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.

L

Peg 1 Peg 2

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

Transverse:

Vibration perpendicular to the X-axis

Model Assumptions

L

Model Assumptions

•Density is assumed constant

= 1

•Initial Deformation is small

Model Assumptions

Tension T is constant and tangent to the curve of the string

T = 1

L

Newton’s Second Law

F = ma

Calculus Prerequisites

T = [ ]1

1 + dydx( )²( ) i + (

dydx

1 + dydx( )²) j |T|

y = f(x) Angle of Inclination

T

y

x

Equational Derivation

x x+x

s

u

x

Equational Derivation

x x+x

s

u

x

Vertical Forces

Horizontal Forces

s

T[ ux (x + x, t)

1 +ux (x, t)( )2]-ux (x, t)

1 +ux (x + x, t)( )2

Vertical Forces:

Get smaller and go to zero

Vertical Forces:

T[ ux (x + x, t)

]-ux (x, t)

1 1

= (s) ²ut² (x,t)

Vertical Forces:

ux

(x + x, t) - ux

(x, t) = s²ut² (x,t)

MassAcceleration

Net Force

ux

(x + x, t) - ux

(x, t)

x

Vertical Forces

= s²ut² (x,t)

x

²ux²

(x, t) = ²ut² (x,t)

Vertical Forces

One dimensional wave equation

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 B C FED 0 ²y ²y ²yx² xx ++ +++ t² t

y yx y =

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] = 0 and [u] = 0

X = 0 X = l

Cauchy Problem

• Infinite String Problem• Initial Conditions• [u] = (x) and [du/dt] (x) t=0 t=00 l=

Multi-Variable Chain Rule example

f(x,y) = xy² + x²

g(x,y) = y sin(x)

h(x) = e

F(x,y) = f(g(x,y),h(x))

x

Let u = g(x,y)

v = h(x)

So F = f(u,v) = uv² + u²

F

u

v

x

y

fu

fv

g

g

x

xh

y

Variable Dependency Diagram

fu

Fx =

gx +

fv

ux

= ((v² + 2u)(y cos(x)) + (2uv)e )x

= (e )² + 2y sin(x) (y cos(x))

+ 2(y sin(x) e e )

x

xx

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 x = (ξ + η)/2

And - ξ + η = 2t t = (η – ξ)/2

Using Multi-Variable Chain Rule

uξ

u t η

u= +

²ut²

t

=ξ

uη

- u +[ ]

Using Algebra to reduce the equation

²ut² ξ²

2 ²uηξ

²u -= ²u

η²+

ηuu

ξ

x

= +

²uηuu

ξx²= +

ux

[ ]

Using Algebra to simplify

²uη²²u²u

ξ²x²= + ηξ

2 ²u+

Substitute What we just found

²ut²

²ux²

=

η²²u²u

ξ²+ ηξ

2²uη²²u²u

ξ²+ ηξ

2 ²u+=

ηξ2 ²u

= ηξ2 ²u

ηξ4 ²u

= 0

We finally come up with

ηξ ²u

= 0

When u = u(ξ, η)

η = x + t

ξ = x - t

Intermission

Can I get a Beer?

Sorry, we don’t

sell to strings here

A String walks into a bar

Can I get a beer?

Again we can’t serve you

because you are a string

I’m afraid not!

And Now Back to the Models Presentation

D’Alemberts Solution

2

2

2

2

xu

tu

tx tx

2

22

2

2

2

2

2nu

nduu

tu

2

22

2

2

2

2

2nu

nduu

xu

2

22

2

2

2

22

2

2

22nu

nduu

nu

nduu

042

u

02

u

)(

ku

)()( cdku

)()(),( cku

)()(),( gfu

)()(),( txgtxftxu

Then unsubstituting

Relabeling in more conventional notation gives

Integrating with respect to Ada

Next integrating with respect to Xi

Infinite String Solution

)()(),( txgtxftxu

0)0,( x

tu

)()0,( xxu

)(')(')0,( xgxfxtu

)(')(' txgtxftu

Which is a cauchy problem

Reasonable initial conditions

0)(')(')0,( xgxfx

tu

So we have

)()()( xxgxf

0)(')(' xgxf

And we have to solve for f and g

)()()()0,( xxgxfxu

)()(),( txgtxftxu

2)()( xcxf

2)()( cxxg

When we solved for f and g, we found

Then when we plug those into U

)()(),( txgtxftxu

2)(

2)( ctxtxc

2)()( txtx

Finite Solution

)()0,( xxu

0)0,( x

tu

0),0( tu0),( tLu

02

)()(),0(

tttu

02

)()(),(

tLtLtLu

Boundary Value Problem

Boundary conditions

0)()(),0( tttu

0)()(),( tLtLtLu

tLb

0)()( bb

)()( bb

)()( LttL

0)()( LttL

bt

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.

0)()2( xLx

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.