# 21.6 Green Functions for First Order Equations

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1 21.6 Green Functions for First Order Equtions Consider the first order inhomogeneous eqution subject to homogeneous initil condition, B[y] y() = 0. The Green function G( ξ) is defined s the solution to L[y] y + p()y = f() for > (21.2) L[G( ξ)] = δ( ξ) subject to G( ξ) = 0. We cn represent the solution to the inhomogeneous problem in Eqution 21.2 s n integrl involving the Green function. To show tht y() = G( ξ)f(ξ) dξ is the solution, we pply the liner opertor L to the integrl. (Assume tht the integrl is uniformly convergent.) L G( ξ)f(ξ) dξ = L[G( ξ)]f(ξ) dξ The integrl lso stisfies the initil condition. B G( ξ)f(ξ) dξ = = = f() = = 0 δ( ξ)f(ξ) dξ B[G( ξ)]f(ξ) dξ (0)f(ξ) dξ Now we consider the qulititive behvior of the Green function. For = ξ, the Green function is simply homogeneous solution of the differentil eqution, however t = ξ we epect some singulr behvior. G ( ξ) will hve Dirc delt function type singulrity. This mens tht G( ξ) will hve jump discontinuity t = ξ. We integrte the differentil eqution on the vnishing intervl (ξ... ξ + ) to determine this jump. G + p()g = δ( ξ) ξ + G(ξ + ξ) G(ξ ξ) + p()g( ξ) d = 1 ξ G(ξ + ξ) G(ξ ξ) = 1 (21.3) The homogeneous solution of the differentil eqution is y h = e p() d Since the Green function stisfies the homogeneous eqution for = ξ, it will be constnt times this homogeneous solution for < ξ nd > ξ. c 1 e p() d < < ξ c 2 e p() d ξ < 661

2 In order to stisfy the homogeneous initil condition G( ξ) = 0, the Green function must vnish on the intervl (... ξ). 0 < < ξ c e p() d ξ < The jump condition, (Eqution 21.3), gives us the constrint G(ξ + ξ) = 1. This determines the constnt in the homogeneous solution for > ξ. 0 < < ξ e ξ p(t) dt ξ < We cn use the Heviside function to write the Green function without using cse sttement. e ξ p(t) dt H( ξ) Clerly the Green function is of little vlue in solving the inhomogeneous differentil eqution in Eqution 21.2, s we cn solve tht problem directly. However, we will encounter first order Green function problems in solving some prtil differentil equtions. Result The first order inhomogeneous differentil eqution with homogeneous initil condition hs the solution L[y] y + p()y = f() for < y() = 0 y = where G( ξ) stisfies the eqution G( ξ)f(ξ) dξ L[G( ξ)] = δ( ξ) for < G( ξ) = 0. The Green function is e ξ pt) dt H( ξ) 21.7 Green Functions for Second Order Equtions Consider the second order inhomogeneous eqution L[y] = y + p()y + q()y = f() for < < b (21.4) subject to the homogeneous boundry conditions B 1 [y] = B 2 [y] = 0. The Green function G( ξ) is defined s the solution to L[G( ξ)] = δ( ξ) subject to B 1 [G] = B 2 [G] = 0. The Green function is useful becuse you cn represent the solution to the inhomogeneous problem in Eqution 21.4 s n integrl involving the Green function. To show tht y() = b G( ξ)f(ξ) dξ 662

