Eigenvectors, Eigenvalues, and Finite Strain

Transcription

1 Eigenvectors, Eigenvalues, and Finite Strain GG303, 013 Lab 9 10/3/13 GG303 1 I Main Topics A Elementar linear algebra relanons B EquaNons for an ellipse C EquaNon of homogeneous deformanon D Eigenvalue/eigenvector equanon E SoluNons for smmetric homogeneous deformanon matrices F SoluNons for general homogeneous deformanon matrices G RotaNons in homogeneous deformanon 10/3/13 GG303 1

2 Eamples of D homogeneous deformanon Note that the smmetr of the displacement fields (or lack thereof) in the eamples corresponds to the smmetr (or lack thereof) in the deformanon gradient matri [F]. What is a simple wa to describe homogeneous deformanon that is geometricall meaningful? What is the geologic relevance? 0 F = F = F = /3/13 GG303 3 II Elementar linear algebra relanons A Inverse [A] - 1 of a real matri A 1 [A][A] - 1 = [A] - 1 [A] = [I], 1 0 where [I] = idennt matri (e.g., [ I ] = ) [A] and [A] must be square nn matrices 3 Inverse [A] - 1 of a matri [ A] = a c b d [ A] 1 = 1 ad bc d c b a = 1 A d c 4 Inverse [A] - 1 of a 33 matri also requires determinant A to be non- zero b a 10/3/13 GG303 4

3 II Elementar linear algebra relanons B Determinant A of a real matri A 1 A number that provides scaling informanon on a square matri Determinant of a matri a A = c b, A = ad bc d 3 Determinant of a 33 matri a b c A = d e f, A = a e f g h i h i b d f g i Akin to: Cross product (an area) Scalar triple product (a volume) + c d e g h 10/3/13 GG303 5 II Elementar linear algebra relanons C Transpose For [ A] = a c b d, A [ ] T = a b c d D Transpose of a matri product a If [ A] = c [ A] [ B] = [ B] T [ A] T = b e and [ B] = d g ae + bg ce + dg ea + gb fa + hb af + bh cf + dh ec + gd fc + hd, A f h, then [ A a ]T = b [ ][ B] = A T ae + bg = af + bh [ ][ B] c e and [ B] T = d f ce + dg cf + dh 10/3/13 GG303 6 T g h This is true for an real nn matrices 3

4 II Elementar linear algebra relanons E RepresentaNon of a dot product using matri mulnplicanon and the matri transpose a b = a,a,a z b,b,b z = a b + a b + a z b z = a a a z b b b z = [ a] T [ b] 10/3/13 GG303 7 III EquaNons for an ellipse A EquaNon of a unit circle 1 + = X X = 1 3 = cosθ = sinθ = X [ ] T [ X] = 1 10/3/13 GG

5 III EquaNons for an ellipse B Ellipse centered at (0,0), aligned along, aes 1 Standard form a + b = 1 General form A + D + F = 0 3 Matri form A 0 0 D = 10/3/13 GG303 9 A D = F A, D, and F are constants here, not matrices III EquaNons for an ellipse B Ellipse centered at (0,0), aligned along, aes 4 Parametric form = acosθ = bsinθ 5 Vector form r = acosθ i + bsinθ j 10/3/13 GG

6 III EquaNons for an ellipse C Ellipse centered at (0,0), arbitrar orientanon 1 General form A + ( B + C) + D + F = 0 provided 4AD > (B+C) Matri form A C B D = A + B C + D = F A, B, C, D, and F are constants here, not matrices 10/3/13 GG III EquaNons for an ellipse D PosiNon vector for an ellipse r = acosθ i + bsinθ j E DerivaNve of posinon vector for an ellipse (dr/dθ) dr dθ = asinθ i + bcosθ j F Dot product of r and dr/dθ r d r dθ = ( b a )sinθ cosθ G The posinon vector and its tangent are perpendicular if and onl if 1 a=b, or θ = 0, or 3 θ = 90 Along aes of ellipse We will use these results shortl So the aes of an ellipse/ellipsoid are perpendicular, and the tangents to an ellipse/ellipsoid at the ends of the aes are perpendicular. Those tangents parallel the aes. 10/3/13 GG

