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### ch11

Course: MATH 260, Spring 2008
School: Gonzaga
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CHAPTER ---------------------------------------------------- 11. ---- Chapter Eleven Section 11.1 1. Since the right hand sides of the ODE and the boundary conditions are all zero, the boundary value problem is homogeneous. 3. The right hand side of the ODE is nonzero. Therefore the boundary value problem is nonhomogeneous. 6. The ODE can also be written as C ww -^&quot; B# C oe ! . Although the second...

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CHAPTER ---------------------------------------------------- 11. ---- Chapter Eleven Section 11.1 1. Since the right hand sides of the ODE and the boundary conditions are all zero, the boundary value problem is homogeneous. 3. The right hand side of the ODE is nonzero. Therefore the boundary value problem is nonhomogeneous. 6. The ODE can also be written as C ww -^" B# C oe ! . Although the second boundary condition has a more general form, the boundary value problem is homogeneous. 7. First assume that - oe ! . The general solution of the ODE is CaBb oe -" B -# . The boundary condition at B oe ! requires that -# oe ! . Imposing the second condition, - " a 1 "b - # oe ! . It follows that -" oe -# oe ! . Hence there are no nontrivial solutions. Suppose that - oe .# . In this case, the general solution of the ODE is CaBb oe -" -9=2 .B -# =382 .B . The first boundary condition requires that -" oe ! . Imposing the second condition, -" a-9=2 .1 .=382 .1b -# a=382 .1 .-9=2 .1b oe ! . -# a>+82 .1 .b oe ! . The two boundary conditions result in Since the only solution of the equation >+82 .1 . oe ! is . oe ! , we have -# oe ! . Hence there are no nontrivial solutions. Let - oe .# , with . ! . Then the general solution of the ODE is CaBb oe -" -9= .B -# =38 .B . Imposing the boundary conditions, we obtain -" oe ! and -" a-9= .1 .=38 .1b -# a=38 .1 .-9= .1b oe ! . -9= .1 oe ! =38 .1 oe ! , For a nontrivial solution of the ODE, we require that =38 .1 .-9= .1 oe ! . Note that which is false. It follows that >+8 .1 oe . . From a plot of 1 >+8 1. and 1. , ________________________________________________________________________ page 720 ---------------------------------------------------- CHAPTER 11. ---- we find that there is a sequence of solutions, ." !()(' , .# "'("' , ; For large values of 8, 1 1 .8 a#8 "b . # Therefore the eigenfunctions are 98 aBb oe =38 .8 B , with corresponding eigenvalues -" !'#!% , -# #(*%\$ , a#8 "b# -8 . % Asymptotically, 8. With - oe ! , the general solution of the ODE is CaBb oe -" B -# . Imposing the two boundary conditions, -" oe ! and #-" -# oe ! . It follows that -" oe -# oe ! . Hence there are no nontrivial solutions. Setting - oe .# , the general solution of the ODE is CaBb oe -" -9=2 .B -# =382 .B . The first boundary condition requires that -# oe ! . Imposing the second condition, -" a-9=2 . .=382 .b -# a=382 . .-9=2 .b oe ! . -" a" . >+82 .b oe ! . The two boundary conditions result in Since . >+82 . ! , it follows that -" oe ! , and there are no nontrivial solutions. Let - oe .# , with . ! . Then the general solution of the ODE is CaBb oe -" -9= .B -# =38 .B . Imposing the boundary conditions, we obtain -# oe ! and ________________________________________________________________________ page 721 ---------------------------------------------------- CHAPTER 11. ---- -" a-9= . .=38 .b -# a=38 . .-9= .b oe ! . For a nontrivial solution of the ODE, we require that -9= . .=38 . oe ! . First note that -9= . oe ! . oe ! or =38 . oe ! . Therefore we find that " . >+8 . oe ! . From a plot of . >+8 . , there is a sequence of solutions, ." !)'!\$ , .# \$%#&' , ; For large 8, Therefore the eigenfunctions are 98 aBb oe -9= .8 B , with corresponding eigenvalues -" !(%!# , -# ""(\$%* , - 8 a8 "b # 1 # . Asymptotically, . 8 a8 "b 1 . 12. First note that T aBb oe " , UaBb oe #B and V aBb oe - . Based on Prob. "" , the integrating factor is a solution of the ODE The differential equation is first order linear, with solution .aBb oe - /B:a B# b. It then follows that the Hermite equation can be written as '/B C w " - /B C oe ! . # . w aBb oe #B .aBb . w # 14. For the Laguerre equation, T aBb oe B , UaBb oe " B and V aBb oe - . Using the result of Prob. "", the integrating factor is a solution of the ODE The general solution of . w aBb oe .aBb is .aBb oe - /B . Therefore the Laguerre equation can be written as ________________________________________________________________________ page 722 B . w aBb oe B .aBb . ---------------------------------------------------- CHAPTER 11. ---- cB /B C w d - /B C oe ! . w 15. For the Chebyshev equation, T aBb oe " B# , UaBb oe B and V aBb oe !# . The integrating factor is a solution of the ODE ^" B# . w aBb oe B .aBb . .. B oe . . " B# The general solution of the resulting ODE is .aBb oe . k" B# k The differential equation is separable, with Recall that the Chebyshev equation is typically defined for kBk Y " . Therefore it can also be written as '" B# C w " w " B# !# C oe !. 16. We consider solutions of the form ?aB >b oe \ aBbX a>b . Substitution into the PDE results in \X ww - \X w 5 \X oe !# \ ww X . Dividing both sides of the equation by \X , we obtain \X ww \X w \ ww X , 5 oe !# \X \X \X that is, ww X ww Xw #\ oe! 5. X X \ Since both sides of the resulting equation are functions of different variables, each must be equal to a constant, say - . Therefore we obtain two ordinary differential equations !# \ ww a- 5 b\ oe ! and X ww -X w -X oe ! . 17a+b. Setting C oe =aBb? , we have C w oe = w ? = ?w and C ww oe = ww ? #= w ? w = ?ww . Substitution into the given ODE results in = ww ? #= w ? w = ?ww #a= w ? = ?w b a" -b=? oe ! . ________________________________________________________________________ page 723 ---------------------------------------------------- CHAPTER 11. ---- Collecting the various terms, = ?ww a#= w #=b? w c= ww #= w a" -b=d? oe ! . The second term on the left vanishes as long as = w oe = . a,b. With =aBb oe /B , the transformed differential equation can be written as Since the boundary conditions are homogeneous, we also have ?a!b oe ?a"b oe ! . It now follows that the eigenfunctions are ?8 oe =38 -8 B , with corresponding eigenvalues Therefore the eigenfunctions for the original problem are 98 aBb oe /B =38 81B , with corresponding eigenvalues " -8 oe " 8# 1 # . a- b. The given equation is a second order constant coefficient differential equation. The characteristic equation is with roots <"# oe " ,, - . <# #< a" -b oe ! , - 8 oe 8# 1 # . ?ww -? oe ! . If - oe ! , then the general solution is C oe -" /B -# B/B . Imposing the two boundary conditions, we find that -" oe -# oe ! , and hence there are no nontrivial solutions. If - ! , then the general solution is C oe -" /B:S" - <B -# /B:S" - <B . It again follows that -" oe -# oe ! , and hence there are no nontrivial solutions. Therefore - ! , and the general solution is C oe -" /B -9= - B -# /B =38 - B . Invoking the boundary conditions, we have -" oe ! and -# /=38- oe ! . For a nontrivial solution, - oe 81 . 19. First write the differential equation as C ww a" -bC w -C oe ! , which is a second order constant coefficient differential equation. The characteristic equation is ________________________________________________________________________ page 724 ---------------------------------------------------- CHAPTER 11. ---- < # a" - b< - oe ! , with roots <" oe " and <# oe - . For - " , the general solution is C oe -" /B -# /-B . Imposing the boundary conditions, we require that -" -# oe ! and -" /" -# /- oe ! . For a nontrivial solution, it follows that /" oe /- , and hence - oe " , which is contrary to the assumption. If - oe " , then the general solution is C oe -" /B -# B/B . The boundary conditions require that -" oe ! and -" /" -# /" oe ! . Hence there are no nontrivial solutions. 