Chapter 7
Partial Differential Equations
7.3 Hyperbolic Equations
7.3-2 D’Alembert’s Method
Consider the one-dimensional wave equation
2
 2u
2 u
=
c
t 2
x 2
0 < x < L, t > 0
(7.3-1)
The boundary and initial conditions required for the solution of the wave equations are
B.C. : u(0,t) = 0
and
u(L,t) = 0, for t  0
I.C. : u(x,0) = f(x)
and
u
(x,0) = g(x), for 0  x  L
t
f(x) and g(x) are known initial position and initial velocity, respectively. Another solution of
the wave equation is given by d’Alembert as
u(x,t) =
1 *
1
[f (x  ct) + f*(x + ct)] +
2
2c

x  ct
x ct
g * ( s)ds
(7.3-2)
where f* and g* denote the odd extension of f(x) and g(x). The reason why the odd extension
is used can be deduced from the Fourier solution of (7.3-1) with g(x) = 0.

u(x,t) =

bn sin(
n 1
n
cn 
x)cos(
t)
L
L
From the trigonometric identity
sin()cos() =
sin(
1
[sin(  ) + sin( + )]
2
n
cn 
1
n
n
x)cos(
t) = { sin[
(x  ct)] + sin[
(x + ct)]}
L
L
2
L
L
u(x,t) =
1
2


n 1
bn { sin[
n
n
(x  ct)] + sin[
(x + ct)]}
L
L
Therefore, u(x,t) has the form
7-26
u(x,t) =
1 *
[f (x  ct) + f*(x + ct)]
2

At t = 0,
u(x,0) = f(x) =

n 1
bn sin(
n
x) = f*(x)
L
The above function is an odd extension of f(x) with period 2L.
The d’Alembert’s solution can be obtained by a transformation of variables so that u(x, t) will
become u(, ) where
 = x + ct,
 = x  ct.
The new independent variables are substituted into equation (7.3-1) by applying the chain
rule
u
u 
u
u 
u
=
+
=c
c
 t

t
  t

u u 
u
u 
u
=
+
=
+

x    x
  x


  u   
  u   
 2u
=
+
=c
 
 
2

  t   t
  t   t
t
 u
u 

c
 c
  c
 

 
2
2
2
 2u
2  u
2  u
2  u
=c
 2c
+c
 2
t 2

 2
Similarly
 2u  2u
 2u
 2u
=
+2
+
 x 2  2

 2
Substitute
2
 2u
 2u
 2u
2 u
and
into
=
c
to obtain
t 2
x 2
t 2
x 2
2
  u 
 2u
 2u
2  u
 2c
= 2c

=0
  =0
   



2
u
= ()  u(, ) =   (  )d + F()

7-27
 u
u 
c
 c

 
 
Hence
u(, ) = F() + G(), where G() =   (  )d
In terms of x and t
u(x, t) = F(x + ct) + G(x  ct)
(7.3-2)
From the initial displacement u(x,0) = f(x)
F(x) + G(x) = f(x)
From the initial velocity
u
(x,0) = g(x)
t
u(, ) = F() + G() 
 = x + ct,
At t = 0,
Hence
dF 
u
dG 
=
+
t
d  t
d  t


= c;
=c
t
t
 = x  ct 
dF
dF
dG
dG
=
= F’(x);
=
= G’(x)
d
dx
dx
d
u
(x,0) = cF’(x)  cG’(x) = g(x)
t
Dividing [cF’(x)  cG’(x) = g(x)] by c and integrating with respect to x

x
xo
F ' ( x )dx 

x
xo
G ' ( x )dx =
1
c

x
g ( s )ds
xo
We obtain
F(x)  G(x)  [F(xo)  G(xo)] =
1
c

x
xo
g ( s )ds
Let k(xo) = F(xo)  G(xo)
F(x)  G(x) = k(xo) +
1
c

x
xo
g ( s )ds
F(x) and G(x) can be solved from the above equation and the initial displacement F(x) + G(x)
= f(x).
7-28
F(x) =
1
1
f(x) +
2
2c

