# In Section 5.3 we considered initial value problems for the linear second order equation. y.a/ C ˇy 0.a/ D k 1 (13.1.4)

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1 678 Chpter 13 Boundry Vlue Problems for Second Order Ordinry Differentil Equtions 13.1 TWO-POINT BOUNDARY VALUE PROBLEMS In Section 5.3 we considered initil vlue problems for the liner second order eqution P./y C P 1./y C P 2./y D F./: (13.1.1) Suppose P, P 1, P 2, nd F re continuous nd P hs no zeros on n open intervl.; b/. From Theorem 5.3.1, if is in.; b/ nd k 1 nd k 2 re rbitrry rel numbers then (13.1.1) hs unique solution on.; b/ such tht y. / D k 1 nd y. / D k 2. Now we consider different problem for (13.1.1). PROBLEM Suppose P, P 1, P 2, nd F re continous nd P hs no zeros on closed intervl Œ; b. Let, ˇ,, nd ı be rel numbers such tht nd let k 1 nd k 2 be rbitrry rel numbers. Find solution of on the closed intervl Œ; b such tht 2 C ˇ2 nd 2 C ı 2 ; (13.1.2) P./y C P 1./y C P 2./y D F./ (13.1.3) y./ C ˇy./ D k 1 (13.1.4) nd y.b/ C ıy.b/ D k 2 : (13.1.5) The ssumptions stted in this problem pply throughout this section nd won t be repeted. Note tht we imposed conditions on P, P 1, P 2, nd F on the closed intervl Œ; b, nd we re interested in solutions of (13.1.3) on the closed intervl. This is different from the sitution considered in Chpter 5, where we imposed conditions on P, P 1, P 2, nd F on the open intervl.; b/ nd we were interested in solutions on the open intervl. There is relly no problem here; we cn lwys etend P, P 1, P 2, nd F to n open intervl.c; d/ (for emple, by defining them to be constnt on.c; d nd Œb; d/) so tht they re continuous nd P hs no zeros on Œc; d. Then we cn pply the theorems from Chpter 5 to the eqution y C P 1./ P./ y C P 2./ P./ y D F./ P./ on.c; d/ to drw conclusions bout solutions of (13.1.3) on Œ; b. We cll nd b boundry points. The conditions (13.1.4) nd (13.1.5) re boundry conditions, nd the problem is two-point boundry vlue problem or, for simplicity, boundry vlue problem. (We used similr terminology in Chpter 12 with different mening; both menings re in common usge.) We require (13.1.2) to insure tht we re imposing sensible condition t ech boundry point. For emple, if 2 C ˇ2 D then D ˇ D, so y./ C ˇy./ D for ll choices of y./ nd y./. Therefore (13.1.4) is n impossible condition if k 1, or no condition t ll if k 1 D. We bbrevite (13.1.1) s Ly D F, where Ly D P./y C P 1./y C P./y; nd we denote B 1.y/ D y./ C ˇy./ nd We combine (13.1.3), (13.1.4), nd (13.1.5) s B 2.y/ D y.b/ C ıy.b/: Ly D F; B 1.y/ D k 1 ; B 2.y/ D k 2 : (13.1.6)

2 Section 13.1 Two-Point Boundry Vlue Problems 679 This boundry vlue problem is homogeneous if F D nd k 1 D k 2 D ; otherwise it s nonhomogeneous. We leve it to you (Eercise 1) to verify tht B 1 nd B 2 re liner opertors; tht is, if c 1 nd c 2 re constnts then B i.c 1 y 1 C c 2 y 2 / D c 1 B i.y 1 / C c 2 B i.y 2 /; i D 1; 2: (13.1.7) The net three emples show tht the question of eistence nd uniqueness for solutions of boundry vlue problems is more complicted thn for initil vlue problems. Emple Consider the boundry vlue problem The generl solution of y C y D 1 is y C y D 1; y./ D ; y.=2/ D : y D 1 C c 1 sin C c 2 cos ; so y./ D if nd only if c 2 D 1 nd y.=2/ D if nd only if c 1 D 1. Therefore y D 1 sin cos is the unique solution of the boundry vlue problem. Emple Consider the boundry vlue problem Agin, the generl solution of y C y D 1 is y C y D 1; y./ D ; y./ D : y D 1 C c 1 sin C c 2 cos ; so y./ D if nd only if c 2 D 1, but y./ D if nd only if c 2 D 1. Therefore the boundry vlue problem hs no solution. Emple Consider the boundry vlue problem y C y D sin 2; y./ D ; y./ D : You cn use the method of undetermined coefficients (Section 5.5) to find tht the generl solution of y C y D sin 2 is sin 2 y D C c 1 sin C c 2 cos : 3 The boundry conditions y./ D nd y./ D both require tht c 2 D, but they don t restrict c 1. Therefore the boundry vlue problem hs infinitely mny solutions where c 1 is rbitrry. sin 2 y D C c 1 sin ; 3 Theorem If 1 nd 2 re solutions of Ly D such tht either B 1. 1/ D B 1. 2/ D or B 2. 1/ D B 2. 2/ D ; then f 1; 2g is linerly dependent. Equivlently; if f 1; 2g is linerly independent, then B / C B / nd B / C B / :

