1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

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488 Answers to Odd-Numbered Exercises



  1. 25(ui+ 1 − 2 ui+ui− 1 )− 25 ui=−25,i= 1 , 2 , 3 , 4 ,5;u 0 =2,u 5 +(u 6 −
    u 4 )/( 2 / 5 )=1. When the equation fori=5 and the boundary condition
    are combined, they become 2u 4 − 3. 4 u 5 =− 1 .4. Solution:u 1 = 1 .382,
    u 2 = 1 .146,u 3 = 1 .057,u 4 = 1 .023,u 5 = 1 .014.

  2. 9(ui+ 1 − 2 ui+ui− 1 )+( 3 / 2 )(ui+ 1 −ui− 1 )−ui=−( 1 / 3 )i,i= 0 , 1 ,2;
    u 3 =1,(u 1 −u− 1 )/( 2 / 3 )=0. Whenu− 1 is eliminated and coefficients
    are collected, the equations to solve are


− 19 u 0 + 18 u 1 = 0 ,
712 u 0 − 19 u 1 + 1012 u 2 =−^13 ,
712 u 1 − 19 u 2 =− 1116.

Solution:u 0 = 0. 795 ,u 1 = 0. 839 ,u 2 = 0 .919.

Section 7.2



  1. Linemof the solution should be exactly the same as linem+1ofTable4.
    3.r= 2 /5, t= 1 /40.


i
m 01 2 3 4
0 00000
1 1 0. 0. 0. 0.
2 1 0. 40. 0. 0.
3 1 0. 48 0. 16 0. 0.
4 1 0. 56 0. 224 0. 064 0.
5 1 0. 6016 0. 2944 0. 1024 0. 0512


  1. t= 1 /32. Remember thatu 4 (m)=u 0 (m)=m t.Allnumbersinthe
    table should be multiplied by t.


i
m 01 2 3 4
0 00 0 0 0
1 1 0001
2 2 1 / 20 1 / 2 2
3 3 11 / 21 3
4 4 7 / 41 7 / 4 4
5 5 5 / 27 / 45 / 2 5


  1. t= 1 /32. All numbers in this table should be multiplied by t.

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