boyce/diprima 9 th ed, ch 10.7: the wave equation: vibrations of an elastic string elementary...
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Boyce/DiPrima 9th ed, Ch 10.7: The Wave Equation: Vibrations of an Elastic StringElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.
A second partial differential equation that occurs frequently in applied mathematics is the wave equation.
Some form of this equation, or generalizations of it, almost inevitably arises in any mathematical analysis of phenomena involving propagation of waves in a continuous medium.
The studies of acoustic waves, water waves, electromagnetic waves, and seismic waves are all based on this equation.
Perhaps the easiest situation to visualize occurs in the investigation of mechanical vibrations.
In this section we focus on the vibrations of an elastic string.
The string may be thought of as a violin string, a guy wire, or possibly an electric power line.
Vibrating String: Assumptions (1 of 5)
Suppose that an elastic string of length L is tightly stretched between two supports at the same horizontal level.
Let the x-axis be chosen to lie along the axis of the string, and let x = 0 and x = L denote the ends of the string.
Suppose that the string is set in motion so that it vibrates in a vertical plane, and let u(x, t) denote the vertical displacement experienced by the string at the point x at time t.
Assume that damping effects, such as air resistance, can be neglected, and that the amplitude of motion is not too large.
Wave Equation (2 of 5)
Under these assumptions, the string vibration is governed by the one-dimensional wave equation, and has the form
The constant coefficient a2 is given by a2 = T /, where T is the tension, is the mass per unit length of the string material.
It follows that the units of a are length/time. It can be shown that a is the velocity of propagation of waves along the string.
See Appendix B for a derivation of the wave equation.
0,0,2 tLxuua ttxx
Wave Equation: Initial and Boundary Conditions (3 of 5)
We assume that the ends of the string remain fixed, and hence
Since the wave equation is of second order with respect to t, it is plausible to prescribe two initial conditions, the initial position of the string, and its initial velocity:
where f and g are a given functions.
In order for these four conditions to be consistent, we require
,0),()0,(),()0,( Lxxgxuxfxu t
0,0),(,0),0( ttLutu
0)()0(,0)()0( LggLff
Wave Equation Problem (4 of 5)
Thus the wave equation problem is
This is an initial value problem with respect to t, and a boundary value problem with respect to x.
Alternatively, it is a boundary value
problem in xt-plane, with one condition
imposed at each point on semi-infinite
sides, and two imposed at each point
on the finite base.
Lxxgxuxfxu
ttLutu
tLxuua
t
ttxx
0),()0,(),()0,(
0,0),(,0),0(
0,0,2
Wave Equation Problems (5 of 5)
The wave equation governs a large number of other wave problems besides the transverse vibrations of an elastic string.
For example, it is only necessary to interpret the function u and the constant a appropriately to have problems dealing with water waves in an ocean, acoustic or electromagnetic waves in the atmosphere, or elastic waves in a solid body.
If more than one space dimension is significant, then we can generalize the wave equation, for example, to two dimensions:
This equation can be used to describe the motion of a thin drumhead, with suitable boundary and initial conditions.
ttyyxx uuua 2
Nonzero Initial Displacement (1 of 9)
Suppose the string is disturbed from its equilibrium position and then released at t = 0 with zero velocity to vibrate freely.
The vertical displacement u(x, t) must then satisfy
where f is a given function describing the configuration of the string at t = 0.
We will use the separation of variables method to obtain solutions of this problem.
Lxxuxfxu
ttLutu
tLxuua
t
ttxx
0,0)0,(),()0,(
0,0),(,0),0(
0,0,2
Separation of Variables Method (2 of 9)
As in Section 10.5, we begin by assuming
Substituting this into our differential equation
we obtain
or
where is a constant, as in Section 10.5.
We next consider the boundary conditions.
