Introduction
• Descriptions
• Statements
A string can be defined as
a rigid body whose
dimensions are small when
compared with its length.
The string in our model will be stretched
between two fixed pegs that are separated
by a distance of length L.
Peg 1
Peg 2
L
Tension (T0) will be the force of the
two pegs pulling on the string.
For our model, we will assume near
constant tension.
Density can be defined as the ratio of the
mass of an object to its volume.
For a string, density is mass per unit length.
In our model, we will also assume
near constant density for the string.
Derivation of the Wave Equation
•
•
•
•
Basic modeling assumptions
Review of Newton’s Law
Calculus prereqs
Equational Derivation
Model Assumptions
Transverse:
L
Vibration perpendicular to the X-axis
Model Assumptions
•Density is assumed constant
=1
•Initial Deformation is small
Model Assumptions
T=1
L
Tension T is constant and tangent to the curve of the
string
Newton’s Second Law
F = ma
Calculus Prerequisites
y
y = f(x)
Angle of Inclination

x
T
=
[(
1
)) i (
dy ²
1 + dx
(
+
dy
dx
|T|
]
)
j
)
dy ²
1 + dx
(
Equational Derivation
u
s
x
x
x+x
Equational Derivation
Horizontal Forces
s
u
Vertical Forces
s
x
x
x+x
Vertical Forces:
[
T
u
x (x + x, t)
u
1 +
x (x + x, t)
(
2
)
-
u
x (x, t)
]
2
u
1 +
x (x, t)
(
Get smaller and go to zero
)
Vertical Forces:
[
T
u
(x + x, t)
x

1
= (s)
-
]
u
x (x, t)
 1
²u
t² (x,t)
Vertical Forces:
u
(x + x, t)
x
-
Net Force
u
(x, t)
x
Mass
Acceleration
=
²u
s t² (x,t)
Vertical Forces
u
(x + x, t)
x
x
-
u
(x, t)
x
=
²u
s t² (x,t)
x
Vertical Forces
One dimensional wave equation
²u
(x, t)
x²
=
²u
t² (x,t)
Solution to the Wave Equation
• Partial Differential equations
• Multivariate Chain rule
• D’Alemberts Solution
• Infinite String Case
• Finite String Case
• Connections with Fourier Analysis
2nd Order Homogeneous Partial
Differential Equation
A
 ²y
 ²y
x² + Bxx +
C
 ²y
 t² +
D
y
x
+
E
y
t
+
Fy
=
0
Classification of P.D.E. types
•
•
•
•
 = B² - AC
Hyperbolic  > 0
Parabolic  = 0
Elliptic
<0
Boundary Value Problem
• Finite String Problem
• Fixed Ends with 0 < x < l
• [u] X = 0= 0 and [u] X = l= 0
Cauchy Problem
• Infinite String Problem
• Initial Conditions
• [u] t=0= 0 (x) and [du/dt] t=0 =  l (x)
Multi-Variable Chain Rule
example
f(x,y) = xy² + x²
g(x,y) = y sin(x)
h(x)
=e
x
F(x,y) = f(g(x,y),h(x))
Let u = g(x,y)
v = h(x)
So F = f(u,v) = uv² + u²
Variable Dependency Diagram
f
u
u
g
x
g
y
F
f
v
x
v
h
x
y
F
f g
f u
=
x
u x + v x
x
= ((v² + 2u)(y cos(x)) + (2uv)e )
= (ex)² + 2y sin(x) (y cos(x))
x x
+ 2(y sin(x) e e )
Multi-Variable Chain Rule for
Second Derivative
Very Similar to that of the first derivative
Our Partial Differential Equation
ξ=x–t
η=x+t
So
ξ + η = 2x
And - ξ + η = 2t
x = (ξ + η)/2
t = (η – ξ)/2
Using Multi-Variable Chain Rule
u =
t
²u = 
t²
t
u
u
+
ξ
η
[
u
- u
+
η
ξ
]
Using Algebra to reduce the
equation
²u
t²
=
²u - 2 ²u +
ξ²
ηξ
²u
η²
u
x
²u
x²
=
=
u
ξ

x
+
u
η
u
[ ξ
+
u
]
η
Using Algebra to simplify
²u
x²
=
²u
ξ²
+
2 ²u
ηξ
+
²u
η²
Substitute What we just found
²u
t²
=
²u
x²
²u
ξ²
2²u
ηξ
+
²u
²u
=
η²
ξ²
2 ²u
ηξ =
4 ²u
ηξ
=
+
2 ²u
ηξ
0
2 ²u
ηξ
+
²u
η²
We finally come up with
²u
ηξ
=
0
When u = u(ξ, η)
ξ=x-t
η=x+t
Intermission
Sorry, Can I get
we don’t a Beer?
sell to
strings
here
A String walks into a bar
Again we can’t
serve you
because you are
a string
Can I
get a
beer?
I’m
afraid
not!
And Now Back
to the Models
Presentation
D’Alemberts Solution
u u

2
2
t
x
2
2
  x t
  x t
u u
u u
 2 2
 2
2
x

nd n
2
2
2
2
u u
u u


2

2
2
2
t

nd n
2
2
2
2
u
u u u
u u
2
 2  2 2
 2
2

nd n

nd n
2
2
2
2
u
4
0

2
u
0

2
2
2
Integrating with respect to Ada
u
 k ( )

Next integrating with respect to Xi
u   k ( )d  c( )
u ( , )  k ( )  c( )
Relabeling in more conventional notation gives
u ( , )  f ( )  g ( )
Then unsubstituting
u ( x, t )  f ( x  t )  g ( x  t )
Infinite String Solution
Which is a cauchy problem
u ( x, t )  f ( x  t )  g ( x  t )
Reasonable initial conditions
u
( x ,0)  0
t
u ( x,0)   ( x)
u
 f ' (x  t)  g' (x  t)
t
u
( x,0)  f ' ( x)  g ' ( x)
t
u
( x,0)  f ' ( x)  g ' ( x)  0
t
u ( x, t )  f ( x  t )  g ( x  t )
u ( x,0)  f ( x)  g ( x)   ( x)
So we have
f ( x)  g ( x)   ( x)
f ' ( x)  g ' ( x)  0
And we have to solve for f and g
When we solved for f and g, we found
c   ( x)
f ( x) 
2
 ( x)  c
g ( x) 
2
Then when we plug those into U
u ( x, t )  f ( x  t )  g ( x  t )
c   (x  t)  (x  t)  c


2
2

 (x  t)   (x  t)
2
Finite Solution
Boundary Value Problem
Boundary conditions
u ( x,0)   ( x)
u
( x ,0)  0
t
u (0, t )  0
u ( L, t )  0
u (0, t ) 
 (t )   (t )
0
2
 (L  t)   (L  t)
u ( L, t ) 
0
2
u (0, t )   (t )   (t )  0
u ( L, t )   ( L  t )   ( L  t )  0
t b
 (b)   (b)  0
 (b)   (b)
b  L t
 ( L  t )   (t  L)
 ( L  t )   (t  L)  0
 ( x  2 L)   ( x)  0
This is a periodic function with period 2L. If the
boundary conditions hold this above is true. This
equation relates to the sin and cos functions.
NEED A CONCLUSION
A Special thanks To
• Dr. Steve Deckelman for all your help and
support
• S.L. Sobolev “Partial Differential Equations
of Mathematical Physics
• Scott A. Banaszynski for use of his
wonderful guitar
Thank you for coming, enjoy the
rest of the presentations.