that u u1 U2 U3 where u f u10 u1L 0 ug 0 u20 oz u2L 0 ug 0 U30 0 U3L B It turns out that L u1XO Gx7xofxodXO u2ltxgt a1 f BG X0O X07 L Existense of Green39s function Green s function of an initialboundary value problem exists only if there is always a unique solution Example 1 u u f u0 uL 0 Green s function exists if L y mr n 12 otherwise u1X 0 and U2X sinX are both solutions for f 0 Example 2 u X uX fX 0 lt X lt L u0 uO 0 Green s function exists for any L gt 0 Math 412 501 Theory of Partial Differential Equations Lecture 46 Review for the final exam Math 412 501 Fall 2006 Sample problems for the final exam Any problem may be altered or replaced by a different one Some possibly useful information o Parseval39s equality for the complex form of the Fourier series on inn 0 7r 0 x Z cneinx gt lfxl2dx 2w 2 lcnlz n7oo 7 n7oo o Fourier sine and cosine transforms of the second derivative Slf lw i f0w 7 wzsmw Cf w 7 no 7 w2Cfw o Laplace39s operator in polar coordinates r70 V2 7 amp l g i amp 7 8r2 r 8r r2 80239 c Any nonzero solution of a regular Sturm Liouville equation p gtq gta gt0 altxltb satisfies the Rayleigh quotient relation b b 2 2 7p gt gt p gt q gt dx aa b 120 dx 3 0 Some table integrals 2 iax X2 2X 2 iax xe dx 7E7 9 C7 ay O 393 I 00 2 7r 2 e70 e quot dx e K4 04 gt 07 B E R 00 04 0 204 eiaMeI x dx W 04 gt 07 B E R 700 Problem 1 Let fx X2 i Find the Fourier series complex form of fx on the interval 7713977139 ii Rewrite the Fourier series of fx in the real form iii Sketch the function to which the Fourier series converges iv Use Parseval39s equality to evaluate 221 n Problem 1 Let fx X2 i Find the Fourier series complex form of fx on the interval 7713977139 CO The required series is E cne X where n7oo cn i fxe i quot dx 27139 W In particular 1 7r 1 7r 2 1X3 7 1271393 71392 mfg 7WfXdX7E7WX Wig iwigTi 3 If n 7 0 then we have to integrate by parts twice To save time we could instead use the table integral 2 2 7 X 2X 2 X e axdxi ltEgiiai3gt e a l C7 ay O According to this integral 1 7 2 firm 1 X2 2X 2 i d i if i 7 InX C 27139 4X8 X 27139 inlnzlin3 8 7r in 1 27reiin7r eimr 7 271n 7 27139 n2 n2 Thus 2 2 7quot 2 1 n inx X N 2 n2 9 39 icoltnltoo n70 II Rewrite the Fourier series of fx in the real form W2 271n 39x 712 00 271n 39x 739x 7lt in e n2 e e oonn0 0 71392 471 7 i 2 cos nx 3 n n1 Thus 00 2 x2 4n2 cosnx iii Sketch the function to which the Fourier series converges The series converges to the 27r periodic function that coincides with fx for 77139 g X g 7139 The sum is continuous and piecewise smooth hence the convergence is uniform The derivative of the sum has jump discontinuities at points 7139 2k7r k E Z The graph is a scaoped curve iv Use Parseval39s equality to evaluate 221 n 4 In our case Parseval39s equality can be written as 00 f7 n 2 ltf fgt Z ll gtnlgt gtnlgt n7oo where gtnx emquot and ltg7hgt 7r gMde in Since an and lt gtn7 gtngt 27139 for all n E Z it can be reduced to an equivalent form lfxl2dx27r Z lcnlz Therefore lt follows that Problem 2 Solve Laplace39s equation in a disk Vzu0 O rlta7 ua0f0 Laplace39s operator in polar coordinates r7 0 82L 1 8H 1 82L v2 7 7 7 i 7 u 8r2 r 8r r2 802 We search for the solution of the boundary value problem as a superposition of solutions ur7 0 hr gt0 with separated variables Solutions with separated variables satisfy periodic boundary conditions 8U 8U uim 07 7w Em and the singular boundary condition iuooi lt oo Separation of variables provides the following solutions Ho 17 unr0 r cos n07 Dnr0 r sin n07 n 1727 A superposition of these solutions is a series 00 n I ur7 0 040 l ZPI r an cos n0 l Br sin n07 where 0407 041 and 3132 are constants