ch4

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CHAPTER 4

_–

ORDINARY AND PARTIAL DIFFERENTIAL EQUATION ORDINARY AND PARTIAL DIFFERENTIAL EQUATION

ORDINARY DIFFERENTIAL EQUATION (ODE) ORDINARY DIFFERENTIAL EQUATION (ODE)

4.1

4.1 INITIAL-VALUE INITIAL-VALUE PROBLEM PROBLEM (IVP)(IVP)

Q1

Q1 Consider the following initial-value problem (IVP)Consider the following initial-value problem (IVP)

2 2 3 3 , , ((00)) 11.. dy dy x x y y yy dx dx    Solve the IVP for

Solve the IVP for 00_{ }_ x x _0..6 06 aannd d hh__0..202_{by using}_{by using Euler’s method}_{Euler’s method and RK4 method.}_{and RK4 method.}

Q2

Q2 Consider the following initial-value problem (IVP)Consider the following initial-value problem (IVP)

2 2 ((11 x x ))dydy xy xy 00, , ((2yy 2)) 55.. dx dx     

Solve the IVP for

Solve the IVP for 22_{ }_ x x _2..3 23 aannd d hh__0..101_{by using Euler’s method}_{by using Eule}_{r’s method and RK4 method.}_{and RK4 method.}

Q3

Q3 Solve the following ordinary differential equationSolve the following ordinary differential equation

x x y y dx dx dy dy 2 2       ,, y y((00))11

with uniform step size

with uniform step size hh00..11 over interval [0, 0.3] by using over interval [0, 0.3] by using Euler’s methodEuler’s method and RK4 and RK4

method. method.

Q4

Q4 Given an initial-value problem (IVP) as followsGiven an initial-value problem (IVP) as follows

0.3 0.3 1 1..22 77 x x, , ((00)) 33.. dy dy y y e e yy dx dx       

Solve the IVP by using

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Q5

Q5 The The concentration concentration of of a a chemical chemical in in a a batch batch reactor reactor can can be be modeled modeled by by the the followingfollowing differential equation differential equation 1 1 2 2 , ( , (00)) 00..88.. 1 1 k k C C dC dC C C d dt t k k C C       

Find a numerical solution for this problem at

Find a numerical solution for this problem at t t __11 s with s with

1

1 11, , 22 0.30.3

k

k   k k  and step length, and step length,

0.5 0.5 h

h by using by using Euler’s method andEuler’s method and classical fourth-order Runge-kutta method. classical fourth-order Runge-kutta method.

Q6

Q6 Velocity of a falling object can be modVelocity of a falling object can be modeled as the following initial-value problem (IVP)eled as the following initial-value problem (IVP)

2 2 , , ((00) ) 2200 d d cc dv dv v v g g vv dt dt   mm    where

where vv velocity of the falling object (m/s), velocity of the falling object (m/s), t t  time (s), time (s), g g  acceleration due to acceleration due to

gravity (9.81 m/s

gravity (9.81 m/s22),), mm mass (kg) and mass (kg) and cc_d_d  drag coefficient (kg/m). Initially, the object drag coefficient (kg/m). Initially, the object

was at

was at v_{v }_2020_{m/s with}_{m/s with} cc_d_d 0.2250.225. Calculate the velocity for a falling 5-k. Calculate the velocity for a falling 5-kg object atg object at

0 (

0 ( 0.10.1) 0.5) 0.5 t

t  s by using Euler’s methods by using Euler’s method..

Q7

Q7 A voltage source,A voltage source, E E t ( ( ))t _{is supplied to an electrical circuit with inductance}_{is supplied to an electrical circuit with inductance} _L_L_{and a}_{and a}

resistance

resistance R R. If the switch is closed at. If the switch is closed at t _{t }_00_{s, the current}_{s, the current (} I I t _{( ))}t _{will satisfy the following}_{will satisfy the following}

initial-value problem (IVP) initial-value problem (IVP)

( ( ) ) ( ( ) ) ( ( )), , ((00) ) 00.. d d L L I I t t RI RI t t E E t t I I dt dt     

Parameter values are given as

Parameter values are given as _L_L5050 H, H, R R  2020  and and E E t ( ( ) t ) 1100 V. Estimate the value V. Estimate the value

of the current at

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4.2

4.2 BOUNDARY-VALUE BOUNDARY-VALUE PROBLEM PROBLEM (BVP)(BVP) Q8

Q8 Given the boundary-value problem (BVP)Given the boundary-value problem (BVP)

4

4 ssiinn

x

x   _ x x __ t t ,, 0 0 __{ }t t _11

with conditions

with conditions x x((00) ) 00 and and x x((11) ) 00. Solve the BVP by using finite difference method. Solve the BVP by using finite difference method

by taking

by taking __{ }t t _{ }hh_0.250.25_..

