### Create a StudySoup account

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

Already have a StudySoup account? Login here

# Boundary-Value Problems 3363

UH

### View Full Document

## 69

## 0

## Popular in Intro to PDE

## Popular in Math

This 17 page Reader was uploaded by Shahmeer Baweja on Saturday November 8, 2014. The Reader belongs to 3363 at University of Houston taught by a professor in Fall. Since its upload, it has received 69 views. For similar materials see Intro to PDE in Math at University of Houston.

## Reviews for Boundary-Value Problems

### 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: 11/08/14

DERIVATION OF THE HEAT DIFFUSION EQUATION IN ONE SPACE DIMENSION PHILIP W WALKER A rod of length L units of length insulated except perhaps at its ends lies along the maxis with its left end at coordinate 0 and its right end at coordinate L Suppose that the mass density p units of mass divided by units of length and thermal conductivity K0 energygtltlength time gtlttemperature and speci c heat c energymassgtlttemperature at each point in the rod depend only on the mcoordinate of the point Let e and Q be as follows The thermal energy density energy length at 75 units of time after the time origin at points with rst coordinate x is em t The heat flux energytime to the right at time 75 through the cross section consisting of points with rst coordinate x is m t A negative value for mt indicates heat flow to the left The heat energy per unit length being generated per unit time inside the rod at time t at points with rst coordinate x is Qx t A negative value for Q indicates a heat sink Suppose that 0 g a g b g L Conservation of thermal energy tells us that the timerateofchange in thermal energy in the section of the rod consisting of points with rst coordinate x satisfying a g x g b is the net heat energy owing per unit time across the boundaries of this section plus the net heat energy being generated internally in the section Thus b b E em wdsv lta tgt agtltb t c2ltxtgtdw Assuming that e and have continuous rst order partial derivatives we have b b b mtdm a mtdm Qmtdm 1 Thus ab gems mt Qzt div 0 Since this is true for each choice of a and b with 0 g a g b g L if Q is continuous and e and have continuous rst order partials it follows that 36 amt axmt Qxt for 0 gm g L andt 2 0 By de nition the temperature u is given by emt cmpmumt for 0 g x g L and t 2 0 So cmpmmt mt 1 Qmt for 0 g x g L and t 2 0 1 2 PHILIP W WALKER According to Fourier s law of heat conduction 3 g0mt K0ma umt for 0 g x g L and t 2 0 m Thus we arrive at the heat diffusion equation in one space dimension 8a 3 8n ltgt cp KO 875 8x 856 If each of c p and K 0 is constant and there are no internal sinks or sources so that Q is zero we have Qfor0 m Landt20 8u 82u Z f lt ltL d gt atmt ax2mt or0m an t0 where K0 s 0p is the thermal diffusivity REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 PHILIP WWALKER Suppose that a and b are real numbers with a lt b each of p q and w is a continuous real valued function with domain a b the function p has a continuous rst deriv ative each of pm and wm is positive for all x in a b each of M1 and N1 is a complex number for i 1 2 andj 1 2 and the quadruples M11 M12 N11 N12 and M21 M22 N21 N22 are linearly independent Suppose that the operator 739 is given by T90 099039 qr whenever go is a twice differentiable complex valued function with domain a b We shall be concerned with nding the complex numbers A and complex valued functions 90 such that 1 73990 Awgo on a b 2 M1190a M12907 N1190b N1295 0 and 3 M2190a M22lt70 N21lt95 N229039b 0 Remark 1 The equation 1 is equivalent to P90 p s0 q Aws0 0 Thus 1 is a regular second order linear homogeneous differential equation Remark 2 The zero function on a b e the function 90 such that gom 0 for all x in ab is always a solution to 1 2 and p De nition 1 The statement that A0 is an eigenvalue for the problem 1 2 and 3 means that there is a complex valued function 90 other than the zero function on ab that satis es 