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17.2 The Heat Equation on [0, L]  625


                               SECTION 17.2        PROBLEMS


                            In each of Problems 1 through 7, write a solution of the  its left end maintained at zero temperature, while its
                            initial-boundary value problem. Graph the twentieth par-  right end is perfectly insulated. The bar has an initial
                                                                                              2
                            tial sum of the series for u(x,t) on the same set of axes for  temperature f (x) = x for 0 ≤ x ≤ 2. Determine the
                            different values of the time.                     temperature distribution u(x,t) and lim t→0 u(x,t).
                                                                           11. Show that the partial differential equation
                                        2
                                  ∂u   ∂ u
                             1.     = k    for 0 < x < L,t > 0,
                                                                                               2
                                  ∂t   ∂x 2                                            ∂u     ∂ u   ∂u
                                u(0,t)= u(L,t) = 0for t ≥ 0,                           ∂t  = k  ∂x  2  + A  ∂x  + Bu
                               u(x,0)= x(L − x) for 0 ≤ x ≤ L
                                                                              can be transformed into a standard heat equation for v
                                        2
                                  ∂u   ∂ u                                    by letting u(x,t) = e αx+βt v(x,t) and choosing α and
                             2.     = 4    for 0 < x < L,t > 0,
                                  ∂t   ∂x 2                                   β appropriately.
                                u(0,t)= u(L,t) = 0for t ≥ 0,               12. Use the idea of Problem 11 to solve
                                       2
                               u(x,0)= x (L − x) for 0 ≤ x ≤ L
                                                                                        2
                                                                                   ∂u  ∂ u   ∂u
                                        2
                                  ∂u   ∂ u                                           =    + 4   + 2u for 0 < x <π,t > 0,
                             3.     = 3    for 0 < x < L,t > 0,                    ∂t  ∂x  2  ∂x
                                  ∂t   ∂x 2
                                                                                u(0,t) = u(π,t) = 0for t ≥ 0,
                                u(0,t)= u(L,t) = 0for t ≥ 0,
                               u(x,0)= L(1 − cos(2πx/L)) for 0 ≤ x ≤ L          u(x,0) = x(π − x) for 0 ≤ x ≤ π.
                                         2
                                   ∂u   ∂ u                                13. Solve
                             4.       =    for 0 < x <π,t > 0,
                                   ∂t   ∂x  2                                             2
                               ∂u       ∂u                                           ∂u  =  ∂ u  + 6 ∂u  for 0 < x < 4,t > 0,
                                  (0,t)=  (π,t) = 0for t ≥ 0,                              2
                               ∂x       ∂x                                           ∂t  ∂x    ∂x
                                u(x,0)= sin(x) for 0 ≤ x ≤ π                      u(0,t) = u(4,t) = 0for t ≥ 0,
                                         2
                                   ∂u   ∂ u                                       u(x,0) = 1for0 ≤ x ≤ 4.
                             5.       = 4   for 0 < x < 2π,t > 0,
                                   ∂t   ∂x  2
                               ∂u       ∂u                                    Graph the twentieth partial sum of the solution for
                                  (0,t)=  (2π,t) = 0for t ≥ 0,                selected times.
                               ∂x       ∂x
                                u(x,0)= x(2π − x) for 0 ≤ x ≤ 2π           14. Solve
                                                                                          2
                                         2
                                   ∂u   ∂ u                                          ∂u  ∂ u   ∂u
                             6.       = 4   for 0 < x < 3,t > 0,                       =    − 6   for 0 < x <π,t > 0,
                                   ∂t   ∂x  2                                        ∂t  ∂x  2  ∂x
                               ∂u       ∂u                                        u(0,t) = u(π,t) = 0for t ≥ 0,
                                  (0,t)=  (3,t) = 0for t ≥ 0,
                               ∂x       ∂x
                                        2
                                u(x,0)= x for 0 ≤ x ≤ 3                           u(x,0) = x(π − x) for 0 ≤ x ≤ π.
                                         2
                                   ∂u   ∂ u                                   Graph the twentieth partial sum of the solution for
                             7.       = 2   for 0 < x < 6,t > 0,              selected times.
                                   ∂t   ∂x  2
                               ∂u       ∂u                                 15. Solve
                                  (0,t)=  (6,t) = 0for t ≥ 0,
                               ∂x       ∂x                                             ∂u    ∂ u
                                                                                              2
                                u(x,0)= e  −x  for 0 ≤ x ≤ 6                             = 16   for 0 < x < 1,t > 0,
                                                                                       ∂t    ∂x  2
                             8. A thin, homogeneous bar of length L has insulated   u(0,t) = 2,u(1,t) = 5for t ≥ 0,
                               ends and initial temperature B, a positive constant.         2
                                                                                    u(x,0) = x for 0 ≤ x ≤ 1.
                               Find the temperature distribution in the bar.
                             9. A thin homogeneous bar of length L has initial tem-  16. Solve
                               perature f (x) = B where the right end x = L is insu-   ∂u   ∂ u
                                                                                             2
                               lated, while the left end is kept at zero temperature.    = k    for 0 < x < L,t > 0,
                                                                                       ∂t   ∂x  2
                               Find the temperature distribution in the bar.
                                                                                    u(0,t) = T,u(L,t) = 0for t ≥ 0,
                            10. A thin, homogeneous bar having thermal diffusivity
                               of 9 and a length of 2 cm has insulated sides and    u(x,0) = x(L − x) for 0 ≤ x ≤ L.
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                                   October 14, 2010  15:25  THM/NEIL   Page-625        27410_17_ch17_p611-640
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