Substituting the series into the boundary condition ua0 1 09 we get 00 n 0 a0 ZPI a an cos n0 l Br sin n0 The boundary condition is satisfied if the right hand side coincides with the Fourier series A0 l Zoilmn cos n0 l Bn sin n0 of the function 1 09 on 77r77r Hence 040A07 ozrf An7 Br zfr39Br7 n12 and ur7 0 A0 21nAn cos n0 B sin n07 A0 1 1 09 d0 Anl ft9cosnt9dt97 7 7r 77r Bnl 05innt9dt97 n1727 7139 77r Bonus Problem 7 Find a Green function implementing the solution of Problem 2 The solution of Problem 2 ur0 A0 i i ltggtnAn cos n0 i Bn sin n07 n1 where 1 7r 1 7 A0 E 77r 1007 An 77r COS n00 1007 1 7r Bni ft90sinnt90dt907 n12 7r 77r It can be rewritten as 7r ur7 0 Cr 00 H00 100 7l39 where 1 1 r n Cr 00 7 i 7 lt7 cos n0 cos n00 i sin n0 sin n00 7r 3 2 7rn1 This is the desired Green function The expression can be simplified CO Cr 00 7 7 ltggtncos n0 cos n00 i sin n0 sin n00 ililtigtnCOSn00 7 27139 7139 a 0 ein9790 eiin9790 i2 2 2 iltrai1ei9790gt iltrai1eii9790gt Cr7 0 00 1 1 ra le iw eo E 17 ra lei990 17 ra le i990 1 a re iw eo E a 7 re 9 90 a 7 re iw eo aZ7r2 1 E a 7 re 9 90a 7 re i9 90 1 327 r2 E 32 7 2arcos0 7 00 1 r2 Poisson s formula Problem 3 Find Green39s function for the boundary value problem dzu W7ufx 0ltXlt17 u0u10 The Green function GXX0 should satisfy 27 7 G 6X 7X07 07X0 17X0 0 It follows that G aequot be for X lt X07 XX 0 cequot de x for X gt X07 where constants a7 b7 7 d may depend on X0 Then 86 aequot 7 be for X lt X07 7X7X0 7 8X cequot 7 de X for X gt X0 The boundary conditions imply that a b and ce de 1 Gluing conditions at X X0 are continuity of the function and jump discontinuity of the first derivative 86 86 CX7X0 lxxo7 CX7X0 lxxo7 7 a X fa xxo77 The two conditions imply that anO i be quot0 cequot0 l de xo7 cequot0 7 de m 7 aequot0 7 be xo 1 Now we have 4 equations to determine 4 quantities a7 b7 c7 d Solution equot0 l e 0 e 3 l ez x0 C 217e2 217e27 equot0 l e 0 bi exo 827m 2e 2717 7 217e2 39 Finally exo 827x0ex six G 21782 for XltX07 X7X 0 exo sixoxex 827x for X gt X 217 e2 039 Observe that GXX0 GX0X Problem 4 Solve the initial boundary value problem for the heat equation 8U 82L E 0ltxlt7r tgt0 ux0 fx 0 lt X lt 7r7 8U u0 t 7 0 am 1 2u7r t 7 0 In the process you will discover a sequence of eigenfunctions and eigenvalues which you should name gtnx and n Describe the n qualitatively eg find an equation for them but do not expect to find their exact numerical values Also do not bother to evaluate normalization integrals for on We search for the solution of the initial boundary value problem as a superposition of solutions ux7 t gtXgt with separated variables of the heat equation that satisfy the boundary conditions We get an equation for g g Ag gf Coe and an eigenvalue problem for j gt WW7 MO 07 W 2M 0 This is a regular Sturm Liouville eigenvalue problem Rayleigh quotient 7w l gt xl2 dx 7 0 0 A 0l l gtxl2 dx Note that 7M l3 gt0 gt 0 7 gt7r gt 7r 2llt gt7rl2 It follows that gt 0 The eigenvalues O lt 1 lt 2 lt are solutions of the equation 1 5 taanIr7 and the corresponding eigenfunctions are gtnx sinxn X Solutions with separated variables urx7 t e A t gtnX e Antsinb rx7 n 1727 A superposition of these solutions is a series 00 7A 00 7A I ux7 t ZPI one t gtnx ZPI one tsin n X7 where c179 are constants Substituting the series into the initial condition ux0 fx we get 00 fx ZPI cn gtnx The initial condition is satisfied if the right hand side coincides with the generalized Fourier series of the function 1 that is if ltf7 ngt 12 Cn 7lt n7 ngt7 I7 77 Problem 5 By the method of your choice solve the wave equation on the half line 82 i 82 0 lt lt lt f lt w w X 007 00 0 subject to 