1 2 and when A A0 When this is the case the statement that go is an eigenfunction corresponding to the eigenvalue A0 means that go is not the zero function on ab and go satis es 1 2 and P when A A0 When A0 is an eigenvalue the eigenspace corresponding to A0 consists of all eigenfunctions corresponding to A0 together with the zero function on a b Remark 3 Suppose that A0 is an eigenvalue for 1 2 and e Since the eigenspace corresponding to A0 is a subspace of the set of all solutions to 1 the eigenspace is either one dimensional or two dimensional 1 REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 2 Example 1 Consider the problem L is a positive number 4 90 A90 on U713 5 900 0 and 6 s0L 0 In order to nd the eigenvalues and eigenfunctions we will consider three cases Case 1 Suppose that A lt 0 Then go satis es 4 only in case g0m C1 cosh Am 1 C2 sinh Am for some pair of numbers C1 and C2 and all m in 0 L Since cosh0 1 and sinh0 0it follows that 5 will also hold only in case C1 p n Thus 4 and 5 hold only in case 7 g0x C2 sinh Am for some number C2 Since sinhz 0 only in case 2 0 and AL gt 0 it follows that 4 5 and 6 hold only in case 7 holds with C2 0 or g0m 0 for all m in 0 L Thus there are no negative eigenvalues Case 2 Suppose that A 0 Then go satis es 4 only in case 8 s0 01 0250 for some pair of numbers C1 and C2 and all x in 0 L Thus it follows that 4 5 and 6 hold only in case 8 holds with C1 C2 0 or g0m 0 for all m in 0 L Thus zero is not an eigenvalue Case 3 Suppose that A gt 0 Then go satis es 4 only in case g0m C1 cos Am 1 C2 sin Am for some pair of numbers C1 and C2 and all m in 0L Since cos0 1 and sin0 0it follows that 5 will also hold only in case C1 5i Thus 4 and 5 hold only in case 9 g0m C2 sin Am for some number C2 Since sinz 0 only in case 2 is an integral multiple of 7r and A gt 0 it follows that 4 5 and 6 hold with 90 different from the zero function only in case 9 holds with C2 74 0 and AL k7r or A k7r L or k7r 2 10 A 0kf for some positive integer k REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 3 As we will see below all eigenvalues of 4 5 and 6 must be real numbers Thus A is an eigenvalue only in case 10 holds and when 10 holds 90 is a corresponding eigenfunction only in case h gom csin for some number c 7E 0 and all m in 0 L From this it follows that each eigenspace is one dimensional Remark 4 Here is a procedure for nding the eigenvalues and eigenfunctions of the problem consisting of 1 2 and For each complex number A let u 2 be a linearly independent pair of solution to Then go satis es 1 only in case gom c1um 1 c2um for all x in a b for some pair of complex numbers c1 c2 Moreover go is different from the zero function only in case at least one of c1 and c2 is different from zero When 90 c1u 1 c2u then go39m c1u3m 1 c2u m for all x in a b so J ltIgtm M J for allm in ab where I UAW UAW J 1 l um um Conditions 2 and 3 together are equivalent to Ms0a Ns0b J 31 ltP39a 9039b where M11 M12 11 M l M21 M22 l and N11 N12 12 N l N21 N22 l 39 Thus 90 will satisfy 1 2 s only in case 90 c1u 1 c222 and wz1m ltb 11121 Dlt 11121 where REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 4 Moreover there will be a solution where at least one of c1 and c2 is not zero only in case A 0 where A detD We have established the following Theorem 1 The eigenvalues of the problem 1 2 and are the zeros of the function A and x A0 0 then go is an eigenfunction corresponding to the eigenvalue A0 only in case 90 C1UA0 1 C21A0 where and at least one of the numbers c1 and c2 is different from zero and C1 0 Dlt0gtlc2llol From this it follows that if A0 is an eigenvalue the corresponding eigenspace is twodimensional when D075lg 8 De nition 2 The statement that the problem 1 2 and 0u2 is selfadjoint means that b b Tf W wheneuer each of f and g is a twice continuously differentiable complea ualued func tion with domain a b and each off and g satis es the conditions P and 1 Duo 3 3 and is onedimensional when Example 2 Consider the problem 4 5 and 6 given in the last example Ilere 