)()(),( tTxXtxu
ttxx uua 2
TXTXa 2
,0
0122
TaT
XX
T
T
aX
X
Boundary Conditions (3 of 9)
Our vibrating string problem is
Substituting u(x,t) = X(x)T(t) into the second of the initial conditions at t = 0, we find that
Similarly, the boundary conditions require X(0) = 0, X(L) = 0:
We therefore have the following boundary value problem in x:
Lxxuxfxu
ttLutu
tLxuua
t
ttxx
0,0)0,(),()0,(
0,0),(,0),0(
0,0,2
0)0(0,0)0()()0,( TLxTxXxut
0)()0(,0 LXXXX
0,0)()(),(,0)()0(),0( ttTLXtLutTXtu
Eigenvalues and Eigenfunctions (4 of 9)
From Section 10.1, the only nontrivial solutions to this boundary value problem are the eigenfunctions
associated with the eigenvalues
With these values for , the solution to the equation
is
where k1, k2 are constants. Since T'(0) = 0, k2 = 0, and hence
,3,2,1,/sin)( nLxnxX n
,3,2,1,/ 222 nLnn
02 TaT
,/sin/cos)( 21 LtankLtanktT
LtanktT /cos)( 1
Fundamental Solutions (5 of 9)
Thus our fundamental solutions have the form
where we neglect arbitrary constants of proportionality.
To satisfy the initial condition
we assume
where the cn are chosen so that the initial condition is satisfied:
,,3,2,1,/cos/sin),( nLtanLxntxun
Lxxfxu 0),()0,(
11
/cos/sin),(),(n
nn
nn LtanLxnctxuctxu
L
nn
n dxLxnxfL
cLxncxfxu0
1
/sin)(2
/sin)()0,(
Solution (6 of 9)
Therefore the solution to the vibrating string problem
is given by
where
L
n dxLxnxfL
c0
/sin)(2
1
/cos/sin),(n
n LtanLxnctxu
Lxxuxfxu
ttLutu
tLxuua
t
ttxx
0,0)0,(),()0,(
0,0),(,0),0(
0,0,2
Natural Frequencies (7 of 9)
Our solution is
For a fixed value of n, the expression
is periodic in time t with period T = 2L/na, and represents a vibratory motion with this period and frequency n a /L.
The quantities a = n a /L, for n = 1, 2, …, are the natural frequencies of the string – that is, the frequencies at which the string will freely vibrate.
LtanLxn /cos/sin
1
/cos/sin),(n
n LtanLxnctxu
Natural Mode (8 of 9)
Our solution is
For a fixed value of n, the factor
represents the displacement pattern occurring in the string as it vibrates with a given frequency.
Each displacement pattern is called a natural mode of vibration and is periodic in the space variable x.
The spatial period 2L/n is called the wavelength of the mode of frequency n a /L, for n = 1, 2, ….
Lxn /sin
1
/cos/sin),(n
n LtanLxnctxu
Graphs of Natural Modes (9 of 9)
Thus the eigenvalues n2 2/L2 of the vibrating string problem are proportional to the squares of the natural frequencies, and the eigenfunctions sin(n x /L) give the natural modes.
The first three natural modes are graphed below.
The total motion of the string u(x,t) is a combination of the natural modes of vibration and is also a periodic function of time with period 2L/a.
Example 1: Vibrating String Problem (1 of 5)
Consider the vibrating string problem of the form
where
300,0)0,(),()0,(
0,0),30(,0),0(
0,300,4
xxuxfxu
ttutu
txuu
t
ttxx
3010,20/)30(
100,10/)(
xx
xxxf
Example 1: Solution (2 of 5)
The solution to our vibrating string problem is
where
Thus
,2,1,3/sin9
30/sin20
30
30
230/sin
1030
2
22
30
10
10
0
nnn
dxxnx
dxxnx
cn
1
30/2cos30/sin),(n
n tnxnctxu
1
2230/2cos30/sin3/sin
9),(
n
tnxnnn
txu
Example 1: Displacement Pattern (3 of 5)
The graphs below of u(x,t) for fixed values of t shows the displacement pattern of the string at different times.
Note that the maximum initial displacement is positive and occurs for x = 10, while at t = 15, a half-period later, the maximum displacement is negative and occurs at x = 20.
The string then retraces its motion and returns to its original configuration at t = 30.
Example 1: Spatial Behavior Over Time (4 of 5)
The graphs below of u(x,t) for fixed values of x shows the behavior of the string at x = 10, 15, and 20, as time advances.
These plots confirm that the motion is periodic with period 30.
Observe also that each interior point on the string is motionless for one-third of each period.
Example 1: Graph of u(x,t) (5 of 5)
A three-dimensional plot of u versus x and t is given below.
Observe that we obtain the previous graphs by intersecting the surface below by planes on which either t or x is constant.
Justification of Solution (1 of 10)
At this stage, the solution to the vibrating string problem
where
is only a formal solution until a rigorous justification of the limiting processes is provided.
