wave

College of Engineering and Computer Science

Mechanical Engineering Department

Engineering Analysis Notes

Larry Caretto December 21, 2011





Solutions to the Wave Equation

The wave equation

The one-dimensional wave equation, shown below, describes the propagation of a disturbance,

u, over space and time. For example, u might be the amplitude of a vibrating string which varies

with space and time.



 2u  2u

 c2 2 [1]

 2t x



The D’Alambert solution and its proof

The D’Alambert solution to equation [1] is written in terms of coordinates ξ and η, defined as

follows:



  x  ct and   x  ct [2]



The D’Alambert solution to the wave equation is written in terms of two arbitrary functions, F(ξ)

and G(η). This gives the following solution.



u  F ( )  G()  F ( x  ct )  G( x  ct ) [3]



To show that this is a solution, we have to rewrite the wave equation in terms of these two

functions. This means that we have to transform the coordinates in the wave equation from (x,t)

to (ξ,η). To start the coordinate transformations we look at the first derivative terms. If we have u

as a function of  and , we can get the time derivatives with respect to t by the following

equation.



u  u  u

  [4]

t t  t 



The D’Alambert solution gives simple expressions for the ξ and η derivatives since F is a function

of ξ only and G is a function of η only.



u  dF ()

 F ()  G ()   F ' ( ) [5]

  d



Here we use the notation F’(ξ) for the first derivative of F with respect to ; we will subsequently

define a second derivative in a similar manner.



dF ( ) d 2 F ( )

F ' ( )  F ' ' ( )  [6]

d d 2





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Wave equation solutions L. S. Caretto, December 21, 2011 Page 2





In a similar fashion we can write



u 

 F ()  G()  dG()  G' () [7]

  d



Finally, from the definitions of ξ = x + ct and η = x - ct, we can write the following partial

derivatives.



 

c  c [8]

t t

Combining this equation with equations [4], [5], and [7] gives the following result.



u

 cF 'cG'  c( F 'G' ) [9]

t

We find the second time derivative by taking the time derivative of this first derivative. We can

simplify the result using the equations above for coordinate transformation, the D’Alambert

solution that u = F + G, and the definitions of F’’ and G’’ as ordinary second derivatives.



 2u  u   u   u  

    c c( F 'G ' )  c c( F 'G ' )  c 2 ( F ' 'G ' ' ) [10]

t 2

t t t  t t  t  



We can repeat this process for the x-derivatives; the main difference is in the partial derivatives of

the new coordinates with respect to x.



 

1 1 [11]

x x

With these relationships, we can obtain the first derivative with respect to x as follows.



u  u  u [ F ( )  G ( )] [ F ( )  G ( )]

   (1)  (1)  F ' ( )  G ' ( ) [12]

x x  x   



The second derivative is obtained by taking the derivative of the first derivative.



 2u  u   u   u  

    (1) ( F 'G ' )  (1) ( F 'G ' )  ( F ' 'G ' ' ) [13]

x 2

x x x  x x  x  



We can now show that the two expressions for the second derivatives in equations [10] and [13]

satisfy the original two-dimensional wave equation in [1].



 2u  2u  2u  2u

 c2 2  c 2 F ' 'G ' ' c2  c 2 F ' 'G ' ' [14]

 2t x t2

x 2







Thus, the D’Alambert solution, u = F(x + ct) + G(x – ct), where F and G are arbitrary functions

satisfies the differential equation. We now have to show how we can use this solution to satisfy

the differential equation and the initial conditions.

Wave equation solutions L. S. Caretto, December 21, 2011 Page 3





The D’Alambert solution with initial conditions

We want to be able to satisfy arbitrary initial conditions on the displacement, u, and its first

derivative, the velocity, at t = 0. These arbitrary initial conditions are written as follows.



u

u (0, x)  f ( x) and  g ( x) [15]

t x 0





In this section we show that the solution to the wave equation [1] with the boundary conditions in

equation [15] can be written as follows.



x  ct

u (t , x)   f ( x  ct )  f ( x  ct ) 

1 1

2  g ( )d

2c x  ct

[16]





In this equation, we have used the D’Alambert solution for the first two terms where the arbitrary

functions F() and G() in the D’Alambert solution, are both the initial condition function f(). We say

that u(t=0,x) = f(x) and u(t,x) has a contribution f(x + ct) and f(x – ct). This means that we use the

same functional form, but the argument to the function f() at later times is computed as x + ct and

x – ct. For example if f(x) = 1 when x = 0 then f(x + ct) would equal 1 when x + ct = 0. (This

would occur at all points where x = -ct.