3 is the solution, we pply the liner opertor L to the integrl. (Assume tht the integrl is uniformly convergent.) b b L G( ξ)f(ξ) dξ = L[G( ξ)]f(ξ) dξ The integrl lso stisfies the boundry conditions. b B i G( ξ)f(ξ) dξ = = b = f() = = 0 b b δ( ξ)f(ξ) dξ B i [G( ξ)]f(ξ) dξ [0]f(ξ) dξ One of the dvntges of using Green functions is tht once you find the Green function for liner opertor nd certin homogeneous boundry conditions, L[G] = δ( ξ) B 1 [G] = B 2 [G] = 0 you cn write the solution for ny inhomogeneity, f(). L[f] = f() B 1 [y] = B 2 [y] = 0 You do not need to do ny etr work to obtin the solution for different inhomogeneous term. Qulittively, wht kind of behvior will the Green function for second order differentil eqution hve? Will it hve delt function singulrity; will it be continuous? To nswer these questions we will first look t the behvior of integrls nd derivtives of δ(). The integrl of δ() is the Heviside function, H(). 0 for < 0 H() = δ(t) dt = 1 for > 0 The integrl of the Heviside function is the rmp function, r(). 0 for < 0 r() = H(t) dt = for > 0 The derivtive of the delt function is zero for = 0. At = 0 it goes from 0 up to +, down to nd then bck up to 0. In Figure 21.2 we see conceptully the behvior of the rmp function, the Heviside function, the delt function, nd the derivtive of the delt function. We write the differentil eqution for the Green function. G ( ξ) + p()g ( ξ) + q() δ( ξ) we see tht only the G ( ξ) term cn hve delt function type singulrity. If one of the other terms hd delt function type singulrity then G ( ξ) would be more singulr thn delt function nd there would be nothing in the right hnd side of the eqution to mtch this kind of singulrity. Anlogous to the progression from delt function to Heviside function to rmp function, we see tht G ( ξ) will hve jump discontinuity nd G( ξ) will be continuous. 663

4 Figure 21.2: r(), H(), δ() nd d d δ() Let y 1 nd y 2 be two linerly independent solutions to the homogeneous eqution, L[y] = 0. Since the Green function stisfies the homogeneous eqution for = ξ, it will be liner combintion of the homogeneous solutions. c 1 y 1 + c 2 y 2 d 1 y 1 + d 2 y 2 for < ξ for > ξ We require tht G( ξ) be continuous. G( ξ) ξ = G( ξ) ξ + We cn write this in terms of the homogeneous solutions. c 1 y 1 (ξ) + c 2 y 2 (ξ) = d 1 y 1 (ξ) + d 2 y 2 (ξ) We integrte L[G( ξ)] = δ( ξ) from ξ to ξ+. ξ + ξ + ξ [G ( ξ) + p()g ( ξ) + q()g( ξ)] d = ξ δ( ξ) d. Since G( ξ) is continuous nd G ( ξ) hs only jump discontinuity two of the terms vnish. ξ + ξ p()g ( ξ) d = 0 nd ξ + ξ + ξ + G ( ξ) d = δ( ξ) d ξ ξ G ( ξ) ξ + = H( ξ) ξ + ξ ξ G (ξ + ξ) G (ξ ξ) = 1 ξ q()g( ξ) d = 0 We write this jump condition in terms of the homogeneous solutions. d 1 y 1(ξ) + d 2 y 2(ξ) c 1 y 1(ξ) c 2 y 2(ξ) = 1 Combined with the two boundry conditions, this gives us totl of four equtions to determine our four constnts, c 1, c 2, d 1, nd d

5 Result The second order inhomogeneous differentil eqution with homogeneous boundry conditions L[y] = y + p()y + q()y = f() for < < b B 1 [y] = B 2 [y] = 0 hs the solution y = b where G( ξ) stisfies the eqution G( ξ)f(ξ) dξ L[G( ξ)] = δ( ξ) for < < b B 1 [G( ξ)] = B 2 [G( ξ)] = 0. G( ξ) is continuous nd G ( ξ) hs jump discontinuity of height 1 t = ξ. Emple Solve the boundry vlue problem y = f() y(0) = y(1) = 0 using Green function. A pir of solutions to the homogeneous eqution re y 1 = 1 nd y 2 =. First note tht only the trivil solution to the homogeneous eqution stisfies the homogeneous boundry conditions. Thus there is unique solution to this problem. The Green function stisfies The Green function hs the form G ( ξ) = δ( ξ) G(0 ξ) = G(1 ξ) = 0. c 1 + c 2 for < ξ d 1 + d 2 for > ξ. Applying the two boundry conditions, we see tht c 1 = 0 nd d 1 = d 2. The Green function now hs the form c for < ξ d( 1) for > ξ. Since the Green function must be continuous, From the jump condition, ξ cξ = d(ξ 1) d = c ξ 1. d d c ξ ξ 1 ( 1) =ξ d d c =ξ = 1 ξ c ξ 1 c = 1 c = ξ 1. Thus the Green function is (ξ 1) for < ξ ξ( 1) for > ξ. The Green function is plotted in Figure 21.3 for vrious vlues of ξ. The solution to y = f() is 665