21. Suppose that - oe ! . In that case the general solution is C oe -" B -# . The boundary conditions require that -" #-# oe ! and -" -# oe ! . We find that -" oe -# oe ! , and hence there are no nontrivial solutions. a+b. Let - oe .# , with . ! . Then the general solution of the ODE is CaBb oe -" -9= .B -# =38 .B . The boundary conditions require that #-" . -# oe ! and -" -9= . -# =38 . oe ! . These equations have a nonzero solution only if #=38 . . -9= . oe ! , which can also be written as #>+8 . . oe ! . Based on the graph, the positive roots of the determinantal equation are ________________________________________________________________________ page 725 ---------------------------------------------------- CHAPTER 11. ---- 1 ." %#(%) , .# (&*'& , ; for large 8, .8 a#8 "b . # Therefore the eigenvalues are -" ")#(\$) , -# &((!(& , ; for large 8, -8 a#8 "b# CaBb oe -" -9=2 .B -# =382 .B . 1# . % a,b. Setting - oe .# ! , the general solution of the ODE is Imposing the boundary conditions, we obtain the equations #-" . -# oe ! and -" -9=2 . -# =382 . oe ! . These equations have a nonzero solution only if #=382 . . -9=2 . oe ! . The latter equation is satisfied only for . oe ! and . oe ,, "*"&! . Hence the only negative eigenvalue is -" oe \$''(\$ . 24. Based on the physical problem, - oe 7=# IM ! . Let - oe .% . The characteristic equation is <% .% oe ! , with roots <"# oe ,, .3 , <\$ oe . and <% oe . . Hence the general solution is CaBb oe -" -9=2 .B -# =382 .B -\$ -9= .B -% =38 .B . a+b. Simply supported on both ends Ca!b oe C ww a!b oe ! C aPb oe C ww aPb oe ! Invoking the boundary conditions, we obtain the system of equations -" - \$ oe ! -" - \$ oe ! -" -9=2 .P -# =382 .P -\$ -9= .P -% =38 .P oe ! # -" . -9=2 .P -# .# =382 .P -\$ .# -9= .P -% .# =38 .P oe ! . The determinantal equation is .% =382 .P =38 .P oe ! . The nonzero roots are .8 oe 81P , 8 oe " # . The first two equations result in -" oe -\$ oe ! . The last two equations, -# =382 81 -% =38 81 oe ! -# =382 81 -% =38 81 oe ! , imply that -# oe ! . Therefore the eigenfunctions are 98 oe =38 .8 B , with corresponding eigenvalues -8 oe 8% 1% P% . ________________________________________________________________________ page 726 ---------------------------------------------------- CHAPTER 11. ---- a,b. Simply supported Ca!b oe C ww a!b oe ! clamped C aPb oe C w aPb oe ! Invoking the boundary conditions, we obtain the system of equations -" - \$ oe ! -" - \$ oe ! -" -9=2 .P -# =382 .P -\$ -9= .P -% =38 .P oe ! -" .=382 .P -# .-9=2 .P -\$ .=38 .P -% .-9= .P oe ! . The determinantal equation is #.\$ =382 .P -9= .P #.\$ -9=2 .P =38 .P oe ! . Based on numerical analysis, ." \$*#''P and .# (!')'P . The first two equations result in -" oe -\$ oe ! . The last two equations, -# =382 .8 P -% =38 .8 P oe ! -# -9=2 .8 P -% -9= .8 P oe ! , imply that -# oe Therefore the eigenfunctions are 98 oe =38 .8 P =382 .8 B =38 .8 B , =382 .8 P =38 .8 P -% . =382 .8 P with corresponding eigenvalues -8 oe .% . 8 a- b. Clamped Ca!b oe C w a!b oe ! free C ww aPb oe C www aPb oe ! Invoking the boundary conditions, we obtain the system of equations ________________________________________________________________________ page 727 ---------------------------------------------------- CHAPTER 11. ---- -" - \$ oe ! . -# . - % oe ! # # # -" . -9=2 .P -# . =382 .P -\$ . -9= .P -% .# =38 .P oe ! -" .\$ =382 .P -# .\$ -9=2 .P -\$ .\$ =38 .P -% .\$ -9= .P oe ! . The determinantal equation is " -9=2 .P -9= .P oe ! . The first two nonzero roots are ." ")(&"P and .# %'*%"P . With -\$ oe -" and -% oe -# , the system of equations reduce to -" a-9=2 .8 P -9= .8 Pb -# a=382 .8 P =38 .8 Pb oe ! -" a=382 .8 P =38 .8 Pb -# a-9=2 .8 P -9= .8 Pb oe ! . Let E8 oe a-9=2 .8 P -9= .8 Pba=382 .8 P =38 .8 Pb . The eigenfunctions are given by 98 aBb oe -9=2 .8 B -9= .8 B E8 a=38 .8 B =382 .8 Bb , with corresponding eigenvalues -8 oe .% . 8 25a+b. Assume that the solution has the form ?aB >b oe \ aBbX a>b . Substitution into the PDE results in I ww \ X oe \X ww . 3 Dividing both sides of the equation by \X , we obtain I \ ww X \X ww oe , 3 \X \X that is, ________________________________________________________________________ page 728 ---------------------------------------------------- CHAPTER 11. ---- \ ww 3 X ww oe . \ I X Since both sides of the resulting equation are functions of different variables, each must be equal to a constant, say - . Therefore we obtain two ordinary differential equations \ ww -\ oe ! and X ww I X oe !. 3 a,b. Given that ?a! >b oe \ a!bX a>b for > ! , it follows that \ a!b oe ! . The second boundary condition can be expressed as From the result in Part a+b, IE\ w aPbX a>b 7\ aPbX ww a>b oe ! , > ! . IE\ w aPbX a>b -7 I \ aPbX a>b oe ! , > ! . 3 Since the condition is to be satisfied for all > ! , we arrive at the boundary condition 7 \ w aPb \ aPb oe ! . 3E a- b. If - oe ! , the general solution of the spatial equation is \ aBb oe -" B -# . The boundary condition require that -" oe -# oe !. Hence there are no nontrivial solutions. If - oe .# ! , then the general solution is \ aBb oe -" -9=2 .B -# =382 .B . The first boundary condition implies that -" oe !. The second boundary condition requires that 7 -# -9=2 .P -# . =382 .P oe ! . 3E The solution is nontrivial only if 3E . 7 Since . >+82 .P ! , there are no nontrivial solutions. . >+82 .P oe Let - oe .# ! . The general solution of the spatial equation is \ aBb oe -" -9= .B -# =38 .B . ________________________________________________________________________ page 729 ---------------------------------------------------- CHAPTER 11. ---- The first boundary condition implies that -" oe !. The second boundary condition requires that 7 -# -9= .P -# . =38 .P oe ! . 3E For a nontrivial solution, it is necessary that 7 -9= .P . =38 .P oe ! , 3E or >+8 .P oe For the case a73EPb oe !& , 3E . 7. we find that ." P "!('* and .# P \$'%\$' . Therefore the eigenfunctions are given by 98 aBb oe =38 .8 B . The corresponding eigenvalues are solutions of -9= -8 P P -8 =38 -8 P oe ! . # The first two eigenvalues are approximated as -" ""&*(P# and -# "\$#('P# . ________________________________________________________________________ page 730 ---------------------------------------------------- CHAPTER 11. ---- Section 11.2 2. Based on the boundary conditions, - ! . The general solution of the ODE is The boundary condition C w a!b oe ! requires that -# oe ! . Imposing the second boundary condition, we find that -" -9=- oe ! . So for a nontrivial solution, - oe a#8 "b1# , 8 oe " # Therefore the eigenfunctions are given by 98 aBb oe 58 -9= a#8 "b1B . # # CaBb oe -" -9=-B -# =38-B . In this problem, <aBb oe " , and the normalization condition is # 58 ( ! " a#8 "b1B "-9= .B oe " . # # It follows that 58 oe # . Therefore the normalized eigenfunctions are 98 aBb oe # -9= a#8 "b1B , 8 oe " # # 3. Based on the boundary conditions, - !. For - oe !, the eigenfunction is 9! aBb oe 5 ! Set 5! oe " . With - ! , the general solution of the ODE is Invoking the boundary conditions, we require that -# oe ! and -" - =38- oe ! . Since - ! , the eigenvalues are -8 oe 8# 1# , 8 oe " # , with corresponding eigenfunctions 98 aBb oe 58 -9= 81B . " CaBb oe -" -9=-B -# =38-B . The normalization condition is # 58 ( -9=# 81B .B oe " . ! # It follows that 58 oe # . Therefore the normalized eigenfunctions are 9! aBb oe " , and 98 aBb oe # -9= 81B , 8 oe " # 4. From Prob. ) in Section """ , the eigenfunctions are 98 aBb oe 58 -9= -8 B , in which -9=-8 -8 =38-8 oe ! . The normalization condition is ________________________________________________________________________ page 731 ---------------------------------------------------- CHAPTER 11. ---- # 58 ( ! " -9=# -8 B .B oe " . -9= -8 =38 -8 -8 . #-8 First note that # ( -9= -8 B .B oe " ! Based on the determinantal equation, -9= -8 =38 -8 -8 " =38# -8 oe # #-8 \$ -9= #-8 oe . % % \$ -9= #-8 # -9= -8 B Therefore # 58 oe and the normalized eigenfunctions are given by 98 aBb oe \$ -9= #-8 . 