Silly String
Our modeling presentation deals with the wave equation for a vibrating string. For our
model, we will use the motion of a string that has been plucked to represent this wave. The first
thing we will do is describe the vibrating string and the physics surrounding it. Next, we will
derive the wave equation using basic modeling assumptions and a decomposition method of
equation derivation. Finally, we will find the solution to the wave equation using partial
differential equations, the multivariate chain rule, D'Alembert's solution, and infinite and finite
string cases.
A string can be defined as a rigid body whose other dimensions are relatively small when
compared with its length. The string for our model will be stretched between two fixed pegs that
are separated by a distance L, where L will represent the length. The force that these two pegs
exert when pulling the string will be the tension. For our model, we will assume constant
tension. Next, density can be defined as the ratio of an object's mass compared to its volume.
For a string though, density will be the mass per unit length. Also we are assuming near constant
density for our model.
Derivation of the Wave Equation
This part of the models presentation will show how to get the one-dimensional wave equation of
a vibrating string. Some steps that will need to be done to calculate the equation are some basic
assumptions, a quick review of Newton’s Second Law of Physics, prerequisite equations, and
then derive the equation.
There are a couple of assumptions that are going to be need for the wave equation model.
One will be to understand that for any point on the string from 0 to L will be traveling in a
traverse motion, or perpendicular to the X-axis. The second assumption will be that we are
going to assume that the density of the string is constant at all times. For this model will assume
that the density is equal to one. Another assumption is that the initial deformation of the string
will be small. The last assumption is that the tension for every point on the string will be tangent
to the string at be constant, which will be equal to one also.
A quick review is need about Newton’s Second Law of Physics, which we will be
applying to last part of are deriving the equation. Newton’s Second Law of Physics is equal to F
= ma, which is equal to force, mass, and acceleration, respectively.
Next for the preparation of the equation some prerequisite will need to be established.
The figure below shows how the vector T equation was created:
y = f(x)
T
θ
Angle of Inclination
T
1
=
dy
dx
1 +
2
dy
dx
i +
dy
dx
1 +
2
j
T
Next a small chunk of the string is needed to calculate the one-dimensional wave equation. As
show in the diagram below:
Horizontal
∆S
Forces
Vertical
Forces
Since all points on the string are going in a vertical motion the horizontal force of the vector T
can be dropped. Hence, from all the information calculated so far the following equation can be
established:
T
∂u
(x + ∆x, t )
∂x
∂u
(x + ∆x, t )
1+
∂x
2
−
∂u
( x, t )
∂x
∂u
(( x, t )
1+
∂x
2
∂u
(x + ∆x, t ) and the ∂u (( x, t ) will
∂x
∂x
get small, and since there squared they will go to zero faster. Plus, the tension T and the density
ρ is equal to one, they can be dropped by the equation. This will then give us the equation:
Also since the initial deformation of the string was small the
∂u
∂u
∂ 2u
(x + ∆x, t ) − (x, t ) = ∆s 2 (x, t )
∂x
∂x
∂t
Net force
mass
acceleration
Hence Newton’s Second Law of Physics is show here. The last thing that is needed to be done
is divide both sides by ∆x, which will then equal the one-dimensional wave equation:
∂ u
∂ u
(
)
(
)
x
,
t
=
,
x
t
2
2
∂x
∂t
2
2
Derivation of two-dimensional wave equation
The next part of our modeling is coming up with the basis for our two-dimensional wave
equation. The model is based on a second order homogeneous partial differential equation in the
form of:
A
∂2 y
∂2 y
∂2 y
∂y
∂y
+
B
+
C
+D +E + Fy = 0
2
2
∂x∂y
∂x
∂t
∂x
∂t
There are three different classifications of partial differential equations. Using the equation
above if ∆ = B*B – AC, and ∆>0 the equation is Hyperbolic, If ∆<0 the equation is Elliptic and
if ∆=0 then the equation is Parabolic. Our equation that we will be working with is going to be
Hyperbolic.
There are also two different types of problems that you can work with. The first is Boundary
Valued Problems. This is a case where there are fixed endpoints, finite problem. The set U
when x=0 is 0, and the set U when x=L is also 0. The second type of problem is the Cauchy
Valued Problem. The Cauchy problem is an infinite string problem where there are no
endpoints.
Now to figure out the two-dimensional wave equation we will need to use the Multi-Variable
Chain Rule. The Multi-Variable Chain Rule for our problem though is very difficult to follow
and understand so we are going to show an example of how it works.
f ( x, y ) = xy 2 + x 2
g ( x, y ) = y sin( x)
h( x) = e x
F ( x, y ) = f ( g ( x, y ), h( x))
Let U = g (x,y) and v = h (x)
That gives:
F = f (u, v) = uv2 + u2
Looking at a Variable Dependency Diagram you will see that:
∂F ∂f ∂g ∂f ∂u
=
+
∂x ∂u ∂x ∂v ∂x
Which is equal to
((v 2 + 2u )( y cos( x)) + (2uv)e x
With substitution this is broken down to
(e x ) 2 + 2 y sin( x)( y cos( x)) + 2( y sin( x)e x e x
That shows the multivariable chain rule for the first derivative. The multivariable chain rule for
the second derivative is very close to that so we will use the same explanation to show it.