3 68 Chpter 13 Boundry Vlue Problems for Second Order Ordinry Differentil Equtions Proof Recll tht B 1. / D./ C ˇ./ nd 2 C ˇ2. Therefore, if B 1. 1/ D B 1. 2/ D then. ; ˇ/ is nontrivil solution of the system This implies tht 1./ C ˇ 1./ D 2./ C ˇ./ D : 1./ 2./ 1./ 2./ D ; so f 1; 2g is linerly dependent, by Theorem We leve it to you to show tht f 1; 2g is linerly dependent if B 2. 1/ D B 2. 2/ D. Theorem The following sttements re equivlenti tht is; they re either ll true or ll flse: () There s fundmentl set f 1; 2g of solutions of Ly D such tht B 1. 1/B 2. 2/ B 1. 2/B 2. 1/ : (13.1.8) (b) If fy 1 ; y 2 g is fundmentl set of solutions of Ly D then B 1.y 1 /B 2.y 2 / B 1.y 2 /B 2.y 1 / : (13.1.9) (c) For ech continuous F nd pir of constnts.k 1 ; k 2 /; the boundry vlue problem Ly D F; B 1.y/ D k 1 ; B 2.y/ D k 2 (d) hs unique solution: The homogeneous boundry vlue problem Ly D ; B 1.y/ D ; B 2.y/ D (13.1.1) (e) hs only the trivil solution y D. The homogeneous eqution Ly D hs linerly independent solutions 1 nd 2 such tht B 1. 1/ D nd B 2. 2/ D : Proof We ll show tht./ H).b/ H).c/ H).d/ H).e/ H)./: () H) (b): Since f 1; 2g is fundmentl set of solutions for Ly D, there re constnts 1, 2, b 1, nd b 2 such tht y 1 D 1 1 C 2 2 ( ) y 2 D b 1 1 C b 2 2: Moreover, ˇˇˇ ˇ 1 2 b 1 b 2 ˇ : ( ) becuse if this determinnt were zero, its rows would be linerly dependent nd therefore fy 1 ; y 2 g would be linerly dependent, contrry to our ssumption tht fy 1 ; y 2 g is fundmentl set of solutions of Ly D. From (13.1.7) nd ( ), B1.y 1 / B 2.y 1 / B 1.y 2 / B 2.y 2 / D 1 2 B1. 1/ B 2. 1/ b 1 b 2 B 1. 2/ B 2. 2/ :

