Block 4 Nonlinear Systems
Lesson 13 – Nonlinear Equations
The World is not Linear
Narrator: Charles Ebeling
University of Dayton
With a Special bonus on quadratic forms
1
The Nonlinear Road Ahead



Lesson 13 – Nonlinear Systems of Equations
Lesson 14 – Methods of Differential Calculus
Lesson 15 – Nonlinear Optimization Models
Gosh, just 3 easy
lessons and I am
through this block.
2
The General Problem
f1 ( x1 , x,..., xn )  0
My problem is that there
may be more than one
solution, and I may not be
able to find them all
analytically (or numerically).
f 2 ( x1 , x,..., xn )  0
:
f n ( x1 , x,..., xn )  0
A civil war
general…
3
Order of difficulty?


Linear
quadratic – 2 variables
ax  bxy  cy  d
2

2
quadratic – more variables
ax  bxy  cy  ez  fxz  gyz  d
2


2
2
other polynomials
other functions; e.g. exponential, logarithmic, rational
4
Our very first (and simplest)
problem…

Simultaneous equations involving quadratics
one linear and one quadratic
4 x 2 solve
 3y2 
16linear equation for one variable, substitute in the
the

to obtain a quadratic in one variable
y7
5 x quadratic

y  7  5x
y
4 x 2  3  7  5 x   16
2
4 x  3  49  70 x  25 x
2
2
  16
79 x 2  210 x  131   x  1 79 x  131  0
x  1, y  2; x  131/ 79, y  102 / 79
(1,2)
x
(1.65,-1.29)
5
Two Quadratics – count them!

Simultaneous equations involving two
quadratics


In general, results in a 4th degree equation in at least
one of unknowns
In general, must solve numerically
3x 2  2 xy  y 2  200
x  xy  2 y  300
2
2
try using solver
6
Solver Solution
3x  2 xy  y  200
2
2
x  xy  2 y  300
2
variable
Eq1
Eq2
x
1.429
3
1
2
y
12.57
1
2
xy
compute
2
-1
14
200
300
x+y
RHS
200
300
7
Sometimes we can eliminate an
equation
Now what? We have a
fourth degree polynomial
to solve!
3 x 2  2 xy  2 y 2  200
 2
2
 x  2 y  100
x 2  100  2 y 2
3 100  2 y 2   2



100  2 y 2 y 
100  2 y
2

100  2 y 2 y  2 y 2  200
200  3 100  2 y 2   2 y 2
2
 2 y 2  50
4
2
y

2
y

50

4
y

200
y
 2500
 

2
2
2
100 y 2  2 y 4  4 y 4  200 y 2  2500
6 y 4  300 y 2  2500  0
8
Look for one of those
mathematical tricks…
6 y 4  300 y 2  2500  0
Let z  y 2
insert trick here
6 z 2  300 z  2500  0
z  10.566, 39.434
y   10.566  3.25 ;
x
y   39.434  6.28
100  2  3.25  8.88
2
x   100  2 y   100  2  6.28  4.596
2
4 solutions!
2
why not use Excel???
9
The Excel Solution…
2
2

3
x

2
xy

2
y
 200

 2
2
x

2
y
 100


x
y
xy
variable 4.597 6.28
Eq1
3
2
2
Eq2
1
2
0
compute
10.88
200
100
RHS
200
100
Which of the 4 solutions will Solver find?
10
Without that pesky cross-product
term

Two equations of the form
a1 x  b1 y  c1
2
2
a2 x  b2 y  c2
2

2
solve by method of addition
see example next slide
11
Our very next problem…
4 x 2  9 y 2  72 (1)
 2
2
3 x  2 y  19 (2)
8 x 2  18 y 2  144
27 x  18 y  171
2
35 x
2
2
 315
multiply eq (1) by 2
multiply eq (2) by 9
add
x 2  9 and x  3
9 y 2  72  4 x 2  72  36  36
y  4 and y  2
2
four solutions: (3,2), (3,-2), (-3,2), (-3,-2)
12
Or use substitution
4 x 2  9 y 2  72 (1)
 2
2
3 x  2 y  19 (2)
2
72

9
y
x2 
4
 72  9 y 2 
2
3

2
y
 19

4


35 2
54  y  19
4
4
2
y   54  19   4
35
four solutions: (3,2), (3,-2),
(-3,2), (-3,-2)
13
A Problem in Economics
A supply curve is given by the following where p is the (selling)
price per unit and q is the quantity of available units.
supply curve: p = (q + 10)2
A corresponding demand curve is given by the following:
demand curve: p = 388 – 16q – q2
Where the two curves intersect is called the equilibrium point and
the resulting values of (p,q) are the price at which consumers will
purchase the same quantity as the producers wish to sell at that
price. It is the price at which stability in the producer-consumer
relationship occurs.
14
The Equilibrium Point
Equilibrium Point
price
600
500
400
300
200
100
0
0
2
4
6
8
10
12
14
quantity
supply
(q + 10)2 = 388 – 16q – q2
2q2 + 36q – 288 = q2 + 18q – 144 = 0
(q + 24) (q-6) = 0; q = -24,6
demand
q = 6, p= 256
15
Homogenous equation
An equation whose terms are all of the same degree in the
variables and equals zero is a homogeneous equation:
 x 2  3 xy  2 y 2  0 (1)
 2
2
2
x