7 IV EquaNon of homogeneous deformanon A [X ] = [F][X] B D d d = C 3D d d = d z z z z z z z d d = a b c d d d = dz z 10/3/13 GG = F F F F F z F F F z F z F z F z z F F z For homogeneous strain, the derivanves are uniform (constants), and d, d can be small or large IV EquaNon of homogeneous deformanon [X ] = [F][X] D CriNcal maner: Understanding the geometr of the deformanon E In homogeneous deformanon, a unit circle transforms to an ellipse (and a sphere to an ellipsoid) F Proof X [ X] T X =1 [ X] = 1 [ X ] = [ F] [ X] Now solve for [X] [ F] 1 [ X ] = [ F] 1 [ F] [ X] = [ I ][ X] = [ X] [ X] = [ F] 1 [ X ] Now solve for [X] T [ X] T = [ F] 1 T [ X ] = [ X ] T [ F] 1 T 10/3/13 GG [ X] T [ X] = [ X ] T [ F] 1 Now subsntute for [X] T and [X] in first equanon T [ F] 1 X [ X ] T [ Smmetric matri] [ X ] = 1 EquaNon of ellipse See slide 11 [ ] = 1 7

8 IV EquaNon of homogeneous deformanon [X ] = [F][X] G [F] transforms a unit circle to a strain ellipse H Strain ellipse geometricall represents [F][X] I [F] - 1 transforms a strain J ellipse back to a unit circle [F] - 1 transforms a unit circle to a reciprocal strain ellipse K [F] - 1 transforms a reciprocal strain ellipse back to a unit circle L Reciprocal strain ellipse geometricall represents [F] - 1 [X] [ F] = a c b d [ F] 1 1 = ad bc d c b a 10/3/13 GG V Eigenvectors and eigenvalues A The eigenvalue matri equanon [A][X] = λ[x] 1 [A] is a (known) square matri (nn) [X] is a non- zero direcnonal eigenvector (n1) 3 λ is a number, an eigenvalue 4 λ[x] is a vector (n1) parallel to [X] 5 [A][X] is a vector (n1) parallel to [X] 10/3/13 GG

9 A The eigenvalue matri equanon [A][X] = λ[x] (cont.) 6 The vectors [[A][X]], λ[x], and [X] share the same direcnon if [X] is an eigenvector 7 If [X] is a unit vector, λ is the length of [A][X] 8 Eigenvectors [X i ] have corresponding eigenvalues [λ i ], and vice- versa 9 In Matlab, [v,d] = eig(a), finds eigenvectors (v) and eigenvalues (d) 10/3/13 GG V Eigenvectors and eigenvalues (cont.) B Eamples 1 IdenNt matri [I] All vectors in the - plane maintain their orientanon when operated on b the idennt matri, so all vectors are eigenvectors, and all vectors maintain their length, so all eigenvalues of [I] equal 1. The eigenvectors are not uniquel determined but could be chosen to be perpendicular. = = 1 10/3/13 GG

10 V Eigenvectors and eigenvalues (cont.) B Eamples (cont.) A matri for rotanons in the plane cosω sinω sinω cosω = λ All non- zero real vectors rotate; a D rotanon matri has no real eigenvectors and hence no real eigenvalues 10/3/13 GG V Eigenvectors and eigenvalues (cont.) B Eamples (cont.) 3 A 3D rotanon matri a The onl unit vector that is not rotated is along the ais of rotanon b The real eigenvector of a 3D rotanon matri gives the orientanon of the ais of rotanon c A rotanon does not change the length of vectors, so the real eigenvalue equals 1 10/3/13 GG