6 Figure 21.3: Plot of G( 0.05),G( 0.25),G( 0.5) nd G( 0.75). y() = ( 1) y() = 1 0 G( ξ)f(ξ) dξ ξf(ξ) dξ (ξ 1)f(ξ) dξ. Emple Solve the boundry vlue problem y = f() y(0) = 1 y(1) = 2. In Emple we sw tht the solution to u = f() u(0) = u(1) = 0 is u() = ( 1) Now we hve to find the solution to ξf(ξ) dξ (ξ 1)f(ξ) dξ. v = 0 v(0) = 1 u(1) = 2. The generl solution is Applying the boundry conditions yields v = c 1 + c 2. v = 1 +. Thus the solution for y is y = ( 1) ξf(ξ) dξ (ξ 1)f( i) dξ. Emple Consider y = y(0) = y(1) = 0. Method 1. Integrting the differentil eqution twice yields y = c 1 + c 2. Applying the boundry conditions, we find tht the solution is y = 1 6 (3 ). 666

7 Method 2. Using the Green function to find the solution, 1 y = ( 1) ξ 2 dξ + (ξ 1)ξ dξ 0 = ( 1) y = 1 6 (3 ). Emple Find the solution to the differentil eqution tht is bounded for ll. The Green function for this problem stisfies y y = sin G ( ξ) δ( ξ). The homogeneous solutions re y 1 = e, nd y 2 = e. The Green function hs the form c 1 e +c 2 e for < ξ d 1 e +d 2 e for > ξ. Since the solution must be bounded for ll, the Green function must lso be bounded. c 2 = d 1 = 0. The Green function now hs the form c e for < ξ d e for > ξ. Requiring tht G( ξ) be continuous gives us the condition c e ξ = d e ξ d = c e 2ξ. G( ξ) hs jump discontinuity of height 1 t = ξ. d d c e2ξ e d =ξ d c e = 1 =ξ c e 2ξ e ξ c e ξ = 1 Thus The Green function is then c = 1 2 e ξ 1 2 e ξ 1 2 e +ξ for < ξ for > ξ 1 2 e ξ. A plot of G( 0) is given in Figure The solution to y y = sin is y() = = e ξ sin ξ dξ sin ξ e ξ dξ + = 1 + cos ( sin y = 1 sin. 2 sin + cos ) 2 sin ξ e +ξ dξ 667

8 Figure 21.4: Plot of G( 0) Green Functions for Sturm-Liouville Problems Consider the problem L[y] = (p()y ) + q()y = f() subject to B 1 [y] = α 1 y() + α 2 y () = 0 B 2 [y] = β 1 y(b) + β 2 y (b) = 0. This is known s Sturm-Liouville problem. Equtions of this type often occur when solving prtil differentil equtions. The Green function ssocited with this problem stisfies L[G( ξ)] = δ( ξ) B 1 [G( ξ)] = B 2 [G( ξ)] = 0. Let y 1 nd y 2 be two non-trivil homogeneous solutions tht stisfy the left nd right boundry conditions, respectively. L[y 1 ] = 0 B 1 [y 1 ] = 0 L[y 2 ] = 0 B 2 [y 2 ] = 0. The Green function stisfies the homogeneous eqution for = ξ nd stisfies the homogeneous boundry conditions. Thus it must hve the following form. c 1 (ξ)y 1 () for ξ c 2 (ξ)y 2 () for ξ b Here c 1 nd c 2 re unknown functions of ξ. The first constrint on c 1 nd c 2 comes from the continuity condition. G(ξ ξ) = G(ξ + ξ) c 1 (ξ)y 1 (ξ) = c 2 (ξ)y 2 (ξ) We write the inhomogeneous eqution in the stndrd form. G ( ξ) + p p G ( ξ) + q δ( ξ) p p The second constrint on c 1 nd c 2 comes from the jump condition. G (ξ + ξ) G (ξ ξ) = 1 p(ξ) c 2 (ξ)y 2(ξ) c 1 (ξ)y 1(ξ) = 1 p(ξ) 668

