Wave equation in 1D (part 1)*

• Derivation of the 1D Wave equation– Vibrations of an elastic string

• Solution by separation of variables– Three steps to a solution

• Several worked examples• Travelling waves

– more on this in a later lecture• d’Alembert’s insightful solution to the 1D

Wave Equation*Kreysig, 8th Edn, Sections 11.2 – 11.4

A plucked stringt = 0, small displacement

a short while latera little while later …

These are snapshots, freezing time, and looking at the vertical displacement of the string:

,...1,0),,( =itxu iEqually, we could fix attention to one point on the string and watch it over time

x

Physical assumptions*1. The mass of the string per unit length is

constant (“homogeneous string”). The string is perfectly elastic and does not offer resistance to bending

2. The tension caused by stretching the string is so large that gravitational effects can be neglected.

3. The string performs small transverse motions –that is, every particle on the string moves strictly vertically and so that the deflection and the slope at every point of the string remain small.

*Kreysig, 8th Edn, page 585

Derivation of the 1D Wave Equation

Consider the transverse vibrations in a string held under tension …….

x δx

β

α1T

2T

),( txu

constant) (a :3assumptionBy

TTT == βα coscos 21

Newton’s Law of motion in the vertical direction

( )

( )2

2

2

2

12

tantan

sinsin

tu

Tx

tuxTT

∂∂∆

=−

∂∂

∆=−

ραβ

ραβ

xxx xu

xu

∆+

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

= βα tantan and Since 2

21tu

Txu

xu

x xxx ∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

∆ ∆+

ρ

ρTc

xuc

tux =

∂∂

=∂∂

→∆ 22

22

2

2

,,0 where:obtain we Letting

Boundary conditions2

22

2

2

xuc

tu

∂∂

=∂∂ :solve To

0),(0),0( == tLutu and :ends Fixed

)(

)()0,(

0

xgtu

xfxu

t

=∂∂

=

=

:conditions Initial

LxLxLLk

LxxLk

xf

xg

≤<−

≤<=

=

2)(2

202

)(

0)(

if

if

:Example

Solution by Separation of Variables

1. Convert the PDE into two separate ODEs

2. Solve the two (well known) ODEs3. Compose the solutions to the two ODEs

into a solution of the original PDE– This uses Fourier series

The first step is to assume that the function of two variables has a very special form: the product of two separate functions, each of one variable, that is:

Step 1: PDE 2 ODEs

)()(),( tGxFtxu = :that Assume

2

2

2

2

2

2

2

2

''

''

dxFdF

dtGdG

GFxu

GFtu

==

=∂∂

=∂∂

and where

:find weating,Differenti

&&

&&GFcGF ''2=&&

kFF

GcG

==''

2

&&

So that

equivalently

0

0''2 =−

=−

GkcG

kFF&&

The two (homogeneous) ODEs are:

Step 2a: boundary conditions

0)(0)0(

0)(0)()(),(0)()0(),0(

==

≡====

LFF

tGtGLFtLutGFtu

:so and interest, no of is and

Step 2b: solving for F0'' =−kFF

00

0

)(02

===+

=+

+=

>=

−

−

BABeAe

BA

BeAexFk

LL

xx

so and

conditionsboundary the ng Applyi

:is solution general the case

µµ

µµ

µ

0)(

:0

==+=

=

babaxxF

k

:conditionsboundary the Applying

case

0sin,0sin)(0)0(

sincos)(:02

=====

+=<−=

pLpLBLFAF

pxBpxAxFpk

so and

find we,conditionsboundary the ng Applyi

is solution the case

We have to solve:

xL

nxF

Lnp

npL

nπ

ππ

sin)( =

=

=

that so

:isthat

Step 2c: solving for G2

2 ⎟⎠⎞

⎜⎝⎛−=−=

Lnpk π

tBtBtG

tGL

cn

nnnnn

n

λλ

πλ

sin*cos)(

)(,

+=

=

are for solutions the Denoting

( ) xL

ntBtBtGxFtxu nnnnnnnπλλ sinsin*cos)()(),( +==

:are Equation Wave1D the for Solutions

As a result of solving for F, we have restricted

These functions are the eigenfunctions of the vibrating string, and the values are called the eigenvalues. The set of the eigenvalues is called the spectrum.