The partial differential equation that we are going to work with is set up in the following matter:
ξ = x – t and η = x + t
So therefore ξ + η = 2x and η - ξ = 2t
Now using the Multi-Variable Chain Rule which we just established we have
∂u
∂u ∂u
=−
+
∂t
∂ξ ∂η
So therefore the partial of that is equal to
∂ 2 u ∂ ∂u ∂u
= [−
+
]
∂t 2 ∂t ∂ξ ∂η
Using algebra to simplify this you get
∂ 2u ∂ 2u
∂ 2u
∂ 2u
=
−
=
2
∂η∂ξ ∂η 2
∂t 2 ∂ξ 2
And now with respect to x
∂u
∂u ∂u
=−
+
∂x
∂ξ ∂η
So therefore the partial of this equation is equal to
∂ 2 u ∂ ∂u ∂u
= [−
+
]
∂t 2 ∂t ∂ξ ∂η
Using algebra to simplify this equation you get
∂ 2u ∂ 2u
∂ 2u
∂ 2u
=
+
2
=
∂η∂ξ ∂η 2
∂x 2 ∂ξ 2
Now if you substitute
∂ 2u
∂ 2u
=
∂x 2
∂t 2
You end up getting that
∂ 2u
4
=0
∂η∂ξ
So we finally come up with
∂ 2u
=0
∂η∂ξ
When ξ = x - t and η = x + t
D’Alambert’s Solution
D’Alambart’s solution is a general solution to the one-dimensional wave equation found earlier.
In order to find D’Alambert’s solution, two substitutions that we will be making will be that
η=x+t
ξ=x–t
In order to find D’Alambert’s solution we need to find the second partial of U with respect to
time and second partial of U with respect to x, which were also found earlier. Substituting the
second partial of U with respect to time and x back into the one-dimension wave equation gives
∂ 2u
∂ 2u ∂ 2u ∂ 2u
∂ 2u ∂ 2u
−
+
=
+
2
2
+
∂ε 2
∂ndε ∂n 2 ∂ε 2
∂ndε ∂n 2
Canceling like terms and simplifying gives
∂ 2u
=0
∂η∂ε
If we integrate with respect to Ada, we get the partial of U with respect to Xi.
∂u
= k (ε )
∂ε
If we integrate then with respect to Xi, and simplify, we get
u (ε ,η ) = k (ε ) + c(η )
Relabeling in more conventional terms gives
u (ε ,η ) = f (η ) + g (ε )
Then substituting gives us D’Alambart’s Solution.
u ( x, t ) = f ( x + t ) + g ( x − t )
Infinite String Problem
Next, we will find a solution to the infinite string problem. First we set some reasonable initial
conditions.
∂u
( x ,0 ) = 0
∂t
u ( x,0) = φ ( x)
Next find the partial of U with respect to time
∂u
= f '(x + t) − g ' (x − t)
∂t
and,
∂u
( x ,0 ) = f ' ( x ) − g ' ( x )
∂t
so,
∂u
( x ,0 ) = f ' ( x ) − g ' ( x ) = 0
∂t
and also,
u ( x, t ) = f ( x + t ) + g ( x − t )
remembering initial conditions
u ( x ,0 ) = f ( x ) + g ( x ) = φ ( x )
which gives us
f ( x ) + g ( x) = φ ( x )
Now using the previous two equations, we have to solve for f and g. When we do solve for f and
g, we get,
f ( x) =
g ( x) =
c + φ ( x)
φ ( x2) − c
2
Now to find the solution to the Infinite String Problem, we need to plug f and g back into U, we
get,
U ( x, t ) =
φ (x + t) + φ (x − t)
2
Which is the solution to the Infinite String Problem.
Infinite String Problem
Next we will find a solution to the finite string problem. In this example we will consider a
string with endpoints 0 and L. For the finite string problem we will start out with the two initial
conditions from the infinite string problem plus two others.
u ( x,0) = φ ( x)
∂u
( x ,0 ) = 0
∂t
u (0, t ) = 0
u ( L, t ) = 0
Using the solution to the infinite string problem, we can consider
u (0, t ) =
u ( L, t ) =
φ (t ) + φ (−t )
2
=0
φ (L + t) + φ (L − t)
2
=0
Now we can simplify these two equations by multiplying each side by two. Now take the
equation u (0,t) and let b stand for t. Now through manipulation, we get the equation
φ (b) = −φ (−b)
Then substitute L-t back in for b and come up with this equation.
φ ( L − t ) = −φ (t − L)
Which equals
φ ( L + t ) − φ (t − L) = 0
Manipulating this equation gives us
φ ( x + 2 L) − φ ( x) = 0
Which is the solution to the finite string problem. Now this is a periodic function with a period
of 2L. Because this function is periodic, it is often modeled using a series of sine and cosine's,
which is more commonly known as the Fourier series.
Silly String
By: Michael Janssen
Jeffery Rogers
Nathaniel Smith
Zebulon Zielie
May 17, 2000
References:
S.L. Sobolev. Partial Differential Equations of Mathematical Physics.
Steve Deckelman. Math Models II Professor