Q9

Q9 Given the boundary-value Given the boundary-value problem (Bproblem (BVP)VP)

2 2 2 2 44 00,, 00 22 d d y y ddyy x x xx d dx x   ddxx       with conditions

with conditions y y((00) ) 00 and and y y((22) ) 11.. Solve the BVP by using finite-difference method Solve the BVP by using finite-difference method

by taking

by taking __ x x__00..55..

Q10

Q10 Solve the boundary-value problem (BVP),Solve the boundary-value problem (BVP), y y     xy xy  

3

y y 

11

xx with conditions with conditions y y((00) ) 11 and

and y y((11) ) 22 where where hh0.250.25 by using finite-difference method. by using finite-difference method.

Q11

Q11 The boundary-value problem (BVP) for the steady-state temperature in a rod of length 2The boundary-value problem (BVP) for the steady-state temperature in a rod of length 2 m is represented as follows m is represented as follows 2 2 0 0 00 2 2 00..1 1 00, , ((00) 2) 2000 0 , , ((22) ) 11000 0 .. d d T T T T TT CC TT CC d d xx    

Approximate the temperature,

Approximate the temperature, T T throughout the rod for throughout the rod for _{ }_ x x _{ }hh_0.50.5_{by using finite-}_{by using}

finite-difference method. difference method.

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Q12

Q12 A heated rod with a uniform heat source can be modeled with the Poisson equation,A heated rod with a uniform heat source can be modeled with the Poisson equation,

2 2 2 2 ( ( )).. d d T T f f xx dx dx   Given the heat source,

Given the heat source, f f xx( ) ( ) 2255 and the boundary conditions, and the boundary conditions, T T x( ( x  0) 40) 40 0 and and

(

( 1100) ) 220000..

T

T xx   Solve for the temperature distribution with Solve for the temperature distribution with h h    xx 2.52.5 by using finite- by using

finite-difference method. difference method.

Q13

Q13 The position of a falling object is governed by the following boundary-value problemThe position of a falling object is governed by the following boundary-value problem (BVP) (BVP) 2 2 2 2 0 0 , , ffoor r 0 0 1122,, d x d x c c ddxx g g t t d dt t  m m ddt t     

where boundary conditions are

where boundary conditions are x x((00) ) 00 and and x x(1(12) 2) 55000.0. Given that the parameter Given that the parameter

values are

values are cc a first-order drag coefficient a first-order drag coefficient (12.5 kg/s),(12.5 kg/s), mm mass of the falling object mass of the falling object

(50 kg)

(50 kg) and and g g  gravitational acceleration gravitational acceleration

2 2

(( 99..881 1 mm//ss ).). Approximate the position of Approximate the position of the falling object,

the falling object, x x(m)(m) for for hh33 by using finite-difference method. by using finite-difference method.

Q14

Q14 A thin rod of length,A thin rod of length, l l _{is moving in the}_{is moving in the}_xy_xy_{-plane. The rod is fixed with a pin on one end}_{-plane. The rod is fixed with a pin on one end}

and a mass at the other end. This system is represented in the form of boundary-value and a mass at the other end. This system is represented in the form of boundary-value problem (BVP) as follows problem (BVP) as follows ( ( ) t t ) g g ( ( ) 0t t ) 0 l l   _ __   __ _{, for}_{, for} 0 0 __{ }t t _00..44,,

where boundary conditions are

where boundary conditions are  ((00) ) 00 and and   (0(0.4.4) ) 11. The parameter values. The parameter values are given as

are given as g g gragravitavitationtional foal force (9.81 rce (9.81 m/sm/s ))22 and and l l 0.90.9 m. Approximate the angle m. Approximate the angle   (in radian) for

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PARTIAL DIFFERENTIAL EQUATION (PDE) PARTIAL DIFFERENTIAL EQUATION (PDE)

