04-The Wave Equation
Transcript of 04-The Wave Equation
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The Wave Equation
Analytic Solution using Separation
of Variables
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The Wave Equation 2
Motion of an Elastic String
Consider an elastic string (e.g. a guitar string) of
length L, fixed at both ends.
This is distorted and at some instant, say t = 0, released
to that it vibrates.
The problem is to calculate the transverse (i.e. vertical)deflection of the string u(x,t) at any point x between the
fixed endpoints and for any time t > 0.
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The Wave Equation 3
Elastic String
u
x x+xO
PQ
T1
T2
L
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The Wave Equation 4
Outline of Solution State modelling assumptions and initial conditions
Derive the (Partial) Differential Equation (PDE),
and associated boundary conditions
Convert PDE into Ordinary Differential Equations
(ODEs) using Separation of Variables Solve ODEs that satisfy the boundary conditions
Use Fourier Series to satisfy the initial conditions
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The Wave Equation 5
Modelling Assumptions
We assume the following:
The mass of the string per unit length (the line density) is
constant i.e. a homogeneous string. This string isperfectly elastic and does not offer any resistance tobending.
The tension caused by stretching the string is so largecompared to the deflection caused by gravity that thelatter can be neglected.
The string performs small transverse motions in a
vertical plane; that is every particle in the string movesstrictly vertically so that the deflection and the slope atevery point of the string always remain small in absolutevalue.
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The Wave Equation 6
The Differential Equation
To obtain the differential equation, we consider the forces
acting on a small element of the string.
Since the string does not offer any resistance tobending, the tension is tangential to the curve of the
string at each point.
Let T1 and T2 be the tensions at the endpoints, P and Q, ofthis portion of string.
Since there is no horizontal motion, the horizontal
component of the string must be constant.Using the notation of the previous diagram, we obtain
(1)constantT)cos(T)cos(T 21 ===
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The Wave Equation 7
The Differential Equation
In the vertical direction, we have two forces:
the vertical components of the tension and
where the minus sign appears because the component at P
is directed downwards.
By Newton's second law, the resultant of these two forces
is equal to the mass, , of the string element, times
the acceleration in the vertical direction , evaluated atsome point between x and
)sin(T1 )sin(T2
x
2
2
t
u
xx +
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The Wave Equation 8
The Differential Equation
Hence
using eq(1), we can divide this by to obtain
(2)
i.e.
2
2
12 t
ux)sin(T)sin(T
=
2
2
1
1
2
2
t
u
T
x
)cos(T
)sin(T
)cos(T
)sin(T
=
2
2
t
u
T
x)tan()tan(
=
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The Wave Equation 9
The Differential Equation
Now tan() and tan() are the slopes of the string at x andx + x respectively:
so and
which gives
i.e.
xx
u)tan(
=xxx
u)tan(
+
=
2
2
xxx t
u
T
x
x
u
x
u
=
+
2
2
xxx t
u
Tx
u
x
u
x
1
=
+
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The Wave Equation 10
The Differential Equation
If we now let x approach zero,
the left hand side of this equation becomes the partial
derivative of with respect to x,
i.e. the second order partial derivative .
The equation therefore can be written as
where
x
u
2
2
x
u
2
2
22
2
t
u
c
1
x
u
=
=
T
c
2
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The Wave Equation 11
The Differential Equation
This is known as the One-Dimensional Wave Equation.
It is a second order partial differential equation whichdescribes how the vibrations evolve in time in one space
dimension.
The notation c2
rather than c for the physical constantT/ has been chosen to indicate that this constant ispositive.
2
2
22
2
t
u
c
1
x
u
=
(3)
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The Wave Equation 12
Boundary ConditionsVibrations in an elastic string are governed by the one-dimensional wave equation
The solution of this equation also has to satisfy certainboundary conditions due to the fact that the two ends of thestring are fixed.
The string is fixed for all times, i.e. for all values of t, at x = 0,and is similarly fixed at x = L .
The boundary conditions are therefore
u(0,t) = 0 and u(L,t) = 0 for all t (4)
2
2
22
2
xuc
tu
=
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The Wave Equation 13
Initial Conditions
The form of the motion of the string will also depend on its
initial deflection profile (i.e. u(x,0) ) and on the initial velocityprofile (i.e. ).
