CHAPTER 8e. DIFFERENTIAL
EQUATIONS
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
by
Dr. Ibrahim A. Assakkaf
Spring 2001
ENCE 203 - Computation Methods in Civil Engineering II
Department of Civil and Environmental Engineering
University of Maryland, College Park
First-order Ordinary Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
!
Galerkin Method
Procedure for Deriving Galerkin
Approximating Polynomial
1. For a given ordinary differential equation,
assume that the solution is a differentiable
function such as given by the following
polynomial:
yˆ = b0 + b1 x + b2 x 2 + L + bn x n
(29)
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 157
1
First-order Ordinary Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
!
Galerkin Method
2. Use the boundary condition of the ordinary
differential equation to evaluate one of the
coefficients of the assumed function or to
provide a condition toward the solution of the
coefficients.
3. Provide an expression for the error e as
e=
dyˆ dy
−
dx dx
(30)
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 158
First-order Ordinary Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
!
Galerkin Method
4. Compute the weight wi for each coefficient,
where
wi =
dyˆ
dbi
(31)
5. The normal equations are computed from
∫ w e dx = 0
x
i
for i = 1,2, L, n
(32)
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 159
2
First-order Ordinary Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 15 – Galerkin Method
Using the Galerkin method with a linear
model ( i.e. ŷ = bo + b1x) and for 0 ≤ x ≥ 1 to
find an approximating function for the
following differential:
dy
= xy such that y = 1 at x = 0
dx
Using this function, estimate y at x = 0.6
and compare with the true value.
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 160
First-order Ordinary Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 15 (cont’d) – Galerkin Method
yˆ = b0 + b1 x
yˆ (0 ) = 1 = b0 + b1 (0) ⇒ b0 = 1
dyˆ dy
e=
−
dx dx
dyˆ
yˆ = 1 + b1 x
= b1
and
dx
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 161
3
First-order Ordinary Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 15 (cont’d) – Galerkin Method
dy
(given by the DE)
= xy
dx
dyˆ dy
e=
−
= b1 − xy = b1 − x(1 − b1 x )
dx dx
dyˆ
=x
w1 =
db1
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 162
First-order Ordinary Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 15 (cont’d) – Galerkin Method
1
1
∫ w e dx = ∫ x[b − x(1 − b x )] dx = 0
1
1
0
1
0
1
∫ [xb − x
1
2
]
+ b1 x 3 dx = 0
0
1
x 2b1 x 3 x 4b1
b 1 b
− +
= 1 − + 1 =0
2
3
4 0 2 3 4
3b1 1
4
= ⇒ b1 =
4 3
9
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ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 163
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First-order Ordinary Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 15 (cont’d) – Galerkin Method
Therefore the approximating function (linear
polynomial) is given by
yˆ = b0 + b1 x = 1 +
4
x
9
4
and yˆ (0.6) = 1 + 9 (0.6) = 1.2667
The true value is given by
y (0.6 ) = e x
2
error (%) =
/2
= e ( 0.6)
2
/2
= 1.1972
1.1972 - 1.2667
×100 = 5.8%
1.1972
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 164
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Many engineering problems require the
solution of higher-order differential
equations.
! For example a second-order differential
equation can be given as
d2y
dy 

= f  x, y , 
2
dx 
dx

(33)
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ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 165
5
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! The function f(.) does not need to
include all the parameters x, y, and
dy/dx.
! Examples of higher-order differential
equations are
dy
d2y
= 2 x such that y = 0 and
= 0 at x = 0
2
dx
dx
d3y
− 2 x − yx + 4 = 0
dx 3
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 166
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! The methods previously introduced can
be used to solve higher-order
differential equations after transforming
them into systems of first-order
differential equations.
! The procedure for transforming the DE’s
for the following example is as follows:
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ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 167
6
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
dy 
d2y

