1. Solve the wave equation utt = c 2uxx on the interval 0

Solution 1
Suppose that the string is set in motion so that it vi
ates in a vertical plane, and let u(x,t) denote
the vertical displacement experienced by the string at the point x at time t. If damping effects,
such as air resistance, are neglected, and if the amplitude of the motion is not too large, then
u(x,t) satisfies the partial differential equation


in the domain 0 < x < L, t > 0. Equation is known as the wave equation and is derived in
Appendix B at the end of the chapter. The constant coefficient a2 appearing in Eq. is given by
where T is the tension (force) in the string, and ρ is the mass per unit length of the string
material. It follows that a has the units of length/time, that is, of velocity. In this it is shown that a
is the velocity of propagation of waves along the string. To describe the motion of the string
completely it is necessary also to specify suitable initial and boundary conditions for the
displacement u(x,t). The ends are assumed to remain fixed, and therefore the boundary
conditions are
Since the differential equation is of second order with respect to t, it is plausible to prescribe two
initial conditions. These are the initial position of the string
educes to that of Wave equation, which can be given by
2
2
2
2
2
x
u
C
t
u





where the values of C
2
are for different cases are given in the following Table.
Case C
2
Lateral Vi
ation of taut String (T/  )
Longitudinal Vi
ation of rod E/ 
Torsional Vi
ation of Rod G/ 
To find the response of different system one may use the variable separation method by using the
following equation.
)()(),( tqxtxu 
( )x is known as the mode shape of the system and ( )q t is known as the time modulation. Now
equation (1) reduces to
2
2
2
2
2
)()(
x
tqC
t
q
x




 

or
2 2
2
2 2
1 1 q
C
x q t
     
   
     
(1)
Since the left side of equation (1) is independent of time t and the right side is independent of x
the equality holds for all values of t and x. Hence each side must be a constant. As the right side
term equals to a constant, it implies that the acceleration term
2
2
q
t
 
 
 
is proportional to
displacement term ( )q t , one may take the proportionality constant equal to 2 to have a simple
harmonic motion in the system. If one takes a positive constant, the response will grow
exponentially and make the system unstable. Hence one may write equation (1) as
2 2
2 2
2 2
1 1 q
C
x q t
     
     
     

Hence, 02
2
2
 q
dt
qd

And
22
2 2
0
x C
  
  


The solution of equation can be written as
1 2( ) sin cosq t C t C t    ...