4.3

4.3 Heat Heat equation equation (explicit (explicit finite-difference finite-difference method)method)

Q15

Q15 Given the heat equationGiven the heat equation

2 2 2 2 0 0..9 9 , , 0 0 11, , 00 u u uu x x t t t t xx           

with the boundary conditions,

with the boundary conditions, u t u t u ((00, , ) )   u t ((11, , ) t ) 11 for for t t 00, and the initial condition,, and the initial condition,

(1 (1 )) ( ,0) ( ,0) x x xx u u x x ee  

 for for 0 0   x x11. Find. Find u xu ( ( ,, 0.x 0.0101)) and and u u x( ( ,, 0.x 0.0202)) by using explicit finite- by using explicit

finite-difference method with

difference method with __{ } x x _{ }hh_0.2.0.2.

Q16

Q16 Given the heat equationGiven the heat equation

,, 0 0 ,, 2 2 0 0 ,, )) ,, (( 2 2 )) ,, (( 2 2 2 2                 t t x x x x t t x x u u t t t t x x u u

with the boundary conditions with the boundary conditions

,, 0 0 )) ,, 2 2 (( )) ,, 0 0 (( t t uu t t  u u

and the initial condition and the initial condition

). ). sin( sin( )) 0 0 ,, (( x x xx u u    Find

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Q17

Q17 Consider the heat conduction equationConsider the heat conduction equation

2 2 2 2 ( ( , , ) ) ( ( , , )), , 0 0 1100, , 00 T T xx tt TT xx tt xx tt t t xx              ,, where

where   is thermal diffusity is thermal diffusity 10,10, since since

2 2 c c    ..

Given the boundary conditions, Given the boundary conditions,

((00, , ) ) 00, , ((1100, , ) ) 110000

T

T tt   TT tt 

and initial condition, and initial condition,

2 2 ( ,0) ( ,0) T T x x  xx ..

By using explicit finite-difference method, find

By using explicit finite-difference method, find T T x( ( ,,0.x 0.05055)5)_and_and _T_{T xx}₍_{( ,, 0.}_0.11₁₁₎₎_{with 5 grid}_{with 5 grid} intervals on the

intervals on the x x coordinate. coordinate.

Q18

Q18 The temperature distributionThe temperature distribution ( u u x ( , , ))x t t of one dimensional silver rod is governed by the heat of one dimensional silver rod is governed by the heat equation equation 2 2 2 2 2 2 u u uu t t xx           with

with   22 is is thermal difthermal diffusity =fusity =1.71.1.71.

Given the initial condition, Given the initial condition,

               .. 4 4 2 2 ,, 4 4 ,, 2 2 0 0 ,, )) 0 0 ,, (( x x x x x x x x x x u u

and boundary conditions, and boundary conditions,

2 2

((0 0 , , ) ) , , ((44, , )) u

u tt   tt uu tt tt .. Find the temperature distribution of the rod with

Find the temperature distribution of the rod with __{ } x x _{ }hh_11_and_and __{ }t t _{ }k k _0.20.2_for_for 0

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4.4

4.4 Wave equation Wave equation (finite-difference (finite-difference method)method)

Q19

Q19 LetLet uu(( x x,,t t )) be the displacement of uniform wire which be the displacement of uniform wire which is fixed at both ends alongis fixed at both ends along x x-axis-axis

at time

at time t t . . The disThe distribution oftribution of uu(( x x,,t t )) is given by the wave equation is given by the wave equation

2 2 2 2 2 2 2 2 4 4 x x u u t t u u           ,, 0 0    x x11,, 0 0   t t 00..55

with the boundary conditions

with the boundary conditions uu((00,,t t ))uu((11,,t t ))00 and the initial conditions and the initial conditions

x x x x u u(( ,,00))sinsin  ,, (( ,,00))00     x x t t u u for

for 00__ xx __11. Solve the wave equation up to level. Solve the wave equation up to level t t 0.20.2

by using finite-difference method with

by using finite-difference method with __ x x __hh__00..2525_and_and __t t __k k __00..11

Q20

Q20 LetLet y y x ( ( , , ))x t t denotes displacement of a vibrating string. If denotes displacement of a vibrating string. If T T is the tension of the string, is the tension of the string,



 is the weight per unit length and is the weight per unit length and g g is acceleration due to gravity, then is acceleration due to gravity, then y y satisfies the satisfies the

equation equation 2 2 22 2 2 22 ,, 00 22 ,, 00 y y Tg Tg yy x x t t t t   xx



_







_

_













..