11 V Eigenvectors and eigenvalues (cont.) B Eamples (cont.) 4 A A = 0 0 = 0 0 = Eigenvalues Eigenvectors = A = 0 0 = = 10/3/13 GG303 1 V Eigenvectors and eigenvalues (cont.) B Eamples (cont.) 5 A = Eigenvalues Eigenvectors A 0.1 = = = A = = = /3/13 GG303 11

12 V Eigenvectors and eigenvalues (cont.) E Geometric meanings of the real matri equanon [A][X] = [B] = 0 1 A 0 ; a [A] - 1 eists b Describes two lines (or 3 planes) that intersect at the origin c X has a unique solunon A = 0 ; a [A] - 1 does not eist b Describes two co- linear lines that that pass through the origin (or three planes that intersect in a line or a plane through the origin) c [X] has no unique solunon n 1 n n 1 n Intersecting lines have non-parallel normals AX = B = 0 n (1) n (1) d 1 =0 n () n () = d =0 A = n (1) * n () - n (1) * n () 0 n 1 n 0 Parallel lines have parallel normals AX = B = 0 n (1) n (1) d 1 =0 n () n () = d =0 A = n (1) * n () - n (1) * n () = 0 n 1 n = 0 10/3/13 GG303 3 V Eigenvectors and eigenvalues (cont.) F AlternaNve form of an eigenvalue equanon 1 [A][X]=λ[X] SubtracNng λ[ix] = λ[x] from both sides ields: [A- Iλ][X]=0 (same form as [A][X]=0) G SoluNon condinons and connecnons with determinants 1 Unique trivial solunon of [X] = 0 if and onl if A- Iλ 0 Eigenvector solunons ([X] 0) if and onl if A- Iλ =0 * See previous slide 10/3/13 GG

13 V Eigenvectors and eigenvalues (cont.) H CharacterisNc equanon: A- Iλ =0 1 The roots of the characterisnc equanon are the eigenvalues 10/3/13 GG303 5 V Eigenvectors and eigenvalues (cont.) H CharacterisNc equanon: A- Iλ =0 (cont.) Eigenvalues of a general matri A = a A Iλ = a I b c d λ = 0 a c b d b c d ( a λ) ( d λ) bc = 0 λ ( a + d)λ + ( ad bc) = 0 ( λ 1,λ = a + d) ± a + d ( ) 4 ad bc ( ) 10/3/13 GG303 6 (a+d) = tr(a) (ad- bc) = A λ 1 + λ = tr( A) λ 1 λ = A 13

14 V Eigenvectors and eigenvalues (cont.) I To solve for eigenvectors, subsntute eigenvalues back into AX= lx and solve for X J See notes of lecture 19 for details of analnc solunon for eigenvectors of D matrices 10/3/13 GG303 7 V Eigenvectors and eigenvalues (cont.) K Matlab solunon: [vec,val] = eig(m) 1 M = matri to solve for vec = matri of unit eigenvectors (in columns) 3 val = matri of eigenvalues (in columns) L Eample:>> [vec,val]=eig([ ; ]) vec = val = /3/13 GG

15 VI SoluNons for smmetric matrices A Eigenvalues of a smmetric matri ( 1 λ 1,λ = a + d) ± ( a + d A = ) 4( ad b ) ( λ 1,λ = a + d) ± ( a + ad + d) 4ad + 4b ( 3 λ 1,λ = a + d) ± ( a ad + d) + 4b ( 4 λ 1,λ = a + d) ± ( a d) + 4b a b b d Replace c b b in eqns. Of slide 6 Radical term cannot be neganve. Eigenvalues are real. 10/3/13 GG303 9 VI SoluNons for smmetric matrices (cont.) B An disnnct eigenvectors (X 1, X ) of a smmetric nn matri are perpendicular (X 1 X = 0) 1a AX 1 =λ 1 X 1 1b AX =λ X AX 1 parallels X 1, AX parallels X (propert of eigenvectors) Dovng AX 1 b X and AX b X 1 can test whether X 1 and X are orthogonal. a X AX 1 = X λ 1 X 1 = λ 1 (X X 1 ) b X 1 AX = X1 λ X = λ (X 1 X ) 10/3/13 GG