4 Section 13.1 Two-Point Boundry Vlue Problems 681 Since the determinnt of product of mtrices is the product of the determinnts of the mtrices, (13.1.8) nd ( ) imply (13.1.9). (b) H) (c): Since fy 1 ; y 2 g is fundmentl set of solutions of Ly D, the generl solution of Ly D F is y D y p C c 1 y 1 C c 2 y 2 ; where c 1 nd c 2 re rbitrry constnts nd y p is prticulr solution of Ly D F. To stisfy the boundry conditions, we must choose c 1 nd c 2 so tht (recll (13.1.7)), which is equivlent to k 1 D B 1.y p / C c 1 B 1.y 1 / C c 2 B 1.y 2 / k 2 D B 2.y p / C c 1 B 2.y 1 / C c 2 B 2.y 2 /; c 1 B 1.y 1 / C c 2 B 1.y 2 / D k 1 B 1.y p / c 1 B 2.y 1 / C c 2 B / D k 2 B 2.y p /: From (13.1.9), this system lwys hs unique solution.c 1 ; c 2 /. (c) H) (d): Obviously, y D is solution of (13.1.1). From (c) with F D nd k 1 D k 2 D, it s the only solution. (d) H) (e): Let fy 1 ; y 2 g be fundmentl system for Ly D nd let 1 D B 1.y 2 /y 1 B 1.y 1 /y 2 nd 2 D B 2.y 2 /y 1 B 2.y 1 /y 2 : Then B 1. 1/ D nd B 2. 2/ D. To see tht 1 nd 2 re linerly independent, note tht 1 1 C 2 2 D 1 ŒB 1.y 2 /y 1 B 1.y 1 /y 2 C 2 ŒB 2.y 2 /y 1 B 2.y 1 /y 2 D ŒB 1.y 2 / 1 C B 2.y 2 / 2 y 1 ŒB 1.y 1 / 1 C B 2.y 1 / 2 y 2 : Therefore, since y 1 nd y 2 re linerly independent, 1 1 C 2 2 D if nd only if B1.y 1 / B 2.y 1 / 1 D : B 1.y 2 / B 2.y 2 / 2 If this system hs nontrivil solution then so does the system B1.y 1 / B 1.y 2 / c1 D B 2.y 1 / B 2.y 2 / c 2 : This nd (13.1.7) imply tht y D c 1 1 C c 2 2 is nontrivil solution of (13.1.1), which contrdicts (d). (e) H) (). Theorem implies tht if B 1. 1/ D nd B 2. 2/ D then B 1. 2/ nd B 2. 1/. This implies (13.1.8), which completes the proof. Emple Solve the boundry vlue problem 2 y 2y C 2y 2 3 D ; y.1/ D 4; y.2/ D 3; ( ) given tht f; 2 g is fundmentl set of solutions of the complementry eqution Solution Using vrition of prmeters (Section 5.7), you cn show tht y p D 3 is solution of the complementry eqution 2 y 2y C 2y D 2 3 D :

5 682 Chpter 13 Boundry Vlue Problems for Second Order Ordinry Differentil Equtions Therefore the solution of ( ) cn be written s y D 3 C c 1 C c 2 2 : Then y D 3 2 C c 1 C 2c 2 ; nd imposing the boundry conditions yields the system c 1 C c 2 D 3 c 1 C 4c 2 D 9; so c 1 D 7 nd c 2 D 4. Therefore is the unique solution of ( ) y D 3 C Emple Solve the boundry vlue problem y 7y C 12y D 4e 2 ; y./ D 3; y.1/ D 5e 2 : Solution From Emple 5.4.1, y p D 2e 2 is prticulr solution of y 7y C 12y D 4e 2 : ( ) Since fe 3 ; e 4 g is fundmentl set for the complementry eqution, we could write the solution of ( ) s y D 2e 2 C c 1 e 3 C c 2 e 4 nd determine c 1 nd c 2 by imposing the boundry conditions. However, this would led to some tedious lgebr, nd the form of the solution would be very unppeling. (Try it!) In this cse it s convenient to use the fundmentl system f 1; 2g mentioned in Theorem (e); tht is, we choose f 1; 2g so tht B 1. 1/ D 1./ D nd B 2. 2/ D 2.1/ D. It is esy to see tht 1 D e 3 e 4 nd 2 D e 3.1/ e 4.1/ stisfy these requirements. Now we write the solution of ( ) s y D 2e 2 C c 1 e 3 e 4 C c 2 e 3.1/ e 4.1/ : Imposing the boundry conditions y./ D 3 nd y.1/ D 5e 2 yields Therefore nd 3 D 2 C c 2 e 4.e 1/ nd 5e 2 D 2e 2 C c 1 e 3.1 e/: y D 2e 2 C c 1 D 3 e.1 e/ ; c 2 D e4 e 1 ; 3 e.1 e/.e3 e 4 / C e4 e 1.e3.1/ e 4.1/ /: Sometimes it s useful to hve formul for the solution of generl boundry problem. Our net theorem ddresses this question.