Suppose a particular string with 2 m long is fixed at both ends. By taking

Suppose a particular string with 2 m long is fixed at both ends. By taking T T 11..55 N,N, 01 01 .. 0 0   

 kg/m and kg/m and g g 1010 m/sm/s22 , use finite-difference method to solve for, use finite-difference method to solve for y y up to up to

second level. second level.

The initial conditions are The initial conditions are

0 0..55 ,, 00 11 (( ,, 00 )) 1 1 00..55 ,, 11 22 x x xx y y xx x x xx



 







 







 





andand 2 2 ( ( ,, 00 ) ) 2 2 .. y y x x x x xx t t       

Perform all calculations with

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Q21

Q21 The air pressureThe air pressure u u x ( ( , , ))x t t in an organ pipe is governed by the wave equationin an organ pipe is governed by the wave equation

2 2 22 2 2 2 2 22 1 1 , , 0 0 , , 00,, u u uu x x l l t t t t   xx            where

where l l _{is the length of the pipe and}_{is the length of the pipe and}   is a physical constant. If the pipe is closed at the is a physical constant. If the pipe is closed at the

end where

end where x x l l , the boundary conditions are, the boundary conditions are

((00, , ) ) 00..9 9 aannd d ( ( , , ) ) 00..9 9 ffoor r 0 0 00..22..

u

u tt __ uu ll tt _{ }__ __tt__

Assume that

Assume that     11, , l l 00..55 and the initial conditions areand the initial conditions are

(

( ,, 00) ) 00..9 9 ccooss ((2 2 ) a) annd d uu( ( ,, 00) ) 0 0 ffoor r 0 0 00..55..

u u xx xx xx xx t t           

Approximate the pressure for the closed-pipe by using finite-difference method with Approximate the pressure for the closed-pipe by using finite-difference method with

0.1 0.1

h

h    xx _and_and _k_k   _t_t_0.1._0.1.

Q22

Q22 The longitudinal vibration of a bar with the The longitudinal vibration of a bar with the length oflength of l l _{m is governed by}_{m is governed by} 2 2 22 2 2 2 2 22 c c x x t t             with

with cc E E  



 , where, where     (( ,, )) x t x t is the axial displacement, is the axial displacement, E E is Young’s modulus andis Young’s modulus and



 is the mass density of the bar. The boundary conditions and the initial conditions are is the mass density of the bar. The boundary conditions and the initial conditions are

given as follows, given as follows, ((00, , ) t t ) ( ( , l l t , ) t ) 00        for for 0 0 __{ }t t _00..0044 0 0 )) 0 0 ,, (( x x    and and x x t t x x      __₍₍ _,,₀₀₎₎ for for 00 xx 2020..

Determine the variation of the axial displacement of the bar by using finite-difference Determine the variation of the axial displacement of the bar by using finite-difference method with the following data:

method with the following data:

6 6

3 30 0 1100 E

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PARTIAL DIFFERENTIAL EQUATION (PDE) PARTIAL DIFFERENTIAL EQUATION (PDE)

4.3

4.3 Heat Heat equation equation (explicit (explicit finite-difference finite-difference method)method)

Q15

Q15 Refer to class noteRefer to class note

Q16 Q16

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Q17 Q17 2 2 2 2 (( ,, )) (( ,, )),, 00 1100,, 00 T T xx tt TT xx tt xx tt t t   xx           

 



2 2 2 2 , , 1 1 ,, 11,, ,, 11,, 2 2 , , 1 1 ,, 11,, ,, 11,, 2 2 , , 11 ,, 11,, ,, 11,, 10 10 2 2 10 10 2 2 10 10 0 0..00555 5 22 0 0..11338 8 22 i i jj ii jj ii jj ii jj ii jj i i jj ii jj ii jj ii jj ii jj i i jj ii jj ii jj ii jj ii jj T T T T t t xx T T TT TT TT TT k k hh T T TT TT TT TT T T TT TT TT TT                                           1 1,, ,, 11,, 0 0..11338 8 T T _i_i__ _j_j 0..202776 6 T T _i_i _j_j 0..1013388T T _i_i__ _jj      , , 11 00..11338 8 1,1, 00..22776 6 ,, 00..113388 1,1, ,, i i jj ii jj ii jj ii jj ii jj T T __  TT__  TT  TT__  TT , , 11 00..11338 8 11,, 00..77224 4 ,, 00..113388 1,,1 0 0..11338 8 00..77224 4 00..113388 i i jj ii j j ii jj ii jj T T TT TT TT A A B B C C                T T 0,00,0 T T 0,10,1 T T 0,20,2 T T 2,02,0 T T 3,03,0 T T 4,04,0 T T 5,05,0 T T 1,11,1 T T 1,21,2 T T 2,12,1 T T 2,22,2 T T 3,13,1 T T 3,23,2 T T 4,14,1 T T 4,24,2 T T 5,15,1 T T 5,25,2

