21.6 Green Functions for First Order Equations
|
|
- Adela Mason
- 5 years ago
- Views:
Transcription
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
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)
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
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationu t = k 2 u x 2 (1) a n sin nπx sin 2 L e k(nπ/l) t f(x) = sin nπx f(x) sin nπx dx (6) 2 L f(x 0 ) sin nπx 0 2 L sin nπx 0 nπx
Chpter 9: Green s functions for time-independent problems Introductory emples One-dimensionl het eqution Consider the one-dimensionl het eqution with boundry conditions nd initil condition We lredy know
More informationPartial Differential Equations
Prtil Differentil Equtions Notes by Robert Piché, Tmpere University of Technology reen s Functions. reen s Function for One-Dimensionl Eqution The reen s function provides complete solution to boundry
More information(4.1) D r v(t) ω(t, v(t))
1.4. Differentil inequlities. Let D r denote the right hnd derivtive of function. If ω(t, u) is sclr function of the sclrs t, u in some open connected set Ω, we sy tht function v(t), t < b, is solution
More informationSturm-Liouville Eigenvalue problem: Let p(x) > 0, q(x) 0, r(x) 0 in I = (a, b). Here we assume b > a. Let X C 2 1
Ch.4. INTEGRAL EQUATIONS AND GREEN S FUNCTIONS Ronld B Guenther nd John W Lee, Prtil Differentil Equtions of Mthemticl Physics nd Integrl Equtions. Hildebrnd, Methods of Applied Mthemtics, second edition
More information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
More informationAQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions
Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic
More informationVariational Techniques for Sturm-Liouville Eigenvalue Problems
Vritionl Techniques for Sturm-Liouville Eigenvlue Problems Vlerie Cormni Deprtment of Mthemtics nd Sttistics University of Nebrsk, Lincoln Lincoln, NE 68588 Emil: vcormni@mth.unl.edu Rolf Ryhm Deprtment
More informationLinearity, linear operators, and self adjoint eigenvalue problems
Linerity, liner opertors, nd self djoint eigenvlue problems 1 Elements of liner lgebr The study of liner prtil differentil equtions utilizes, unsurprisingly, mny concepts from liner lgebr nd liner ordinry
More informationGreen s functions. f(t) =
Consider the 2nd order liner inhomogeneous ODE Green s functions d 2 u 2 + k(t)du + p(t)u(t) = f(t). Of course, in prctice we ll only del with the two prticulr types of 2nd order ODEs we discussed lst
More information(9) P (x)u + Q(x)u + R(x)u =0
STURM-LIOUVILLE THEORY 7 2. Second order liner ordinry differentil equtions 2.1. Recll some sic results. A second order liner ordinry differentil eqution (ODE) hs the form (9) P (x)u + Q(x)u + R(x)u =0
More informationSTURM-LIOUVILLE BOUNDARY VALUE PROBLEMS
STURM-LIOUVILLE BOUNDARY VALUE PROBLEMS Throughout, we let [, b] be bounded intervl in R. C 2 ([, b]) denotes the spce of functions with derivtives of second order continuous up to the endpoints. Cc 2
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationConsequently, the temperature must be the same at each point in the cross section at x. Let:
HW 2 Comments: L1-3. Derive the het eqution for n inhomogeneous rod where the therml coefficients used in the derivtion of the het eqution for homogeneous rod now become functions of position x in the
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationGreen function and Eigenfunctions
Green function nd Eigenfunctions Let L e regulr Sturm-Liouville opertor on n intervl (, ) together with regulr oundry conditions. We denote y, φ ( n, x ) the eigenvlues nd corresponding normlized eigenfunctions
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationSTEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.
STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t
More informationMath Fall 2006 Sample problems for the final exam: Solutions
Mth 42-5 Fll 26 Smple problems for the finl exm: Solutions Any problem my be ltered or replced by different one! Some possibly useful informtion Prsevl s equlity for the complex form of the Fourier series
More information1 1D heat and wave equations on a finite interval
1 1D het nd wve equtions on finite intervl In this section we consider generl method of seprtion of vribles nd its pplictions to solving het eqution nd wve eqution on finite intervl ( 1, 2. Since by trnsltion
More informationChapter 28. Fourier Series An Eigenvalue Problem.
Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationApplied Partial Differential Equations with Fourier Series and Boundary Value Problems 5th Edition Richard Haberman
Applied Prtil Differentil Equtions with Fourier Series nd Boundry Vlue Problems 5th Edition Richrd Hbermn Person Eduction Limited Edinburgh Gte Hrlow Essex CM20 2JE Englnd nd Associted Compnies throughout
More information1 2-D Second Order Equations: Separation of Variables
Chpter 12 PDEs in Rectngles 1 2-D Second Order Equtions: Seprtion of Vribles 1. A second order liner prtil differentil eqution in two vribles x nd y is A 2 u x + B 2 u 2 x y + C 2 u y + D u 2 x + E u +
More informationHeat flux and total heat
Het flux nd totl het John McCun Mrch 14, 2017 1 Introduction Yesterdy (if I remember correctly) Ms. Prsd sked me question bout the condition of insulted boundry for the 1D het eqution, nd (bsed on glnce
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More information7 Green s Functions and Nonhomogeneous Problems
7 Green s Functions nd Nonhomogeneous Problems The young theoreticl physicists of genertion or two erlier subscribed to the belief tht: If you hven t done something importnt by ge 3, you never will. Obviously,
More information1.1. Linear Constant Coefficient Equations. Remark: A differential equation is an equation
1 1.1. Liner Constnt Coefficient Equtions Section Objective(s): Overview of Differentil Equtions. Liner Differentil Equtions. Solving Liner Differentil Equtions. The Initil Vlue Problem. 1.1.1. Overview
More informationGreen s function. Green s function. Green s function. Green s function. Green s function. Green s functions. Classical case (recall)
Green s functions 3. G(t, τ) nd its derivtives G (k) t (t, τ), (k =,..., n 2) re continuous in the squre t, τ t with respect to both vribles, George Green (4 July 793 3 My 84) In 828 Green privtely published
More information1.9 C 2 inner variations
46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for
More informationMath 115 ( ) Yum-Tong Siu 1. Lagrange Multipliers and Variational Problems with Constraints. F (x,y,y )dx
Mth 5 2006-2007) Yum-Tong Siu Lgrnge Multipliers nd Vritionl Problems with Constrints Integrl Constrints. Consider the vritionl problem of finding the extremls for the functionl J[y] = F x,y,y )dx with
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationHOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016
HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M
More informationFundamental Theorem of Calculus
Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under
More informationThe problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.
ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationg i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f
1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where
More informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06-073 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More information221B Lecture Notes WKB Method
Clssicl Limit B Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using
More information440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam
440-2 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP
More informationAMATH 731: Applied Functional Analysis Fall Some basics of integral equations
AMATH 731: Applied Functionl Anlysis Fll 2009 1 Introduction Some bsics of integrl equtions An integrl eqution is n eqution in which the unknown function u(t) ppers under n integrl sign, e.g., K(t, s)u(s)
More informationM344 - ADVANCED ENGINEERING MATHEMATICS
M3 - ADVANCED ENGINEERING MATHEMATICS Lecture 18: Lplce s Eqution, Anltic nd Numericl Solution Our emple of n elliptic prtil differentil eqution is Lplce s eqution, lso clled the Diffusion Eqution. If
More informationBoundary-value problems
226 Chpter 10 Boundry-vlue problems The initil-vlue problem is chrcterized by the imposition of uxiliry dt t single point: if the eqution is of the nth order, the n otherwise rbitrry constnts in its solution
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationAnalytical Methods Exam: Preparatory Exercises
Anlyticl Methods Exm: Preprtory Exercises Question. Wht does it men tht (X, F, µ) is mesure spce? Show tht µ is monotone, tht is: if E F re mesurble sets then µ(e) µ(f). Question. Discuss if ech of the
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More informationLecture Solution of a System of Linear Equation
ChE Lecture Notes, Dept. of Chemicl Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updted /) Lecture 8- - Solution of System of Liner Eqution 8. Why is it importnt to e le to solve system of liner
More informationTHE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem
More informationBest Approximation in the 2-norm
Jim Lmbers MAT 77 Fll Semester 1-11 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the -norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationLecture 6: Singular Integrals, Open Quadrature rules, and Gauss Quadrature
Lecture notes on Vritionl nd Approximte Methods in Applied Mthemtics - A Peirce UBC Lecture 6: Singulr Integrls, Open Qudrture rules, nd Guss Qudrture (Compiled 6 August 7) In this lecture we discuss the
More informationMapping the delta function and other Radon measures
Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support
More informationMath Theory of Partial Differential Equations Lecture 2-9: Sturm-Liouville eigenvalue problems (continued).
