ch4
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Sem II 14/15
BEE 31602
CHAPTER 4 – ORDINARY ORDINARY AND PARTIAL DIFFERENTIAL EQUATION
ORDINARY DIFFERENTIAL EQUATION (ODE)
4.1
INITIAL-VALUE PROBLEM (IVP)
Q1
Consider the following initial-value problem (IVP)
dy dx Solve the IVP for
Q2
3 x2 y, y(0) 1.
0 x 0.6 and h 0.2 by
using Euler’s method and method and RK4 method.
Consider the following initial-value problem (IVP)
(1 x2 ) Solve the IVP for
Q3
dy dx
xy 0, y(2) 5.
2 x 2.3 and h 0.1 by
using Euler’s Euler’s method and method and RK4 method.
Solve the following ordinary differential equation dy dx
with uniform step size
h
y
0.1 over
x ,
2
y (0)
1
interval [0, 0.3] by using Euler’s method and RK4
method.
Q4
Given an initial-value problem (IVP) as follows
dy dx
1.2 y
7e 0.3 x , y(0) 3.
Solve the IVP by using Euler’s method at
x 0 (0.5) 2.
Sem II 14/15
BEE 31602
Q5
The concentration of a chemical in a batch reactor can be modeled by the following differential equation dC
k1C
1 k2C
dt
,
(0) 0.8. C (0
Find a numerical solution for this problem at h
Q6
0.5 by
with
k1
1, k 2
0.3 and step length,
using Euler’s method and classical and classical fourth-order Runge-kutta method.
Velocity of a falling object can be modeled mod eled as the following initial-value problem (IVP)
dv dt where
v
was at t
cd m
v2
g
, v(0) 20
velocity of the falling object (m/s),
gravity (9.81 m/s2),
Q7
t 1 s
v 20 m/s
m
with
mass (kg) and
cd
0.225 .
cd
t time
(s),
g
acceleration due to
drag coefficient (kg/m). Initially, the object
Calculate the velocity for a falling 5-kg 5-k g object at
method. 0 ( 0.1) 0.1) 0.5 s by using Euler’s method.
A voltage source, E (t ) is supplied to an electrical circuit with inductance L and a resistance R . If the switch is closed at t 0 s, the current I (t ) will satisfy the following initial-value problem (IVP) L
d dt
I (t ) RI (t ) E (t ) ,
I (0) 0.
Parameter values are given as L 50 H, R 20 and E (t ) 10 V. Estimate the value of the current at t 0 (1) 5 s by using Euler’s method.
Sem II 14/15
BEE 31602
4.2
BOUNDARY-VALUE PROBLEM (BVP)
Q8
Given the boundary-value problem (BVP) x 4 x sin t , 0 t 1
with conditions x(0) 0 and by taking t h 0.25 .
Q9
(1)
x
0 . Solve the BVP by using finite difference method
Given the boundary-value problem (BVP) (BVP)
d2y dx 2 with conditions y (0) by taking x 0.5.
Q10
dy dx
x
0,
0 x 2
0 and y (2) 1. Solve the BVP by using finite-difference method
Solve the boundary-value problem (BVP), y xy 3 y 11x with conditions and y (1) 2 where h 0.25 by using finite-difference method.
Q11
4
y(0)
1
The boundary-value problem (BVP) for the steady-state temperature in a rod of length 2 m is represented as follows 2
d T dx
2
0.1T
0,
T (0)
20 200 0 C,
T ( 2)
Approximate the temperature, T throughout the rod for difference method.
100 0 C .
x h 0.5 by
using finite-
Sem II 14/15
BEE 31602
Q12
A heated rod with a uniform heat source can be modeled with the Poisson equation,
d 2T
dx 2
f ( x) .
Given the heat source, f ( x) 25 and the boundary conditions, T ( x 0) 40 40 and T ( x 10) 200 . Solve for the temperature distribution with h x 2.5 by using finitedifference method.
Q13
The position of a falling object is governed by the following boundary-value problem (BVP) 2
d x dt 2
c dx
g
m dt
0
,
for 0 t 12 ,
where boundary conditions are x ( 0 ) 0 and x (12) (12) 500. Given that the parameter values are c a first-order drag coefficient (12.5 kg/s), m mass of the falling object
(50 kg) and g gravitational acceleration ( 9.81 m/s2 ). Approximate the position of the falling object, x (m) for h 3 by using finite-difference method.