G(x) =
1
1
f(x) 
2
2c

x
xo
x
xo
g ( s )ds +
1
k(xo)
2
g ( s )ds 
1
k(xo)
2
Replacing x by x + ct for F(x) and x by x  ct for G(x), we obtain
G(x  ct) =
F(x + ct) =
1
1
f(x + ct) +
2
2c

G(x  ct) =
1
1
f(x  ct) 
2
2c

1
1
f(x  ct) +
2
2c

xo
x ct
g ( s)ds 
x ct
xo
x ct
xo
g ( s )ds +
1
k(xo)
2
g ( s )ds 
1
k(xo)
2
1
k(xo)
2
The final solution is then
u(x, t) = F(x + ct) + G(x  ct)
u(x, t) =
1
1
[f(x  ct) + f(x + ct)] +
2
2c

x ct
x ct
g ( s)ds
When the initial velocity is zero, d’Alembert solution is simply
u(x, t) =
1
[f(x  ct) + f(x + ct)]
2
Geometrically, the above solution of the wave equation is an average of two waves traveling
in opposite directions with shapes determined from the initial displacement.
Example 7.3-2. ____________________________________
An infinite string is subjected to the initial displacement
f(x) =
0.02
1  9x2
Find an expression for the subsequent motion of the string if it is released from rest. The
tension is 20 N and the mass per unit length is 510-4 kg/m.
Solution
7-29
u(x, t) =
c=
1
1
1
0.02
0.02
[f(x  ct) + f(x + ct)] =
+
2
2
2 1  9( x  ct )
2 1  9( x  ct ) 2

=

u(x, t) =
20
= 200 m/s
5  10  4
1
1
0.02
0.02
+
2
2 1  9( x  200t )
2 1  9( x  200t ) 2
Table 7.3-2 lists the Matlab program to plot the wave motion in Figure 7.3-4.
__________ Table 7.3-2 Matlab program to plot u(x,t) at various time ___________
% Plot u for example 7.3-2 at various t
%
x=-20:.02:20;
%
% Label the time t for each displacement u, the character vector ax hold the data
%
ax='t=0.00t=0.01t=0.02t=0.04t=0.06t=0.08';tv=.01*[0 1 2 4 6 8];
%
% Set y-coordinate from -1 to 1
%
x1=[0 0];y1=[-0.02 0.02];x2=[-20 20];y2=[0 0];
for i=1:6;
t=tv(i);xpt2=9*(x+200*t).^2;xmt2=9*(x-200*t).^2;
%
% Extract the time from ax, label axi is used for x-axis label for the time
%
ib=1+(i-1)*6;ie=ib+5;
axi=ax(ib:ie);
u=0.5*(.02./(1+xpt2)+.02./(1+xmt2));
%
% Divide the plot window into 3 rows and 2 columns using subplot command
%
subplot(3,2,i),plot(x,u,x1,y1,x2,y2)
xlabel(axi);ylabel('u')
end
7-30
0
0
0
t=0.00
10
-10
0
t=0.02
10
-10
0
t=0.06
Figure 7.3-4. Plot of u(x, t) =
10
20
-10
0
t=0.01
10
20
-10
0
t=0.04
10
20
-10
0
t=0.08
10
20
0
-0.02
-20
0.02
20
0
-0.02
-20
-0.02
-20
0.02
20
0
-0.02
-20
0.02
u
-10
u
u
-0.02
-20
0.02
u
0.02
0
u
u
0.02
-0.02
-20
1
1
0.02
0.02
+
at various t.
2
2 1  9( x  200t )
2 1  9( x  200t ) 2
Example 7.3-3 _______________________________
u(x,t)
0.1
u(x,0) = f(x)
0
1
 3
 10 x
f ( x)  
3(1  x )

 20
0 x
1
3
1
 x 1
3
x
Figure 7.3-5 Initial shape of the string in Example 7.3-2
Figure 7.3-2 shows the initial displacement u(x,t) of a string stretched along the x-axis
between x = 0 and x = 1. The string is free to vibrate in a fixed plane through the x-axis.
a) Use d’Alembert’s solution to determine the shape of the string at times t =
is released from rest, given that c = 1/.
b) Determine the first time when the string returns to its initial shape.
7-31
2

and
if it
3
3
Solution
a) Since the string is released from rest,
u
(x,0) = g(x) = 0
t
The shape of the string at any time t is given from d’Alembert’s solution by
u(x,t) =
1 *
[f (x  ct) + f*(x + ct)]
2
f*(x) is the odd extension of the original string and is shown over the interval  1  x  1 in
Figure 7.3-6a.
At t =

