# Advanced Modern Engineering Mathematics by G. James

By G. James

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**Example text**

Deﬁne p(t) = g ◦ τ −1 t ; that is, p(x, t) is deﬁned by mapping back x ∈ Ω(t) to corresponding point in Ω and evaluating g at the mapped-back point. Then d+ d+ (p(t) ◦ τ t ) t=0 = g ◦ τ −1 t ◦ τt dt dt δp = δm p − δφ · ∇p(0) = −δφ · ∇p(0). δm p = t=0 = 0, (19) Thus, when p(t) is “moving along” with the deformation, the material derivative vanishes. Next example illustrates the opposite situation. Example 2. Let f : Rd → R. Deﬁne p(t) = f |Ω(t) . Then d+ d+ (p(t) ◦ τ t ) t=0 = f |Ω(t) ◦ τ t t=0 dt dt d+ τt = ∇f |Ω(0) · = ∇f |Ω(0) · δφ = δφ · ∇p(0), dt t=0 δm p = (20) δp = δm p − δφ · ∇p(0) = 0.

Holds for all (˜ v01,1 , w Notice again the diﬀerence in the notation for the inﬁnitesimal rotation ˜ ki,j and the notation for the inﬁnitesimal rotation solution test functions w ˜ ki,j . Also notice that all the intermediate boundary terms on the functions ω right-hand side of the equation (10) cancel out in the formulation (15) due to the kinematic and dynamics contact conditions. Solution to the problem (15) is not unique. Namely, since only the deriva˜ appears in the weak formulation, the solution will be determined tive of ω ˜ 0 .

Therefore, H describes the elastic properties of the rod and the geometry of the cross-section. The equation (8) is a condition that requires that the middle line is approximately inextensible and that allowable deformations of the cross-section are approximately orthogonal to the middle line. This condition has to be included in the solution space for the weak formulation of the problem (5)–(8) (pure traction problem for a single curved rod). Thus, introduce the space ˜ +t×w ˜ =0 . ˜ ∈ H 1 (0, )6 : v V = (˜ v, w) (9) ˜ ∈ V is called a weak solution of the problem (5)–(8) if Function (˜ u, ω) ˜f ·˜ ˜ )−˜ ˜ ˜ ( )·˜ ˜ (0)·˜ vds+˜ q( )·w( q(0)·w(0)+ p v( )−p v(0) ˜ ·w ˜ ds = QHQT ω 0 0 (10) ˜ ∈ V (notice the diﬀerence in the notation between ω ˜ holds for all (˜ v, w) ˜ and w).