Proposal of Finite Difference

LisanEldeen Galhoud

C̊

≤ ≤4 ,0≤C̊ ≤4

Question1: Question1 y let the boundary temperatures be 0 For the square 0 x on the horizontal and 50  on  on the vertical edges find the temperatures at the interior points of square grid with 0C h=1. 4    

3

Solution:

U13

As we see the square their similarity so the unknown reduced to be:

UU == UU == UU = U = U UU == UU = U

U23

U33

                      

   C    0    52

                      

U12

U22

U32

U11

U21

U31

   0    5

1

0 1

2

3

4

0C

   

At new unknown find U where: 0C

   

4 1

3 U1

1

-4

1

U1

U2

  .U=0                         

                        

    C     0     52

1

U3

U4

U3

U1

U2

U1

    0     5

1

U+,j  U−,j  U,j+  U,j−  4U,j = 0 4U4U  U2U UU ==500 4U4  22U  2U==500 [1]

0 1

2

3

0C

   

4

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=42 14 10 10− 500  0 2 20 4 2 14 500 ∴ =25=U =U =U =U     =18. 7 5=U =U    =31. 2 5=U =U   =25=U Question 2:

For the grid in figure below compute the potential at the four internal  points by finite difference method if the boundary values on the edges are:

=0  ℎ9,ℎℎℎ, ,27  27  ℎ  .

∇²U=0

3

2

U12

U22

U11

U21

Solution:

The region defined by Laplace equation so the solution will be:

1

0 1

1

²

U=

1

-4

1

  .U=0

1

First find the boundary points values:

 AtUlo=1wer, boundar y poi n t s def i n ed by U= U =2 =8 At left all points are zeros.

[2]

2

3

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 27 AtUth=26 e upper, boundary poi n t e s def i n ed by U= U =46  U=279 U =18 , U =9 At the right boundary points defined by

U13=-26 U23=-46

U02=0

Finding the unknown function depending on equation:

U32=-9

U4U+,jUU−,Uj U=1,j+ U,j− 4U,j =0 4U4U UU UU =26 =188 4U U4 U1 =469 − 1 1 0 =10 1 410 40 1114 265526  ∴ =2 ,  =2 ,  =11,  =16

U12

U22

U11

U21

U01=0

U31=18

U10=1

U20=8

Question 3:

∇U=12xy on square of figure below ,where h=0.5.

Solve the mixed boundary value problem for Poisson equation:

y

Solution: 1

AtUri=3(0. ght edge5)poi=0.n3ts75,defUined=3by U=3

du/dy=6x

First we find the boundary points:

At the upper edge points are unknown too and defined by derivative du/dy=6x so we imagine

[3]

U=0

U=6xy

U=0

U=3y

1.5

x

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the region to be extended above the first row (corresponding to y=1.5).

1.5 1

U13

U23

U=3

1 ²

U=

1

-4

1

U12

  .U=h²f(x,y)      0     =      U

U22

U=0.375 U11

U21

1

U=0

For Poisson equation solution is:

At4the internalpoint=s: ((0U.5),U. 12)(0.5×0.5) =0.75 4At the boundar  y poi =3.nts: ((U0.5)0. 3 75=1. 1 25 , U )   4   =1. 5    4   =0     ≈ 2ℎ =6⇒ = 3  ≈ 2ℎ =6⇒ = 6 4U4U UU UU =0.=1.17255 4U4U 2UU2UU =1. 5 =6 =41 41 10 01− 1.0.17255   20 02 1 4 14 1.65 [4]

1.5

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∴ =0.0769 ,  =0.191 , =0.8665,  =1.18121 Question 4: Solve the Laplace equation in the region and for the boundary values shown in figure below using indicated grid the sloping portion of the boundary is y=4.5-x.

y U=0 3 U=x²-1.5x       0 2     =       U

Solution:

1

First we find the boundary points values:

U12

U22

U11

U21

U=9-3y

x

0 1

2

3

U=3x

At left boundary all points are zeros.

AtAt lroigwerht boundar boundaryy:Uthe =93 points ar(1e): U=6 =3, U =6 PoiFornunknown ts at slope:poiUnts=2.U5, U,U, U=1 solution is: At upper boundary all points are zeros.

1

U=0

3

U=1

²

U=

1

-4

1

       0 2      =        U

  .U=0

U12

U22

U11

U21

U=2.5

U=6

1

1

x

For unknown poi n t U , u se equat i o n:   2           ≈ ℎ (( )  ( )  ( )  (  )  ()  ) 0

1

2

U=3

U=2

3

from the grid at this point we find:

B

a=b=1/2, p=q=1 so the solution is:

bh

Q 

ah

qh

ph

[5]

P

A

LisanEldeen Galhoud

4/3 {3/2 3/24 4/3}.U=0 4U4U UU UU =3=0 4U12UU2U U 2U =12 =14 = 41 14 10 10 −  30   0120421 12 1214 ∴ =2 ,  =4 ,  =1,  =2 The system of equations:

Question 5:

PDE:    =0 0<<1, 0<<2 & ℎ==0. 5 (, 0) = , (, 2) = ( 2), (0, ) =, (1, ) =(1)  ={1 1 4 1}.U=ℎ (, ) 1 4UU 4U UU=(0=0.5). (4) 3. (0. 5) U 4U =3.75

Use finite difference method to approximate the solution to the elliptic

Solution:

U(x,2)=(x-2)²

2

1.5

              ²

   y   =     )    y  ,     0 1     (     U

              ²

    )     1      y     (   =     )    y  ,     1     (     U

U13

U12

0.5 U11

0 [6]

0.5 U(x,0)=x ²

1

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[]=[41 41 01 ][ 0.025 ]  0 1 4 3.75 ∴U =0, U =0.25,U =1 Question 6:

∇² UU,, =0=3onontthheerliegfhtt, U, U=x=xon 1theonlowerthe upper edge, . 1 ( ) 0= = U U g i v es U =   . ,      2h Sionmthilearrligyht,Uedge,=Usoth3at ftrhoemequatthe condi t i o n ions are 4U4U UU UU =1=0.25 4U4U UU UU =1 =3. 5 =41 42 01 00− 0.125  00 10 2 4 14 6.15 ∴ =0.25 ,  =0 ,  =0.75,  =2

Solve the mixed boundary value problem for the Laplace equation ( U=0) in the rectangle in figure shown and the  boundary conditions: U=x²-1

1

       0      =      x

        /      u

Solution:

U11

U21

U=x²

U=-0.75

1.5

U=0

1

Ua1

[7]

U01

U11

U=0.25

U21

U=1

U31

1.5

Ub1