,
3
u(x,t) =
1 *
[f (x  1/3) + f*(x + 1/3)]
2
The graph of f*(x + 1/3) is obtained by translating the graph of f*(x) to the left by 1/3 unit.
The graph of f*(x  1/3) is obtained by translating the graph of f*(x) to the right by 1/3 unit.

The shape of the string at time t =
is obtained by averaging the graphs of f*(x  1/3) and
3
f*(x + 1/3). We restrict the graph to the interval  1  x  1 which is shown in Figure 7.3-6b.
At t =
2
,
3
u(x,t) =
1 *
[f (x  2/3) + f*(x + 2/3)]
2
Similar procedures are followed to obtain the shape of the string at t =
2
as shown in
3
0.1
0.1
0.05
0.05
f*(t=0)
f*(t=0)
Figure 7.3-6b.
0
-0.05
-0.1
-1
0
-0.05
0
1
-0.1
-1
2
x
0
1
x
Figure 7.3-6a Initial shape of the string with odd extension
7-32
2
0.1
0.05
0.05
f*(x+2/3)
f*(x+1/3)
0.1
0
-0.05
-0.1
-1
0
-0.05
0
1
-0.1
-1
2
0
0.1
0.1
0.05
0.05
0
-0.05
-0.1
-1
1
2
1
2
-0.05
0
1
-0.1
-1
2
0
x
0.1
0.1
0.05
0.05
u(x,t=2*pi/3)
u(x,t=pi/3)
2
0
x
0
-0.05
-0.1
-1
1
x
f*(x-2/3)
f*(x-1/3)
x
0
1
0
-0.05
-0.1
-1
2
0
x
x
Figure 7.3-6b Shapes of the string at times t =
2

and
3
3
b) Determine the first time when the string returns to its initial shape.
The string returns to its initial shape when
1 *
[f (x  t/) + f*(x + t/)] = f*(x)
2
Since f*(x) is 2-periodic, when t/ = 2 the string returns to its initial shape.
We now will show that the simple algorithm
uin 1 = uin1 + uin1  uin 1
matches the d’Alembert solution
u(x,t) =
1 *
[f (x + ct) + f*(x  ct)] = F(x + ct) + G(x  ct)
2
We need u in that represents u value at x = xi = ix and at t = tn = nt.
7-33
Since c
t
= 1  ct = x, we have
x
ct = ctn = nct = nx
Therefore
u in = F(xi + ctn) + G(xi  ctn)
u in = F(ix + nx) + G(ix  nx)
u in = F[(i + n)x)] + G[(i  n)x)]
From the above expression, we obtain
uin 1 = F{[(i + (n+1)x)]} + G{[(i  (n+1)x)]}
If the formula uin 1 = uin1 + uin1  uin 1 matches the d’Alembert solution we will have
uin1 + uin1  uin 1 = F{[(i + (n+1)x)]} + G{[(i  (n+1)x)]}
From the expression u in = F[(i + n)x)] + G[(i  n)x)], we can write similarly
uin1 = F[(i + 1 + n)x)] + G[(i + 1  n)x)]
uin1 = F[(i  1 + n)x)] + G[(i  1  n)x)]
uin 1 = F[(i + n  1)x)] + G[(i  n + 1)x)]
Since both F and G are linear function of x, F(a) + F(b) = F(a + b) and G(a) + G(b) = G(a +
b)
uin1 + uin1  uin 1 = F[(i + 1 + n)x) + (i  1 + n)x)  (i + n  1)x]
+ G[(i + 1  n)x) + (i  1  n)  (i  n + 1)x)]
uin1 + uin1  uin 1 = F{[(i + (n+1)x)]} + G{[(i  (n+1)x)]} = uin 1
Thus, the solution to the wave equation is given exactly by
uin 1 = uin1 + uin1  uin 1
7-34