9 Now we hve system of equtions to determine c 1 nd c 2. We solve this system with Krmer s rule. c 1 (ξ)y 1 (ξ) c 2 (ξ)y 2 (ξ) = 0 c 1 (ξ)y 1(ξ) c 2 (ξ)y 2(ξ) = 1 p(ξ) y 2 (ξ) c 1 (ξ) = p(ξ)( W (ξ)) c y 1 (ξ) 2(ξ) = p(ξ)( W (ξ)) Here W () is the Wronskin of y 1 () nd y 2 (). The Green function is y1()y 2(ξ) p(ξ)w (ξ) The solution of the Sturm-Liouville problem is Result The problem y = for ξ y 2()y 1(ξ) p(ξ)w (ξ) for ξ b. b G( ξ)f(ξ) dξ. L[y] = (p()y ) + q()y = f() subject to B 1 [y] = α 1 y() + α 2 y () = 0 B 2 [y] = β 1 y(b) + β 2 y (b) = 0. hs the Green function y1 )y 2 ξ) pξ)w ξ) y 2 )y 1 ξ) pξ)w ξ) for ξ for ξ b where y 1 nd y 2 re non-trivil homogeneous solutions tht stisfy B 1 [y 1 ] = B 2 [y 2 ] = 0, nd W () is the Wronskin of y 1 nd y 2. Emple Consider the eqution y y = f() y(0) = y(1) = 0. A set of solutions to the homogeneous eqution is {e e }. Equivlently, one could use the set {cosh sinh }. Note tht sinh stisfies the left boundry condition nd sinh( 1) stisfies the right boundry condition. The Wronskin of these two homogeneous solutions is W () = sinh sinh( 1) cosh cosh( 1) = sinh cosh( 1) cosh sinh( 1) = 1 2 [sinh(2 1) + sinh(1)] 1 [sinh(2 1) sinh(1)] 2 = sinh(1). The Green function for the problem is then sinh sinh(ξ 1) sinh(1) for 0 ξ sinh( 1) sinh ξ sinh(1) for ξ

10 The solution to the problem is y = sinh( 1) sinh(1) 0 sinh(ξ)f(ξ) dξ + sinh() sinh(1) 1 sinh(ξ 1)f(ξ) dξ Initil Vlue Problems Consider L[y] = y + p()y + q()y = f() subject the the initil conditions for < < b y() = γ 1 y () = γ 2. The solution is y = u + v where u + p()u + q()u = f() u() = 0 u () = 0 nd Since the Wronskin v + p()v + q()v = 0 v() = γ 1 v () = γ 2. W () = c ep p() d is non-vnishing, the solutions of the differentil eqution for v re linerly independent. Thus there is unique solution for v tht stisfies the initil conditions. The Green function for u stisfies G ( ξ) + p()g ( ξ) + q() δ( ξ) G( ξ) = 0 G ( ξ) = 0. The continuity nd jump conditions re G(ξ ξ) = G(ξ + ξ) G (ξ ξ) + 1 = G (ξ + ξ). Let u 1 nd u 2 be two linerly independent solutions of the differentil eqution. For < ξ, G( ξ) is liner combintion of these solutions. Since the Wronskin is non-vnishing, only the trivil solution stisfies the homogeneous initil conditions. The Green function must be 0 for < ξ u ξ () for > ξ where u ξ () is the liner combintion of u 1 nd u 2 tht stisfies u ξ (ξ) = 0 u ξ(ξ) = 1. Note tht the non-vnishing Wronskin ensures unique solution for u ξ. We cn write the Green function in the form H( ξ)u ξ (). This is known s the cusl solution. The solution for u is u = = = b b G( ξ)f(ξ) dξ H( ξ)u ξ ()f(ξ) dξ u ξ ()f(ξ) dξ 670