= f  x, y , 
2
dx 
dx

becomes
dy2
= f (x, y, y2 )
dx
dy1
where y1 = y2
= y2
dx
dy dy
y2 = 1 =
dx dx
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 168
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Higher-order Differential Equations
dy1
= f1 (x, y1 , y2 , L, yn )
dx
dy2
= f1 (x, y1 , y2 , L , yn )
dx
M
(34)
dyn
= f n (x, y1 , y2 , L , yn )
dx
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 169
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Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example – Simply Supported Beam
w = 2 k/ft
x
10 ft
d 2 y M (x )
=
dx 2
EI
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 170
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example (cont’d) – Simply Supported
Beam
M
w = 2 k/ft
x
10 kips
V
x
∑ M = −M + 10 x − 2 x 2  = 0
− M + 10 x − x 2 = 0
M = 10 x − x 2
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ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 171
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Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example – Simply Supported Beam
Let
dy
= rotation
dx
d 2 y dθ M (x )
=
=
dx 2 dx
EI
θ=
Hence
dθ 10 x − x 2
=
dx
EI
dy
=θ
dx
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 172
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example (cont’d) – Simply Supported Beam
Assume that EI = 3600 kip/ft2 and at x = 0,
y = 0 and θ = - 0.02314, and h = 0.1.
Using basic Euler’s method for example,
the following equations result:
θ i +1 = θ i + fθ (xi , yi ,θ i ) h
yi +1 = yi + f y (xi , yi ,θ i ) h
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 173
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Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example (cont’d) – Simply Supported Beam
First Iteration (i = 0)
x0 = 0, θ 0 = −0.02314, and y = 0
fθ (xi ,θ i , yi ) = fθ (x0 ,θ 0 i , y0 ) =
dθ
dx
x = x0
θ =θ 0
y = y0
=
10 x − x 2
EI
10(0) − (0 )
=0
3600
θ1 = θ 0 + fθ (0,−0.0231481,0) h
fθ (0,−0.0231481,0 ) =
2
θ1 = −0.0231481 + 0 = −0.0231481
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 174
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example (cont’d) – Simply Supported
Beam
dy
= θ0
dx
f y (0,−0.0231481,0 ) = θ 0 = −0.0231481
f y (xi ,θ i , yi ) = f y (x0 ,θ 0 , y0 ) h =
y1 = y0 + f y (x0 ,θ 0 , y0 ) h
= y0 + f y (0,−0.0231481,0 )
y1 = 0 − 0.0231481(0.1) = −0.00231481
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 175
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Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example (cont’d) – Simply Supported Beam
Second Iteration (i = 1)
x1 = 1.1, θ1 = −0.02314, and y = −0.00231481
fθ (xi ,θ i , yi ) = fθ (x1 ,θ1i , y1 ) =
dθ
dx
x = x1
θ =θ1
y = y1
=
10 x − x 2
EI
10(0.1) − (0.1)
= 0.00025
fθ (0.1,−0.0231481,−0.00231481) =
3600
θ 2 = θ1 + fθ (0.1,−0.0231481,−0.00231481)h
2
θ 2 = −0.0231481 + 0.000275(0.1) = −0.0231206
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 176
Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example (cont’d) – Simply Supported
Beam
dy
= θ1
dx
f y (0.1,−0.0231206,0 ) = θ1 = −0.0231206
f y (xi ,θ i , yi ) = f y (x1 ,θ1 , y1 ) h =
y2 = y1 + f y (x1 ,θ1 , y1 ) h
= y1 + f y (0,−0.0231206,0 )
y2 = −0.00231481 − 0.0231206(0.1) = −0.0046269
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 177
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Higher-order Differential
Equations
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example (cont’d) – Simply Supported Beam
The exact solution is
given by

