Download Numerical Methods for Generalized Zakharov System

MA3264: Mathematical Modeling
Weizhu Bao
Department of Mathematics
& Center for Computational Science and Engineering
National University of Singapore
Email: [email protected]
URL: http://www.math.nus.edu.sg/~bao
Chapter 1
Introduction
Mathematical modeling
– Aims:
• Convert real-world problems into mathematical equations through
proper assumptions and physical laws
• Apply mathematics to solve real-life problems
• Provide new problems for mathematicians
– History:
• Started by the Egyptians and other ancient civilizations
• Fairly recent named as mathematical modeling & a branch of applied
and computational mathematics – modeling, analysis & simulation
• Rapid development in 20th centuries, especially after the computer
• Mathematical modeling contest (MCM) – undergraduates & high school
Chapter 1
Introduction
– Wide applications in applied sciences
• In physics --- Newton’s laws of motion, quantum physics, particle physics,
nuclear physics, plasma physics, ……..
• In chemistry --- chemical reaction, mixing problems, first principle
calculation, …….
• In engineering --- mechanical engineering (fluid flow, aircraft, Boeing 777,
…), electrical engineering (semiconductor, power transport, …), civil
engineering (building safety, dam analysis), ……
• In materials sciences – fluid-structure interaction, new materials, quantum
dots, …….
• In biology --- cell motion, cell population, plant population, ……
• In social sciences --- population model, traffic flow, president election poll,
casino, gambling, ……
• ……..
Dynamics of soliton in quantum physics
Wave interaction in plasma physics
Wave interaction in particle physics
Vortex-pair dynamics in superfluidity
Vortex-dipole dynamics in superfluidity
Vortex lattice dynamics in superfluidity
Vortex lattice dynamics in BEC
A simple model—A saving certificate
The problem: Suppose you deposit S$10,000 into DBS bank as a fixed
deposit. If the interest is accumulated monthly at 1% and paid at the end of
each month, how much money is in the account after 10 years?
Solution:
– Let S(n) be the amount in the account after nth month
– Mathematical relation
S (n  1)  S (n )  0.01S (n )  1.01S ( n),
n  0,1, 2,
– The result
S (n)  1.01S (n  1)  1.012 S (n  2)  1.01n S (0), n  0,1,2,
S (120)  1.01120 S (0)  3.3004 *10,000  33,004
A simple model
Related question: If the interest is accumulated yearly at 12% and paid at
the end of each year, how much money is in the account after 10 years?
The solution:
– Let S(n) be the amount in the account after nth year
– Mathematical relation
S (n  1)  S (n )  0.12 S (n )  1.12 S (n ),
– The result
n  0,1, 2,
S (n)  1.12S (n  1)  1.122 S (n  2)  1.12n S (0), n  0,1,2,
S (10)  1.1210 S (0)  3.1058*10,000  31,058  33,004
Exercise question: If the year interest rate is at 12% and the interest is
accumulated daily or instantly, how much money is in the account after 10 years,
respectively????
33,195 ( 33,201 )
Another example – Mortgaging a home
The problem: Suppose you want to buy a condo at $800,000 and you can
pay a down payment at $160,000. You find a mortgage with a monthly interest
rate at 0.3%. If you want to pay in 30 years, what is your monthly payment? If
you can pay $4,000 a month, how long do you need to pay?
The solution:
– Let S(n) be the amount due in the mortgage after nth month
– Mathematic relation for the first part::
S (n  1)  S (n )  0.003S (n )  x  1.003S (n )  x,
S (0)  640,000,
n  0,1, 2,
S (360)  0  x  ???
 U ( n)  S ( n)   x
U (n  1)  1.003U (n )  S (n  1)  1.003S (n )  0.003 x    1000 / 3
Another example – Mortgaging a home
U ( n )  1.003U ( n  1) 
 1.003n U (0), n  0,1, 2,...
– The result for the first part: U (360)  0  1000 x / 3  1.003360 (640,000  1000 x / 3)
– Payment information
640,000 
1000
1
[1 
]x
360
3
1.003