Lcnn /πλ =],...,[ 1 nλλ

Remembering that there are several ways to change the spectrum of a string – change the material, the tension, and the length

ρ/2 Tc =

Step 3: Initial conditions & Fourier

g(x)tu

xfxu

t

=∂∂

=

=0

)()0,(

:conditions initial theapply to have still We

xdxL

nxfL

B

xfxL

nBxu

L

n

nn

π

π

sin)(2

)(sin)0,(

0

1

∫

∑

=

==∞

=

so and

( )

∫

∫

∑

∑

=

=

=

⎥⎦

⎤⎢⎣

⎡+−==

∂∂

∞

=

=

∞

==

L

n

L

nn

nnn

tnnnnnnn

t

xdxL

nxgcn

B

xdxL

nxgL

B

xL

nB

xL

ntBtBxgtu

0

0

1

010

sin)(2*

sin)(2*

sin*

sincos*sin)(

ππ

πλ

πλ

πλλλλ

so and

( )∑∞

=

+=1

sinsin*cos),(n

nnnn xL

ntBtBtxu πλλ is found have wesolution The

0)(,0* ≡= xgBn case, the often is as if, that Note

An alternative expression∑∞

=

=≡1

sincos),(,0)(n

n xL

ntL

ncBtxuxg ππ that so that Suppose

⎥⎦⎤

⎢⎣⎡ ++−= )(sin)(sin

21sincos ctx

Lnctx

Lnx

Lnt

Lnc ππππ

Lxff

ctxfctxftxu

odd

oddodd

2)(

))()((21),(

period with of extensionperiodic odd the is

++−=

Remembering formulae for produces of sines and cosines:

Substituting this in the formula for u(x,t) we see that:

Later we will interpret these as travelling waves moving in opposite directions (the sign of +/-c)

Recall of the general solution to the 1D Wave Equation

2

22

2

2

xuc

tu

∂∂

=∂∂ :solve To

0),(0),0( === tLutu and :ends fixed conditionsBoundary

)(

)()0,(

0

xgtu

xfxu

t

=∂∂

=

=

:conditions Initial

( )∑∞

=

+=1

sinsin*cos),(n

nnnn xL

ntBtBtxu πλλ : Solution

xdxL

nxfL

BL

nπsin)(2

0∫= ∫=

L

n xdxL

nxgcn

B0

sin)(2* ππ

Example 1: plucked string*

0)(2

)(22

02

)(

≡

⎪⎩

⎪⎨

⎧

<<−

<<=

xg

LxLxLLk

LxxLk

xf if

if

nBxg n all for , Since 0*,0)( =≡

*Kreysig, 8th Edn, page 593

HLT consult or relationsity orthogonalapply either for solve To nB

k0 L

Fourier coefficients

2212 8)1(

πnkB n

n−−=

6-p545Kreysig,ority,orthogonal HLT, Using

We need to extend the string to a periodic function

We know that the solution is a sum of sin terms, so choose an odd function periodic extension:

⎥⎦⎤

⎢⎣⎡ −+−= ...5cos5sin

513cos3sin

31cossin

118),( 2222 t

Lcx

Lt

Lcx

Lt

Lcx

Lktxu ππππππ

π

d’Alembert’s elegant solution to the 1D Wave Equation

1) (Eqn :solve To 2

22

2

2

xuc

tu

∂∂

=∂∂

( ) ( ) ( )zzvzvvxx

xzzvxvzvxzvxx

zvxzxvx

uuuuzuuvuuuuu

uuzuvuu

ctxzctxv

++=

+++=+=+=+=

−=+=

2

:Rule Chain the Applying

:variablesnew two Introduce

( )zzvzvvtt uuucut

+−= 22 derive can we, forSimilarly

02

=∂∂

∂=

zvuuvz

:find we1, Eqn in values these Inserting

D’Alembert continued

)()()()(),()(

).(

)()(

.)()(

0

ctxctxzvtxuvv

z

zdvvhu

vvvh

vhuzu

v

vz

−++=+=

+=

==

∫

ψφψφφ

ψ

ψ

:that concludemay we, of function a is integral the Since

function unknown some for

:gives to respect withgIntegratin of function unknown some is where

gives to respect with gIntegratin

Jean Le Rond d’Alembert

“D'Alembert was always surrounded by controversy. ... he was a lightning rod which drew sparks from all the foes of the philosophes. ... Unfortunately he carried this... pugnacity into his scientific research and once he had entered a controversy, he argued his cause with vigour and stubbornness. He closed his mind to the possibility that he might be wrong...”

Jean was the illegitimate son of a cavalry officer who was out of the country at his birth in 1717. His “socialite” mother left him on the steps of St Jean le Rond – hence his name. He invented the name d’Alembert for college entry.

He developed into one of the leading French mathematicians of the 18th Century, makingparticular contributions to mechanics, which is odd, since he believed mechanics is essentially metaphysics, and he rejected experimentation. His genes shaped his personality:

Wave propagation

• Displacement of a string• Compression waves• Electrical waves• Water waves• Acoustic waves

The last couple of lectures will be dedicated to variants to the wave equation