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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.
2
2 2 2
2
x
u c t
u
[1]
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
t
u
t t
u [4]
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.
F
d
dF F G
u [5]
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.
2
2 ( ) ''( )
d
d F F d
dF F [6]
In a similar fashion we can write
G
d
dG F G
u [7]
Finally, from the definitions of ξ = x + ct and η = x - ct, we can write the following partial
derivatives.
c t
c t
[8]
Combining this equation with equations [4], [5], and [7] gives the following result.
cF ' cG ' c ( F ' G ') t
u
[9]
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.
2 2
2 c c F G c cF G c F G t
u
t t
u
t t
u
t t
u
[10]
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.
x x
[11]
With these relationships, we can obtain the first derivative with respect to x as follows.
[ ( ) ( )]
[ ( ) ( )]
F G
u F G F G
x
u
x x
u
[12]
The second derivative is obtained by taking the derivative of the first derivative.
2 (^1 ) ^ ( ' ')^ (^1 ) (^ ' ')^ ( '' '')
2 F G F G F G x
u
x x
u
x x
u
x x
u
[13]
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].
2 2
2 2 2 2
2
2
2 2 2
2 c F G x
u c F G c t
u
x
u c t
u
[14]
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.
We have used the usual notation, g’, to indicate an ordinary first derivative.
d
dg g ' [20]
In a similar fashion we can take the second-order space derivative of the integral by starting with
the first derivative.
gx ct g x ct x
g g x ct g x ct
x
g g x ct x
x ct g x ct x
x ct g d x
x ct
x ct
[21]
We then take the second derivative from the result above.
2
g x ct g x ct x ct
g x ct
x
x ct
x ct
g x ct
x
x ct
gx ct g x ct x
g d x x
g d x
x ct
xct
x ct
xct
[22]
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.
^ ^
xct
xct
x ct
x ct
g d x
g d c g x ct g x ct c t
2 2 2 2
2 [23]
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.
0
0
g d f x f x f x c
u x f x f x
x
x
[24]
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.
cg x ct cg x ct c
cf x ct c f x ct
cg x ct cg x ct x ct c
f x ct
t
x ct
x ct
f x ct
t
x ct
g d t c t
f x ct
t
f x ct
t
u xt
xct
xct
[25]
Setting t = 0 in this equation shows that g(x) is the initial velocity.
0
cg x cg x g x c
cf x c f x t
uxt
t
[26]
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.
otherwise
x x
x x
g x allx f x
0
( ) 0 ( ) [27]
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.
-5 -4 -3 -2 -1 0 1 2 3 4 5
distance, x
time, t
t = 0 ct= ct = 2 ct = 3 x+ct x+ct x-ct x-ct
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.
Equation [32] shows that we have two separate differential equations, each of which has a known
general solution. These equations and their general solutions are shown below. 2
( ) 0 ( ) sin( ) cos( )
2
2
X x X x A x B x
dx
d X x
[33]
() 0 () sin( ) cos( )
2
2 Tt T t C ct D ct dt
dTt [34]
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 ) A sin( 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 A sin( 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 A sin( 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
L
n L n or
[38]
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 = nx/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 Bn.
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
^
1
( ,) sin( ) cos( )sin( ) n
uxt An nct Bn nct nx [39]
(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.
^
1 1
( , 0 ) ( ) sin( 0 ) cos( 0 )sin( ) sin( ) n
n n n
ux f x An nc Bn nc nx B x [40]
We can obtain an equation for Bn by using the orthogonality relationships for integrals of the sine.
If we multiply both sides by sin(mx/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.
L
m n
L
n
L
n
n
L
dx L
m x dx B L
n x
L
m x B
dx L
n x
L
m x dx B L
m x f x
0
2
1 0
0 0 1
sin sin sin
( )sin sin sin
[41]
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
integral^3 in equation [41] gives the following result.
L
L
L
m dx L
m x f x L dx L
m x
dx L
m x f x
B 0
0
2
(^0 2) ( )sin
sin
( )sin
[42]
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.
^
1
cos( ) sin( )sin( )
n
n cAn nct Bn nct nx t
uxt
[43]
Using this equation for the initial condition on velocity as g(x) gives.
^ ^
1 1
( , 0 ) ( ) cos( 0 ) sin( 0 )sin( ) sin( ) n
n n n n
x g x n cAn nc Bn nc nx cA x t
u [44]
The eigenvalue expansion to obtain the coefficients An 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:
sin 2 4
sin 2 4
sin 0 0
2 L
L
m L
m
L L
L
m x
m
x L dx L
m x
L L
^
1
( , ) sin sin n
n L
n x
L
n ct u xt A
[53]
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
cos
cos 2
n
n L
n x ct
L
n x ct uxt A
[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
1
sin
sin 2
cos
cos 2
n
n
n
n
L
n x ct
L
n x ct B
L
n x ct
L
n x ct uxt A
[58]
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 1
sin sin 2
cos cos 2
n
n n
n L
n x
L
n x B L
n x
L
n x u x A
[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 1
sin sin sin 2
n
n n
n L
n x B L
n x
L
n x ux f x B
[60]
Equation [60] tells us that
1
( ) sin n
n L
n x f x B
so that
1 1
( ) sin
( ) sin n
n n
n L
n x ct and f x ct B L
n x ct f x ct B
[61]
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
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.
-
-0.
0
0.
1
-2 -1 0 1 2
x
f(x)
Initial Conditions Periodic Extension
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.
-
-0.
0
0.
1
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 x
f(x)
(^) f(x) f(x+ct)/ f(x - ct)/ ct = 0.
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 =
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.