1

0.138

0.724

0.138

0.138 =

= T

T

i,j+1i,j+1

(A)

T

i-1,ji-1,j

(B)

T

i,ji,j

(C)

T

i+1,ji+1,j

x x tt 0 0 2 2 44 6 6 88 1010 0.055 0.055 0.11 0.11 T T 1,01,0

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Q18 Q18 100 100 0.11 0.11 0.055 0.055 0,0 0,0 T T T T _1,0_1,0 T T _2,0_2,0 T T _3,0_3,0 T T 4,04,0 T T 5,05,0 0,1 0,1 T T 0,2 0,2 T T 1, 1,11 T T 1,2 1,2 T T 2,1 2,1 T T 2,2 2,2 T T 3, 3,11 T T 3,2 3,2 T T _T_T_4,2_4,2 4,1 4,1 T T 5,2 5,2 T T 5, 5,11 T T tt x x 0 0 5.104 5.104 17.104 17.104 37.104 37.104 65.10465.104 0 0 6.056 6.056 18.208 18.208 38.208 38.208 66.05666.056 100 100 0 0 2 2 4 4 6 6 8 8 1010 0 4 16 36 64 100 0 4 16 36 64 100

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4.4

4.4 Wave equation Wave equation (finite-difference (finite-difference method)method)

Q19

Q19 Refer to class noteRefer to class note

Q20 Q20 ₂₂ ,, 00 22,, 00 2 2 2 2 2 2                 t t x x x x y y Tg Tg t t y y   where

where T T 11..55,,   00..0101 and and g g 1010..

2 2 2 2 2 2 2 2 1500 1500 x x y y t t y y           2 2 ,, 1 1 ,, ,, 1 1 2 2 1 1 ,, ,, 1 1 ,, )) (( 2 2 1500 1500 )) (( 2 2 h h y y y y y y k k y y y y y y_ii_j j _ii_j j _ii_j j _ii _j j  _ii_j j  _ii _j j       2 2 ,, 1 1 ,, ,, 1 1 2 2 1 1 ,, ,, 1 1 ,, )) 5 5 .. 0 0 (( 2 2 1500 1500 )) 01 01 .. 0 0 (( 2 2 _ii_j j _ii_j j _ii _j j _ii_j j _ii _j j j j

ii y y y y y y y y yy

y y             )) 2 2 (( 6 6 .. 0 0 2 2 _,, _,, ₁₁ ₁₁_,, _,, ₁₁_,, 1 1 ,, j j ii j j ii j j ii j j ii j j ii jj

y y __   __  __   __ j j ii j j ii j j ii j j ii j j ii j j

y y,, 11 00..66(( 11,, 22 ,,  11,, )) ,, 11 22 ,, 1 1 ,, ,, 1 1 ,, ,, 1 1 1 1 ,, j j 00..66 ii j j 00..88 ii j j 00..66 ii j j  ii j j

ii y y y y y y yy

y

y --- --- (1)(1)

(D) (D)

y

i,j-1i,j-1

)) 2 2 (( 2 2 1 1 ,, 1 1 ,,       __   x x x x k k y y y y_ii_j_j _ii_j_j )) 2 2 (( 02 02 .. 0 0 1 1 ,, 1 1 ,,   y y   x x xx y yii j j ii j j )) 2 2 (( 02 02 .. 0 0 1 1 ,, 1 1 ,,   y y   x x xx y y_ii_j_j _ii_j_j --- --- (2)(2)

1

0.6

0.8

0.6

0.6 =

= yy

i,j+1i,j+1

(A)

yy

i-1,ji-1,j

(B)

y

i,ji,j

(C)

(C) y

y

i+1,ji+1,j

-1

(13)

Substitute eqn. (2) into eqn. (1): Substitute eqn. (2) into eqn. (1):

