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406    CHAPTER 12  Vector Integral Calculus

                                 This is the flux of K∇u across  , multiplied by the length of the time interval. But, the change
                                 in temperature at (x, y, z) over  t is approximately (∂u/∂t) t so the resulting heat loss in M is
                                                             ⎛              ⎞
                                                                      ∂u

                                                                    μρ   dV    t.
                                                             ⎝              ⎠
                                                                      ∂t
                                                                 M
                                 Assuming no heat sources or sinks in M (which occur, for example, during chemical reactions
                                 or radioactive decay), the change in heat energy in M over  t must equal the heat exchange
                                 across  :
                                                                        ⎛              ⎞

                                                                                 ∂u


                                                       (K∇u) · ndσ  t =  ⎝     μρ   dV  ⎠   t.
                                                                                 ∂t

                                                                            M
                                 Therefore,
                                                                                ∂u

                                                           (K∇u) · ndσ =     μρ    dV.
                                                                                ∂t
                                                                          M
                                 Now use the divergence theorem to convert the surface integral to a triple integral:

                                                          (K∇u) · ndσ =    ∇· (K∇u)dV.

                                                                        M
                                 Substitute this into the preceding equation to obtain
                                                                ∂u

                                                              μρ   −∇ · (K∇u) dV = 0.
                                                                 ∂t
                                                         M
                                 Now   is an arbitrary closed surface within the medium. If the integrand in the last equation
                                 were, say, positive at some point P 0 , then it would be positive throughout some (perhaps very
                                 small) sphere about P 0 , and we could choose   as this sphere. But then the triple integral over M
                                 of a positive quantity would be positive, not zero, a contradiction. The same conclusion follows
                                 if this integrand were negative at some P 0 .
                                    Vanishing of the last integral for every closed surface   in the medium therefore forces the
                                 integrand to be identically zero:
                                                                ∂u
                                                             μρ   −∇ · (K∇u) = 0.
                                                                ∂t
                                 This is the partial differential equation
                                                                  ∂u
                                                               μρ   =∇ · (K∇u)
                                                                  ∂t
                                 for the temperature function. This equation is called the heat equation.
                                    We can expand
                                                            ∂u      ∂u     ∂u

                                            ∇· (K∇u) =∇ · K    i + K  j + K  k
                                                            ∂x      ∂y     ∂z
                                                       ∂     ∂u     ∂     ∂u     ∂     ∂u
                                                     =     K     +     K     +     K
                                                       ∂x    ∂x    ∂y    ∂y    ∂z    ∂z
                                                                                    2
                                                                                         2
                                                                                               2
                                                       ∂K ∂u   ∂K ∂u   ∂K ∂u       ∂ u  ∂ u   ∂ u
                                                     =       +       +       + K      +    +
                                                       ∂x ∂x   ∂y ∂y   ∂z ∂z      ∂x  2  ∂y  2  ∂z 2
                                                                   2
                                                     =∇K ·∇u + K∇ u.


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                                   October 14, 2010  14:53  THM/NEIL   Page-406        27410_12_ch12_p367-424
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