6. As shown in Prob. ", the normalized eigenfunctions are 98 aBb oe # =38 Based on Eq. a\$%b, with <aBb oe " , the coefficients in the eigenfunction expansion are given by -7 oe ( 0 aBb97 aBb.B " a#8 "b1B , 8 oe " # # oe #( =38 " ! oe a#7 "b1 ## ! a#7 "b1B .B # . Therefore we obtain the formal expansion "oe ## _ " a#8 "b1B " =38 . 1 8 oe " #8 " # ________________________________________________________________________ page 732 ---------------------------------------------------- CHAPTER 11. ---- 8. We consider the normalized eigenfunctions 98 aBb oe # =38 a#8 "b1B , 8 oe " # # Based on Eq. a\$%b, with <aBb oe " , the coefficients in the eigenfunction expansion are given by -7 oe ( 0 aBb97 aBb.B " oe #( oe ! "# ## a#7 "b1 "" -9= . a#7 "b1 % ! =38 a#7 "b1B .B # Therefore we obtain the formal expansion 0 aBb oe ## _ a#8 "b1 a#8 "b1B " "" -9= . =38 1 8oe" % # 9. The normalized eigenfunctions are Based on Eq. a\$%b, with <aBb oe " , the coefficients in the eigenfunction expansion are given by -7 oe ( 0 aBb97 aBb.B " " a#7 "b1B a#7 "b1B .B #( =38 .B # # ! "# ) 71 71 oe # # '=38 # -9= # " . a#7 "b 1 98 aBb oe # =38 a#8 "b1B , 8 oe " # # oe #( ! "# #B =38 Therefore the formal expansion of the given function is 0 aBb oe ) _ =38 8#1 -9= 8#1 a#8 "b1B " =38 . # 1# 8 oe " a#8 "b # # -9= -8 B 11. From Prob. %, the normalized eigenfunctions are given by 98 aBb oe \$ -9= #-8 , ________________________________________________________________________ page 733 ---------------------------------------------------- CHAPTER 11. ---- in which the eigenvalues satisfy -9=-8 -8 =38-8 oe !. Based on Eq. a\$%b, the coefficients in the eigenfunction expansion are given by -7 oe ( 0 aBb97 aBb.B " ! oe oe # ^# -9= -7 " -7 !7 \$ -9= #-7 # ( B -9= -7 B.B " ! , in which !7 oe " =38# -7 12. The normalized eigenfunctions are given by 98 aBb oe in which the eigenvalues satisfy -9=-8 -8 =38-8 oe !. Based on Eq. a\$%b, the coefficients in the eigenfunction expansion are given by -7 oe ( 0 aBb97 aBb.B " ! \$ -9= #-8 # -9= -8 B , oe \$ -9= #-7 -7 !7 # oe # ^" -9= -7 ( a" Bb-9= -7 B.B " ! , in which !7 oe " =38# -7 13. We consider the normalized eigenfunctions 98 aBb oe in which the eigenvalues satisfy -9=-8 -8 =38-8 oe !. The coefficients in the eigenfunction expansion are given by \$ -9= #-8 # -9= -8 B , ________________________________________________________________________ page 734 ---------------------------------------------------- CHAPTER 11. ---- -8 oe ( 0 aBb98 aBb.B " ! oe oe in which !8 oe " =38# -8 # =38^-8 # , -8 !8 \$ -9= #-8 # ( "# ! -9= -8 B.B 15. The differential equation can be written as with :aBb oe " B# and ; aBb oe " . The boundary conditions are homogeneous and separated. Hence the BVP is self-adjoint. 16. Since the boundary conditions are not separated, the inner product is computed: Given ? and @, sufficiently smooth and satisfying the boundary conditions, aPc?d @b oe ( c? ww @ ?@d.B " ! " ! " ! ^" B# C w ` w C oe ! , oe ? w @ ( c? w @ w ?@d.B oe c? w @ ?@ w d a? Pc@db " ! Based on the given boundary conditions, ? w a"b@a"b ? w a!b@a!b oe ?a!b@a"b #?a"b@a!b ?a"b@ w a"b ?a!b@ w a!b oe ?a"b@a!b #?a!b@a"b c? w @ ?@ w d oe ?a"b@a!b ?a!b@a"b , " ! Since the BVP is not self-adjoint. 18. The differential equation can be written as w with :aBb oe " , ; aBb oe ! , and <aBb oe " . The boundary conditions are homogeneous and separated. Hence the BVP is self-adjoint. cC w d oe - C , ________________________________________________________________________ page 735 ---------------------------------------------------- CHAPTER 11. ---- 19. If +# oe ! , then ? w a"b@a"b ?a"b@ w a"b oe ,# w ,# ? a"b@ w a"b ? w a"b@ w a"b oe ! , ," ," and since ?a!b oe @a!b oe ! , If ,# oe ! , then ?a"b oe @a"b oe ! implies that Furthermore, ? w a!b@a!b ?a!b@ w a!b oe ! . ? w a"b@a"b ?a"b@ w a"b oe ! . +# w +# ? a!b@ w a!b ? w a!b@ w a!b oe ! . +" +" ? w a!b@a!b ?a!b@ w a!b oe Clearly, the results are also true if +# oe ,# oe ! . 20. Suppose that 9" aBb and 9# aBb are linearly independent eigenfunctions associated with an eigenvalue - . The Wronskian is given by Each of the eigenfunctions satisfies the boundary condition +" Ca!b +# C w a!b oe ! . If either +" oe ! or +# oe ! , then clearly [ a9" , 9# ba!b oe ! . On the other hand, if +# is not equal to zero, then [ a9" , 9# ba!b oe 9" a!b9#w a!b 9# a!b9"w a!b +" +" oe 9" a!b9# a!b 9# a!b9" a!b +# +# oe !. [ a9" , 9# baBb oe 9" aBb9#w aBb 9# aBb9"w aBb. By Theorem \$\$# , [ a9" , 9# baBb oe ! for all ! Y B Y " . Based on Theorem \$\$\$ , 9" aBb and 9# aBb must be linearly dependent. Hence - must be a simple eigenvalue. 22. We consider the operator PcCd oe c:aBbC w d ; aBbC w on the interval ! B " , together with the boundary conditions +" Ca!b +# C w a!b oe ! , ," C a"b ,# C w a"b oe ! . Let ? oe 9 3< and @ oe 0 3( . If ? and @ both satisfy the boundary conditions, then the real and imaginary parts also satisfy the same boundary conditions. Using the inner product a? @b oe ( ?aBb@aBb.B , " ! ________________________________________________________________________ page 736 ---------------------------------------------------- CHAPTER 11. ---- w aPc?d @b oe ( c:aBb? w d @ ; aBb?@`.B " ! " w oe ( ~ c:aBba9 w 3< w bd @ ; aBb?@TM.B ! " ! " oe :aBba9 w 3< w b@ ( e:aBba9 w 3< w b @ w ; aBb?@f.B ! Integrating by parts, again, w w w w w w ( e:aBba9 3< b@ f.B oe a9 3<b:aBb@ ( ~c:aBb@ d ?TM.B . " " ! " ! ! Collecting the boundary terms, :aBbca9 w 3< w b@ a9 3<b@ w d oe :aBbca9 w 3< w ba0 3(b a9 3<ba0 w 3( w bd " ! " ! . The real part is given by :aBbca9 w 0 < w (b a90 w <( w bd oe :aBbca9 w 0 90 w b a< w ( <( w bd " ! w w " ! w " ! " ! oe :aBbc9 0 90 d :aBbc< ( <( w d Since 9 , < , 0 and ( satisfy the boundary conditions, it follows that :aBbca9 w 0 < w (b a90 w <( w bd oe ! . " ! Similarly, the imaginary part also vanishes. That is, :aBbca< w 0 <0 w b a9 w ( 9( w bd oe ! . " ! Therefore w aPc?d @b oe ( ~ c:aBb@ w d ? ; aBb?@TM.B " The result follows from the fact that a? Pc@db oe a? Pc@db . 24. Based on the physical problem, - oe T IM ! . Let - oe .# . The characteristic equation is <% .# <# oe ! , with roots <"# oe ! , <\$ oe .3 and <% oe .3 . Hence the general solution is CaBb oe -" -# B -\$ -9= .B -% =38 .B . oe a P c @ d ?b oe a ? P c@ d b . ! ________________________________________________________________________ page 737 ---------------------------------------------------- CHAPTER 11. ---- a+b. Simply supported on both ends Ca!b oe C ww a!b oe ! C aPb oe C ww aPb oe ! Invoking the boundary conditions, we obtain the system of equations -" - \$ oe ! -\$ oe ! -\$ -9= .P -% =38 .P oe ! -" -# P -\$ -9= .P -% =38 .P oe ! . The determinantal equation is =38 .P oe ! . The nonzero roots are .8 oe 81P , 8 oe " # . Therefore the eigenfunctions are 98 oe =38 .8 B , with corresponding eigenvalues -8 oe 8# 1# P# . Hence the smallest eigenvalue is -" oe 1# P# . a,b. Simply supported Ca!b oe C ww a!b oe ! clamped C aPb oe C w aPb oe ! Invoking the boundary conditions, we obtain the system of equations -" - \$ oe ! -\$ oe ! -# -\$ .=38 .P -% .-9= .P oe ! -" -# P -\$ -9= .P -% =38 .P oe ! . The determinantal equation is .P -9= .P =38 .P oe ! . It follows that the eigenfunctions are given by 98 aBb oe =38-8 B S-8 -9=-8 P<B , and the eigenvalues satisfy the equation P-8 -9= -8 P =38 -8 P oe ! . The smallest eigenvalue is estimated as -" a%%*\$%b# P# a- b. Clamped Ca!b oe C w a!b oe ! clamped C aPb oe C w aPb oe ! Invoking the boundary conditions, we obtain the system of equations -" - \$ oe ! -# .- % oe ! -" -# P -\$ -9= .P -% =38 .P oe ! -# -\$ .=38 .P -% .-9= .P oe ! . The determinantal equation is # #-9= .P oe .P =38 .P It follows that the eigenfunctions are given by ________________________________________________________________________ page 738 ---------------------------------------------------- CHAPTER 11. ---- 98 aBb oe " -9=-8 B , and the eigenvalues satisfy the equation # #-9= -8 P oe -8 P =38 -8 P The smallest eigenvalue is -" oe a#1b# P# 26. As shown is Prob. #& , the general solution is CaBb oe -" -# B -\$ -9= .B -% =38 .B . -# oe ! -" - \$ oe ! -# .