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Solved Problems





               Exercise 114. Consider a semi-infinite string, x > 0. For all time the end of
               the string is displaced according to u(0, t) = f(t). Find the motion of the string,
               u(x, t) with the method of characteristics. The wave speed is c.

               Solution: The problem is

                                              2
                                      u − c u    xx  = 0,    x > 0,    −∞ < t < ∞,
                                        tt
                                                       u(0, t) = f(t).

               Because the left end of the string is being displaced, there will only be right-moving

               waves. The solution has the form

                                                    u(x, t) = F(x − ct).


               We substitute this into the boundary condition.

                                                       F(−ct) = f(t)


                                                                      ξ
                                                      F(ξ) = f     −
                                                                      c
                                                    u(x, t) = f(t − x/c)




               Exercise 115. Sketch the solution to the wave equation:

                                                                                Z  x+ct
                                      1                                       1
                          u(x, t) =     (u(x + ct, 0) + u(x − ct, 0)) +                 u (τ, 0) dτ,
                                                                                          t
                                      2                                      2c   x−ct
               for various values of t corresponding to the initial conditions:

                 (a) u(x, 0) = 0,        u (x, 0) = sin ωx       where ω is a constant,
                                           t
                                                      
                                                      1       for 0 < x < 1
                                                      
                (b) u(x, 0) = 0,         u (x, 0) =      −1 for − 1 < x < 0
                                           t
                                                      
                                                         0     for |x| > 1.
                                                      
               Solution:

                 (a)

                                                                                    Z  x+ct
                                         1                                       1
                              u(x, t) =    (u(x + ct, 0) + u(x − ct, 0)) +                 u (τ, 0) dτ
                                                                                             t
                                         2                                      2c   x−ct
                                                             1  Z  x+ct
                                                 u(x, t) =              sin(ωτ) dτ
                                                             2c  x−ct
                                                               sin(ωx) sin(ωct)
                                                   u(x, t) =
                                                                       ωc


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