11 Now we hve the solution for y, y = v + u ξ ()f(ξ) dξ. Result The solution of the problem is y + p()y + q()y = f() y() = γ 1 y () = γ 2 y = y h + y ξ ()f(ξ) dξ where y h is the combintion of the homogeneous solutions of the eqution tht stisfy the initil conditions nd y ξ () is the liner combintion of homogeneous solutions tht stisfy y ξ (ξ) = 0, yξ (ξ) = Problems with Unmied Boundry Conditions Consider L[y] = y + p()y + q()y = f() subject the the unmied boundry conditions for < < b The solution is y = u + v where nd α 1 y() + α 2 y () = γ 1 β 1 y(b) + β 2 y (b) = γ 2. u + p()u + q()u = f() α 1 u() + α 2 u () = 0 β 1 u(b) + β 2 u (b) = 0 v + p()v + q()v = 0 α 1 v() + α 2 v () = γ 1 β 1 v(b) + β 2 v (b) = γ 2. The problem for v my hve no solution, unique solution or n infinite number of solutions. We consider only the cse tht there is unique solution for v. In this cse the homogeneous eqution subject to homogeneous boundry conditions hs only the trivil solution. The Green function for u stisfies The continuity nd jump conditions re G ( ξ) + p()g ( ξ) + q() δ( ξ) α 1 G( ξ) + α 2 G ( ξ) = 0 β 1 G(b ξ) + β 2 G (b ξ) = 0. G(ξ ξ) = G(ξ + ξ) G (ξ ξ) + 1 = G (ξ + ξ). Let u 1 nd u 2 be two solutions of the homogeneous eqution tht stisfy the left nd right boundry conditions, respectively. The non-vnishing of the Wronskin ensures tht these solutions eist. Let W () denote the Wronskin of u 1 nd u 2. Since the homogeneous eqution with homogeneous boundry conditions hs only the trivil solution, W () is nonzero on [ b]. The Green function hs the form c 1 u 1 for < ξ c 2 u 2 for > ξ. 671

12 The continuity nd jump conditions for Green function gives us the equtions Using Krmer s rule, the solution is Thus the Green function is The solution for u is c 1 u 1 (ξ) c 2 u 2 (ξ) = 0 c 1 u 1(ξ) c 2 u 2(ξ) = 1. c 1 = u 2(ξ) W (ξ) u = u1()u 2(ξ) W (ξ) c 2 = u 1(ξ) W (ξ). for < ξ u 1(ξ)u 2() W (ξ) for > ξ. b G( ξ)f(ξ) dξ. Thus if there is unique solution for v, the solution for y is y = v + Result Consider the problem b G( ξ)f(ξ) dξ. y + p()y + q()y = f() α 1 y() + α 2 y () = γ 1 β 1 y(b) + β 2 y (b) = γ 2. If the homogeneous differentil eqution subject to the inhomogeneous boundry conditions hs the unique solution y h, then the problem hs the unique solution where y = y h + b u1 )u 2 ξ) W ξ) u 1 ξ)u 2 ) W ξ) G( ξ)f(ξ) dξ for < ξ for > ξ u 1 nd u 2 re solutions of the homogeneous differentil eqution tht stisfy the left nd right boundry conditions, respectively, nd W () is the Wronskin of u 1 nd u Problems with Mied Boundry Conditions Consider L[y] = y + p()y + q()y = f() for < < b subject the the mied boundry conditions B 1 [y] = α 11 y() + α 12 y () + β 11 y(b) + β 12 y (b) = γ 1 B 2 [y] = α 21 y() + α 22 y () + β 21 y(b) + β 22 y (b) = γ 2. The solution is y = u + v where u + p()u + q()u = f() B 1 [u] = 0 B 2 [u] = 0 nd v + p()v + q()v = 0 B 1 [v] = γ 1 B 2 [v] = γ

13 The problem for v my hve no solution, unique solution or n infinite number of solutions. Agin we consider only the cse tht there is unique solution for v. In this cse the homogeneous eqution subject to homogeneous boundry conditions hs only the trivil solution. Let y 1 nd y 2 be two solutions of the homogeneous eqution tht stisfy the boundry conditions B 1 [y 1 ] = 0 nd B 2 [y 2 ] = 0. Since the completely homogeneous problem hs no solutions, we know tht B 1 [y 2 ] nd B 2 [y 1 ] re nonzero. The solution for v hs the form Applying the two boundry conditions yields v = c 1 y 1 + c 2 y 2. v = γ 2 B 2 [y 1 ] y 1 + γ 1 B 1 [y 2 ] y 2. The Green function for u stisfies G ( ξ) + p()g ( ξ) + q() δ( ξ) B 1 [G] = 0 B 2 [G] = 0. The continuity nd jump conditions re G(ξ ξ) = G(ξ + ξ) G (ξ ξ) + 1 = G (ξ + ξ). We write the Green function s the sum of the cusl solution nd the two homogeneous solutions H( ξ)y ξ () + c 1 y 1 () + c 2 y 2 () With this form, the continuity nd jump conditions re utomticlly stisfied. Applying the boundry conditions yields B 1 [G] = B 1 [H( ξ)y ξ ] + c 2 B 1 [y 2 ] = 0 B 2 [G] = B 2 [H( ξ)y ξ ] + c 1 B 2 [y 1 ] = 0 B 1 [G] = β 11 y ξ (b) + β 12 y ξ(b) + c 2 B 1 [y 2 ] = 0 B 2 [G] = β 21 y ξ (b) + β 22 y ξ(b) + c 1 B 2 [y 1 ] = 0 H( ξ)y ξ () β 21y ξ (b) + β 22 y ξ (b) B 2 [y 1 ] y 1 () β 11y ξ (b) + β 12 yξ (b) y 2 (). B 1 [y 2 ] Note tht the Green function is well defined since B 2 [y 1 ] nd B 1 [y 2 ] re nonzero. The solution for u is u = b G( ξ)f(ξ) dξ. Thus if there is unique solution for v, the solution for y is y = b G( ξ)f(ξ) dξ + γ 2 B 2 [y 1 ] y 1 + γ 1 B 1 [y 2 ] y