8U uioir 07 uixio M ax70 gx Bonus Problem 6 Solve Problem 5 by a distinctly different method Fourier39s method In view of the boundary condition let us apply the Fourier sine transform with respect to X to both sides of the equation 82L 8 5 iwi 5 iwi Let 2 00 Um t 5 01M a uix t Sm dx 7quot 0 Then 82L aZU 82L 2 W7 LIO7 t u17wZUw7 t 7w2Uw7 t Hence 82U W w2Uw t lf w 7 0 then the general solution of the latter equation is Uw7 t acoswt l bsinwt where a aw b bw Applying the Fourier sine transform to the initial conditions we obtain Uw PM w70 Giwii where F f G g It follows that aw Fw bw Gww Now it remains to apply the inverse Fourier sine transform Cw sin wt sin wx dw uxt ltFwcoswt 0 where 2 0 2 0 Fw7 l X0SII 1uJX0dX07 Cw7 gX0SIan0dX0 7139 0 0 7T D39Alembert39s method Define fX and gX for negative X to be the odd extensions of the functions given for positive X ie f7X 7fX and g7X 7gX for all X gt 0 By d39Alembert39s formula the function uX t ax t fX7 t 3 tgX0 dXo 1 2 2 is the solution of the wave equation that satisfies the initial conditions 8U uX7 0 fX7 at X7 0 gX on the entire line Since f and g are odd functions it follows that uX7 t is also odd as a function of X As a consequence u07 t O for all t Math 412 December 67 2005 Green s Function R Gil7 P Sirneon7 A Stoker7 and T Strong Texas AampM University We are given a second order non hornogenous ODE with hornogenous boundary condi tions and ask for the Greens function In order to do this7 we make use of several identities of Green7s functions and the Dirac Delta 77function g 4y we lt1 110 0 yL 0 The relevant equation for the Greens function is below Our justi cation for being able to write such an odd looking equation is given later in this document Notice we solve the equation in two regions7 each corresponding to either side of the Dirac Delta irnpulse g4G67z 2 Gzzlt Acos 2x B sin 2x x lt z 3 Gzzgt Ccos 2x 7 L D sin 2x 7 L x gt z 4 When z lt z the boundary condition implies that A 0 When z gt z the boundary condition implies that C 0 We can evaluate the integral of Eqn L 61 L 0 W4GdzO 672d2 5 1 6 dx 1 dx ff The Greens function is continuous in the neighborhood of Z7 and in fact everywhere on the interval If it weren7t7 then would be worse behaved than 6 7 z7 which is a clear sign that G isn7t such The reason the G terrn itself does not contribute to the integral is that7 since 6s7z only contributes in the interval zez 76 where E is an arbitrarily small number such that E E R7 we can reduce the integration interval to such On this interval7 by our stipulation that G is continuous7 G will not vary since 6 is in nitely small The continuity condition and the condition we just derived result in a linear system of order 2 We write this in matrix forrn7 and then seek the determinant 1223E Eiiigtltfigtltgt lt7 After computing the determinant and simplifying with a trigonornetric identity7 the result A 7 sin 2L 8 Using Crarner7s rule7 we have the following gt gt w 2sin 2L 1 sin 22 0 7 7 sin 22 lt lt i l 2smlt2Lgt 10 l 2 Therefore7 our nal Green7s function is 7 sin 22 7 L sin 2x Gxzlt Sin 2L x lt z 11 cm zgt 22311222 L x gt z 12 However7 it is clear that for some values of L7 A 0 may vanish In this case7 our Green7s function diverges This becomes clear upon considering the original ODE7s hornogenous equation7 which when solved with the boundary data7 results in the series a Its no coincidence that setting A 0 results in the same series Then7 the solution to the ODE is non unique7 and in our case there are in nitely many solutions No single Green7s function could hope to represent all this


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