These will be functions of x only so we write
(5)
The problem will therefore be solved if we can find a solution
of the one-dimensional wave equation (3) which also
satisfies the boundary conditions (4) and the initial
conditions (5).
0tt
u
=
)x(gt
u,)x(f)0,x(u
0t
=
==
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The Wave Equation 14
Method of Solution
Outline
1. The first step is to apply the method ofseparation of
variables to the partial differential equation to obtaintwo ordinary differential equation.
2. The second step is to determine a solution of the two
ordinary differential equations that satisfy the boundaryconditions.
3. Finally, using Fourier Series, we shall compose these
solutions to get a solution which satisfies the initialconditions.
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The Wave Equation 15
We first assume that the solution of the wave equation
u(x,t) which is a function which depends on both x and t,
can be written as the product of a function of x only and afunction of t only.
(6)
We then substitute this back into the wave equation.
Clearly we need to differentiate with respect to x and t.
We use the dot notation to indicate differentiation withrespect to t and the dash notation to denote differentiation
with respect to x.
1. Separations of Variables
u(x,t) = F(x)G(t)
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The Wave Equation 16
1. Separations of Variables
and
[ ]
)t(G)x(F
dt)t(Gd)x(F
)t(G)x(Ftt
u
2
2
2
2
2
2
&&
=
=
=
[ ]
)t(G)x(F
)t(Gdx
)x(Fd
)t(G)x(Fxx
u
2
2
2
2
2
2
=
=
=
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The Wave Equation 17
1. Separations of Variables
Inserting these back into the wave equation we obtain
Dividing both sides by c2F(x)G(t) we obtain
The left hand side of the equation depends only on the
variable t whereas the right hand side depends only on the
variable x. The only way that this can occur is if each sideequals an arbitrary constant, say k.
i.e.
)t(G)x(Fc)t(G)x(F 2 =&&
)x(F
)x(F
)t(Gc
)t(G
2
=
&&
k)x(F
)x(F
)t(Gc
)t(G
2=
=
&&
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The Wave Equation 18
1. Separations of Variables
Therefore we have two equations
(7)
and
(8)
2
2
G(t)k G(t) c kG(t) 0c G(t) = =
&&&&
0)x(kF)x(Fk)x(F
)x(F==
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The Wave Equation 19
2. Satisfying the Boundary Conditions
We shall now determine the functions F(x) and G(t) so that
satisfies the boundary conditions (4), i.e.
and for all t
Considering the first of these equations,
Clearly either F(0) = 0 or G(t) = 0 for all t. The latter solution
means that u(x,t) would be identically equal to zero which is of
no interest, so we must have that F(0) = 0.
Similarly for the second equation, we must have F(L) = 0.
The boundary conditions therefore imply the following
conditions on F(x): (9)
0)t(G)0(F)t,0(u == 0)t(G)L(F)t,L(u ==
0)t(G)0(F)t,0(u ==
0)L(F0)0(F ==
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The Wave Equation 20
2. Satisfying the Boundary Conditions
We now consider the ordinary differential equation (8)
There are three possible possibilities for the value of k.
Firstly k could be zero,
secondly k could be a positive number
thirdly k could be a negative number.
0)x(kF)x(F =
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The Wave Equation 21
2. Satisfying the Boundary Conditions
For k = 0, the equation states that
which has a general solution of the form
where a and b are constants.
If this is the case, then
andand hence F(x) is identically equal to zero.
This means that u(x,t) is also identically equal to zero
which is uninteresting.
0)x(F =
bax)x(F +=
0b0)0(F ==
0a0)L(F ==
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The Wave Equation 22
2. Satisfying the Boundary Conditions
If k > 0, then
where
is some real number.
This means that the general solution of eq(8) is
where A and B are constants.
The boundary condition
whileHence the only solution is the uninteresting solutionu(x,t) = 0.
2k =
xx BeAe)x(F +=
AB0)0(F ==
0A0)L(F ==
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The Wave Equation 23
2. Satisfying the Boundary Conditions
The final possibility is that k < 0.
In this case we can write for some real number p.