Q14
A thin rod of length, l is moving in the xy-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 problem (BVP) as follows (t )
g l
(t ) 0 ,
for
0 t 0.4 ,
where boundary conditions are (0) 0 and (0.4 (0.4)) 1 . The parameter values gravitation tional al force force (9.81 m/s2 ) and l 0.9 m. Approximate the angle are given as g gravita (in radian) for h 0.1 by using finite-difference method.
Sem II 14/15
BEE 31602
PARTIAL DIFFERENTIAL EQUATION (PDE)
4.3
Heat equation (explicit finite-difference method)
Q15
Given the heat equation u t
2
0.9
u x
2
, 0 x 1,
with the boundary conditions, u(0, t )
t
u (1, t ) 1 for
( ,0) e (1 ) for 0 x 1 . Find u( x, 0.01 0.01)) and difference method with x h 0.2. u x
Q16
x
0
x
t 0 ,
and the initial condition,
( , 0.02 0.02)) by using explicit finite-
u x
Given the heat equation u( x, t )
2
t
2
u( x, t ) x
2
, 0 x 2,
t
0,
with the boundary conditions u (0, t )
u (2, t )
0,
and the initial condition ( ,0)
u x
sin( x).
Find u ( x, 0.3 0.3) by using explicit finite-difference method with
x 0.5 and t 0.3.
Sem II 14/15
BEE 31602
Q17
Consider the heat conduction equation t
where
T ( x, t )
is thermal diffusity
2
x
2
T ( x, t ),
10, since
0 x 10,
c
2
t 0 ,
.
Given the boundary conditions, T (0, t )
0,
T (10, t )
100
and initial condition, T ( x,0)
x
2
.
By using explicit finite-difference method, find intervals on the x coordinate.
Q18
T ( x, 0.05 0.055) 5) and
0.11)) with 5 grid T ( x, 0.11
The temperature distribution u( x, t ) of one dimensional silver rod is governed by the heat equation u
2
t
with
2
2
u x
2
is thermal diffusity diffusity =1.71. =1.71.
Given the initial condition, 0 x 2, x, 4 x, 2 x 4.
( ,0)
u x
and boundary conditions, u (0 , t )
t,
u (4, t )
t
2
.
Find the temperature distribution of the rod with 0 t 0.4 by using explicit finite-difference method.
x h 1
and
t k 0.2
for
Sem II 14/15
BEE 31602
4.4
Wave equation (finite-difference method)
Q19
Let u ( x, t ) be the displacement of uniform wire which is fixed at both ends along at time t . The distribution distribution of u( x, t ) is given by the wave equation 2
u t
2
( , 0)
sin x ,
u t
4
x
2
,
( x, 0) 0 for
0 x 1 , 0 t 0.5
0
u (1, t ) x
by using finite-difference method with
Q20
-axis
2
u
with the boundary conditions u(0, t ) u x
x
1.
0 and the initial conditions
Solve the wave equation up to level
t 0.2
x h 0.25 and t k 0.1
Let y ( x, t ) denotes displacement of a vibrating string. If T is the tension of the string, is the weight per unit length and g is acceleration due to gravity, then y satisfies the equation
2 y Tg 2 y t 2 x 2
,
0 x 2 , t 0 .
Suppose a particular string with 2 m long is fixed at both ends. By taking T 1.5 N, 2 0.01 kg/m and g 10 m/s , use finite-difference method to solve for y up to second level.
The initial conditions are
0.5 x , 0 x 1 1 0 . 5 , 1 2 x x
y ( x , 0 )
Perform all calculations with
h x 0.5 m
and
and
y t
( x , 0 ) x 2 2 x .
k t 0.01 s.
Sem II 14/15
BEE 31602
Q21
The air pressure
u ( x, t )
in an organ pipe is governed by the wave equation 2
u t
2
2
1 u
2
x
2
0 x l ,
,
t 0,
where l is the length of the pipe and is a physical constant. If the pipe is closed at the end where x l , the boundary conditions are
u (0, t ) 0.9
Assume that
and
u(l, t) 0.9
for
0 t 0. 2.
1, l 0.5 and the initial conditions are
u ( x, 0)
0.9
u
cos (2 x) an and
t
( x, 0) 0
for
0
x
0.5.
Approximate the pressure for the closed-pipe by using finite-difference method with h x 0.1 and k t 0.1.
Q22
The longitudinal vibration of a bar with the length of l m is governed by c
2
2
x
with
c
E
, where
2
2
2
t
( x, t ) is the axial displacement,
E
is Young’s modulus and
is
the mass density of the bar. The boundary conditions and the initial conditions are given as follows,
(0, t ) (l , t ) 0 for ( x,0) 0 and
( x,0) t
0 t 0.04
x
for 0 x 20.