739go go So if each of f and g has a continuous second derivative and each satis es 5 and 6 0Lltnquotgty UL f f t UL 1 0 0L 1 fa 60Lf030Lfv g Thus the problem 4 5 and 6 is selfadjoint The following theorem gives a straightforward test for selfadjointness REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 5 Theorem 2 The problem 1 2 and is self adjoint only in case pbM 1 01M paN 1 01N where M and N are given by 11 and Remember that u is a continuous positive Valued function de ned on a b De nition 3 Whenever each of f and g is a piecewise continuous complea ualued function de ned on ab the inner product of f and g is denoted by lt fg gt and is de ned by lt fg gt bfmuJxdx De nition 4 Suppose that each of f and g is a piecewise continuous complex ualued function de ned on a b The statement that f and g are orthogonal means that lt ggt0 Suppose that goko gokO1 gokO2 is a sequence of piecewise continuous complex ualued functions de ned on ab The statement that goko gok01 gok02 is or thogonal means that lt 9013903 gt2 0 wheneueri 7E j Theorem 3 If the problem 1 2 and is self adjoint then all eigenvalues are real and eigenfunctions corresponding to different eigenvalues are orthogonal Proof Suppose that A is an eigenvalue and go is a corresponding eigenfunction Then b b b lts0s0gt ltltpltpgtf ws0 fTs0 f lt9 b b goAuJgoA go uJAltgogogt Since go is continuous and not the zero function it follows that lt go go gt75 0 Thus A A showing that A is real Suppose now that A and a are eigenvalues A 7E a go is an eigenfunction corresponding to A and 1b is an eeigenfunction corresponding to a lts0 gt ltAs0 gtfbws0fbTltgtE b b b fs0T fltpuw fruw ltltp gt Thus lt 907 gt 07 and since A 7E a it follows that lt go1b gt 0 I REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 6 Theorem 4 Suppose that the problem 1 2 and F is self adjoint There will be in nitely many eigenvalues and they can be arranged in a nondecreasing sequence Ako k01 k02 with llt gtoo When necessary the GramSchmidt process can be used to convert a linearly inde pendent sequence into an orthogonal one The following theorem gives the process for a linearly independent pair Theorem 5 Suppose that A0 is an eigenvalue and the corresponding eigenspace is two dimensional An orthogonal basis for this eigenspace is oz where ozu0 and lt uOoz gt To v ltozozgt De nition 5 Suppose that the problem 1 2 and N is self adjoint A proper listing of eigenvalues and eigenfunctions for the problem consists of a non decreasing sequence of eigenvalues Ako k01 k02 in which each eigenualue is listed exactly the number of times that is the dimension of the corresponding eigenspace and an orthogonal sequence of eigenfunctions goko gok01 gok02 in which goj is an eigenfunction corresponding toJforj kg kg 1 1 kg 1 2 Theorem 6 If the problem 1 2 and P is self adjoint then there is a proper listing of eigenvalues and eigenfunctions for the problem De nition 6 The statement that the problem 1 2 and 4 is a Sturm Liouville problem means that the conditions 2 and b are equivalent to ones of the form 0 and 0 M11ltP M12lt70 N21lt95 N2295 where each of M11 M12 N21 and N22 is real at least of M11 and M12 is not zero and at least one on N21 and N22 is not zero Theorem 7 All Sturm Liouuille problems are self adjoint and have eigenspaces that are all one dimensional De nition 7 Suppose that the problem 1 2 and is self adjoint and k quotkO and gok quotkO is a proper listing of eigenvalues and eigenfunctions When f is a function that is piecewise continuous on a b the expansion of f in terms of gok quotkO is the sequence offunctions SnfL k0 given by lt gt Snm go fol all x in ab and n l 0l 0 1 l 0 1 2 Theorem 8 Suppose that the problem 1 2 and p e is self adjoint and k quotkO and gok quotkO is a proper listing of eigenfunctions and eigenvalues and suppose that REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 7 f is a function that is piecewise continuous on ab It follows that ck quotkO is a sequence of numbers and f EOE CkS0k kzko with convergence in the mean ie