The final term in equation [16] is the integral of the initial derivative condition that is integrated

over the dummy variable, . The first two terms have just been shown to satisfy the differential

equation. (Recall that any function of x + ct or x – ct satisfied the differential equation.) In order

to show that the integral term in equation [16] satisfies the differential equation we need to use

the general formula for differentiation under the integral sign.



b( y )

 b a

y  ( x)dx  y (b)  y (a) [17]

a( y)



We can apply this general result to compute the space and time derivatives for the wave

equation. First find the time derivatives. Using the general rule above, we find the following

result.



x  ct

  ( x  ct ) ( x  ct )

t  g ()d  t

g ( x  ct ) 

t

g ( x  ct )

x  ct

[18]

 cg ( x  ct )  (c) g ( x  ct )  cg ( x  ct )  g ( x  ct )



The second derivative of the integral is just the derivative of the first derivative, which becomes



x  ct x  ct

2   

t 2  g ( )d 

x  ct

ctg ( )d  t cg ( x  ct )  cg ( x  ct )

t t x 

[19]

  ( x  ct ) g ( x  ct )  ( x  ct ) g ( x  ct ) 

 c    c g ' ( x  ct )  g ' ( x  ct )

2



 t  ( x  ct ) t  ( x  ct ) 

Wave equation solutions L. S. Caretto, December 21, 2011 Page 4





We have used the usual notation, g’, to indicate an ordinary first derivative.



dg  

g '    [20]

d

In a similar fashion we can take the second-order space derivative of the integral by starting with

the first derivative.



x  ct

  ( x  ct )  ( x  ct ) g ( )

ctg ( )d  x g ( x  ct )  x g ( x  ct )  x

x x  [21]

g ( )

 (1) g ( x  ct )  (1) g ( x  ct )   g ( x  ct )  g ( x  ct )

x



We then take the second derivative from the result above.



x  ct x  ct

2   

2 

x x  ct

g ( )d  ctg ( )d  x g ( x  ct )  g ( x  ct )

x x x  [22]

( x  ct ) g ( x  ct ) ( x  ct ) g ( x  ct )

  )  (1) g ' ( x  ct )  (1) g ' ( x  ct )

x  ( x  ct ) x ( x  ct )



If we substitute equations [19] and [22] into the original two-dimensional wave equation in [1] we

see that the integral term satisfies the differential equation.



x  ct x  ct

2 2

 g ( )d  c g ' ( x  ct )  g ' ( x  ct )  c  g ( )d

2 2

[23]

t 2 x  ct

x 2 x  ct





We can now show that the integral is added to the solution to satisfy the initial conditions. First

consider the condition that u(0,x) = f(x). Setting t = 0 in the proposed solution gives.



x0

u (0, x)   f ( x  0)  f ( x  0)  0g ( )d  2  f ( x)  f ( x)  0  f ( x)

1 1 1

[24]

2 2c x 



Here we use the result that the value of a definite integral is zero when both the lower and upper

limits are the same. Taking the time derivative of the solution and using equation [18] for the first

derivative of the integral in the solution gives the following result.



x  ct

u ( x, t ) 1  f ( x  ct ) f ( x  ct )  1 

t

 

2 t



t   2c t  g ( )d

 x  ct





1   ( x  ct ) f ( x  ct )  ( x  ct ) f ( x  ct )  1

 cg ( x  ct )  cg ( x  ct )

[25]

 

2  t

  ( x  ct ) t  ( x  ct )  2c





1

cf ' ( x  ct )  (c) f ' ( x  ct )  1 cg ( x  ct )  cg ( x  ct )

2 2c

Setting t = 0 in this equation shows that g(x) is the initial velocity.

Wave equation solutions L. S. Caretto, December 21, 2011 Page 5





u ( x, t )

 cf ' ( x)  (c) f ' ( x)  cg ( x)  cg ( x)  g ( x)

1 1

[26]

t t  0 2 2c



Thus the solution proposed in equation [16] satisfies the wave equation and the initial conditions.