- % oe ! -\$ -9= .P -% =38 .P oe ! . For a nontrivial solution, it is necessary that -9= .P oe ! . We find that -# oe -% oe ! , and hence the eigenfunctions are given by The corresponding eigenvalues are -8 oe a#8 "b# 1# %P# , 8 oe " # . The smallest eigenvalue is -" oe 1# %P# 98 aBb oe " -9=-8 B . Imposing the boundary conditions, we obtain the system of equations ________________________________________________________________________ page 739 ---------------------------------------------------- CHAPTER 11. ---- Section 11.3 4. The eigensystem of the associated homogeneous problem is given in Prob. "" of Section ""# . The normalized eigenfunctions are 98 aBb oe " =38# -8 # -9= -8 B , in which the eigenvalues satisfy -9=-8 -8 =38-8 oe !. Rewrite the given differential equation as C ww oe #C B . Since . oe # -8 , the formal solution of the nonhomogeneous problem is CaBb oe " _ -8 98 aBb, - # 8oe" 8 in which -8 oe ( 0 aBb98 aBb.B " ! oe oe " =38# -8 # #^# -9= -8 " ! ( B -9= -8 B .B " -8 " =38# -8 . Therefore we obtain the formal expansion CaBb oe # " _ #^# -9= -8 " -9= -8 B -8 a-8 #b^" =38# -8 . 8oe" 5. The solution follows that in Prob. " , except that the coefficients are given by -8 oe ( 0 aBb98 aBb.B " oe #( oe% ! "# ! # =38a81#b 8# 1 # #B =38 81B .B #( . " "# a# #Bb =38 81B .B Therefore the formal solution is CaBb oe ) " _ =38a81#b =38 81B . 8# 1# a8# 1# #b 8oe" ________________________________________________________________________ page 740 ---------------------------------------------------- CHAPTER 11. ---- 6. The differential equation can be written as C ww oe .C 0 aBb. Note that ; aBb oe ! and <aBb oe " . As shown in Prob. " in Section ""# , the normalized eigenfunctions are with associated eigenvalues -8 oe a#8 "b# 1# % . Based on Theorem ""\$" , the formal solution is given by CaBb oe # " _ 98 aBb oe # =38 a#8 "bB , # -8 a#8 "bB =38 , a-8 .b # 8oe" " as long as . -8 The coefficients in the series expansion are computed as -8 oe #( 0 aBb=38 ! a#8 "bB .B . # 7. As shown in Prob. " in Section ""# , the normalized eigenfunctions are 98 aBb oe # -9= a#8 "bB , # with associated eigenvalues -8 oe a#8 "b# 1# % . Based on Theorem ""\$" , the formal solution is given by CaBb oe # " _ -8 a#8 "bB -9= , a-8 .b # 8oe" " as long as . -8 The coefficients in the series expansion are computed as a#8 "bB -8 oe #( 0 aBb-9= .B . # ! # -9= -8 B 9. The normalized eigenfunctions are 98 aBb oe The eigenvalues satisfy -9=-8 -8 =38-8 oe !. Based on Theorem ""\$" , the formal solution is given by CaBb oe # " _ 8 oe " a-8 " =38# -8 . .b" =38# -8 -8 -9= -8 B , as long as . -8 The coefficients in the series expansion are computed as ________________________________________________________________________ page 741 ---------------------------------------------------- CHAPTER 11. ---- # ( 0 aBb-9= -8 B .B . " ! -8 oe " =38# -8 13. The differential equation can be written as C ww oe 1# C -9= 1B + . Note that . oe 1# and 0 aBb oe -9= 1B + . Furthermore, . oe 1# is an eigenvalue corresponding to the eigenfunction 9" aBb oe # =38 1B . A solution exists only if 0 aBb and 9" aBb are orthogonal. Since ( a-9= 1B +b=38 1B .B oe #+1 , " ! there exists a solution as long as + oe ! . In that case, the ODE is The complementary solution is C- aBb oe -" -9= 1B -# =38 1B . A particular solution is ] aBb oe EB -9= 1B FB =38 1B . Using the method of undetermined coefficients, we find that E oe ! and F oe "#1 . Therefore the general solution is B CaBb oe -" -9= 1B -# =38 1B =38 1B . #1 The boundary conditions require that -" oe ! . Hence the solution of the boundary value problem is B CaBb oe -# =38 1B =38 1B . #1 15. Let CaBb oe 9" aBb 9# aBb . It follows that PcCd oe Pc9" d Pc9# d oe 0 aBb. Also, +" Ca!b +# C w a!b oe +" 9" a!b +" 9# a!b +# 9"w a!b +# 9#w a!b oe +" 9" a!b +# 9"w a!b +" 9# a!b +# 9#w a!b oe !. C ww 1# C oe 1B -9= . Similarly, the boundary condition at B oe " is satisfied as well. 16. The complementary solution is C- aBb oe -" -9= 1B -# =38 1B . A particular solution is ] aBb oe E FB . Using the method of undetermined coefficients, we find that E oe ! and F oe " . Therefore the general solution is CaBb oe -" -9= 1B -# =38 1B B . CaBb oe -9= 1B -# =38 1B B . Imposing the boundary conditions, we find that -" oe " . Therefore the solution of the BVP is Now attempt to solve the problem as shown in Prob. "& . Let BVP-1 be given by ________________________________________________________________________ page 742 ---------------------------------------------------- CHAPTER 11. ---- ? ww 1# ? oe 1# B , ?a!b oe ! , ?a"b oe ! The general solution of the ODE is ?aBb oe -" -9= 1B -# =38 1B B . The boundary conditions require that -" oe ! and -" " oe ! . We find that BVP-1 has no solution. Let BVP-2 be given by @ ww 1# @ oe ! , @ a !b oe " , @ a " b oe ! The general solution of the ODE is @aBb oe -" -9= 1B -# =38 1B . Imposing the boundary conditions, we obtain -" oe " and -" oe ! . Thus BVP-2 has no solution. 17. Setting CaBb oe ?aBb @aBb, substitution results in ? ww @ ww :aBbc? w @ w d ; aBbc? @d oe ? ww :aBb? w ; aBb? @ ww :aBb@ w ; aBb@ ? ww :aBb? w ; aBb? oe c@ ww :aBb@ w ; aBb@d 1aBb oe a+ ,b:aBb a+ ,bB ; aBb + ; aBb Since the left hand side of the equation is zero, Furthermore, ?a!b oe Ca!b @a!b oe ! and ?a"b oe Ca"b @a"b oe ! . The simplest function having the assumed properties is @aBb oe a, +bB + . In this case, 20. The associated homogeneous PDE is ?> oe ?BB , ! B " , with ?B a! >b oe !, ?B a" >b ?a" >b oe ! and ?aB !b oe " B. Applying the method of separation of variables, we obtain the eigenvalue problem \ ww -\ oe ! , with boundary conditions \ w a!b oe ! and \ w a"b \ a"b oe !. It was shown in Prob. % , in Section ""# , that the normalized eigenfunctions are 98 aBb oe where -9=-8 -8 =38-8 oe !. We assume a solution of the form ?aB >b oe ",8 a>b98 aBb . _ 8 oe" " =38# -8 # -9= -8 B , ________________________________________________________________________ page 743 ---------------------------------------------------- CHAPTER 11. ---- Substitution into the given PDE results in w ww ",8 a>b98 aBb oe ",8 a>b98 aBb /> _ _ 8 oe" 8 oe" _ oe "-8 ,8 a>b98 aBb /> , 8 oe" that is, w "c,8 a>b -8 ,8 a>bd98 aBb oe /> _ 8 oe" We now note that "oe" _ 8 oe" -8 " # =38-8 =38# -8 98 aBb . Therefore / > oe ""8 /> 98 aBb , _ 8 oe" in which "8 oe # =38-8 "-8 " =38# -8 . Combining these results, w ",8 a>b -8 ,8 a>b "8 /> `98 aBb oe ! _ 8 oe" Since the resulting equation is valid for ! B " , it follows that w ,8 a>b -8 ,8 a>b oe "8 /> , 8 oe " # . Prior to solving the sequence of ODEs, we establish the initial conditions. These are obtained from the expansion ?aB !b oe " B oe "!8 98 aBb , _ in which !8 oe # ^" -9=-8 "-8 " =38# -8 That is, ,8 a!b oe !8 . Therefore the solutions of the first order ODEs are ,8 a>b oe "8 ^/> /-8 > !8 /-8 > , 8 oe " # . a - 8 "b 8 oe" Hence the solution of the boundary value problem is ________________________________________________________________________ page 744 ---------------------------------------------------- CHAPTER 11. ---- ?aB >b oe " _ 8 oe" "8 ^/> /-8 > !8 /-8 > --98 aBb a - 8 "b 21. Based on the boundary conditions, the normalized eigenfunctions are given by 98 aBb oe # =38 81B , _ with associated eigenvalues -8 oe 8# 1# We now assume a solution of the form ?aB >b oe ",8 a>b98 aBb . 