The document discusses numerical methods for solving mathematical problems. It begins by defining numerical methods as algorithms used to obtain numerical solutions when an analytical solution does not exist or is difficult to obtain. It then provides the Navier-Stokes equations as an example of equations that require numerical methods to solve. The document concludes by discussing the importance of numerical accuracy and computation time for numerical methods.

1. Mathematical Modeling and Engineering
Problem Solving

2. What are NUMERICAL METHODS?
Why do we need them?

3. Numerical Methods:
Algorithms that are used to obtain numerical solutions of
a mathematical problem.
Why do we need them?
1. No analytical solution exists,
2. An analytical solution is difficult to obtain
or not practical.

4. Navier and Stokes Equations
For a viscous flow, the relationships between the normal/shear stresses and the rate
of deformation (velocity field variation) can be determined by making a simple
assumption. That is, the stresses are linearly related to the rate of deformation
(Newtonian fluid). The proportional constant for the relation is the dynamic
viscosity of the fluid (m). Based on this, Navier and Stokes derived the famous
Navier-Stokes equations:
2 2 2
2 2 2
2 2 2
2 2 2
2 2 2
2 2 2
x
y
z
x
y
z
u u u u P u u u
u v w g
t x y z x y z
v v v v P v v v
u v w g
t x y z x y z
w w w w P w w w
u v w g
t x y z x y z
  m
  m
  m
         
                    
         
                    
         
                   


5. 5
Engineering Simulations
Finite element analysis (FEA) and product design services
Computational Fluid Dynamics (CFD)
Molecular Dynamics
Particle Physics
Earthquake simulations
Development of new products and performance improvement of existing
products
Benefits of Simulations
Cost savings by minimizing material usage.
Increased speed to market through reduced product development time.
Optimized structural performance with thorough analysis
Eliminate expensive trial-and-error.

6. Basic Needs in the Numerical Methods:
– Practical:
Can be computed in a reasonable amount of time.
– Accurate:
• Good approximate to the true value,
• Information about the approximation error
(Bounds, error order,… ).

7. Every part in this book
requires some
mathematical background

8. Computers are great tools,
however, without fundamental understanding
of engineering problems, they will be useless.

9. 9
The engineering problem-solving process
Newton’s 2nd law
Conservation laws

10. 10
Newton’s 2nd law of Motion
• “The time rate change of momentum of a body is equal
to the resulting force acting on it.”
• Formulated as F = m.a
F = net force acting on the body
m = mass of the object (kg)
a = its acceleration (m/s2)
• Some complex models may require more sophisticated
mathematical techniques than simple algebra
– Example, modeling of a falling parachutist:
FU = Force due to air resistance = -cv (c = drag
coefficient)
FD = Force due to gravity = mg
UD FFF 

11. m
cvmg
dt
dv
cvF
mgF
FFF
m
F
dt
dv
U
D
UD






v
m
c
g
dt
dv

• This is a first order ordinary
differential equation. We would like to
solve for v (velocity).
• It can not be solved using algebraic
manipulation
• Analytical Solution:
If the parachutist is initially at rest
(v=0 at t=0), using calculus dv/dt can
be solved to give the result:
 tmc
e
c
gm
tv )/(
1)( 

Independent
variableDependent
variable
ParametersForcing
function

12. Analytical Solution
 tmc
e
c
gm
tv )/(
1)( 

t (sec.) V (m/s)
0 0
2 16.40
4 27.77
8 41.10
10 44.87
12 47.49
∞ 53.39
If v(t) could not be solved analytically, then
we need to use a numerical method to solve
it
g = 9.8 m/s2 c =12.5 kg/s
m = 68.1 kg

13. 13
)(
)()(
lim........
)()(
1
1
0
1
1
i
ii
ii
t
ii
ii
tv
m
c
g
tt
tvtv
t
v
dt
dv
tt
tvtv
t
v
dt
dv

















))](([)()( 11 iii
tttv
m
c
gtvtv ii  
This equation can be rearranged to yield
∆t = 2 sec
To minimize the error, use a smaller step size, ∆t
No problem, if you use a computer!
Numerical Solution
t (sec.) V (m/s)
0 0
2 19.60
4 32.00
8 44.82
10 47.97
12 49.96
∞ 53.39

14. t (sec.) V (m/s)
0 0
2 19.60
4 32.00
8 44.82
10 47.97
12 49.96
∞ 53.39
t (sec.) V (m/s)
0 0
2 16.40
4 27.77
8 41.10
10 44.87
12 47.49
∞ 53.39
m=68.1 kg c=12.5 kg/s
g=9.8 m/s
 tmc
e
c
gm
tv )/(
1)( 
 ttv
m
c
gtvtv iii
 )]([)()( 1
∆t = 2 sec
Analytical
t (sec.) V (m/s)
0 0
2 17.06
4 28.67
8 41.95
10 45.60
12 48.09
∞ 53.39
∆t = 0.5 sec
t (sec.) V (m/s)
0 0
2 16.41
4 27.83
8 41.13
10 44.90
12 47.51
∞ 53.39
∆t = 0.01 sec
CONCLUSION: If you want to minimize
the error, use a smaller step size, ∆t
Numerical solutionvs.

15. Comparison of numerical and analytical solutions
Larger step size  less accurate result
Smaller step size  more steps (longer computing time)

17. Derive the mathematical equations that describes the fluid system below

18. • The rate of change in liquid stored in the tank is equal to the flow in
minus flow out.
• The resistance R may be written as
• Rearranging equation (2)
oi qq
dt
dh
C 
(1)
0q
h
dQ
dH
R  (2)
R
h
q 0 (3)

19. • Substitute qo in equation (3)
• After simplifying above equation
oi qq
dt
dh
C  (1)
R
h
q 0 (3)
R
h
q
dt
dh
C i 
iRqh
dt
dh
RC 

21.    0M-M outin
SweatOutAirFecesSkinUrineMetabolismInAirDrinkFood 
MetabolismInAirFoodSweatOutAirFecesSkinUrineDrink 
L2.13.005.012.04.02.035.04.1Drink 

22. 3out,3out,2in,1 AvQQ 
2
33
out,3
out,2in,1
3 m333.3
m/s6
/sm20/sm40





v
QQ
A

25. Special Problem 1
Schematic of a cart-pendulum system
Derive the equations of motion for the two-degree of freedom system.
In this system…….
 It requires two coordinates, x and .
 It requires two equations of motion:
1. The linear motion of the system.
2. The rotation of the pendulum.