n1iI1Oltf nCkS0kf nCkS0k gt20 kzko kzko only in case lt gt ck i forkk0h0 1k0 2 lt 901979076 gt P If f is piecewise smooth and lt gt c i forkk0h0 1k0 2 lt 90199019 gt then TL nlggo 2 lt igtsokltxgt rltagt rlta gt for each x with a lt x lt b If f has a continuous second derivative satis es the boundary conditions and 3 and lt gt S nm k 2 fol all x in ab andn l 0l 0 1 l 0 1 2 0 then S nconuerges uniformly to f on a b Theorem 9 The Rayleigh Quotient Suppose that the problem 1 2 and is self adjoint that A is an eigenvalue and that go is a corresponding eigenfunction It follows that pltagtso39ltagtWgt pltbgtso39ltbgtW fltpW M Li ltm2wgt Thus s qx 0 for all x in ab and the boundary conditions Pj and 0 imply pago agoa pbgo bgob 2 0 it follows that A 2 0 If it is also true that the non zero constant functions fail to satisfy either 2 or 3 then LEW gt 0 A and it follows that A gt 0 Theorem 10 Suppose that the problem 1 2 and is self adjoint If each Mj and each Nj is a real number then each eigenualue has a corresponding real valued eigenfunction REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 8 Remark 5 In the special case where 739 is given by T90 907 wm 1 the problem 1 2 and w is self adjoint A is an eigenvalue and go is a corresponding real valued eigenfunction the Rayleigh Quotient becomes iomo www aEwr Ewe Remark 6 In the special case where 739 is given by 73990 S0 and wm 1 equation 1 is equivalent to 90 Ago 0 and we will let the linearly independent pair of solutions u 2 be given by cosh Am when A lt 0 1 A um when A 0 cos Am when A gt 0 and sinh Am when A lt 0 um x when A 0 sin Am when A gt 0 With this de nition of u 2 note that h V Am sinh V Am cp C05 J h A lt 0 A l TA s1nh Am TA cosh Am w en ltIgtm1 T when A0 and cos Am sin Am Ina i AsinxAm Acos Am Example 3 Consider the problem J whenA gt 0 90 Ar 0n0L 900 0 and goL 0 This is a SturmLiouville problem so it is selfadjoint If A is an eigenvalue and go is a corresponding real valued eigenfunction then Rayleigh Quotient ww o w mM4wr A L fo r2 7 L fo r 2 L lo 902 Thus all eigenvalues are nonnegative The nonzero constant functions do not satisfy the boundary conditions so all eigenvalues are positive REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 9 The boundary conditions are equivalent to Ml 55ltgt l Vi 5539lti3gt l 2 l gl where D 181xXf8 SsifL ffxL 1 0 DO 2 i cos XL sinXL J and A0 detD sin xXL From this we see that A is an eigenvalue only in case 67 2 XL k7r or A T for some positive integer k Note that SO only in case C1 Z From this it follows that go is an eigenfunction corresponding to the eigenvalue If2 only in case Imrm g0m C2 s1n T for all x in 0 L and some number c2 7E 0 Based on these observations it follows that a proper listing of eigenvalues and eigenfunctions for this problem is k quot1 and g0k quot1 where is 2 is Akz fork123 and g0kmsin fork123 and0 x L Computation shows that L L Q0k2 fOI39 123 0 REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 10 Example 4 Consider the problem 90 A90 0n0L 900 0 and s0 L 0 This is a SturmLiouville problem so it is selfadjoint If A is an eigenvalue and go is a corresponding real valued eigenfunction then Rayleigh Quotient 0 900 o s0L foLs0 2 A so f0Ls02 L f0 902 TL fo W2 Thus all eigenvalues are nonnegative The nonzero constant functions do satisfy the boundary conditions so all that we can conclude at this point is that all eigenvalues are nonnegative The boundary conditions are equivalent to Ml sf39ltgt l Vi sf39lti3gt l 2 l til 01 00 M00andN01 where When A gt 0 D 811xA8f Ssi L ffxL DO 2 l ExAsinxAL gcos L l and AA det DA Asin AL From this we see that A is a positive eigenvalue only in case k 2 AL k7r or A for some positive integer k Note that when k is a positive integer lmr 2 0 E Df 0 1k SO REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 11 only in case C2 From this it follows that go is an eigenfunction corresponding to the eigenvalue 39f2 when k is a positive integer only in case x ccos L Imrm for some number c 7E 0 and all m in 0 L 39WmnAQ and Thus zero is