Elastic Wave Lecture Notes, ECE471 University of Illinois at Urbana-Champaign W. C. Chew Fall, 1991 1 Elastic Wave Class Note 1, ECE471, U. of Illinois W. C. Chew Fall, 1991 Derivation of the Elastic Wave Equation (optional read- ing) The elastic eave equation governs the propagation of the waves in solids. We shall illustrate its derivation as follows: the waves in a solid cause perturbation of the particles in the solid. The particles are displaced from their equilibrium position. The elasticity of the solid will provide the restoring force for the displaced particles. Hence, the study of the balance of these forces will lead to the elastic wave equation. The displacement of the particles in a solid from their equilibrium position causes a displacement field u(x; t) where u is the displacement of the particle at position x at time t. Here, x is a position vector in three dimensions. Usually, we use r for position vector, but we use x here so that indicial notation can be used conveniently. In indicial notation, x1, x2, and x3 refer to x, y, z respectively. The displacement field u(x; t) will stretch and compress distances between particles. For instance, particles at x and x + ±x are ±x apart at equilibrium. But under a perturbation by u(x; t), the charge in their separation is given by ±u(x; t) = u(x + ±x; t) ¡ u(x; t) (1) By using Taylor’s series expansion, the above becomes ±u(x; t) ' ±x ¢ ru(x; t) + O(±x2) (2) or in indicial notation, ±ui ' @jui±xj (3) where @ u = @ui . This change in separation ±u can be decomposed in to a sym- j i @xj metric and an antisymmetric part as follows: symmetric antisymmetric 1 z }| { 1 z }| { ±u = (@ u + @ u ) ±x + (@ u ¡ @ u ) ±x (4) i 2 j i i j j 2 j i i j j Using indicial notation, it can be shown that [(r £ u) £ ±x]i = ²ijk(r £ u)j±xk = ²ijk²jlm@lum±xk (5) From the identity that ²ijk²jlm = ¡²jik²jlm = ±im±kl ¡ ±il±km (6) 1 we deduce that [(r £ u) £ ±x]i = (@kui ¡ @iuk)±xk (7) Hence, stretch rotation z }| { 1 z }| { 1 ±u = e ±x + [(r £ u) £ ±x] = (e¯ ¢ ±x) + [r £ u) £ ±x] (8) i ij j 2 i i 2 i where we have defined 1 e = (@ u + @ u )) (9) ij 2 j i i j The first term in (8) results in a change in distance between the particles, while the second term, which corresponds to a rotation, has a higher order effect. This can be shown easily as follows: The perturbed distance between the particles at x and x + ±x is now ±x + ±u. The length square of this distance is (using (8)) (±x + ±u)2 =» ±x ¢ ±x + 2±x ¢ ±u + O(±u2) = j±xj2 + 2±x ¢ e¯ ¢ ±x + O(±u2) (10) assuming that ±u ¿ ±x, since u is small, and ±u is even smaller. The second term in (8) vanished in (10) because it is orthogonal to ±x. The above analysis shows that the stretch in the distance between the particles is determined to first order by the first term in (8). The tensor e¯ describes how the particles in a solid are stretched in the presence of a displacement field: it is called the strain tensor. This strain produced by the displacement field will produce stresses in the solid. Stress in a solid is described by a stress tensor T¯ . Given a surface 4S in the body of solid with a unit normaln