8 oe" Substitution into the given PDE results in w ww ",8 a>b98 aBb oe ",8 a>b98 aBb " k" #Bk _ _ 8 oe" 8 oe" _ oe "-8 ,8 a>b98 aBb " k" #Bk, 8 oe" that is, w "c,8 a>b -8 ,8 a>bd98 aBb oe " k" #Bk _ 8 oe" It was shown in Prob. & that " k" #Bk oe "% _ 8 oe" # =38a81#b 8# 1# 98 aBb . Substituting on the right hand side and collecting terms, we obtain w ",8 a>b -8 ,8 a>b % _ 8 oe" # =38a81#b 8# 1 # --98 aBb oe ! Since the resulting equation is valid for ! B " , it follows that w ,8 a>b 8# 1# ,8 a>b oe % # =38a81#b 8# 1# , 8 oe " # . Based on the given initial condition, we also have ,8 a!b oe ! , for 8 oe " # . The solutions of the first order ODEs are ,8 a>b oe % # =38a81#b 8% 1% S" /8 1 > <, 8 oe " # . # # Hence the solution of the boundary value problem is ________________________________________________________________________ page 745 ---------------------------------------------------- CHAPTER 11. ---- ) _ =38a81#b # # ?aB >b oe % " S" /8 1 > <=38 81B % 1 8 oe" 8 23a+b. Let ?aB >b be a solution of the boundary value problem and @aBb be a solution of the related BVP. Substituting for ?aB >b oe AaB >b @aBb, we have <aBb?> oe <aBbA> and Hence AaB >b is a solution of the homogeneous PDE The required boundary conditions are c:aBb?B dB ; aBb? J aBb oe c:aBbAB dB ; aBbA c:aBb@ w d ; aBb@ J aBb oe c:aBbAB dB ; aBbA J aBb J aBb oe c:aBbAB dB ; aBbA . w <aBbA> oe c:aBbAB dB ; aBbA . Aa! >b oe ?a! >b @a!b oe ! , Aa" >b oe ?a" >b @a"b oe ! . The associated initial condition is AaB !b oe ?aB !b @aBb oe 0 aBb @aBb. a,b. Let @aBb be a solution of the ODE w and satisfying the boundary conditions @ w a!b 2" @a!b oe X" , @ w a"b 2# @a"b oe X# . If AaB >b oe ?aB >b @aBb, then it is easy to show the A satisfies the PDE and initial condition given in Part a+b. Furthermore, AB a! >b 2" Aa! >b oe ?B a! >b @ w a!b 2" ?a! >b 2" @a!b oe ?B a! >b 2" ?a! >b @ w a!b 2" @a!b oe !. c:aBb@ w d ; aBb@ oe J aBb , Similarly, the other boundary condition is also homogeneous. ________________________________________________________________________ page 746 ---------------------------------------------------- CHAPTER 11. ---- 25. In this problem, J aBb oe 1# -9= 1B . First find a solution of the boundary value problem The general solution is @aBb oe EB F -9= 1B . Imposing the initial conditions, the solution of the related BVP is @aBb oe -9= 1B . Now let AaB >b oe ?aB >b -9= 1B . It follows that AaB >b satisfies the homogeneous boundary value problem, and the initial condition AaB !b oe -9=a\$1B#b -9= 1B a -9= 1Bb oe -9=a\$1B#b . We now seek solutions of the homogeneous problem of the form AaB >b oe ",8 a>b98 aBb , _ 8 oe" @ ww oe 1# -9= 1B , @ w a!b oe ! , @a"b oe " . in which 98 aBb oe # -9= a#8 "b1B# are the normalized eigenfunctions of the homogeneous problem and -8 oe a#8 "b# 1# % , with 8 oe " # . Substitution into the PDE for A, we have w ww ",8 a>b98 aBb oe ",8 a>b98 aBb _ _ 8 oe" 8 oe" _ oe "-8 ,8 a>b98 aBb . 8 oe" Since the latter equation is valid for ! B " , it follows that with ,8 a>b oe ,8 a!b/B:a -8 >b. Hence _ 8 oe" w ,8 a>b -8 ,8 a>b oe ! , 8 oe " # , AaB >b oe ",8 a!b/B:a -8 >b98 aBb . Imposing the initial condition, we require that _ It is evident that all of the coefficients are zero, except for ,# a!b oe "# . Therefore AaB >b oe /B:^ *1# >%-9= \$1B , # # ",8 a!b-9= a#8 "b1B oe -9= \$1B . # # 8 oe" and the solution of the original BVP is ?aB >b oe /B:^ *1# >%-9= \$1B -9= 1B . # ________________________________________________________________________ page 747 ---------------------------------------------------- CHAPTER 11. ---- 26a+b. Let ?aB >b oe \ aBbX a>b. Substituting into the homogeneous form of a3b, <aBb\X ww oe c:aBb\ w d X ; aBb\X . w Now divide both sides of the resulting equation by \X to obtain X ww c:aBb\ w d w ; aBb oe oe -. X <aBb\ <aBb w It follows that Since the boundary conditions a33b are valid for all > ! , we also have \ w a!b 2" \ a!b oe ! , \ w a"b 2# \ a"b oe ! . a,b. Let -8 and 98 aBb denote the eigenvalues and eigenfunctions of the BVP in Part a+b. Assume a solution, of the PDE a3b, of the form ?aB >b oe ",8 a>b98 aBb _ 8 oe" c:aBb\ w d ; aBb\ oe -<aBb\ . Substituting into a3b, ww w w <aBb ",8 a>b98 oe ",8 a>b~c:aBb98 d ; aBb98 TM J aB >b _ _ 8 oe" oe ",8 a>bc -8 <aBb98 d J aB >b . 8 oe" 8 oe" _ Rearranging the terms, ww <aBb "c,8 a>b -8 ,8 a>bd98 oe J aB >b , _ 8 oe" or ww "c,8 a>b -8 ,8 a>bd98 oe _ 8 oe" J a B >b . <aBb Now expand the right hand side in terms of the eigenfunctions. That is, write _ J a B >b oe "#8 a>b98 aBb , <aBb 8 oe" in which ________________________________________________________________________ page 748 ---------------------------------------------------- CHAPTER 11. ---- #8 a>b oe ( <aBb " ! " ! oe ( J aB >b98 aBb.B , 8 oe " # . J a B >b 98 aBb.B <aBb Combining these results, we have _ 8 oe" ww "c,8 a>b -8 ,8 a>b #8 a>bd98 oe ! . It follows that w In order to solve this sequence of ODEs, we require initial conditions ,8 a!b and ,8 a!b Note that w ?aB !b oe ",8 a!b98 aBb and ?> aB !b oe ",8 a!b98 aBb . _ _ 8 oe" 8 oe" ww ,8 a>b -8 ,8 a>b oe #8 a>b , 8 oe " # . Based on the given initial conditions, w 0 aBb oe ",8 a!b98 aBb and 1aBb oe ",8 a!b98 aBb . _ _ 8 oe" 8 oe" w Hence ,8 a!b oe !8 and ,8 a!b oe "8 , the expansion coefficients for 0 aBb and 1aBb in terms of the eigenfunctions, 98 aBb . 27a+b. Since the eigenvectors are orthogonal, they form a basis. Given any vector b , b oe ",3 0 a3b 8 3oe" Taking the inner product, with 0 a4b , of both sides of the equation, we have ^b 0 a4b oe ,4 ^0 a4b 0 a4b a,b. Consider solutions of the form x oe "+3 0 a3b 8 3oe" Substituting into Eq. a3b, and using the above form of b , 8 8 3oe" 3oe" "+3 A0 a3b ". +3 0 a3b oe ",3 0 a3b 8 3oe" It follows that ________________________________________________________________________ page 749 ---------------------------------------------------- CHAPTER 11. ---- "c+3 -3 . +3 ,3 d0 a3b oe 0 8 3oe" Since the eigenvectors are linearly independent, +3 -3 . +3 ,3 oe ! , for 3 oe " # 8 . That is, +3 oe ,3 a-3 .b , 3 oe " # 8 . xoe" 8 Assuming that the eigenvectors are normalized, the solution is given by ab 0 a3b b a3b 0 , -3 . 3oe" as long as . is not equal to one of the eigenvalues. 29. First write the ODE as C ww C oe 0 aBb . A fundamental set of solutions of the homogeneous equation is given by The Wronskian is equal to [ c-9= B =38 Bd oe " . Applying the method of variation of parameters, a particular solution is ] aBb oe C" aBb?" aBb C# aBb?# aBb , B C" oe -9= B and C# oe =38 B . in which ?" aBb oe ( =38a=b0 a=b.= and ?# aBb oe ( -9=a=b0 a=b.= . B ! ! Therefore the general solution is C oe 9aBb oe -" -9= B -# =38 B -9= B( =38a=b0 a=b.= =38 B( -9=a=b0 a=b.=. B B ! ! Imposing the boundary conditions, we must have -" oe ! and -# =38 " -9= "( =38a=b0 a=b.= =38 "( -9=a=b0 a=b.= oe ! . " " ! ! It follows that -# oe and " " ( =38a" =b0 a=b.= , =38 " ! ________________________________________________________________________ page 750 ---------------------------------------------------- CHAPTER 11. ---- B =38 B " 9aBb oe ( =38a" =b0 a=b.= ( =38aB =b0 a=b.=. =38 " ! ! Using standard identities, =38 B =38a" =b =38 " =38aB =b oe =38 = =38a" Bb . =38 B =38a" =b =38 = =38a" Bb =38aB =b oe . =38 " =38 " B ! " =38 = =38a" Bb =38 B =38a" =b 0 a=b.= ( 0 a=b.= =38 " =38 " B Therefore Splitting up the first integral, we obtain 9aBb oe ( oe ( KaB =b0 a=b.= , " ! in which K aB =b oe =38 ==38a"Bb =38 " =38 B=38a"=b =38 " !Y=YB B Y = Y " 31. The general solution of the homogeneous problem is By inspection, it is easy to see that C" aBb oe " satisfies the BC C w a!b oe ! and that the function C# aBb oe " B satisfies the BC C a"b oe ! . The Wronskian of these solutions is [ cC" C# d oe " . Based on Prob. \$! , with :aBb oe " , the Green's function is given by K aB =b oe oe a" Bb a " =b !Y=YB B Y = Y " C oe -" -# B . Therefore the solution of the given BVP is 9aBb oe ( a" Bb0 a=b.= ( a" =b0 a=b.= . B " ! B 32. The general solution of the homogeneous problem is We find that C" aBb oe B satisfies the BC Ca!b oe ! . Imposing the boundary condition C oe -" -# B . ________________________________________________________________________ page 751 ---------------------------------------------------- CHAPTER 11. ---- Ca"b C w a"b oe ! , we must have -" #-# oe ! . Hence choose C# aBb oe # B . The Wronskian of these solutions is [ cC" C# d oe # . Based on Prob. \$! , with :aBb oe " , the Green's function is given by K aB =b oe oe =aB #b# Ba= #b# !Y=YB B Y = Y " Therefore the solution of the given BVP is " B " " 9aBb oe ( =aB #b0 a=b.