Mth 412-501 Theory of Prtil Differentil Equtions Lecture 2-9: Sturm-Liouville eigenvlue problems (continued). Regulr Sturm-Liouville eigenvlue problem: d ( p dφ ) + qφ + λσφ = 0 ( < x < b), dx dx β 1 φ()
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationAMS 212A Applied Mathematical Methods I Lecture 06 Copyright by Hongyun Wang, UCSC. ( ), v (that is, 1 ( ) L i
AMS A Applied Mthemticl Methods I Lecture 6 Copyright y Hongyun Wng, UCSC Recp of Lecture 5 Clssifiction of oundry conditions Dirichlet eumnn Mixed Adjoint opertor, self-djoint opertor Sturm-Liouville
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationNotes on the Eigenfunction Method for solving differential equations
Notes on the Eigenfunction Metho for solving ifferentil equtions Reminer: Wereconsieringtheinfinite-imensionlHilbertspceL 2 ([, b] of ll squre-integrble functions over the intervl [, b] (ie, b f(x 2
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl Won-Kwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
More informationSTURM-LIOUVILLE THEORY, VARIATIONAL APPROACH
STURM-LIOUVILLE THEORY, VARIATIONAL APPROACH XIAO-BIAO LIN. Qudrtic functionl nd the Euler-Jcobi Eqution The purpose of this note is to study the Sturm-Liouville problem. We use the vritionl problem s
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationCalculus of Variations
Clculus of Vritions Com S 477/577 Notes) Yn-Bin Ji Dec 4, 2017 1 Introduction A functionl ssigns rel number to ech function or curve) in some clss. One might sy tht functionl is function of nother function
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n nti-derivtive is not esily recognizble, then we re in
More informationModule 9: The Method of Green s Functions
Module 9: The Method of Green s Functions The method of Green s functions is n importnt technique for solving oundry vlue nd, initil nd oundry vlue prolems for prtil differentil equtions. In this module,
More information( ) 2. ( ) is the Fourier transform of! ( x). ( ) ( ) ( ) = Ae i kx"#t ( ) = 1 2" ( )"( x,t) PC 3101 Quantum Mechanics Section 1
1. 1D Schrödinger Eqution G chpters 3-4. 1.1 the Free Prticle V 0 "( x,t) i = 2 t 2m x,t = Ae i kxt "( x,t) x 2 where = k 2 2m. Normliztion must hppen: 2 x,t = 1 Here, however: " A 2 dx " " As this integrl
More informationOrthogonal functions
Orthogonl functions Given rel vrible over the intervl (, b nd set of rel or complex functions U n (ξ, n =, 2,..., which re squre integrble nd orthonorml b U n(ξu m (ξdξ = δ n,m ( if the set of of functions
More informationChapter 4. Additional Variational Concepts
Chpter 4 Additionl Vritionl Concepts 137 In the previous chpter we considered clculus o vrition problems which hd ixed boundry conditions. Tht is, in one dimension the end point conditions were speciied.