Therefore the differential equation for F(x) takes the form
which is the simple harmonic motion equation which has
general solution
(A,B constants)
2pk =
0)x(Fp)x(F 2 =+
)pxsin(B)pxcos(A)x(F +=
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The Wave Equation 24
2. Satisfying the Boundary Conditions
The boundary condition F(0) = 0 implies that A = 0,
while F(L) = 0 implies that Bsin(pL) = 0.
For this to occur either B = 0 or . The former result would again mean that F(x) and hence
u(x,t) is identically equal to zero. Consequently we must
take which implies that pL is an integermultiple of.
Therefore(10)
where n is an integer.
0)pLsin( =
0)pLsin( =
L
n
p
=
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The Wave Equation 25
3. Finding the General Solution
We therefore obtain infinitely many solutions for the
function F(x) depending on the value of the integer n
(n = 1, 2, 3, ) (11)
[Note that we could also include the negative integers but
we would obtain essentially the same solutions since
sin(-) = - sin() and so the minus sign is just incorporatedinto the constant Bn.]
== x
L
nsinB)x(F)x(F nn
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The Wave Equation 26
3. Finding the General Solution
We can now solve the second ordinary differential equation
0)t(kGc)t(G 2 =&&
subject to the restriction that 22
L
npk
==
i.e. 0)t(GL
cn)t(G
2
=
+&&
Once again, this is the simple harmonic equation with
an infinite number of solutions depending on the integer n,
n n n
cn cnG(t) G (t) C cos t D sin t
L L
= = +
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The Wave Equation 27
3. Finding the General Solution
The function u(x,t) is the product of the functions F(x)
and G(t)
So we obtain infinitely many solutions of the partialdifferential equation which satisfy the boundary
conditions (4).
n n n n
n n n
cn cn nu (x,t) C cos t D sin t B sin x
L L L
cn cn nu (x,t) C cos t D sin t sin x
L L L
= +
= +
(where in the second line we have simply absorbed the constant Bn
into
the other constants Cn and Dn).
(12)i.e.
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The Wave Equation 28
These functions are called eigenfunctions or
characteristic functions
The values are called the characteristic
frequencies of the vibrating string.
The most general solution of the partial differential
equation satisfying the boundary conditions will be a sum
over all possible eigenfunctions, so we obtain:
3. Finding the General Solution
L
cn
=
=
+
==
1n
nn
1n
n xL
nsint
L
cnsinDt
L
cncosC)t,x(u)t,x(u
(13)
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The Wave Equation 29
4. Satisfying the Initial Conditions
The initial conditions that we need to satisfy are given by
equations (5),
i.e
From equation (13) we see that
that is
t 0
uu(x,0) f(x) and g(x)
t =
= =
( ) ( )n n nn 1 n 1
nf(x) u(x,0) u (x,0) C cos 0 D sin 0 sin x
L
= =
= = = +
=
=
1n
n x
L
nsinC)x(f (14)
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The Wave Equation 30
4. Satisfying the Initial Conditions
The function f(x), which is only defined for , can
however be written as a Fourier sine series given by
These two expressions for the function f(x) are clearly thesame if we identify the constants Cn as the Fourier sine
coefficients, bn given by
0 x L
=
=
1n
n xL
nsinb)x(f (15)
dxxL
nsin)x(f
L
2bC
L
0
nn
== (n = 1, 2, 3 ) (16)
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The Wave Equation 31
4. Satisfying the Initial Conditions
The second initial condition that needs to be satisfied is
Since u(x,t) is given by
the first partial derivative with respect to the variable t can
be calculated to be
)x(gt
u
0t =
=
=
+
=
1n
nn xL
nsintL
cnsinDtL
cncosC)t,x(u
=
+
=
1n
nn xL
nsint
L
cncos
L
cnDt
L
cnsin
L
cnC
t
u (17)
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The Wave Equation 32
4. Satisfying the Initial Conditions
Therefore
We must again choose the constants Dn so that the right
hand side coincides with the Fourier sine series for the
function .
This will be the case if
==
=
1n
n
0t
x
L
nsin
L
cnD
t
u(18)
t
u
dxxL
nsin)x(g
L
2
L
cnD
L
0
n
=
(19)
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The Wave Equation 33
4. Satisfying the Initial Conditions
Hence
Subject to these conditions on the constants Cn and Dn, the
function defined in equation (13) is a complete solution tothe problem.