Propagation of initial conditions

Consider the following initial conditions giving g = 0 and u a simple triangular profile initially.



x  1 1  x  0



g ( x)  0 all x f ( x)   1  x 0  x 1 [27]

0

 otherwise



With g(t) everywhere zero, equation [16] tells us that u(t,x) = [f(x+ct) + f(x-ct)]/2. This solution

results in the following profiles of the initial conditions at later times as shown in the figure below.









t=0

ct=1

ct = 2

time, t









ct = 3

x+ct

x+ct

x-ct

x-ct









-5 -4 -3 -2 -1 0 1 2 3 4 5

distance, x





The solutions at later times are computed from equation [16] in the following way. The argument

is easier to understand if we talk about the initial condition as f(ξ) or f(η) where ξ = x + ct and η =

x – ct. Of course, f is a single function, as defined in equation [27], such that f(ξ) or f(η) is zero

except where its argument lies between -1 and +1. Between these points this function has the

triangular shape defined in equation [27] and shown as the initial conditions in the figure above.



At a point (t,x) where the time, t, = a/c, the value of ξ = x + ct = x + a. Since the original initial

condition is zero at all points except where -1 ≤ ξ ≤ 1, f(x + a) will be zero except for the region

where -1 ≤ x + a ≤ 1, that is the region where -1-a ≤ x ≤ 1-a. For example, when a = ct = 2, f(x +

ct) will be zero only in the region -3 ≤ x ≤ -1. For ct = 2, then, the f(x+ct) term will occur in this

region. Similarly, when ct = a, the solution for f(η) = f(x – ct) will be zero at all points outside the

region -1 ≤ η ≤ 1, which corresponds to -1 ≤ x – a ≤ 1 or a–1 ≤ x ≤ 1+a. Thus when a = ct = 2 the

u(x,t) term resulting from f(x – ct) will reflect the initial condition between x = 1 and x = 3.

Wave equation solutions L. S. Caretto, December 21, 2011 Page 6





Because the solution u(t,x) = [f(x+ct) + f(x-ct)]/2, the magnitude of the initial condition will be cut in

half in each component.



Solution by separation of variables

Consider the wave equation with the following set of initial and boundary conditions.



 2u 1  2u

  0 0  x  L, t  0

t 2 c 2 x 2 [28]

u

u(0, t )  u( L, t )  0 u(0, t )  f ( x ) (0, t )  g ( x )

t

Here, the boundary conditions give u = 0 at x = 0, x = L. The initial conditions on u and its first

derivative with respect to time are both known functions of x. These functions are typically called

the initial displacement and the initial velocity, respectively.



The solution of the wave equation by separation of variables proceeds in a manner similar to the

solution of other partial differential equations. We postulate a solution that is the product of two

functions, X(x) a function of x only and T(t) a function of time only. With this assumption, our

solution becomes.



u(x,t) = X(x)T(t) [29]



We do not know, in advance, if this solution will work. However, we assume that it will and we

substitute it for u in equation [28]. Since X(x) is a function of x only and T(t) is a function of t only,

we obtain the following result when we substitute equation [29] into equation [28].



1  2u  2u 1  2 X ( x )T (t )  2 X ( x )T (t ) X ( x )  2T (t ) 2 X ( x)

 2  2   2  T (t )  0 [30]

c 2 t 2 x c t 2 x 2 c t 2 x 2



If we divide the final equation through by the product X(x)T(t), and move the x derivative to the

other side of the equal sign, we obtain the following result.



1  2 X ( x) 1  2T ( t )

 2 [31]

X ( x ) x 2 c T (t ) t 2



The left hand side of equation [31] is a function of x only; the right hand side is a function of y

only. The only way that this can be correct is if both sides equal a constant. This also shows that

the separation of variables solution works. In order to simply the solution, we choose the

1 2

constant to be equal to . This gives us two ordinary differential equations to solve.



1  2 X ( x) 1  2T (t )

 2   2 [32]

X ( x ) x 2

c T (t ) t 2









The choice of – for the constant as opposed to just plain  comes from experience. Choosing

1 2



the constant to have this form now gives a more convenient result later. If we chose the constant

to be simply , we would obtain the same result, but the expression of the constant would be

awkward.