1 1 ,, ,, 1 1 ,, ,, 1 1 1 1 ,, j j 00..66 ii j j 00..88 ii j j 00..66 ii j j  ii j j

ii y y y y y y yy

y y 1 1,, ,, 11,, ,, 1 1 0 0..6 6 y y _i_i___j_j 00..8 8 y y _i_i _j_j 0..6 06 y y _i_i___j_j ( ( y y _i_i _jj__ 0..0002 2 ( 2x x x( 2))))x        )) 2 2 (( 02 02 .. 0 0 6 6 .. 0 0 8 8 .. 0 0 6 6 .. 0 0 2

2 y y_ii_,,_j_j__₁₁  y y_ii__₁₁_,,_j_j  y y_ii_,,_j_j  y y_ii__₁₁_,,_j_j  x x xx

)) 2 2 (( 01 01 .. 0 0 3 3 .. 0 0 4 4 .. 0 0 3 3 .. 0 0 ₁₁_,, _,, ₁₁_,, 1 1

,,   y y  y y  y y  x x xx

y y_ii_j_j _ii _j_j _ii_j_j _ii _j_j 0.02 0.02 0.01 0.01 0,2 0,2 y y tt x x 0 0 0.243 0.243 0.340 0.340 0.243 0.243 00 0 0 0.148 0.148 0.064 0.064 0.148 0.148 00 0 0 0.5 0.5 1.0 1.0 1.5 1.5 2.02.0 0 0 0.25 0.25 0.5 0.5 0.25 0.25 00 0,1 0,1 y y 0,0 0,0 y y 2,2 2,2 y y 3,2 3,2 y y 4,2 4,2

y

1, 1,11 y

y y y_2,1_2,1

3,1 3,1

y

y y y_4,1_4,1

1,0 1,0

y

_2,0_2,0

3,0 3,0

y

4,04,0 1,2 1,2 y y

1

0.3

0.4

0.3

0.4

0.3

0.3 =

= yy

i,j+1i,j+1

(A)

yy

i-1,ji-1,j

(B)

yy

i,ji,j

(C)

(C) y

y

i+1,ji+1,j

0.01

(14)

Q21 Q21 2 2 22 22 22 2 2 22 22 22 22 1 1 , , wherwhere e 11 u u uu uu uu t t __ xx   tt xx                , , 11 ,, ,, 11 11,, ,, 11,, 2 2 22 , , 11 ,, ,, 11 11,, ,, 11,, , , 1 1 11, , 11, , , , 11 2 2 22 ((00..11) ) ((00..11)) 2 2 22 i i jj ii jj ii jj ii jj ii jj ii jj i i j j i i j j i i j j i j i j i i j j i i jj i i j j i i j j i i j j i i jj u u uu uu uu uu uu u u uu uu uu uu uu u u u u u u u u A A B B C C                               

      (Calculator formula)(Calculator formula)

Representation in molecule graph (calculating level 2): Representation in molecule graph (calculating level 2):

,, 11 ,, 11 ,, 11 ,, 11 G Giivveen n (( , 0, 0)) 00 0 0 2(0.1) 2(0.1) --- --- (1(1)) t t i i j j i i jj i i j j i i jj u u xx u u uu u u uu               Substitute (1) into Substitute (1) into uui i , 1 , j j 1   uui i j 11, , j   uui i j 1, 1, j uui i , 1, jj1:: , , 1 1 11, , 11, , , , 11 , , 1 1 11, , 11,, , , 1 1 11,, 11,, 2 2 0 0..5 5 00..5 5 00..5 5 00..55 i i j j i i j j i i j j i i jj i i j j i i j j i i jj i i jj ii jj ii jj u u uu uu uu u u u u uu u u uu uu AA BB                        

    (Calculator formula)(Calculator formula)

Representation in molecule graph (calculating level 1): Representation in molecule graph (calculating level 1):

1

1 =

= u

u

,j+1 ,j+1

(A)

u

i-1,ji-1,j

(C)

(C) u

u

i,j-1i,j-1

(B)

(B) u

u

i+1,ji+1,j

1

1 –

–

1

0.5

0.5 (A)

(A)

u

i-1,ji-1,j

(B)

(B) u

u

i+1,ji+1,j

0.5

0.5 =

(15)

0.2 0.2 0.1 0.1 0 0

Thus, pressure of the closed pipe is given

Thus, pressure of the closed pipe is given as follows:as follows:

0, 0, 22 0.9 0.9 u u   1,2 1,2 0.397 0.397 u u   2,2 2,2 0.086 0.086 u u   3,2 3,2 0.086 0.086 u u    4,2 4,2 0.397 0.397 u u    5, 5, 22 0.9 0.9 u u    0,1 0,1 0.9 0.9 u u   1,1 1,1 0.589 0.589 u u   2,1 2,1 0.225 0.225 u u   3,1 3,1 0.225 0.225 u u    4,1 4,1 0.589 0.589 u u    5,1 5,1 0.9 0.9 u u    0, 0, 00 0.9 0.9 u u   1,0 1,0 0.729 0.729 u u   2,0 2,0 0.278 0.278 u u   3,0 3,0 0.278 0.278 u u    4,0 4,0 0.728 0.728 u u    5,0 5,0 0.9 0.9 u u    Q22 Q22 GivenGiven 22 22 22 22 22 22 2 2 22 22 22,, 00 2200 ,, 00 c c cc xx tt x x t t t t xx                       with

with cc E E     6 6 2 2 10 10 63 6366 .. 11 1133 __   c c , , 11 ,, ,, 11 66 11,, ,, 11,, 2 2 22 2 2 22 113 113.63.636 6 1010 i i jj ii jj ii jj ii jj ii jj ii jj k k hh   __     __ __    __    , , 11 ,, ,, 11 66 11,, ,, 11,, 2 2 22 2 2 22 113 113.63.636 6 1010 0 0..0022 55 i i jj ii jj ii jj ii jj ii jj ii jj   __     __ __    __    )) 2 2 (( 18 1822 .. 81 8188 ,, 1 1 2 2 _,, _,, ₁₁ ₁₁_,, _,, ₁₁_,, 1 1 ,, j j ii j j ii j j ii j j ii j j ii jj ii                    1 1 ,, ,, 1 1 ,, ,, 1 1 1 1 ,, j j 18181818..181822 ii j j 36343634..364364 ii j j 18181818..181822 ii j j  ii j j ii           --- --- (1)(1) D D C C B B A A__ __ __  18181818..181822 36343634..363644 18181818..181822

(D)

(D) ϕ

ϕ

i,j-1 i,j-1

1

1 1,818.182

1,818.182

--

3,634.364

1,818.182

ϕ

i,j+1 i,j+1

(A)

ϕ

i-1,j i-1,j

(B)

ϕ

i,j i,j

(C)

ϕ

i+1,j i+1,j

-1

=

(16)

x x t t x x      __₍₍ _,,₀₀₎₎ ,, 11 ,, 11 ,, 11 ,, 11 2 2 22((00..0022)) i i jj ii jj ii jj ii jj x x xx k k                 ,, 11 ,, 11 0.040.04 i i j j i i jj xx             --- --- (2)(2)

Substitute eqn. (2) into eqn. (1): Substitute eqn. (2) into eqn. (1):

)) 04 04 .. 0 0 (( 18 1822 .. 1818 1818 36 3644 .. 3634 3634 18 1822 .. 1818 1818 ₁₁_,, _,, ₁₁_,, _,, ₁₁ 1 1 ,, j j ii j j ii j j ii j j ii j j xx ii                    x x j j ii j j ii j j ii j j ii,, 11 909909..090911  11,, 18171817..181822  ,, 909909..091091  11,, 00..0202   x x C C B B A A 18171817..181822 909909..090911 00..0202 09 0911 .. 90 9099 __ __ __    

1

909.091

909.091 -1,817.182

-1,817.182

909.091

= =

0.02 0.02 x

x

(A)

ϕ

i-1,j i-1,j

(B)

ϕ

i,j i,j

(C)

ϕ

i+1,j i+1,j

ϕ

i,j+1 i,j+1

(17)

3,2 3,2



0.04 0.04 0.02 0.02 0,2 0,2



tt x x 0 0 0.1 0.1 0. 0. 2 2 0.3 0.3 00 0 0 5 5 10 10 1515 0 0 0 0 0 0 0 0 00 0,1 0,1



0,0 0,0



2, 2, 22



4, 4, 22



1, 1,11



2,12,1



_3,1_3,1



_4,1_4,1 1,0 1,0



2,02,0 3,0 3,0



4,04,0 1,2 1,2



0 0 0.2 0.2 0.4 0.4 0.6 0.6 00