L
n
0
2 nD g(x)sin x dx
cn L
=
(20)
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The Wave Equation 34
For the sake of convenience, we demonstrate it only for
the case when the initial velocity profile g(x) is identically
zero (although the procedure can be used moregenerally however).
Applying the trigonometric product formula for cosines
and sines
Further Simplification
=
= 1n n xLn
sintL
cn
cosC)t,x(u (21)
( ) ( )
+
+
=
ctx
L
nsinctx
L
nsin
2
1x
L
nsint
L
cncos
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The Wave Equation 35
Further Simplification
We may rewrite the solution (21) as
The expressions in each of the two series are the Fourier sine
components for the initial displacement profile f(x) but with the
variable x replaced by (x ct) in the first case and (x + ct) in
the second.
Thus we may write
where is the odd periodic extension of f(x) with period 2L.
++
=
=
=
)ctx(LnsinC
21)ctx(
LnsinC
21)t,x(u
1n
n
1n
n (22)
[ ])ctx(f~)ctx(f~2
1)t,x(u ++=
f(x)%
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The Wave Equation 36
Further Simplification
The graph of the function )ctx(f~
is obtained from that of
)x(f by shifting it ct units to the right. This means that )ctx(f
~
represents a wave with profile f(x) travelling to the right with
velocity c units.
Similarly represents a wave with profile f(x)travelling to the left with velocity c units.
)ctx(f
~
+
Equation (22) therefore gives a solution that is a
superposition of these two travelling waves.
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The Wave Equation 37
Further Simplification
The graph of the function is obtained from that of
f(x) by shifting it ct units to the right.
This means that represents a wave with profilef(x) travelling to the right with velocity c units.
Similarly represents a wave with profile f(x)
travelling to the left with velocity c units. Equation (22) therefore gives a solution that is a
superposition of these two travelling waves.
For completeness, we may write out fully, the solution (13)
)ctx(f~
)ctx(f~
)ctx(f~
+
L L
n 10 0
2 n cn 2 n cn nu(x,t) f(x)sin x dx cos t g(x)sin x dx sin t sin x
L L L cn L L L
=
= +
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The Wave Equation 38
ExampleAn elastic string is plucked so that its motion is governed by
the one-dimensional wave equation
Solve the wave equation subject to the following boundary and initial
conditions:
i)
ii)
iii)
2
22
2
2
x
u
ct
u
=
metre1atandorigintheatfixedendpointsi.e.0)t,1(u
0)t,0(u
==
x
0 x 0.510u(x,0) (initial displacement profile)1 x
0.5 x 110
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The Wave Equation 39
Solution
The initial displacement profile, f(x), is shown below:
The general solution of the wave equation with fixed boundary
conditions is given by
where in this case, L = 1, and Dn = 0 since the initial velocity profile
is identically zero.
so where
x
u
O0.5 1.0
=
=
+
==
1n
nn
1n
n xL
nsint
L
cnsinDt
L
cncosC)t,x(u)t,x(u
( ) ( )
=
=1n
n xnsintcncosC)t,x(u
=T
c 2
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The Wave Equation 40
Solution
To satisfy the initial condition for u(x,t), we have to choose the
constants Cn to be the Fourier sine coefficients of the function f(x).
( )
( )
L
n
0
1
0
0.5 1
0 0.5
10.5
2 2 2 20 0.5
2 2
2C f(x)sin n x dxL
2f(x)sin n x dx
1 1
x 1 x2 sin(n x)dx 2 sin(n x)dx
10 10
1 x cos(n x)x cos(n x) sin(n x) sin(n x)2 2
10n 10n10n 10n
2 nsin
25n
=
=
= +
= + + +
=
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Solution
The complete set of solutions in this case is therefore
2 2n 1
2 2n 1
2
2 nu(x, t) sin cos(cn t)sin(n x)
25n
2 1 n
sin cos(cn t)sin(n x)25 n
cos(3c t)sin(3 x)
cos(c t)sin( x)2 9
cos(5c t)sin(5 x) cos(7c t)sin(7 x)5+
25 49
=
=
=
=
=
K