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