6 Section 13.1 Two-Point Boundry Vlue Problems 683 Theorem Suppose the homogeneous boundry vlue problem Ly D ; B 1.y/ D ; B 2.y/ D ( ) hs only the trivil solution: Let y 1 nd y 2 be linerly independent solutions of Ly D such tht B 1.y 1 / D nd B 2.y 2 / D ; nd let W D y 1 y 2 y 1 y 2: Then the unique solution of is y./ D y 1./ Ly D F; B 1.y/ D ; B 2.y/ D ( ) Z P.t/W.t/ dt C y 2./ dt: ( ) P.t/W.t/ Proof where In Section 5.7 we sw tht if y D u 1 y 1 C u 2 y 2 ( ) u 1 y 1 C u 2 y 2 D u 1 y 1 C u 2 y 2 D F; then Ly D F. Solving for u 1 nd u 2 yields which hold if u 1./ D u 1 D Fy 2 P W nd u 2 D Fy 1 P W ; dt nd P.t/W.t/ u 2./ D Z P.t/W.t/ dt: This nd ( ) show tht ( ) is solution of Ly D F. Differentiting ( ) yields Z y./ D y1./ P.t/W.t/ dt C y 2./ dt: ( ) P.t/W.t/ (Verify.) From ( ) nd ( ), B 1.y/ D B 1.y 1 / P.t/W.t/ dt D becuse B 1.y 1 / D, nd B 2.y/ D B 2.y 2 / P.t/W.t/ dt D becuse B 2.y 2 / D. Hence, y stisfies ( ). This completes the proof. We cn rewrite ( ) s where y D 8 y 1.t/y 2./ ˆ< P G.; t/ D.t/W.t/ y 1./y 2.t/ ˆ: P.t/W.t/ G.; t/f.t/ dt; (13.1.2) ; t ; : ; t b:

7 684 Chpter 13 Boundry Vlue Problems for Second Order Ordinry Differentil Equtions This is the Green s function for the boundry vlue problem ( ). The Green s function is relted to the boundry vlue problem ( ) in much the sme wy tht the inverse of squre mtri A is relted to the liner lgebric system y D A; just s we substitute the given vector y into the formul D A 1 y to solve y D A, we substitute the given function F into the formul (13.1.2) to obtin the solution of ( ). The nlogy goes further: just s A 1 eists if nd only if A D hs only the trivil solution, the boundry vlue problem ( ) hs Green s function if nd only the homogeneous boundry vlue problem ( ) hs only the trivil solution. We leve it to you (Eercise 32) to show tht the ssumptions of Theorem imply tht the unique solution of the boundry vlue problem is y./ D Ly D F; B 1.y/ D k 1 ; B 2.y/ D k 2 Emple Solve the boundry vlue problem G.; t/f.t/ dt C k 2 B 2.y 1 / y 1 C k 1 B 1.y 2 / y 2: y C y D F./: y./ C y./ D ; y./ y./ D ; ( ) nd find the Green s function for this problem. Solution Here B 1.y/ D y./ C y./ nd B 2.y/ D y./ y./: Let f 1; 2g D fcos ; sin g, which is fundmentl set of solutions of y C y D. Then nd Therefore B 1. 1/ D.cos sin /ˇˇD D 1 B 2. 1/ D.cos C sin /ˇˇD D 1 B 1. 2/ D.sin C cos /ˇˇD D 1 B 2. 2/ D.sin cos /ˇˇD D 1: B 1. 1/B 2. 2/ B 1. 2/B 2. 1/ D 2; so Theorem implies tht ( ) hs unique solution. Let nd y 1 D B 1. 2/ 1 B 1. 1/ 2 D cos sin y 2 D B 2. 2/ 1 B 2. 1/ 2 D cos C sin : Then B 1.y 1 / D, B 2.y 2 / D, nd the Wronskin of fy 1 ; y 2 g is W./ D cos sin cos C sin ˇ sin cos sin C cos ˇ D 2: Since P D 1, ( ) yields the solution y./ D Z cos sin 2 cos C sin C 2 Z F.t/.cos t C sin t/ dt F.t/.cos t sin t/ dt:

8 Section 13.1 Two-Point Boundry Vlue Problems 685 The Green s function is 8.cos t sin t/.cos C sin / ˆ< ; t ; G.; t/ D 2.cos sin /.cos t C sin t/ ˆ: ; t : 2 We ll now consider the sitution not covered by Theorem Theorem Suppose the homogeneous boundry vlue problem Ly D ; B 1.y/ D ; B 2.y/ D ( ) hs nontrivil solution y 1 ; nd let y 2 be ny solution of Ly D tht isn t constnt multiple of y 1 : Let W D y 1 y2 y 1 y 2: If dt D ; ( ) P.t/W.t/ then the homogeneous boundry vlue problem Ly D F; B 1.y/ D ; B 2.y/ D ( ) hs infinitely mny solutions; ll of the form y D y p C c 1 y 1 ; where nd c 1 is constnt: If y p D y 1./ then ( ) hs no solution: Proof P.t/W.t/ dt C y 2./ dt ; P.t/W.t/ Z P.t/W.t/ dt From the proof of Theorem , y p is prticulr solution of Ly D F, nd y p./ D y 1./ P.t/W.t/ dt C y 2./ Therefore the generl solution of ( ) is of the form where c 1 nd c 2 re constnts. Then y D y p C c 1 y 1 C c 2 y 2 ; Z P.t/W.t/ dt: B 1.y/ D B 1.y p C c 1 y 1 C c 2 y 2 / D B 1.y p / C c 1 B 1.y 1 / C c 2 B 1 y 2 D B 1.y 1 / D c 2 B 1.y 2 / P.t/W.t/ dt C c 1B 1.y 1 / C c 2 B 1.y 2 / Since B 1.y 1 / D, Theorem implies tht B 1.y 2 / ; hence, B 1.y/ D if nd only if c 2 D. Therefore y D y p C c 1 y 1 nd B 2.y/ D B 2.y p C c 1 y 1 / D B 2.y 2 / D B 2.y 2 / P.t/W.t/ dt; P.t/W.t/ dt C c 1B 2.y 1 / since B 2.y 1 / D. From Theorem , B 2.y 2 / (since B 2.y 1 D ). Therefore Ly D if nd only if ( ) holds. This completes the proof.

9 686 Chpter 13 Boundry Vlue Problems for Second Order Ordinry Differentil Equtions Emple Applying Theorem to the boundry vlue problem y C y D F./; y./ D ; y./ D ( ) eplins the Emples nd The complementry eqution y C y D hs the liner independent solutions y 1 D sin nd y 2 D cos, nd y 1 stisfies both boundry conditions. Since P D 1 nd W D ˇ sin cos cos sin ˇ D 1; ( ) reduces to Z F./ sin d D : From Emple , F./ D 1 nd Z F./ sin d D Z sin d D 2; so Theorem implies tht ( ) hs no solution. In Emple , nd Z F./ D sin 2 D 2 sin cos Z F./ sin d D 2 sin 2 cos d D 2 3 sin3 ˇ so Theorem implies tht ( ) hs infinitely mny solutions, differing by constnt multiples of y 1./ D sin. D ; 13.1 Eercises 1. Verify tht B 1 nd B 2 re liner opertors; tht is, if c 1 nd c 2 re constnts then B i.c 1 y 1 C c 2 y 2 / D c 1 B i.y 1 / C c 2 B i.y 2 /; i D 1; 2: In Eercises 2 7 solve the boundry vlue problem. 2. y y D, y./ D 2, y.1/ D 1 3. y D 2 3, y./ D, y.1/ y.1/ D 4. y y D, y./ C y./ D 3, y.1/ y.1/ D 2 5. y C 4y D 1, y./ D 3, y.=2/ C y.=2/ D 7 6. y 2y C y D 2e, y./ 2y./ D 3, y.1/ C y.1/ D 6e 7. y 7y C 12y D 4e 2, y./ C y./ D 8, y.1/ D 7e 2 (see Emple ) 8. Stte condition on F such tht the boundry vlue problem y D F./; y./ D ; y.1/ y.1/ D hs solution, nd find ll solutions.