16 B DisNnct eigenvectors (X 1, X ) of a smmetric matri are perpendicular (X 1 X = 0) (cont.) The material below shows X 1 AX = X AX 1 for the D case: 3a 1 1 a b b d = 1 1 a + b b + d = a + b 1 1 +b 1 + d 1 3b a b b d 1 1 = a 1 + b 1 b 1 + d 1 = a + b 1 1 +b 1 + d 1 The sums on the right sides are scalars, but the ordering of the terms in the sums look like the elements of transposed matrices 10/3/13 GG B DisNnct eigenvectors (X 1, X ) of a smmetric 33 matri are perpendicular (X 1 X = 0) (cont.) The material below shows X 1 AX = X AX 1 for the 3D case: 3c 3d 1 1 z 1 z a b c b d e c e f a b c b d e c e f z 1 1 z 1 = = 1 1 z 1 z a + b + cz b + d + ez = c + e + fz a 1 + b 1 + c 1 z +b 1 + d 1 + e 1 z +cz 1 + ez 1 + fz 1 z a 1 + b 1 + cz 1 a 1 + b 1 + cz 1 b 1 + d 1 + ez 1 = +b 1 + d 1 + ez 1 c 1 + e 1 + fz 1 +c 1 z + e 1 z + fz 1 z Again, the sums on the right sides are scalars, but the ordering of the terms in the sums look like the elements of transposed matrices 10/3/13 GG

17 B DisNnct eigenvectors (X 1, X ) of a smmetric nn matri are perpendicular (X 1 X = 0) (cont.) The D and 3D results suggest matri transposes could test whether X 1 AX = X AX 1 in general X 1 AX = [ X 1 ] T [ A] [ X ] X AX 1 = [ X ] T [ A] [ X 1 ] = [ X ] T [ A] [ X 1 ] T = [ A] [ X 1 ] [ X ] T T = [ X 1 ] T [ A] T [ X ] T [ ] T [ A] T [ X ] [ ] T [ A] [ X ] = X 1 = X 1 T 10/3/13 GG T Are these equal? The transpose of a scalar is the same scalar This step and the net invoke [BC] T = [C] T [B] T [ ] T X [ ] T = X If [A] is smmetric, [A] T = [A] Yes! B DisNnct eigenvectors (X 1, X ) of a smmetric nn matri are perpendicular (cont.) Since the lez sides of (a) and (b) are equal, the right sides must be equal too. Hence, 4 λ 1 (X X 1 ) =λ (X 1 X ) Now subtract the right side of (4) from the lez 5 (λ 1 λ )(X X 1 ) =0 The eigenvalues generall are different, so λ 1 λ 0. This means for (5) to hold that X X 1 =0. The eigenvectors (X 1, X ) of a smmetric nn matri are perpendicular (or can be chosen to be perpendicular) We can pick reference frames with orthogonal aes to simplif problems and gain insight into their solunons 10/3/13 GG

18 VI SoluNons for smmetric matrices (cont.) C Maimum and minimum squared lengths Set derivanve of squared lengths to zero X X = ( AX) ( AX) = L f d( X X ) = X d X dθ dθ + d X dθ X = 0 X d X dθ = 0 dx X dθ = 0 D PosiNon vectors (X ) with maimum and minimum (squared) lengths are those that are perpendicular to tangent vectors (dx ) along ellipse 10/3/13 GG VI SoluNons for smmetric matrices (cont.) E AX=λX F Since eigenvectors of smmetric matrices are mutuall perpendicular, so too are the parallel transformed vectors λx G At the point idennfied b the transformed vector λx, the other eigenvector(s) is (are) perpendicular and hence must parallel dx and be tangent to the ellipse 10/3/13 GG