Wave equation solutions L. S. Caretto, December 21, 2011 Page 7





Equation [32] shows that we have two separate differential equations, each of which has a known

2

general solution. These equations and their general solutions are shown below.



d 2 X ( x)

2

 2 X ( x)  0  X ( x)  A sin( x)  B cos(x) [33]

dx



d 2T (t )

2

 2T (t )  0  T (t )  C sin(ct )  D cos(ct ) [34]

dt



From the solutions in equations [33] and [34], we can write the general solution for u(x,y) =

X(x)T(t) as follows.



u( x, t )  Asin(x )  B cos(x )C sin(ct )  D cos(ct ) [35]



We now apply the boundary conditions shown with the original equation [28] to evaluate the

constants A, B, C, and D. If we substitute the boundary condition that u = 0 at x = 0 into equation

[35], get the following result.



u(0, t )  0  Asin( 0)  B cos( 0)C sin(ct )  D cos(ct ) [36]



Because sin(0) = 0 and cos(0) = 1, equation [36] will be satisfied for all y only if B = 0. Thus, we

set B = 0. Next we apply the solution in equation [35] (with B = 0) to the boundary condition at x

= L.



u( L, t )  0  Asin(L)C sin(ct )  D cos(ct ) [37]



Equation [37] can only be satisfied if the sine term is zero. This will be true only if L is an

integral times . If n denotes an integer, we must have



n

L  n or  [38]

L

Since any integral value of n gives a solution to the original differential equations, with the

boundary conditions that u = 0 at the two boundaries considered so far, the most general solution

is one that is a sum of all possible solutions, each multiplied by a different constant. In the

general solution for one value of n, which we can now write as Asin(nx)[Csin(nct) + Dcos(nct)],

with n = nx/L, we can write the product of two constants, AC, as the single constant, An., which

may be different for each value of n. Similarly we can write the product AD as the constant B n.

Again, this constant can be different for different values of n. The general solution which is a sum

of all solutions with different values of n is written as follows





u( x, t )   An sin(n ct )  Bn cos(n ct )sin(n x ) [39]

n 1





(We start with n = 1 since n = 0 gives zero for the eigenfunction.) We can use eigenfunction

expansions to get the initial conditions on displacement and velocity.





2

As usual, you can confirm that this solution satisfies the differential equation by substituting the

solution into the differential equation.

Wave equation solutions L. S. Caretto, December 21, 2011 Page 8





 

u( x,0)  f ( x )   An sin(n c0)  Bn cos(n c0)sin(n x )   Bn sin(n x ) [40]

n 1 n 1





We can obtain an equation for Bn by using the orthogonality relationships for integrals of the sine.

If we multiply both sides by sin(mx/L), where m is another integer, and integrate from a lower

limit of zero to an upper limit of L, we get the following result.



 mx  

 mx   nx 

L L



 f ( x ) sin  dx    Bn sin  sin dx

0  L  0 n 1  L   L 

[41]

 L

 mx   nx  2  mx 

L

   Bn sin  sin dx  Bm  sin  dx

n 1 0  L   L  0  L 



In the second row of equation [41] we can reverse the order of summation and integration

because these operations commute. We then recognize that the integrals in the summation all

vanish unless m = n, leaving only this integral to evaluate. Solving for Bm and evaluating the last

3

integral in equation [41] gives the following result.



 mx 

L



 f ( x ) sin

 L 

dx

2

L

 mx 

Bm  0

  f ( x ) sin dx [42]

2  m x   L 

L

L0

 sin  L dx

0  



For any initial displacement, then, we can perform the integral on the right hand side of equation

[42] to compute the values of Bm and substitute the result into equation [39]. We find the value of

the constants Am in a similar manner by fitting the condition for the initial velocity. First we take

the time derivative of our solution for u.



u( x, t ) 

  n cAn cos(n ct )  Bn sin(n ct )sin(n x ) [43]

t n 1





Using this equation for the initial condition on velocity as g(x) gives.



u  

( x,0)  g ( x )   n cAn cos(n c0)  Bn sin(n c0)sin(n x )   n cAn sin(n x ) [44]

t n 1 n 1





The eigenvalue expansion to obtain the coefficients A n proceeds in exactly the same manner as

the expansion used to find Bn. The only differences are in the use of g(x) rather than f(x) and the

presence of the factor of λnc multiplying An. Accounting for these differences, we can infer the

equation for An from equation [42] for Bm.