14 Result Consider the problem y + p()y + q()y = f() B 1 [y] = α 11 y() + α 12 y () + β 11 y(b) + β 12 y (b) = γ 1 B 2 [y] = α 21 y() + α 22 y () + β 21 y(b) + β 22 y (b) = γ 2. If the homogeneous differentil eqution subject to the homogeneous boundry conditions hs no solution, then the problem hs the unique solution where y = b G( ξ)f(ξ) dξ + γ 2 B 2 [y 1 ] y 1 + γ 1 B 1 [y 2 ] y 2 H( ξ)y ξ () β 21y ξ (b) + β 22 yξ (b) y 1 () B 2 [y 1 ] β 11y ξ (b) + β 12 yξ (b) y 2 () B 1 [y 2 ] y 1 nd y 2 re solutions of the homogeneous differentil eqution tht stisfy the first nd second boundry conditions, respectively, nd y ξ () is the solution of the homogeneous eqution tht stisfies y ξ (ξ) = 0, yξ (ξ) = Green Functions for Higher Order Problems Consider the n th order differentil eqution L[y] = y (n) + p n 1 ()y (n 1) + + p 1 ()y + p 0 y = f() on < < b subject to the n independent boundry conditions where the boundry conditions re of the form k=0 B j [y] = γ j n 1 n 1 B[y] α k y (k) () + β k y (k) (b). We ssume tht the coefficient functions in the differentil eqution re continuous on [ b]. The solution is y = u + v where u nd v stisfy k=0 L[u] = f() with B j [u] = 0 nd L[v] = 0 with B j [v] = γ j From Result , we know tht if the completely homogeneous problem L[w] = 0 with B j [w] = 0 hs only the trivil solution, then the solution for y eists nd is unique. We will construct this solution using Green functions. 674

15 First we consider the problem for v. Let {y 1... y n } be set of linerly independent solutions. The solution for v hs the form v = c 1 y c n y n where the constnts re determined by the mtri eqution B 1 [y 1 ] B 1 [y 2 ] B 1 [y n ] B 2 [y 1 ] B 2 [y 2 ] B 2 [y n ] B n [y 1 ] B n [y 2 ] B n [y n ] c 1 γ 1 c 2. = γ 2.. c n γ n To solve the problem for u we consider the Green function stisfying L[G( ξ)] = δ( ξ) with B j [G] = 0. Let y ξ () be the liner combintion of the homogeneous solutions tht stisfy the conditions The cusl solution is then y ξ (ξ) = 0 y ξ(ξ) = 0. =. y (n 2) ξ (ξ) = 0 y (n 1) ξ (ξ) = 1. y c () = H( ξ)y ξ (). The Green function hs the form H( ξ)y ξ () + d 1 y 1 () + + d n y n () The constnts re determined by the mtri eqution B 1 [y 1 ] B 1 [y 2 ] B 1 [y n ] B 2 [y 1 ] B 2 [y 2 ] B 2 [y n ] B n [y 1 ] B n [y 2 ] B n [y n ] d 1 B 1 [H( ξ)y ξ ()] d 2. = B 2 [H( ξ)y ξ ()].. d n B n [H( ξ)y ξ ()] The solution for u then is u = b G( ξ)f(ξ) dξ. 675

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