By Lizhen Ji

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**Example text**

Notice that, in the case of a coupled source and sink of equal strength, measured by a 2π exchange flux, and distant (2h) from each other: lim h→0 f (z) = − lim h→0 2h 1 2h ln (z + h) 1 =− . 59) Let M be a positive constant, and consider a coupled source and sink with a 2πC(h) exchange flux, with C(h) to be defined presently. 60) which, as a result of h being a real and positive value, has the following properties: lim Fh (z) = − lim h→0 h→0 ln(z + h) − ln(z − h) 2h lim (2hC(h))). 61) Now, choose C(h), such that lim (2hC(h)) = M.

36) 4η where η is the dynamic fluid viscosity, r2 = x2 + y 2 , R is the cylinder’s radius, and dp P = . 37) dz The field v does not depend upon either z or θ, so that v(r) = v(r)k. 38) (the maximum velocity at the center of the tube). 39) Continuum Mathematical Models 15 we find that (curl v)(x, y) = −2yi + 2xj = 2k × (xi + yj). 40) 2-D Green’s Formula This well-known result reads, as follows: ∂Q ∂P − ∂x ∂y (x, y) dxdy = Ω P (x, y)dx + Q(x, y) dy. 41) C This formula, as well as the 3-D formulae known as Green’s identities, are proved in Appendix G.

76) Ω Suppose also that there are mass sources inside Ω which are described by some scalar field S(x, t), so S(x, t) dV = M˙ S (t). 77) Ω Here, M˙ S (t) stands for the time rate of change for the mass MS as generated by the distributed sources. Now, the net outflux of mass through σ, the closed surface boundary of Ω, is given by n(x) · (ρv)(x, t) dσ. 79) Ω for differentiable j = ρv. Since M (t) = MS (t) + Mv (t), then S(x, t) dV − M˙ s (t)+M˙ v (t) = Ω ∇·(ρv)(x, t) dV = Ω d dt ρ(x, t) dV. 80) Ω If the region Ω, also called the control volume in fluid dynamics, does not change with time t, then ∂ρ + ∇ · (ρv) − S (x, t) dV = 0.