10 Section 13.1 Two-Point Boundry Vlue Problems () Stte condition on nd b such tht the boundry vlue problem y C y D F./; y./ D ; y.b/ D (A) hs unique solution for every continuous F, nd find the solution by the method used to prove Theorem (b) In the cse where nd b don t stisfy the condition you gve for (), stte necessry nd sufficient on F such tht (A) hs solution, nd find ll solutions by the method used to prove Theorem Follow the instructions in Eercise 9 for the boundry vlue problem y C y D F./; y./ D ; y.b/ D : 11. Follow the instructions in Eercise 9 for the boundry vlue problem y C y D F./; y./ D ; y.b/ D : In Eercises find formul for the solution of the boundry problem by the method used to prove Theorem Assume tht < b. 12. y y D F./, y./ D, y.b/ D 13. y y D F./, y./ D, y.b/ D 14. y y D F./, y./ D, y.b/ D 15. y y D F./, y./ y./ D, y.b/ C y.b/ D In Eercises find ll vlues of! such tht boundry problem hs unique solution, nd find the solution by the method used to prove Theorem For other vlues of!, find conditions on F such tht the problem hs solution, nd find ll solutions by the method used to prove Theorem y C! 2 y D F./, y./ D, y./ D 17. y C! 2 y D F./, y./ D, y./ D 18. y C! 2 y D F./, y./ D, y./ D 19. y C! 2 y D F./, y./ D, y./ D 2. Let f 1; 2g be fundmentl set of solutions of Ly D. Given tht the homogeneous boundry vlue problem Ly D ; B 1.y/ D ; B 2.y/ D hs nontrivil solution, epress it eplicity in terms of 1 nd If the boundry vlue problem hs solution for every continuous F, then find the Green s function for the problem nd use it to write n eplicit formul for the solution. Otherwise, if the boundry vlue problem does not hve solution for every continuous F, find necessry nd sufficient condition on F for the problem to hve solution, nd find ll solutions. Assume tht < b. () y D F./, y./ D, y.b/ D (b) y D F./, y./ D, y.b/ D (c) y D F./, y./ D, y.b/ D (d) y D F./, y./ D, y.b/ D

11 688 Chpter 13 Boundry Vlue Problems for Second Order Ordinry Differentil Equtions 22. Find the Green s function for the boundry vlue problem y D F./; y./ 2y./ D ; y.1/ C 2y.1/ D : (A) Then use the Green s function to solve (A) with () F./ D 1, (b) F./ D, nd (c) F./ D Find the Green s function for the boundry vlue problem 2 y C y C. 2 1=4/y D F./; y.=2/ D ; y./ D ; (A) given tht y 1./ D cos p nd y 2./ D sin p re solutions of the complementry eqution. Then use the Green s function to solve (A) with () F./ D 3=2 nd (b) F./ D 5= Find the Green s function for the boundry vlue problem 2 y 2y C 2y D F./; y.1/ D ; y.2/ D ; (A) given tht f; 2 g is fundmentl set of solutions of the complementry eqution. Then use the Green s function to solve (A) with () F./ D 2 3 nd (b) F./ D Find the Green s function for the boundry vlue problem 2 y C y y D F./; y.1/ 2y.1/ D ; y.2/ D ; (A) given tht f; 1=g is fundmentl set of solutions of the complementry eqution. Then use the Green s function to solve (A) with () F./ D 1, (b) F./ D 2, nd (c) F./ D 3. In Eercises 26 3 find necessry nd sufficient conditions on, ˇ,, nd ı for the boundry vlue problem to hve unique solution for every continuous F, nd find the Green s function. 26. y D F./, y./ C ˇy./ D, y.1/ C ıy.1/ D 27. y C y D F./, y./ C ˇy./ D, y./ C ıy./ D 28. y C y D F./, y./ C ˇy./ D, y.=2/ C ıy.=2/ D 29. y 2y C 2y D F./, y./ C ˇy./ D, y./ C ıy./ D 3. y 2y C 2y D F./, y./ C ˇy./ D, y.=2/ C ıy.=2/ D 31. Find necessry nd sufficient conditions on, ˇ,, nd ı for the boundry vlue problem y y D F./; y./ C ˇy./ D ; y.b/ C ıy.b/ D (A) to hve unique solution for every continuous F, nd find the Green s function for (A). Assume tht < b. 32. Show tht the ssumptions of Theorem imply tht the unique solution of Ly D F; B 1.y/ D k 1 ; B 2.y/ D f 2 is y D G.; t/f.t/ dt C k 2 B 2.y 1 /y 1 C k 1 B 1.y 2 / y 2:

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