x  2909.7($)
• Total payment = 2909.7*360=1,047,500
Months n
0
1
2
3
4
Amount owned 640,000 639010.3 638020 637020 636020
Premium paid
0
989.7
1982.4
2978
3976.7
Interest paid
0
1920
3837
5751.1
7662.1
60
120
180
240
300
5
12
635020 627930 575040 497300 404250 292870 159570
360
0
4978.3
12074
64956
142700 235750 347120 480430 639980
9570.2
22842
109630 206460 287990 351200 392480 407510
Another example – Mortgaging a home
– Mathematical relation for the second part:
S (n  1)  S (n )  0.003S (n )  4000  1.003S (n )  4000,
S (0)  640,000, S ( n )  0  n  ???
n  0,1, 2,
U ( n)  S ( n)  1000  4000 / 3
– The result for the second part:
U ( n )  1.003U ( n  1) 
 1.003n U (0), n  0,1, 2,...
U ( n )  0  1000  4000 / 3  1.003n (640,000  1000  4000 / 3)
4000 / 3
1.003 
4000 / 3  640
n
 n  218.3 (months)
Another example--Optimization of Profit
Economics problems:
– Marco-economics: economic policy
– Micro-economics: profit of a company
An example: optimization of profit
Consider an idealized company:
– Object of the management: to produce the best possible dividend for
the shareholders.
– Assumption: The bigger the capital invested in the company, the bigger
will be the profit (the net income)
Optimization of profit
Two strategies to spend the profit:
– Short term management: The total profit is paid out as a dividend to
the shareholders in each year. The company does not grow and
shareholders get the same profit in each year.
– Long term management: The total profit is divided into two parts.
One part is paid out as a dividend to the shareholders and the other part is
to re-invest annually in the company so that the subsequent profits in
future years will increase.
Question: What part of the profit must be paid out annually as a dividend
so that the total yield for the shareholders over a given period of years is a
maximum ???
Optimization of profit
Variables:
t: time
– u(t): the total capital invested in the company in time t
– w(t): total dividend in the period [0,t] to the shareholders
Parameters:
– k: constant fraction of the profit which will be re-invest ( 0  k  1 )
– a: profit rate ( profit per time per capital investment)
Assumptions
– The capital and profit are continuous and the process of re-investment
and dividends is also continuous. (Normally the capital and profit will be
calculated at the end of the financial year of the company!!!)
– The profit is directly proportional to the capital invested.
Optimization of profit
Balance equation:
 rate of change   production rate  loss rate 
 of quantity    of quantity   of quantity 

 
 

Consider the time interval [t , t  t ]
– Profit: a u (t ) t
u (t  t )  u (t )  k a u (t ) t
– Change of the investment:
– Change rate: u(t  t )  u (t )  k a u (t )
t
– Rate of change: du (t )  lim u (t  t )  u (t )  k a u (t )
dt
t 0
t
Optimization of profit
– Dividend paid to shareholders:
(1  k )a u (t ) t
– Change of the total dividend: w(t  t )  w(t )  (1  k ) a u(t ) t
w(t  t )  w(t )
– Change rate:
 (1  k ) a u (t )
t
– Rate of change: dw(t )  lim w(t  t )  w(t )  (1  k ) a u(t )
Mathematical model:
dt
t 0
du (t )
 k a u (t ),
u (0)   ,
dt
dw(t )
 (1  k ) a u (t ),
w(0)  0.
dt
t
Optimization of profit
Solution
u (t )   e a k t ,
t  0,
 (1  k ) a k t
(e  1)

w(t )   k

a t
for
0  k  1,
for k  0.
Interpretation
– If k=0:all profit is paid to shareholders, total dividend increases linearly,
total investment doesn’t change & the company doesn’t grow!!!
– If k=1: all profit is re-invest, total dividend is zero, total investment
increases exponentially & the company grows in the fastest way.
– If 0<k<1: both total investment & dividend increase
Optimization of profit
Central issue: Given a period of time [0,T], how must k be chosen so
that the total dividend over the period [0,T] is a maximum?
Total dividend:
w(k ; T ) 
(1  k ) a k T

(e
 1)  (   )(e a k T  1)
k
k
Question: Find k in [0,1] such that w(k;T) to be maximum?
For simplicity, introduce new variables:
x aT k
&
y
w(k ; T )

Find x in [0,aT], such that y to be maximum
y
aT x x
(e  1)
x
Optimization of profit
Find the derivative of y:
dy
aT
aT  x x
1
  2 (e x  1) 
e  aTe x [
 x 2 (1  x  e  x )]
dx
x
x
aT
1 1 x x2
x
 aTe [
    ...]
aT 2 3! 4!
Different cases:
– If a T=2: y is a decreasing function of x with the maximum of y at x=0
– If a T<2: y is a decreasing function of x with the maximum of y at x=0
– If a T>2: y increases and then decreases & attains its maximum at x*
which is the root of
x2
1 x 
 e x
aT
Optimization of profit
Interpretation:
If a T<=2: then k=0 produces the largest total dividend over
the period of T years, which means that all the profit is paid out
as a dividend. It does not pay to re-invest money in the company
because either a or T or both are too small. In this case, the
maximum profit is a  T
If a T >2: there exists a unique number k=x*/a T such that k u(t)
must be re-invested. In this case, the maximum profit is
(1  k ) a k T
x*
( x*) 2
(e
 1) with k 
, 1 x * 
 e  x*
k
aT
aT