More informationKinematic Waves. These are waves which result from the conservation equation. t + I = 0. (2)
Introduction Kinemtic Wves These re wves which result from the conservtion eqution E t + I = 0 (1) where E represents sclr density field nd I, its outer flux. The one-dimensionl form of (1) is E t + I
More information1 E3102: a study guide and review, Version 1.0
1 E3102: study guide nd review, Version 1.0 Here is list of subjects tht I think we ve covered in clss (your milege my vry). If you understnd nd cn do the bsic problems in this guide you should be in very
More informationSession Trimester 2. Module Code: MATH08001 MATHEMATICS FOR DESIGN
School of Science & Sport Pisley Cmpus Session 05-6 Trimester Module Code: MATH0800 MATHEMATICS FOR DESIGN Dte: 0 th My 06 Time: 0.00.00 Instructions to Cndidtes:. Answer ALL questions in Section A. Section
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationMath 211A Homework. Edward Burkard. = tan (2x + z)
Mth A Homework Ewr Burkr Eercises 5-C Eercise 8 Show tht the utonomous system: 5 Plne Autonomous Systems = e sin 3y + sin cos + e z, y = sin ( + 3y, z = tn ( + z hs n unstble criticl point t = y = z =
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More informationProblem set 1: Solutions Math 207B, Winter 2016
Problem set 1: Solutions Mth 27B, Winter 216 1. Define f : R 2 R by f(,) = nd f(x,y) = xy3 x 2 +y 6 if (x,y) (,). ()Show tht thedirectionl derivtives of f t (,)exist inevery direction. Wht is its Gâteux
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationLecture 13 - Linking E, ϕ, and ρ
Lecture 13 - Linking E, ϕ, nd ρ A Puzzle... Inner-Surfce Chrge Density A positive point chrge q is locted off-center inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on
More information7. Indefinite Integrals
7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find
More informationMath 520 Final Exam Topic Outline Sections 1 3 (Xiao/Dumas/Liaw) Spring 2008
Mth 520 Finl Exm Topic Outline Sections 1 3 (Xio/Dums/Liw) Spring 2008 The finl exm will be held on Tuesdy, My 13, 2-5pm in 117 McMilln Wht will be covered The finl exm will cover the mteril from ll of
More informationA HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction
Ttr Mt. Mth. Publ. 44 (29), 159 168 DOI: 1.2478/v1127-9-56-z t m Mthemticl Publictions A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES Miloslv Duchoň Peter Mličký ABSTRACT. We present Helly
More informationSection 3.3: Fredholm Integral Equations
Section 3.3: Fredholm Integrl Equtions Suppose tht k : [, b] [, b] R nd g : [, b] R re given functions nd tht we wish to find n f : [, b] R tht stisfies f(x) = g(x) + k(x, y) f(y) dy. () Eqution () is
More informationSpace Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More information221A Lecture Notes WKB Method
A Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using ψ x, t = e
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationHomework 11. Andrew Ma November 30, sin x (1+x) (1+x)
Homewor Andrew M November 3, 4 Problem 9 Clim: Pf: + + d = d = sin b +b + sin (+) d sin (+) d using integrtion by prts. By pplying + d = lim b sin b +b + sin (+) d. Since limits to both sides, lim b sin
More informationTopic 1 Notes Jeremy Orloff
Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble
More informationChapter Five - Eigenvalues, Eigenfunctions, and All That
Chpter Five - Eigenvlues, Eigenfunctions, n All Tht The prtil ifferentil eqution methos escrie in the previous chpter is specil cse of more generl setting in which we hve n eqution of the form L 1 xux,tl
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationThe Dirac distribution
A DIRAC DISTRIBUTION A The Dirc distribution A Definition of the Dirc distribution The Dirc distribution δx cn be introduced by three equivlent wys Dirc [] defined it by reltions δx dx, δx if x The distribution
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More information