While such a justification is beyond our scope, we discuss certain features of the argument here.
1
/cos/sin),(n
n LtanLxnctxu
L
n dxLxnxfL
c0
/sin)(2
Partial Derivatives of Formal Solution (2 of 10)
It is tempting to try to justify the solution by substituting
into the equation, and boundary and initial conditions.
However, upon formally computing uxx, for example, we have
Due to the n2 factor in numerator, the series may not converge.
This may not necessarily mean that the series for u(x,t) is incorrect, but that it may not be used to calculate uxx and utt.
1
/cos/sin),(n
n LtanLxnctxu
1
2
/cos/sin),(n
nxx LtanLxnL
nctxu
Comparison of Formal Solutions (3 of 10)
A basic difference between solutions of the wave equation
and the heat equation
is the presence of the negative exponential terms in the latter, which approach zero rapidly and ensures the convergence of the series solution and its derivatives.
In contrast, series solutions of the wave equation contain only oscillatory terms that do not decay with increasing n.
1
/cos/sin),(n
n LtanLxnctxu
1
)/( /sin),(2
n
tLnn Lxnectxu
Alternate Way of Validating Solution (4 of 10)
There is an alternative way to validate our solution
indirectly. We will also gain additional information about the structure of the solution.
We will first show that this solution is equivalent to
where h is the odd periodic extension of f:
1
/cos/sin),(n
n LtanLxnctxu
2)()(),( atxhatxhtxu
)()2(,0),(
0),()( xhLxh
xLxf
Lxxfxh
Alternate Expression for Solution (5 of 10)
Since h is the odd extension of f, it has the Fourier sine series
Then using trigonometric identities
we obtain
Adding these equations, we obtain
L
nn
n dxLxnxfL
cLxncxh0
1
/sin)(2
,/sin)(
BABABA sincoscossin)sin(
1
1
/sin/cos/cos/sin)(
/sin/cos/cos/sin)(
nn
nn
LtanLxnLtanLxncatxh
LtanLxnLtanLxncatxh
2)()(/cos/sin),(1
atxhatxhLtanLxnctxun
n
Continuity of f (6 of 10)
Thus
where
Then u(x,t) is continuous for 0 < x < L, t > 0, provided that h is continuous on the interval (-, ). This requires f to be continuous on the original interval [0, L].
Also, recall the compatibility conditions in the vibrating string problem require f (0) = f (L) = 0.
Thus h (0) = h (L) = h (-L) = 0 also.
0,,2)()(),( tLxLatxhatxhtxu
)()2(,0),(
0),()( xhLxh
xLxf
Lxxfxh
Continuity of f ' and f '' (7 of 10)
We have
where
Note that u is twice continuously differentiable with respect to either variable in 0 < x < L, t > 0, provided h is continuously twice differentiable on (-, ). This requires f ' and f '' to be continuous on [0, L].
0,,2)()(),( tLxLatxhatxhtxu
)()2(,0),(
0),()( xhLxh
xLxf
Lxxfxh
Endpoint Requirements for f '' (8 of 10)
We have
where
Assume h is twice continuously differentiable on (-, ). Since h'' is the odd extension of f '', we must have f ''(0) = 0 and f ''(L) = 0.
However, since h' is the even extension of f ', no further conditions are required on f '.
0,,2)()(),( tLxLatxhatxhtxu
)()2(,0),(
0),()( xhLxh
xLxf
Lxxfxh
Solution to Wave Equation (9 of 10)
We have
where
Provided that all of these conditions are met, uxx and utt can be computed by the above formulas for u and h.
It can then be shown that these derivatives satisfy the wave equation, and the boundary and initial conditions are satisfied.
Thus u(x,t) is a solution to the vibrating string problem, where
0,,2)()(),( tLxLatxhatxhtxu
)()2(,0),(
0),()( xhLxh
xLxf
Lxxfxh
L
nn
n dxLxnxfL
cLtanLxnctxu0
1
/sin)(2
,/cos/sin),(
Effects of Initial Discontinuities (10 of 10)
If some of the continuity requirements are not met, then u is not differentiable at some points in the semi-infinite strip 0 < x < L, and t > 0, and thus u is a solution of the wave equation only in a somewhat restricted sense.
An important physical consequence of this observation is that if there are any discontinuities present in the initial data, then they will be preserved in the solution u(x,t) for all time.
