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3.5
                                                   ∆t =1/20
        3
                                            2.5
       2.5
        2
                 0.5
                                                      0.5
                                                                        1.5
                                                                1
       3.5 0   ∆t =1/10    1       1.5      3.5 3 2 0  1.4 Cauchy Problems  11
                                            3.5
               ∆t =1/40                            ∆t =1/80
        3                                    3
       2.5                                  2.5
        2                                    2
         0       0.5       1       1.5        0       0.5       1       1.5
         FIGURE 1.4. Convergence of the forward Euler approximations as applied to
         problem (1.10) on page 6.
         coordinate. We want to derive a solution of this problem using the method
         of characteristics. The characteristics of (1.20)–(1.21) are curves in the
         x–t-plane defined as follows: For a given x 0 ∈ R, consider the ordinary
         differential equation
                            dx(t)
                                  = a x(t),t ,  t > 0,
                              dt                                    (1.22)
                             x(0) = x 0 .

         The solution x = x(t) of this problem defines a curve  x(t),t ,t ≥ 0
         starting in (x 0 , 0) at t = 0; see Fig. 1.5.
           Now we want to consider u along the characteristic; i.e. we want to study

         the evolution of u x(t),t . By differentiating u with respect to t,weget
                                              dx(t)
                           d
                             u x(t),t = u t + u x
                           dt                  dt
                                     = u t + a(x, t)u x =0,
         where we have used the definition of x(t) given by (1.22) and the differential
         equation (1.20). Since

                                  d
                                    u x(t),t =0,
                                  dt
         the solution u of (1.20)–(1.21) is constant along the characteristic. Hence

                                u x(t),t = u(x 0 , 0)
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