= ( Ba= #b0 a=b.= . # ! # B 34. The general solution of the homogeneous problem is By inspection, it is easy to see that C" aBb oe B satisfies the BC Ca!b oe ! and that the function C# aBb oe " satisfies the BC C w a"b oe ! . The Wronskian of these solutions is [ cC" C# d oe " . Based on Prob. \$! , with :aBb oe " , the Green's function is given by K aB =b oe oe = B !Y=YB B Y = Y " C oe -" -# B . Therefore the solution of the given BVP is 9aBb oe ( =0 a=b.= ( B0 a=b.= . B " ! B 35a+b. We proceed to show that if the expression given by Eq. a3@b is substituted into the integral of Eq. a333b, then the result is the solution of the nonhomogeneous problem. As long as we can interchange the summation and integration, C oe 9aBb oe ( KaB = .b0 a=b.= " oe" 93 aBb " ( 0 a=b93 a=b.= . - . ! 8oe" 3 ! _ Note that ( 0 a=b93 a=b.= oe -3 . " ! Therefore C oe 9aBb oe " _ -3 93 aBb , -3 . 8oe" ________________________________________________________________________ page 752 ---------------------------------------------------- CHAPTER 11. ---- as given by Eq. a"\$b in the text. It is assumed that the eigenfunctions are normalized and -3 . . a,b. For any fixed value of B, KaB = .b is a function of = and the parameter . . With appropriate assumptions on K , we can write the eigenfunction expansion KaB = .b oe "+3 aB .b93 a=b . _ 3oe" Since the eigenfunctions are orthonormal with respect to <aBb, " ! ( KaB = .b<a=b93 a=b.= oe +3 aB .b . C3 aBb oe ( KaB = .b<a=b93 a=b.= . " ! Now let Based on the association 0 aBb oe <aBb93 aBb , it is evident that PcC3 d oe . <aBbC3 aBb <aBb93 aBb . C3 aBb oe " ,35 95 aBb . _ 5oe" In order to evaluate the left hand side, we consider the eigenfunction expansion It follows that PcC3 d oe " ,35 Pc95 d _ 5oe" _ 5oe" oe " ,35 -5 <aBb95 aBb . Therefore <aBb " ,35 -5 95 aBb oe . <aBb" ,35 95 aBb <aBb93 aBb , _ _ 5oe" 5oe" and since <aBb ! , 5oe" " ,35 -5 95 aBb oe . " ,35 95 aBb 93 aBb . _ _ 5oe" Rearranging the terms, we find that ________________________________________________________________________ page 753 ---------------------------------------------------- CHAPTER 11. ---- 93 aBb oe " ,35 a-5 .b95 aBb . _ 5oe" Since the eigenfunctions are linearly independent, ,35 a-5 .b oe \$35 , and thus C3 aBb oe " _ " \$35 95 aBb oe 93 aBb . - . -3 . 5oe" 5 + 3 aB . b oe " 93 aBb , -3 . 93 aBb93 a=b . -3 . 3oe" _ We conclude that which verifies that K aB = .b oe " 36. First note that . # C.=# oe ! for = B . On the interval ! = B , the solution of the ODE is C" a=b oe -" -# = . Given that C a!b oe ! , we have C" a=b oe -# = . On the interval B = " , the solution is C# a=b oe ." .# = . Imposing the condition C a"b oe ! , we have C# a=b oe ." a" =b. Assuming continuity of the solution, at = oe B , which gives -# oe ." a" BbB . Next, integrate both sides of the given ODE over an infinitesimal interval containing = oe B ( It follows that B B .# C .= oe ( \$ a= Bb.= oe " . .=# B -# B oe ." a" Bb , B and hence -# a ." b oe " . Solving for the two coefficients, we obtain -# oe " B and ." oe B . Therefore the solution of the BVP is given by C a =b oe oe =a" Bb Ba" =b !Y=YB B Y = Y ", C w aB b C w aB b oe " , which is identical to the Green's function in Prob. #). ________________________________________________________________________ page 754 ---------------------------------------------------- CHAPTER 11. ---- Section 11.4 1. Let 98 aBb oe N! ^-8 B be the eigenfunctions of the singular problem Let 9aBb be a solution of the given BVP, and set _ 8oe! aB C w b oe -BC , ! B " , C C w bounded as B p ! , C a"b oe ! . w 9aBb oe " ,8 98 aBb . ab Then Substituting ab, we obtain _ 8oe! aB 9 w b oe . B9 0 aBb 0 aBb oe . B9 B . B w _ _ in which the -8 are the expansion coefficients of 0 aBbB for B ! . That is, -8 oe " " 0 aBb 98 aBb.B ( B # m98 aBbm ! B " " oe ( 0 aBb98 aBb.B . m98 aBbm# ! " ,8 -8 B 98 aBb oe .B " ,8 98 aBb B " -8 98 aBb , 8oe! 8oe! It follows that if B ! , " c-8 ,8 a-8 .bd98 aBb oe ! . _ 8oe! As long as . -8 , linear independence of the eigenfunctions implies that -8 , 8 oe " # . ,8 oe -8 . Therefore a formal solution is given by 9aBb oe " _ in which -8 are the positive roots of N! aBb oe ! . -8 N! S-8 B<, -8 . 8oe! ________________________________________________________________________ page 755 ---------------------------------------------------- CHAPTER 11. ---- 3a+b. Setting > oe - B , it follows that .C .C .#C .#C oe and oe- # . .B .> .B# .> . > .C 5 # - oe - > C , .> .> > . .C 5# OE> oe >C .> .> > The given ODE can be expressed as or An equivalent form is given by ># which is known as a Bessel equation of order 5 . A bounded solution is N5 a>b . .C .C > ^># 5 # C oe ! , .> .> a,b. N5 S- B< satisfies the boundary condition at B oe ! . Imposing the other boundary condition, it is necessary that N5 S- < oe ! . Therefore the eigenvalues are given by -8 , 8 oe " # , where -8 are the positive zeroes of N5 aBb . The eigenfunctions of the BVP are 98 aBb oe N5 ^-8 B a- b. The BVP is a singular Sturm-Liouville problem with PcCd oe aB C w b w 5# C and <aBb oe " . B " We note that -8 ( B 98 aBb97 aBb.B oe ( Pc98 d 97 aBb.B " ! oe ( 98 aBbPc97 d.B " ! " ! ! oe -7 ( B 98 aBb97 aBb.B Therefore a-8 -7 b( B 98 aBb97 aBb.B oe ! " ! ________________________________________________________________________ page 756 ---------------------------------------------------- CHAPTER 11. ---- So for 8 7 , we have -8 -7 and ( B 98 aBb97 aBb.B oe ! " ! a. b. Consider the expansion Multiplying both sides of equation by B 94 aBb and integrating from ! to " , and using the orthogonality of the eigenfunction, ( B 0 aBb94 aBb.B oe " +8 ( B 94 aBb98 aBb.B " _ " ! 0 aBb oe " +8 98 aBb _ 8oe! oe +4 ( B 94 aBb94 aBb.B " ! 8oe! ! Therefore +4 oe ( B 0 aBb94 aBb.B( Bc94 aBbd# .B , 4 oe " # " " ! ! a/b. Let 9aBb be a solution of the given BVP, and set _ where 98 aBb oe N5 ^-8 B. Then 9aBb oe " ,8 98 aBb , 8oe! ab Substituting ab, we obtain _ 8oe! Pc9d oe . B9 0 aBb 0 aBb oe . B9 B . B _ _ in which the -8 are the expansion coefficients of 0 aBbB for B ! . That is, -8 oe " " 0 aBb 98 aBb.B ( B # m98 aBbm ! B " " oe ( 0 aBbN5 S-8 B<.B . mN5 ^-8 Bm# ! " ,8 -8 B 98 aBb oe .B " ,8 98 aBb B " -8 98 aBb , 8oe! 8oe! ________________________________________________________________________ page 757 ---------------------------------------------------- CHAPTER 11. ---- It follows that if B ! , " c-8 ,8 a-8 .bdN5 S-8 B< oe ! . _ 8oe! As long as . -8 , linear independence of the eigenfunctions implies that -8 , 8 oe " # . ,8 oe -8 . Therefore a formal solution is given by 9aBb oe " _ -8 N5 S-8 B<. -8 . 8oe! 5a+b. Setting - oe !# in Prob. "& of Section """ , the Chebyshev equation can also be written as '" B# C w " oe w C. " B# Note that :aBb oe " B# , ; aBb oe ! , and <aBb oe "" B# , hence both boundary points are singular. a,b. Observe that :a" &b oe #& &# and :a " &b oe #& &# It follows that if ?aBb and @aBb satisfy the boundary conditions a333b, then & ! lim :a" &bc? w a" &b@a" &b ?a" &b@ w a" &bd oe ! and a- b. For 8 ! , Therefore Eq. a"(b is satisfied and the boundary value problem is self-adjoint. 8# ( " X! aBb X8 aBb .B oe ( X! aBb PcX8 d.B " " B# " " " " & ! lim :a " &bc? w a " &b@a " &b ?a " &b@ w a " &bd oe ! . oe ( PcX! d X8 aBb.B oe !, since PcX! d oe ! X! oe ! . Otherwise, ________________________________________________________________________ page 758 ---------------------------------------------------- CHAPTER 11. ---- " X8 aBb X7 aBb 8 ( .B oe ( PcX8 d X7 aBb.B " " B# " # " oe ( X8 aBbPcX7 d.B " oe 7# ( Therefore ^8# 7# ( So for 8 7 , ( " X8 aBb X7 aBb .B " " B# " X8 aBb X7 aBb .B oe ! " " B# " X8 aBb X7 aBb .B oe ! " " B# " ________________________________________________________________________ page 759 ---------------------------------------------------- CHAPTER 11. ---- Section 11.5 3. The equations relating to this problem are given by Eqs. a#b to a"(b in the text. Based on the boundary conditions, the eigenfunctions are 98 aBb oe N! a-8 <b and the associated eigenvalues -" -# are the positive zeroes of N! a-b . The general solution has the form ?a< >b oe " c-8 N! a-8 <b -9= -8 +> 58 N! a-8 <b =38 -8 +>d . _ 8oe" The initial conditions require that ?a< !b oe " -8 N! a-8 <b oe 0 a<b _ 8oe" and ?