In contrast, in heat conduction problems, initial discontinuities are instantly smoothed out.
See text for more details.
Vibrating String Problem for f = 0 (1 of 6)
Suppose the string is set in motion from its equilibrium position with a given velocity.
Then the vertical displacement u(x, t) must satisfy
where g is the initial velocity at the point x of the string.
We will use the separation of variables method to obtain solutions of this problem.
Lxxgxuxu
ttLutu
tLxuua
t
ttxx
0),()0,(,0)0,(
0,0),(,0),0(
0,0,2
Separation of Variables Method (2 of 6)
As shown previously for the wave equation, assuming
leads us to the two ordinary differential equations
The boundary conditions require X(0) = 0, X(L) = 0, and thus
The only nontrivial solutions to this boundary value problem are the eigenvalues and eigenfunctions
Then T(t) satisfies
)()(),( tTxXtxu
0,0 2 TaTXX
0)()0(,0 LXXXX
,3,2,1,/sin)(,/ 222 nLxnxXLn nn
0/ 2222 TLnaT
Boundary Conditions (3 of 6)
Recall that the initial conditions are
Substituting u(x,t) = X(x)T(t) into the first of these conditions,
Therefore T(t) satisfies
with solution
where k1, k2 are constants.
Since T(0) = 0, it follows that k1 = 0, and hence
0)0(0,0)0()()0,( TLxTxXxu
Lxxgxuxu t 0),()0,(,0)0,(
0)0(,0/ 2222 TTLnaT
,/sin/cos)( 21 LtankLtanktT
LtanktT /sin)( 1
Fundamental Solutions (4 of 6)
Thus our fundamental solutions have the form
where we neglect arbitrary constants of proportionality.
To satisfy the initial condition
we assume
where the kn are chosen so that the initial condition is satisfied.
,,3,2,1,/sin/sin),( nLtanLxntxun
Lxxgxut 0),()0,(
11
/sin/sin),(),(n
nn
nn LtanLxnktxuktxu
Initial Condition (5 of 6)
Thus
where the kn are chosen so that the initial condition is satisfied:
Hence
or
11
/sin/sin),(),(n
nn
nn LtanLxnktxuktxu
1
/sin)()0,(n
nt LxnkL
anxgxu
L
n dxLxnxgL
kL
an0
/sin)(2
L
n dxLxnxgan
k0
/sin)(2
Solution (6 of 6)
Therefore the solution to the vibrating string problem
is given by
where
1
/sin/sin),(n
n LtanLxnctxu
Lxxgxuxu
ttLutu
tLxuua
t
ttxx
0),()0,(,0)0,(
0,0),(,0),0(
0,0,2
L
n dxLxnxgan
c0
/sin)(2
General Problem for Elastic String (1 of 3)
Suppose the string is set in motion from a general initial position with a given velocity.
Then the vertical displacement u(x, t) must satisfy
where f is the given initial position and g is the initial velocity at the point x of the string.
We could use separation of variables to obtain the solution.
However, it is important to note that we can solve this problem by adding together two solutions that we obtained earlier.
Lxxgxuxfxu
ttLutu
tLxuua
t
ttxx
0),()0,(),()0,(
0,0),(,0),0(
0,0,2
Separate Problems (2 of 3)
Let v(x,t) satisfy
and let w(x,t) satisfy
Then u(x,t) = v(x,t) + w(x,t) satisfies the general problem
Lxxvxfxv
ttLvtv
tLxvva
t
ttxx
0,0)0,(),()0,(
0,0),(,0),0(
0,0,2
Lxxgxwxw
ttLwtw
tLxwwa
t
ttxx
0),()0,(,0)0,(
0,0),(,0),0(
0,0,2
Lxxgxuxfxu
ttLutu
tLxuua
t
ttxx
0),()0,(),()0,(
0,0),(,0),0(
0,0,2
Superposition (3 of 3)
Then u(x,t) = v(x,t) + w(x,t) satisfies the general problem
where
This is another use of the principle of superposition.
L
nn
n
L
nn
n
dxLxnxgan
kLtanLxnktxw
dxLxnxfL
cLtanLxnctxv
01
01
/sin)(2
,/sin/sin),(
/sin)(2
,/cos/sin),(
Lxxgxuxfxu
ttLutu
tLxuua
t
ttxx
0),()0,(),()0,(
0,0),(,0),0(
0,0,2