Determine the variation of the axial displacement of the bar by using finite-difference method with the following data: 6 E 30 10 , 0.264 , l 20 m, x h 5 and
t k 0.02.
BEE 31602
PARTIAL DIFFERENTIAL EQUATION (PDE)
4.3
Heat equation (explicit finite-difference method)
Q15
Refer to class note
Q16
Sem II 14/15
Sem II 14/15
BEE 31602
Q17 t 0.11
0.055
T 0,2 0,2
T 1,2 1,2
T 2,2 2,2
T 3,2 3,2
T 4,2 4,2
T 5,2 5,2
T 0,1 0,1
T 1,1 1,1
T 2,1 2,1
T 3,1 3,1
T 4,1 4,1
T 5,1 5,1
T 2,0 2,0
T 3,0 3,0
T 4,0 4,0
T 5,0 5,0
T 0,0 0,0 T 1,0 1,0
2
0 t
T ( x, t )
T t
Ti , j 1 Ti, j k Ti , j 1 Ti, j 0.055
Ti , j 1 Ti, j
x 4
10
10
T ( x, t ),
2
0 x 10,
T
x
2
Ti1, j 2Ti, j Ti1, j h2 Ti1, j 2Ti, j Ti1, j 2
0.138
T
i 1, j
2
2Ti, j Ti1, j
0.138Ti
1
0.138Ti
0.138Ti
1,
0.138 A 0.724B 0.138C
Ti , j
1
t 0
2
Ti, j
10
8
2
x
10
6
0.276Ti, j
0.138T i 1, j
1, j
0.276Ti , j
0.138Ti
j
0.724Ti, j
0.138Ti
1, j
1, j
1,
j
Ti , j
1 = T i,j+1 i,j+1
0.138 (A)
0.724 T i-1,j i-1,j
(B)
0.138 T i,j i,j
(C)
T i+1,j i+1,j
Sem II 14/15
BEE 31602
t
T 0,2
T 1,2
T 2,2
T 3,2
6.056
18.208
38.208
66.056
100
T 2,1
T 3,1 3,1
T 4,1
T 5,1 5,1
5.104
17.104
37.104
65.104
100
T 1,0
T 2,0
T 3,0
T 4,2
T 5,2
0.11 0
T 1,1 1,1
T 0,1
0.055 0 T 0,0
0
2 0
Q18
4 4
6 16
T 4,0
8 36
T 5,0
10 64
100
x
Sem II 14/15
BEE 31602
4.4
Wave equation (finite-difference method)
Q19
Refer to class note
Q20
2
2
y
2
Tg y
t 2
1500
2
x 1
2 yi , j (k )
1
T
1.5 ,
0.01 and g
10 .
1
2
yi , j
1
2
2 yi , j
(0.01) yi , j
0 x 2 , t 0 where
,
y
t
y i , j
2
2
y
y i , j
x
yi , j
1
2
2 yi , j
1500
1500
yi
yi
1
0.6( yi
1
0.6 yi
yi , j
yi , j
1, j
yi
2 yi , j
yi
1, j
1, j
2
1, j
2 yi , j
yi
1, j
2 yi , j
yi
1, j
2
(0.5)
0.6( yi
2 yi , j ( h)
1
yi , j
1, j
0.8 yi , j
1, j
)
1, j
) yi , j
0.6 yi
1, j
1
2 yi , j
yi , j
1
------ (1)
1 = yi,j+1
0.8
0.6 (A)
(B)
yi-1,j
0.6 yi,j
-1 (D) yi,j-1
yi , j
1
yi , j
1
2k yi , j
1
yi , j
x( x 2)
1
0.02 x( x 2)
1
yi , j
yi , j
1
0.02 x( x 2)
------ (2)
(C) yi+1,j
Sem II 14/15
BEE 31602
Substitute eqn. (2) into eqn. (1): yi , j
1
0.6 yi
0.6 yi
1
0.6 yi
1
0.3 yi
2 yi , j
yi , j
1, j
1, j
0.8 yi , j
0.8 yi , j
0.6 yi
0.6 yi
1, j
1, j
0.8 yi , j
0.6 yi
1, j
0.4 yi , j
0.3 yi
1, j
yi , j
( yi , j
1
1
0.02x (x 2) 2))
1, j
0.02 x( x 2)
1, j
0.01 x( x 2)
1 = yi,j+1
0.3 (A)
0.4 yi-1,j
(B)
0.01 x ( x -2) -2)
0.3 (C) yi+1,j
yi,j
t
y1,2
y2,2
0.148
0.064
y1,1 1,1
0 y0,0
y0,2
y3,2
y4,2
0.148
0
y2,1