> a< !b oe " +-8 58 N! a-8 <b oe 1a<b . _ 8oe" The coefficients -8 and 58 are obtained from the respective eigenfunction expansions. That is, -8 oe and " " 58 oe ( <1a<bN! a-8 <b.< , +-8 mN! a-8 <bm# ! " " ( <0 a<bN! a-8 <b.< mN! a-8 <bm# ! in which mN! a-8 <bm# oe ( <cN! a-8 <bd# .< " ! for 8 oe " # . 8. A more general equation was considered in Prob. #\$ of Section "!& . Assuming a solution of the form ?a< >b oe V a<bX a>b, substitution into the PDE results in !# "V ww X " w V X oe VX w . < Dividing both sides of the equation by the factor VX , we obtain ________________________________________________________________________ page 760 ---------------------------------------------------- CHAPTER 11. ---- V ww " Vw Xw oe # . V < V ! X Since both sides of the resulting differential equation depend on different variables, each side must be equal to a constant, say -# . That is, V ww " Vw Xw oe # oe -# . V < V ! X It follows that X w !# -# X oe ! , and V ww " Vw oe -# , V < V which can be written as <# V ww < V w -# <# V oe ! . Introducing the variable 0 oe -< , the last equation can be expressed as 0# V ww 0 V w 0# V oe ! , which is the Bessel equation of order zero. The temporal equation has solutions which are multiples of X a>b oe /B:a !# -# >b. The general solution of the Bessel equation is V a<b oe ," N! a-8 <b ,# ]! a-8 <b . Since the steady state temperature will be zero, all solutions must be bounded, and hence we set ,# oe ! . Furthermore, the boundary condition ?a" >b oe ! requires that V a"b oe ! and hence N! a-b oe ! . It follows that the eigenfunctions are 98 aBb oe N! a-8 <b , with the associated eigenvalues -" -# , which are the positive zeroes of N! a-b . Therefore # the fundamental solutions of the PDE are ?8 a< >b oe N! a-8 <b/B:a !# -8 >b, and the general solution has the form # ?a< >b oe " -8 N! a-8 <b/B:^ !# -8 > . _ 8oe" The initial condition requires that ?a< !b oe " -8 N! a-8 <b oe 0 a<b. _ 8oe" The coefficients in the general solution are obtained from the eigenfunction expansion of 0 a<b. That is, " " -8 oe ( <0 a<bN! a-8 <b.< , mN! a-8 <bm# ! in which mN! a-8 <bm# oe ( <cN! a-8 <bd# .< " ! a8 oe " # b. ________________________________________________________________________ page 761 ---------------------------------------------------- CHAPTER 11. ---- Section 11.6 1. The sine expansion of 0 aBb oe " , on ! B " , is given by 0 aBb oe # " _ " -9= 71 =38 71B , 71 7oe" with partial sums W8 aBb oe # " 8 " -9= 71 =38 71B . 71 7oe" The mean square error in this problem is V8 oe ( k" W8 aBbk# .B . " ! Several values are shown in the Table : 8 V8 & !!'( "! !!% "& !!#' #! !!# Further numerical calculation shows that V8 !!# for 8 #" . 0 aBb oe # " _ 3a+b. The sine expansion of 0 aBb oe Ba" Bb , on ! B " , is given by " -9= 71 =38 71B , 71 7oe" with partial sums W8 aBb oe % " 8 a, - b. The mean square error in this problem is " ! " -9= 71 =38 71B . 7\$ 1\$ 7oe" V8 oe ( k Ba" Bb W8 aBbk# .B . ________________________________________________________________________ page 762 ---------------------------------------------------- CHAPTER 11. ---- We find that V" oe !!!!!%) . The graphs of 0 aBb and W" aBb are plotted below : 6a+b. The function is bounded on intervals not containing B oe ! , so for & ! , "# .B oe # #& ( 0 aBb.B oe ( B " " & & Hence the improper integral is evaluated as "# .B oe # ( 0 aBb.B oe lim ( B " " ! &! & On the other hand, 0 # aBb oe "B for B ! , and " # " & & " ( 0 aBb.B oe ( B .B oe 68& Therefore the improper integral does not exist. a,b. Since 0 # aBb " , it is evident that the Riemann integral of 0 # aBb exists. Let ________________________________________________________________________ page 763 ---------------------------------------------------- CHAPTER 11. ---- TR oe e! oe B" B# BR " oe "f be a partition of c! "d into equal subintervals. We can always choose a rational point, 03 , in each of the subintervals so that the Riemann sum V a0" 0# 0R b oe " 0 a08 b R 8oe" " oe ". R Likewise, can always choose an irrational point, (3 , in each of the subintervals so that the Riemann sum V a(" (# (R b oe " 0 a(8 b R 8oe" It follows that 0 aBb is not Riemann integrable. " oe ". R 8. With T! aBb oe " and T" aBb oe B, the normalization conditions are satisfied. Using the usual inner product on c " "d, and hence the polynomials are also orthogonal Let T# aBb oe +# B# +" B +! . The normalization condition requires that +# +" +! oe " . For orthogonality, we need # # ( ^+# B +" B +! .B oe ! and ( B^+# B +" B +! .B oe ! . " " " " ( T! aBbT" aBb.B oe ! " " It follows that +# oe \$# , +" oe ! and +! oe "# . Hence T# aBb oe a\$B# "b# . Now let T\$ aBb oe +\$ B\$ +# B# +" B +! . The coefficients must be chosen so that +\$ +# +" +! oe " and the orthogonality conditions ( T3 aBbT4 aBb.B oe ! a3 4b " " are satisfied. Solution of the resulting algebraic equations leads to +\$ oe &# , +# oe ! , +1 oe \$# and +! oe ! . Therefore T\$ aBb oe a&B\$ \$Bb# . 11. The implied sequence of coefficients is +8 oe " , 8 " . Since the limit of these coefficients is not zero, the series cannot be an eigenfunction expansion. 13. Consider the eigenfunction expansion 0 aBb oe "+3 93 aBb _ 3oe" Formally, ________________________________________________________________________ page 764 ---------------------------------------------------- CHAPTER 11. ---- # # 0 aBb oe "+3 93 aBb # "+3 +4 93 aBb94 aBb _ # 3oe" 34 Integrating term-by-term, # # ( <aBb0 aBb.B oe "( +3 <aBb93 aBb.B # "( +3 +4 <aBb93 aBb94 aBb.B " _ # " " ! oe 3oe" ! _ " # # "+3 ( 93 aBb.B , ! 3oe" 34 ! since the eigenfunctions are orthogonal. Assuming that they are also normalized, # ( <aBb0 aBb.B oe "+3 . " _ # ! 3oe" ________________________________________________________________________ page 765
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Johns Hopkins - CIVENG - 560.202
Table of ContentsChapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 1 145 242 302 396 504 591 632 666 714 786Engineering Mechanics - DynamicsChapter 12Problem 12-1 A truck t
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5.5mA5.0mA4.5mA4.0mA3.5mA3.0mA2.5mA2.0mA1.5mA1.0mA 0s 0.05us 0.10us 0.15us . I(U1:C5) 0.20us 0.25us 0.30us 0.35us 0.40us 0.45us 0.50us Time 0.55us 0.60us 0.65us 0.70us 0.75us 0.80us 0.85us 0.90us 0.95us 1.00usBasic Current Source
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POLS 102 Test 1 Jan. 15th-17th a. Politics- deciding who gets what, when, and how b. political science- the study of politics; who governs, for what ends and by what meansc. government- organization extending to the whole society that can legitimat
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Vanderbilt - MSE - 150
PRINT NAME _ INTRODUCTION TO MATERIAL SCIENCE MSE-150 QUIZ 1 I PLEDGE MY HONOR THAT I HAVE NEITHER GIVEN OR RECEIVED AID ON THIS EXAMINATION. I FURTHER PLEDGE THAT, IF I HAVE REASON TO SUSPECT THAT THE CODE HAS BEEN VIOLATED, I HAVE TAKEN OR WILL TAK
Vanderbilt - MSE - 150
PRINT NAME _ INTRODUCTION TO MATERIAL SCIENCE MSE-150 QUIZ 2 I PLEDGE MY HONOR THAT I HAVE NEITHER GIVEN OR RECEIVED AID ON THIS EXAMINATION. I FURTHER PLEDGE THAT, IF I HAVE REASON TO SUSPECT THAT THE CODE HAS BEEN VIOLATED, I HAVE TAKEN OR WILL TAK
Vanderbilt - MSE - 150
Introduction The purpose of this experiment was to compare the mechanical properties of metals, ceramics, polymers, and composites. Also, to demonstrate the effect of temperature on the toughness of body centered cubic and face centered cubic metals.