an eigenvalue Note that 0 D C1 only in case C2 0so go is an eigenfunction corresponding to the eigenvalue zero only in case M c for some number c 7E 0 and all m in 0 L Based on these observations it follows that a proper listing of eigenvalues and eigenfunctions for this problem is k quot0 and g0k quot0 where A020 g00x1for0 m L k7r 2 Akz p fork123 and g0kmcos fork123 and0 x L Computation shows that L lt 9007900 gt 9002 L and 0 L L K 0k2 fOI39 23 Example 5 Consider the problem 90 Ago on LL g0L and D W l E I H II REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 12 This is a not a SturmLiouville problem however it is selfadjoint If A is an eigen value and go is a corresponding real valued eigenfunction then Rayleigh Quotient so L s0L ML 9003 ffLltszgt2 so fLLs02 ltp L s03 s0 L ltPL Jim A so fLLs02 7 A L f L sf2 all f L W2 Thus all eigenvalues are nonnegative The nonzero constant functions do satisfy the boundary conditions so all that we can conclude at this point is that all eigenvalues are nonnegative The boundary conditions are equivalent to N W M M 90 N l s0 L s0 L 0 where M gandN1o 1 0 0 1 When A gt 0 DWZH K cosAL sinAL J 1 0 Py cosAL sinxAL Asin AL Acos AL 0 1 Asin AL Acos AL So 0 2 sin L DW l 2AsinxAL 0 l and AA detDA 4xAsin2 t From this we see that A is a positive eigenvalue only in case 67 2 AL k7r or A T for some positive integer k When k is a positive integer Imr 2 0 0 D lt L gt l o o l so the eigenspace corresponding to is two dimensional and a corresponding linearly independent pair of eigenfunctions is u v where pM cos and vm sin L L Computation shows that lt u 391 gt 0 this pair is already orthogonal REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 13 When A 0 Thus zero is an eigenvalue Note that Dlt gtlZll3l only in case C2 0 so go is an eigenfunction corresponding to the eigenvalue zero only in case s0v 0 for some number c 7E 0 and all m in L L Thus a proper listing of eigenvalues and eigenfunctions for this problem is k quot0 and g0k quot0 where A020 g00m1for L m L k7r 2 2k1 A219 for C 123 k m k m g02k1m cosT and g02km s1nT for k123 and0 x L Computation shows that L lt S00 00 gt Q002 2L and L L g0k2Lfor 123 L Example 6 Consider the problem 90 s0 0n01 900ltgt 0 0 and g01 0 This is a SturmLiouville problem so it is selfadjoint and all eigenspaces are one dimensional If A is an eigenvalue and go is a corresponding real valued eigenfunction then Rayleigh Quotient A Thus all eigenvalues are nonnegative The nonzero constant functions do not satisfy the boundary conditions so all eigenvalues are positive REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 14 31 The boundary conditions are equivalent to M19003 Ns01 11 s0 0 s0 1 M 1ao1aw1221 WW1 11 1 1 smzm 1 Dw 1 1oSX 1 AA det DA sin A 1 Acos T From this we see that A is an eigenvalue only in case A p2 where p is a positive number such that sinp1pcosp 0 or such that tan p p For each positive integer k there is exactly one solution to this equation between k7r S and k7r 1 d See page 201 of the text Newton s method Look it up can be used to approximate the zeros of f wheren f p sin p 1 pcos p Note that when A is an eigenvalue then W Z 1 coslxA b 1 Z 1 coslxA 1Li X Dltgt1Z1131 C1 X02 p T From this it follows that go is a corresponding eigenfunction only in case g0m ccos Am 1 sin Am SO only in case for all x in 0 1 and some number c 7E 0 Based on these observations it follows that a proper listing of eigenvalues and eigenfunctions for this problem is Ak quot1 and g0k quot1 where Ak pi pk is the kth positive number such that sinp1pcosp 0 REGULAR TWOPOINT BOUNDARY VALUE PROBLEMS FALL 2011 15 and 1 3 39 3A g0km cos kmks1n km for all x in 01 Numerical approximations for the rst three eigenvalues are as follows 79 Pk Ale 1 20288 41159 2 49132 24139 39 3 79787 63659 DEPARTMENT OF MATHEMATICS UNIVERSITY OF HOUSTON Email address pwalkermathuhedu

### 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."

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### 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.