y3,1
y4,1
0.243
0.340
0.243
y1,0
y2,0
y3,0
0.02 0 y0,1
0.01
0
0.5 0
1.0 0.25
1.5 0.5
0
y4,0 2.0
0.25 0.2 5
0
x
Sem II 14/15
BEE 31602
2
u
Q21
t
ui , j
1
2
2ui , j
(0.1) 1
2ui , j
1
ui
ui , j
ui , j
2
1
u
2
x
ui, j
2
1
ui, j
1, j
ui
, where where 1 ui
2
2
ui
1, j
u i, j
1
(0.1)
2
2u i, j
1, j
t
2u i, j u i
1, j
u 2
2
u x
2
1, j
ui
1, j
formula) A B C (Calculator
1
Representation in molecule graph (calculating level 2):
1 = u ,j+1
1
1
(A)
(B) ui+1,j
ui-1,j – 1 1
(C) ui,j-1 Given
ut ( x, 0) 0) 0
ui , j
1
ui , j
1
2(0.1) ui , j
1
0 ui , j
Substitute (1) into ui , j ui , j
2ui , j
ui , j
1
1
1
ui
1, j
ui
1, j
ui
1, j
1, j
0.5ui
ui
------------------ --- (1) (1)
1
ui
1
ui, j
1, j
ui
1, j
ui, j 1 :
1
1, j
0.5ui
1, j
0.5 A 0. 0.5B (Calculator formula)
Representation in molecule graph (calculating level 1):
1 = ui,j+1
0.5 (A)
0.5 ui-1,j
(B) ui+1,j
Sem II 14/15
BEE 31602
Thus, pressure of the closed pipe is given as follows:
u
0.2
u
0.1 0
Q22
0.9
0.397
u
1,1
0.9
0.589
u
1,0
0.9
0.729
Given
c
2
x
2
u
u
u
0.589
u
u
u
0.278
t
2
t
2
c
2
2
x
2
1
2i , j
i , j
1
k2 i , j
1
2i , j
i , j
1
2
0.02
i , j
i , j
i , j
113.63 113.636 6 10
113.63 113.636 6 10
1
2 i , j
1
1818 .182 182 i
6
1
6
1,818 818.182 182( i
1, j
i
1, j
2i , j
0.9
0 x 20 , t 0 with
,
c
i , j
0.9
5,0
2
4,0
0.728
0.9
5,1
2
4,1
3,0
0.278
2
0.225
2,0
5, 2
3,1
0.225
u
4,2
0.397
0.086
2,1
u
3,2
0.086
u
u
2,2
u
0, 0
u
1,2
0,1
u
u
0, 2
i 1, 1, j
2
113 .636 636 10 113
c
E
6
h2 i
1, j
2i , j
i
1, 1, j
2
5 1, j
3634.364 364 i , j
2 i , j
i
1, j
1818.182 182 i
) 1, j
i , j
1
------ (1)
182 A 3634 .364 364 B 1818.182 182C D 1818.182
1 ϕ i,j+1 =
1,818.182 (A)
-3,634.364 ϕ i-1,j
(B)
ϕ i,j
-1 (D) ϕ i,j-1
1,818.182 (C)
ϕ i+1,j
Sem II 14/15
BEE 31602
( x,0)
x
t
i , j
1
i , j
1
2k
i , j
1
x
i , j
i , j
1
i , j 1
2(0.02)
1
0.04 x
x
------ (2)
Substitute eqn. (2) into eqn. (1): 1
1818.182 182 i
1
909 909.091 091 i
i , j
i , j
1, j
1, j
3634 .364 364 i , j
1817.182 182 i , j
1818.182 182 i
909 909.091 091 i
1, j
1, j
( i , j
1
0.04x)
0.02 x
909 909 .091 091 A 1817 .182 182 B 909 909.091 091 C 0.02 x
1 =
909.091 (A)
ϕ i-1,j
ϕ i,j+1
-1,817.182 (B)
ϕ i,j
0.02 x
909.091 (C)
ϕ i+1,j
Sem II 14/15
BEE 31602
t
0,2
1,2
2, 2
3,2
4, 2
0.04 0
0.2
0.4
0.6
0
0,1
1,1 1,1
2,1
3,1
4,1
0
0.1
0. 2
0.3
0
2,0
3,0
4,0
0.02
0,0 0
1,0 5
0
10 0
0
15 0
0
x
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