ˆ, the stress in the solid will exert a force on this surface 4S. This force acting on a surface, known as traction, is given by T =n ˆ ¢ T¯ 4S (11) nˆ T ∆S S V Figure 1: Hence, if we know the traction on the surface S of a volume V, the total force acting on the body is given by I ZZZ nˆ ¢ T dS = r ¢ T dV (12) S V where the second equality follows from Gauss’ theorem, assuming that T is defined as a continuous function of space. This force caused by stresses in the solid, must be 2 balanced by other forces acting on the body, e.g, the inertial force and body forces. Hence ZZZ 2 Z Z @ u ¯ ½ 2 dV = r ¢ T dV + fdV (13) V @t v V where ½ is the mass density and f is a force density, e.g., due to some externally applied sources in the body. The left hand side is the inertial force while the right hand side is the total applied force on the body. Since (13) holds true for a arbitrary volume V , we have @2u ½ = r ¢ T¯ + f (14) @t2 as our equation of motion. Since the stress force, the first term on the right hand side of (14), is caused by strains in the solid, T¯ should be a function of e¯. Under the assumption of small perturbation, T should be linearly independent on e¯. The most general linear relationship between the second rank tensor is Tij = Cijklekl (15) The above is the constitutive relation for a solid. Cijkl is a fourth rank tensor. For an isotropic medium, Cijkl should be independent of any coordinate rotation. The most general form for a fourth rank tensor that is independent of coordinate rotation is [see Exercise 1] Cijkl = ¸±ij±kl + ¹1±ik±jl + ¹2±il±jk (16) Furthermore, since ekl is symmetric, Cijkl = Cijlk. Therefore, Cijkl = ¸±ij±kl + ¹(±jk±il + ±il±ik) (17) Consequently, in an isotropic medium, the constitutive relation is characterized by two constants ¸ and ¹ known as Lam´econstant. Note that as a consequence of (17), Tij = Tji in (15). Hence, both the strain and the stress tensors are symmetric tensors. Using (17) in (15), we have after using (9) that Tij = ¸±ijell + ¹(eij + eji) = ¸±ij@lul + ¹(@jui + @iuj): (18) Then (r ¢ T¯ )j = @iTij = @j¸@lul + @i¹@jui + @i¹@iuj = ¸@i@lul + (@lul)@j¸ + ¹@j@iui + (@jui)@i¹ + @i¹@iuj = (¸ + ¹)[rr ¢ u]j + (r ¢ ¹ru)j + (r ¢ u)(r¸)j + [(ru) ¢ r¹]j(19) If ¹ and ¸ are constants of positions, we have r ¢ T¯ = (¸ + ¹)rr ¢ u + ¹r2u (20) Using (20) in (14), we have @2u ½ = (¸ + ¹)rr ¢ u + ¹r2u + f (21) @t2 3 which is the elastic wave equation for homogeneous and isotropic media. If ¹, which is also the shear modulus, is zero, taking the divergence of (21), and defining θ = r ¢ u, we have @2θ ½ = ¸r2θ + r ¢ f (22) @T 2 which is the acoustic wave equation for homogenous media, and ¸ is the bulk mod- ulus. Exercise 1 1.(a) Under coordinate rotations, show that the fourth rank tensor Cijkl transforms as 0 Cijkl = Tii0 Tjj0 Tkk0 Tll0 Ci0j0k0l0 (b) Show that if 0 Ci0j0k0l0 = ¸1±i0j0 ±k0l0 + ¹1±i0k0 ±j0l0 + ¹2±i0l0 ±j0k0 then under coordinate rotation, Cijkl = ¸1±ij±kl + ¹1±ik±jl + ¹2±il±jk In other words, Cijkl remains the same under coordinate rotation, i.e., it is an isotropic tensor. (c) Proof that the form in (b) is the only form for isotropic fourth rank tensor (difficult). References [1] Aki, K. and P. G. Richards, Quantitative Seismology, Freeman, New York, 1980. [2] Hudson, J. A., The Excitation and Propagation of Elastic Waves, Combridge University Press, Combridge, 1980. [3] Archenbach, J. D.,Wave Propagation in Elastic Solids, North Holland, Ams- terdam, 1973. 