3
xy

y
 13 (2)

 x  y  x  2 y   0 or x  y, x  2 y
2 x 2  3xy  y 2  13 and x  y
2 y 2  3 y 2  y 2  13; y 2  13 / 4; x  y   13 / 2
2 x 2  3xy  y 2  13 and x  2 y
8 y 2  6 y 2  y 2  13; y  1, x  2
16
How about a cubic and a quadratic?
 y  x3  1

2
 y  2  x
3
2
3
2
set x  1  2  x or x  x  1  0
Solve numerically for x (using Excel): x = .7549
then y = .75493 + 1 = 1.4302
check the 2nd equation:
1.4302 =?= 2 - .75492 = 1.4301
17
Linear Eq + Exponential Eq
 x  4 y  0
 y2
2e  2 y   4 x  0
x  4y
2e
y2

(x,y) = (0,0) ; (x,y) = (4.7096, 1.1774)
(x,y) = (-4.7096, -1.1774)
 2 y   16 y  0

4y e  4  0
y2
As you can
now see, there
are 3
solutions.
4 y  0 or y  0; x  0
e  4  0 or e  4
y2
y2
y 2  ln(4) or y   ln(4)  1.17741
x  4(1.17741)  4.70964
18
Taking advantage of the
structure
40
 30 y  10 z  0
2
x yz
40
 30 x  20 z  0
2
xy z
40
 10 x  20 y  0
2
xyz
Can I
solve
them,
please!
Gosh, 3
equations
in 3
unknowns
19
I taught him
how to do this.
An Old Indian Trick
 40

 2  30 y  10 z  0   x 
 x yz

+ 10xz – 20yz = 0
 40

x = 2y
 2  30 x  20 z  0    y 
 xy z

 40

 10 x  20 y  0   z 

2
 xyz

- 30xy + 10xz = 0
z = 3y
20
Take it Home…
40
40
 10 x  20 y 
 10  2 y   20 y  0
2
2
xyz
 2 y  y 3 y 
40
1
5
 40 y  0 or y 
y  .5609776
4
18 y
18
x  1.1219551; z  1.6829321
x = 2y
z = 3y
21
Look, combining the
matrix development
with quadratic
expressions. What a
marvelous idea!
Quadratic Forms
Much ado about matrices
and those troublesome
quadratic terms
22
Let’s start with an example
 x1 
 2 1
A
;x   

 3 5
 x2 
 2 1  x1 
t
x Ax   x1 x2  
 x 
3
5

 2 
 x1 
  2 x1  3x2 x1  5 x2   
 x2 
 2 x12  4 x1 x2  5 x22
xtAx is called a quadratic form
23
The Matrix A is not unique
However, there is unique symmetrical matrix
 x1 
 2 2
A
;x   

2 5
 x2 
 2 2   x1 
t
2
2
x Ax   x1 x2  

2
x

4
x
x

5
x
1
1 2
2
 x 
2
5

 2 
24
The Vector-Matrix Product
Form
 a11 ... a1n   x1 

 
t
x Ax   x1 ,..., xn   :
.
:  : 
 an1 ... an n   xn 



   a j1 x j
 j 1
n
n
a
j 1
 x1 
 
...  a jn x j   : 
j 1
 x 
 n
n
j2
xj
n
n
n
j 1
j 1
j 1
 x1  a j1 x j  x2  a j 2 x j  ...  xn  a jn x j
  a jk x j xk   a jj x    a jk  akj  x j xk
n
n
j 1 k 1
n
j 1
n
2
j
n
j  k k 1
25
The Vector-Matrix Product
Form
If A is a diagonal matrix:
 a11 ... 0   x1 
n


 
x t Ax   x1 ,..., xn   :
.
:   :    a jj x 2j
 0 ... an n   xn  j 1


26
The 2nd degree polynomial in n
variables
 a11 ... a1n 
 x1 
b 


 
 
x   : ; A   :
.
: ; b   : 
x 
b 


a
...
a
nn 
 n
 n
 n1
p(x1, x2, …,xn ) = a0 + bt x + xt A x
where a0 is a scalar
27
Next – on to the calculus
Typical engineering
management students –
unable to control their
enthusiasm.
28