3

Using a standard integral table, and the fact that the sine of zero is zero and the sine of m is

zero for integer m, we find the following result:

L

2  mx  x  2mx   2mL  L

L

L L L

 sin  L dx   2  4m sin L  0  2  4m sin L   2

0       

Wave equation solutions L. S. Caretto, December 21, 2011 Page 9





L

L  nx 

nc 

g ( x) sin 

 L 

dx

2

L

 nx 

An  0

L

 nx 



nc 

g ( x) sin 

 L 

dx [45]



 sin 2 

 L 

dx 0



0



For any given initial conditions f(x) and g(x), we can find the values of A n and Bn by performing

the integrals in equations [42] and [45]. We can then substitute these results into equation [39],

rewritten below with the substitution of nπ/L for λ.





  nct   nct   nx 

u( x, t )    An sin   Bn cos  sin  [46]

n 1   L   L   L 



We can first consider the case where the initial velocity is zero at all points. According to

equation [45] this will give all values of An = 0 so that our equation becomes.





 nct   nx 

u( x, t )   Bn cos  sin  [47]

n 1  L   L 



We can show that this form is equivalent to the D’Alembert solution by using the trigonometric

identities for the sine of the sum and difference of two angles. These formulae are written below.

(Note that we can obtain the second formula from the first one by setting y = –y in the first

equation and using the facts that sin(–y) = –sin(y) and cos(–y) = cos(y).)



sin(x + y) = sin x cos y + sin y cos x [48]



sin(x – y) = sin x cos y – sin y cos x [49]



Adding these two equations gives



sin(x + y) + sin(x – y) = 2 sin x cos y [50]



We can apply this result to equation [47] writing the product of the two trigonometric functions as

follows.



 nx   nct   nx nct   nx nct 

2 sin  cos   sin    sin   [51]

 L   L   L L   L L 



Simplifying the result and substituting it into equation [47] gives.



1    n ( x  ct )   n ( x  ct ) 

u ( x, t )   Bn sin L   sin L 

2 n 1     

[52]





Equation [52] gives us the form of the D’Alembert solution with the x + ct and x – ct dependence

of two functions.



In a similar fashion we can consider the case where the displacement is zero giving the following

equation.

Wave equation solutions L. S. Caretto, December 21, 2011 Page 10







 nct   nx 

u ( x, t )   An sin   sin   [53]

n 1  L   L 



As before we can manipulate the terms using trigonometric relations for the cosine of the sum

and difference of two angles. These are



cos(x + y) = cos x cos y – sin y sin x [54]



cos(x – y) = cos x cos y + sin y sin x [55]



Subtracting equation [54] from equation [55] gives



cos(x – y) – cos(x + y) = 2 sin x sin y [56]



We can apply this result to equation [53] writing the product of the two trigonometric functions as

follows.



1    n ( x  ct )   n ( x  ct ) 

u ( x, t )   An cos L   cos L 

2 n 1     

[57]





We could have done the same operations with the general equation in which both f(x) and g(x)

are nonzero to give.



1    n ( x  ct )   n ( x  ct ) 

u ( x, t )   An cos L   cos L 

2 n1     

[58]

1    n ( x  ct )   n ( x  ct ) 

  Bn sin   sin 

2 n1   L   L 



Although the equation was solved for the region 0 ≤ x ≤ L, we know that the periodic functions of

sine and cosine extend beyond this region. In fact, they will produce periodic extensions of the

original solution for t = 0 which is shown below.



1    nx   nx  1    nx   nx 

u( x,0)     L   L   2  Bn sin L   sin L 

2 n 1

An cos   cos

 n 1     

[59]





The cosine terms cancel in this equation leaving only the sine terms. In the equation below, we

recognize that the initial condition, u(x,0) is simply f(x).



1    nx   nx    nx 

u( x,0)  f ( x )   Bn sin L   sin L    Bn sin L 

2 n 1      n 1  

[60]







 n x 

Equation [60] tells us that f ( x )   Bn sin   so that

n 1  L 



 n ( x  ct )  

 n ( x  ct ) 

f ( x  ct )   Bn sin  and f ( x  ct )   Bn sin  [61]

n 1  L  n 1  L 

Wave equation solutions L. S. Caretto, December 21, 2011 Page 11





For the case where the initial velocity, g(x), is zero, which is given in equation [52], we see that

the separation of variables solution reduces to the D’Alembert solution that u(x,t) = [f(x + ct) – f( x

– ct)]/2.



Because the initial condition, f(x) is expressed in terms of sine eigenfunctions, we have the same

results as a Fourier series in terms of sines. A Fourier sine series gives an odd periodic

extension of the initial conditions in regions beyond the boundary. This is illustrated in the figure

below for a region 0 ≤ x ≤ 1, and an initial condition that is a triangular peak, with a height of one,

at the midpoint of the region.