x3  1
− 0.0231481
θ =  5 x 3 − 
3  3600

 5x3 x4  1
y = 
− 
− 0.0231481 x
 3 12  3600
x (ft)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
2
3
4
5
6
7
8
9
10
dθ
dx
0
0.0002750
0.0005444
0.0008083
0.0010667
0.0013194
0.0015667
0.0018083
0.0020444
0.0022750
0.0025000
0.0044444
0.0058333
0.0066667
0.0069444
0.0066667
0.0058333
0.0044444
0.0025000
0.0000000
dy
dx
-0.0231481
-0.0231481
-0.0231206
-0.0230662
-0.0229853
-0.0228787
-0.0227467
-0.0225900
-0.0224092
-0.0222048
-0.0219773
-0.0185564
-0.0134412
-0.0071870
-0.0003495
0.0065158
0.0128533
0.0181075
0.0217227
0.0231436
θ=
y (ft)
0
-0.0023148
-0.0046296
-0.0069417
-0.0092483
-0.0115468
-0.0138347
-0.0161094
-0.0183684
-0.0206093
-0.0228298
-0.0434304
-0.0598018
-0.0704996
-0.0746349
-0.0718744
-0.0624403
-0.0471104
-0.0272179
-0.0046518
Exact θ
-0.0231481
-0.0231343
-0.0230933
-0.0230256
-0.0229318
-0.0228125
-0.0226681
-0.0224993
-0.0223066
-0.0220906
-0.0218518
-0.0183333
-0.0131481
-0.0068518
0.0000000
0.0068519
0.0131482
0.0183334
0.0218519
0.0231482
Exact y (ft)
0
-0.0023143
-0.0046260
-0.0069321
-0.0092302
-0.0115176
-0.0137919
-0.0160504
-0.0182909
-0.0205110
-0.0227083
-0.0429629
-0.0588193
-0.0688887
-0.0723377
-0.0688886
-0.0588191
-0.0429626
-0.0227079
0.0000000
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 178
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! The previous methods for solving
ordinary differential equations can be
used to solve higher-order differential
equations as discussed earlier.
! These methods require the initial
conditions to be at the same x value.
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 179
12
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! For example, the initial conditions in
Example 15 were given as
x0 = 0
y0 = 0
θ 0 = −0.0231481
! The rotationθ is not usually known.
! However, in Ex. 15 it is known that at x
= 0, y = 0 and at x = 10, y = 0
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 180
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! The two conditions have different x
values. Therefore, they cannot be used
in solving the differential equation based
on the previous methods.
! In this case, we are dealing with a
boundary-value problem.
! Boundary-value problems can be solved
numerically using Shooting and Finitedifference methods.
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 181
13
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Shooting Method
– The shooting method is a trial-and-error
method that uses any of the previously
introduced methods for solving differential
equations.
– The shooting method is based on
converting the boundary-value problem
into an equivalent initial-value problem.
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 182
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Finite-Difference Method
Recall the finite-difference expressions for
approximating the first and second
derivatives based on two-step method
f (xi +1 ) − f (xi )
2h
f (xi −1 ) − 2(xi ) + f (xi +1 )
f ′′(xi ) ≈
h2
f ′(xi ) ≈
(35)
(36)
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 183
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Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Finite-Difference Method
– In this method, the derivatives in the
differential equation are replaced by finitedifference equations (formulas).
– Then, the resulting DE, which is now in a
finite-difference form, is used at some
interior points using a selected step size h
and the boundary conditions.
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 184
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Finite-Difference Method
– Each use of the finite-difference equation
results in a linear equation in terms of the
unknown solutions at the selected interior
points.
– In this case, we get a system of linear
equations that need to be solved
simultaneously to obtain the solution at the
interior points.
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 185
15
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 – Simply Supported Beam
The simply supported beam is subjected to
distributed loading w as shown in the
figure. Numerically, find the bending
moment M at each node along its span.
w = 2 k/ft
x
10 ft
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 186
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 – Simply Supported Beam
B.C.' s :
at x = 0, M = 0
and at x = 10, M = 0
M2
M1
2.5
M3
2.5
M4
2.5
M5
2.5
10 ft
x
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 187
16
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 (cont’d) – Simply
Supported Beam
The governing differential equation in this
case is given by
d 2M
= w = distributed load
dx 2
or
d 2M
= -2
dx 2
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 188
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 (cont’d) – Simply
f (x
f ′′(x ) ≈
Supported Beam
i
i −1
) − 2(xi ) + f (xi +1 )
h2
(Eq. 36)
The original differential equation can be
converted to a finite-difference equation
(see Eq. 36) as follows:
d 2 M M i −1 − 2 M i + M i +1
≈
dx 2
h2
M i −1 − 2 M i + M i +1
= −2
h2
(37)
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 189
17
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 (cont’d) – Simply
Supported Beam
h, is case, is equal to 2.5 ft, therefore Eq. 37
can be simplified as follows.
M i −1 − 2M i + M i +1
= −2
(2.5)2
or
M i −1 − 2M i + M i +1 = −2(6.25)
M i −1 − 2M i + M i +1 = −12.5
(38)
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 190
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 (cont’d) – Simply
Supported Beam
Application of Eq. 38 at the interior nodes
2, 3, and 4, yields a set of three
simultaneous equations as follows:
M
M
M
M
M
At Node 2:
2
1
2.5
0
3
2.5
4
2.5
5
2.5
10 ft
x
M 1 − 2M 2 + M 3 = −12.5
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 191
18
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 (cont’d) – Simply
Supported Beam
M
M
M
2
1
At Node 3:
2.5
M4
3
2.5
2.5
M5
2.5
10 ft
x
M 2 − 2M 3 + M 4 = −12.5
M2
M1
At Node 4
2.5
M3
2.5
M4
M5
2.5
2.5
10 ft
0
x
M 3 − 2M 4 + M 5 = −12.5
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 192
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 (cont’d) – Simply
Supported Beam
− 2M 2 + M 3
= −12.5
M 2 − 2M 3 + M 4 = −12.5
M 3 − 2M 4 = −12.5
0   M 2  − 12.5
− 2 1
 1 − 2 1   M  = − 12.5

  3 

 0
1 − 2  M 4  − 12.5
(39)
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 193
19
Boundary-Value Problems
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
! Example 16 (cont’d) – Simply
Supported Beam
Equation 39 yields the solution as follows:
M2
M1
2.5
M3
2.5
M4
2.5
10 ft
x
M5
2.5
 M 2  18.75 
 M  = 25.00 ft - kips

 3 
 M 4  18.75 
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 194
• A. J. Clark School of Engineering • Department of Civil and Environmental Engineering
☺
I wish you good luck
in your final ☺
Exams
© Assakkaf
ENCE 203 – CHAPTER 8e. DIFFERENTIAL EQUATIONS
Slide No. 195
20