Vanderbilt - MSE - 150
PRINT NAME _ INTRODUCTION TO MATERIAL SCIENCE MSE-150 QUIZ 3 I PLEDGE MY HONOR THAT I HAVE NEITHER GIVEN OR RECEIVED AID ON THIS EXAMINATION. I FURTHER PLEDGE THAT, IF I HAVE REASON TO SUSPECT THAT THE CODE HAS BEEN VIOLATED, I HAVE TAKEN OR WILL TAK
Vanderbilt - MSE - 150
PRINT NAME _ INTRODUCTION TO MATERIAL SCIENCE MSE-150 QUIZ 4 I PLEDGE MY HONOR THAT I HAVE NEITHER GIVEN OR RECEIVED AID ON THIS EXAMINATION. I FURTHER PLEDGE THAT, IF I HAVE REASON TO SUSPECT THAT THE CODE HAS BEEN VIOLATED, I HAVE TAKEN OR WILL TAK
Vanderbilt - MSE - 150
PRINT NAME _ INTRODUCTION TO MATERIAL SCIENCE MSE-150 QUIZ 5 I PLEDGE MY HONOR THAT I HAVE NEITHER GIVEN OR RECEIVED AID ON THIS EXAMINATION. I FURTHER PLEDGE THAT, IF I HAVE REASON TO SUSPECT THAT THE CODE HAS BEEN VIOLATED, I HAVE TAKEN OR WILL TAK
Vanderbilt - MSE - 150
PRINT NAME _ INTRODUCTION TO MATERIAL SCIENCE MSE-150 QUIZ 6 I PLEDGE MY HONOR THAT I HAVE NEITHER GIVEN OR RECEIVED AID ON THIS EXAMINATION. I FURTHER PLEDGE THAT, IF I HAVE REASON TO SUSPECT THAT THE CODE HAS BEEN VIOLATED, I HAVE TAKEN OR WILL TAK
Vanderbilt - MSE - 150
Travis Owen Section 1 Page 1 Introduction The main objectives of this lab were to compare the mechanical properties of metals, ceramics, polymers, and composites and to demonstrate the effect of temperature on the toughness of body centered cubic and
Vanderbilt - MSE - 150
Intro The Purpose of this experiment was to work a material such as brass so that the microstructure was revealed. The microstructure was revealed by thermal-mechanical processing and the relationship between hardness and grain size was determined. T
Vanderbilt - MSE - 150
Introduction The purpose of this lab was to compare metals and semiconductors electrical properties. Also, to examine how conductivity alters under the influence of a temperature change. This was done using digital multimeters and solution baths set
WVU - POLS - 250
1.) The Tokugawa Shogunate (1603-1867)- was established by Tokugawa Ieayasu following the successful military campaign that unified Japan under his rule. IT was a military dynasty of hereditary Shoguns which ruled japan. - the shogun ruled the centra
Wilkes - FYF - 101
Decision Making&quot;People usually do not lack resources, they lack control over their resources.&quot;Achieverstend to be decisive. They make effective decisions that move them closer to their goals- and they do it consistently.When confronted by
Wilkes - SOC - 101
Sociology 101 EXAM 1 Study Guide1. Review class notes, attendance quizzes and be sure that you can provide a satisfactory answer for each of the learning objectives from lecture. 2. Be able to define the key terms in each chapter summary 3. Know th
Millersville - COMM - 100
Everyone has a some kind of relationship weather it is with a family, friend, romantic, teacher, or peer. There has been a point in that relationship where we have wanted to improve it. Relationship listening is a key point in improving our relations
Millersville - COMM - 100
Nate Campbell MLADeVito, Joeseph. The Interpersonal Communication Book. USA: Allyn &amp; Bacon, 2003.Maxwell-Gunter AFB. 23 Jun 1996. John Kline. 24 Mar 2008 [http:/www.au.af.mil/au/awc/awcgate/kline-listen/b10ch4.htm].Woods, Julia. Communication i
Millersville - COMM - 100
People -Ken Griffey Jr. -Brett FavrePlaces -Canada -LondonActivities -Baseball -Running -Football -Basketball -Milsim-Hank Aaron-California-MLB Stadiums-Weight TrainingEvents -Baseball Games -Football Games -Road Races(Running) -Rock Con
Millersville - CSCI - 161
/* * To change this template, choose Tools | Templates * and open the template in the editor. */ package windows; import javax.swing.*; / For JFrame import java.awt.*; / For Color and Container import java.util.Scanner; /* * * @author */ public class
Millersville - CSCI - 161
45 60 0.0 65 70 1.3 55 75 0.4 -500 -500 -500 78 89 0 72 91 0 45 75 0 -3 34 0 -500 -500 -500 30 35 5.9 -10 0 0 -20 -10 5.4 45 60 0.0 65 70 1.3 55 75 0.4 78 89 0 72 91 0 -3 34 0 30 35 5.9 -10 0 0 -20 -10 5.4 45 60 0.0 78 89 0 72 91 0 45 75 0 -3 34 0
IUP - GOVT - 101
Nate Campbell GOVT101.01 When I look at the President's Budget for the 2007 fiscal year I see two major driving factors. Although Social Security is the largest chunk of money, the President wants to continue his worldwide push against terrorism, alo
Millersville - GEOG - 300
In 1887, The East End of London was no different than any other area. The majority of the population was literate; crime rates were generally stable and it was seen as a moral suburb. Then, all of a sudden, in the fall of 1888 that all changed. The a
Millersville - GEOG - 300
In the sixteenth and seventeenth centuries, Southwark was seen as the more disrespectable section of the London area. It was not always this way however. Southwark was founded in the thirteenth century and was a nice place to live, although always a
Millersville - COMM - 100
Gene Ellis Nate Campbell General Purpose: To Inform Specific Purpose: At the end of my speech the audience will be more informed about Performance Enhancement. INTRODUCTION Attention Grabber I. Hi everyone, my name is Nate and I am a drug addict. The
Millersville - COMM - 100
Everyone knows about the problem with steroids in sports, baseball in particular. Unfortunately we live in a performance enhanced world, and sports are the least of our worries. Performance Enhancers are common in Hollywood, and even in the workplace
Millersville - COMM - 100
Chicago Tribune. 21 November 2006. Chicago Tribune. 13 Feb 2008[http:/featuresblogs.chicagotribune.com/features_julieshealthclub/2006/11/caffeine_abuse_.html]Epstein, David. &quot;The Drugs.&quot; Sports Illustrated 12 March 2007: pg65.Hintz, RL. &quot; Curren
Millersville - CSCI - 161
/* * To change this template, choose Tools | Templates * and open the template in the editor. */ package rm; import javax.swing.*; / For JFrame import java.awt.*; / For Color and Container import java.util.Scanner; import java.text.DecimalFormat; /*