4 Elastic Wave Class Note 2, ECE471, U. of Illinois W. C. Chew Fall, 1991 Solution of the Elastic Wave Equation—A Succinct Derivation The elastic wave equation is (¸ + 2¹)rr ¢ u ¡ ¹r £ (r £ u) ¡ ½u¨ = ¡f(x; t): (1) By Fourier transform, Z 1 1 u(x; t); = d!e¡i!tu(x;!) (2) 2¼ ¡1 (1) becomes (¸ + 2¹)rr ¢ u ¡ ¹r £ (r £ u) + !2½u = ¡f (3) where u = u(x;!), and f = f(x;!) now. By taking r£ of the above equation, and defining Ω = r £ u, the rotation of u, we have ¹r £ (r £ Ω) ¡ !2½Ω = r £ f (4) Since r ¢ Ω = r ¢ r £ u = 0, the above is just 1 r2Ω + k2Ω = ¡ r £ f (5) s ¹ q 2 2 2 2 ¹ where ks = ! ½=¹ = ! =cs and cs = ½ . The solution to the above equation is Z 1 ¡ ¢ Ω(x;!) = dx0g (x ¡ x0)r0 £ f x0;! (6) ¹ s 0 0 iksjx¡x j 0 where gs(x ¡ x ) = e =4¼jx ¡ x j. Taking the divergence of (3), and defining θ = r ¢ u, the dilational part of u, we have (¸ + 2¹)r2θ + !2ρθ = ¡r ¢ f: (7) The solution to the above is Z 1 ¡ ¢ θ(x;!) = dx0g (x ¡ x0)r0 ¢ f(f x0;! (8) ¸ + 2¹ c 1 p 2 2 2 2 0 where kc = ! ½=(¸ + 2¹) = ! =cc , cc = (¸ + 2¹)=½, and gc(x ¡ x ) = 0 eikcjx¡x j=4¼jx ¡ x0j. From (3), we deduce that f 1 1 u(x;!) = ¡ 2 + 2 r £ Ω ¡ 2 rθ (9) ¹ks ks kc Then, using (6) and (8), we have Z f 1 u(x;!) = ¡ + r £ dx0g (x ¡ x0)r0 £ f(x0;!) ¡ ¹k2 ¹k2 s s s Z 1 0 0 0 0 2 r dx gc(x ¡ x )r ¢ f(x ;!) (10) (¸ + 2¹)kc Using integration by parts, and the fact that r0g(x ¡ x0) = ¡rg(x ¡ x0), the above Z f 1 u(x;!) = ¡ + r £ r £ dx0g (x ¡ x0)f(x0;!) ¹k2 ¹k2 s s s Z 1 0 0 0 ¡ 2 rr dx gc(x ¡ x )f(x ;!) (11) (¸ + 2¹)kc 2 2 0 0 Using r £ r £ A = (rr ¡ r )A, and the fact that r gs((x ¡ x ) ¡ ±(x ¡ x ), the above becomes µ ¶ Z rr 1 u(x;!) = I + ¢ dx0g (x ¡ x0)f(x0;!) k2 ¹ s s Z rr 1 0 0 0 ¡ 2 ¢ dx gc(x ¡ x )f(x ;!) (12) kc ¸ + 2¹ Time-Domain Solution If f(x; t) =x ˆj±(x)±(t), then f(x;!) =x ˆj±(x) and µ ¶ rr eiksr rr eikcr u(x;!) = I + 2 ¢ xˆj ¡ 2 ¢ xˆj (13) ks 4¼¹r kc 4¼(¸ + 2¹)r or in indicial notation µ ¶ eiksr 1 eiksr eikcr ui(x;!) = ±ij + @i@j 2 ¡ 2 (14) 4¼¹r 4¼r ¹ks (¸ + 2¹)kc p p Since ks = !=cs = ! ½=¹ , kc = !=cc = ! ½=(¸ + 2¹), the above can be inverse Fourier transformed to yield ·µ ¶ µ ¶ µ ¶ µ ¶¸ ±(t ¡ r=cs) 1 r r r r ui(x; t) = ±ij ¡ @i@j t ¡ u t ¡ ¡ t ¡ u t ¡ 4¼¹r 4¼r½ cs cs cc cc (15) @2 Writing @i@j = (@ir)(@jr) @r2 , the above becomes · µ ¶ µ ¶¸ ±(t ¡ r=cs) 1 r r ui(x; t) = [±ij ¡ (@ir)(@jr)] ¡ (@ir)(@jr) 3 tu t ¡ ¡ tu t ¡ 4¼¹r 2¼½r cs cc ±(t ¡ r=c ) + (@ r)(@ r) c ; (16) i j 4¼(¸ + 2¹)r 2 Consequently, if f(x; t) is a general body force, by convolutional theorem, r fj(x; t ¡ ) u (x; t) = [± ¡ (@ r)(@ r)] cs i ij i j 4¼¹r Z 1 r=cs +(@ir)(@jr) 3 ¿fj(x; t ¡ ¿)d¿ 2¼½r r=cc r fj(x; t ¡ ) +(@ r)(@ r) cc (17) i j 4¼(¸ + 2¹)r Solution of the Elastic Wave Equation—Fourier-Laplace Transform The elastic wave equation (¸ + ¹)rr ¢ u + ¹r2u ¡ ½u¨ = ¡f (18) can be solved by Fourier-Laplace transform.