1

0.5 Initial

f(x)









Conditions

0 Periodic

Extension

-0.5

-1

-2 -1 0 1 2

x



Although the initial conditions are defined only for 0 ≤ x ≤ 1, the sine expansion gives the odd

periodic extensions of the initial conditions beyond the actual region. These extensions are the

basis for the f(x + ct) and f(x – ct) terms in the region for large times. The chart below shows the

propagation of the initial conditions shown above at a time where ct = 0.4.



1



0.5

f(x)

f(x)









0 f(x+ct)/2

f(x - ct)/2

-0.5 ct = 0.4



-1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x



At this point, the triangular initial conditions which are nonzero between x = 0.4 and x = 0.6 have

just reached the boundaries of the region 0 ≤ x ≤ 1. However, as the propagation of the actual

initial conditions is about the leave the region, the periodic extensions are about to enter the

region. Thus we see that the periodic extension centered at x = 1.5 has a component f(x + ct)

that is now centered at 1.1 and is about to propagate into the actual region from the right.

Similarly, the periodic extension at centered at x = –0.5 has a component f(x + ct) that is now

centered at x = –0.1, and is about to enter the region from the left.



When these periodic extensions enter the region, the two solution components will have positive

and negative components that will cancel. The effect of this is shown in the three charts below

for ct = 0.45. The top two charts show the plots of f(x+ct) and f(x-ct), respectively. In the bottom

chart, the expression for the wave form, u(x,t) = [f(x+ct) + f(x-ct)]/2, is plotted.



At the left side of the top chart for f(x+ct) we see the periodic extension of the initial condition that

was initially centered at x = 1.5, whose triangular base extended from 1.4 to 1.6. (Since t = 0

initially, the value of x + ct = 1.4 at the start of this triangle when t = 0. When ct = 0.45, x + ct =

Wave equation solutions L. S. Caretto, December 21, 2011 Page 12





1.4 corresponds to x + 0.45 = 1.4 or x = 0.95. Thus, the start of the triangle which was at x = 1.4

when t = 0 will be located at 0.95 when ct = 0.45. A similar analysis will show that there is a

periodic extension of the initial condition, originally centered at x = –0.5, whose base extends

from –0.6 to –0.4. The part of the triangle that was at x – ct = –0.4 when t = 0 will be located at

an x coordinate given by the equation x – 0.45 = –0.05. This is where we see the start of the

triangle in the plot of f(x-ct).



1





0.5

f(x+ct)









ct = 0.450

0





-0.5

0 0.2 0.4 0.6 0.8 1

x

1



0.5

f(x-ct)









ct = 0.450

0



-0.5

0 0.2 0.4 0.6 0.8 1

x

1



0.5

u(x,t)









ct = 0.450

0



-0.5

0 0.2 0.4 0.6 0.8 1

x





The net solution, u(x,t) = [f(x+ct) + f(x-ct)]/2 is zero between x = 0.15 and 0.85 where both f(x+ct)

and f(x-ct) are zero. Between x = 0.85 and x = 0.95, f(x+ct) is zero, and f(x-ct) is positive, so

u(x,t) is simply f(x-ct)/2. However, between x = 0.95 and x = 1, f(x+ct) is negative, and f(x-ct) is

positive, so u(x,t) = [f(x+ct) + f(x-ct)]/2 has both a positive and negative contribution, which makes

the slope of the wave form more strongly negative.



A similar effect occurs between x = 0.05 and x = 0.15, f(x-ct) is zero, and f(x+ct) is positive, so

u(x,t) is simply f(x+ct)/2. However, between x = 0 and x = 0.05, f(x-ct) is negative, and f(x+ct) is

positive, so u(x,t) = [f(x+ct) + f(x-ct)]/2 has both a positive and negative contribution, which makes

the slope of the wave form more strongly positive.

Wave equation solutions L. S. Caretto, December 21, 2011 Page 13





In summary, the periodic extensions that enter the region from either side cancel part of the initial

conditions that have propagated from the center point of the region, giving a steeper slope to the

start and finish of the solution for u(x,t). When ct = 0.5, the periodic extensions will exactly cancel

the components propagating from x = 0.5 giving a zero solution in the entire region 0 ≤ x ≤ 1.