19 VI SoluNons for smmetric matrices (cont.) H Recall that posinon vectors (X ) with maimum and minimum (squared) lengths are those that are perpendicular to tangent vectors (dx ) along ellipse. Hence, the smallest and largest transformed vectors λx for a smmetric matri give the minimum and maimum distances to an ellipse from its center and the direcnons of the ellipse aes. I The λ values are the principal stretches associated with a smmetric [F] matri J These conclusions etend to three dimensions and ellipsoids 10/3/13 GG VII SoluNons for general homogeneous deformanon matrices A Eigenvalues 1 Start with the defininon of quadra&c elonganon Q, which is a scalar Epress using dot products 3 Clear the denominator. Dot products and Q are scalars. L f L = Q 0 X X X X = Q X X = ( X X )Q 10/3/13 GG

20 VII SoluNons for general homogeneous deformanon matrices A Eigenvalues 4 Replace X with [FX] 5 Re- arrange both sides 6 Both sides of this equanon lead off with [X] T, which cannot be a zero vector, so it can be dropped from both sides to ield an eigenvector equanon 7 [F T F] is smmetric: [F T F] T =[F T F] 8 The eigenvalues of [F T F] are the principal quadranc elonganons Q = (L f /L 0 ) 9 The eigenvalues of [F T F] 1/ are the principal stretches S = (L f /L 0 ) [ F] [ X] nn n1 T X n1 X X = T T F nn F nnt F nn [ F] [ X] nn n1 = F nn X n1 ( X X )Q = X n1 T X Q n1t 11 [ X] Q n1 11 X n1 = Q X n1 "[ A] [ X] = λ[ X]" X n1 10/3/13 GG VII SoluNons for general homogeneous deformanon matrices B Special Case: [F] is smmetric 1 [F T F] = [F ] because F = F T The principal stretches (S) again are the square roots of the principal quadranc elonganons (Q) (i.e., the square roots of the eigenvalues of [F ]) 3 The principal stretches (S) also are the eigenvalues of [F ], directl 4 The direcnons of the principal stretches (S) are the eigenvectors of [F ], and of [F T F] = [F ]! 5 The aes of the principal (greatest and least) strain do not rotate F T F [ X] = Q[ X] F [ X] = Q[ X] Q = L f L ; S = L f 0 L 0 [ F] [ X] = S[ X] Q = S 10/3/13 GG

21 VIII RotaNons in homogeneous deformanon A Just gevng the size and shape of the strain (stretch) ellipse is not enough. Need to consider points on the ellipse B F=VR (which R?) 1 R = rotanon matri V = stretch matri C F=RU (which U? R?) 1 U = stretch matri R = rotanon matri D The choices narrow if the stretch matrices are smmetric 10/3/13 GG VIII RotaNons in homogeneous deformanon E If an ellipse is transformed to a unit circle, the aes of the ellipse are transformed too. F In the diagram, the aes of the ellipses do not maintain their orientanon when the ellipse is transformed back to a unit circle G If F is not smmetric, the aes of the red ellipse and the retro- deformed (black) aes will have a different absolute orientanon H The transformanon from the the retro- deformed (black) aes to the the orientanon of the principal aes gives the rotanon of the aes 10/3/13 GG

22 VIII RotaNons in homogeneous deformanon I We know how to find the principal stretch magnitudes: the are the square roots of the eigenvalues of the smmetric matri [ [F T ][F] ] J The eigenvectors of [ [F T ][F] ] give the direcnon of the principal stretch aes. The rotanon describes the orientanon difference between the unit eigenvectors of [ [F T ][F] ] (the principal strain (stretch) aes) their retro- deformed counterparts {inv(f)*eigenvectors of [ [F T ][F] ] } 10/3/13 GG [F] = [R][U] F = [ F] [ X] [ X] [ X ] = [ F] [ X]; [ F] = [ R] [ U ] [ F] = ; F [ U ] = [ F] T [ F] [ R] = [ F] [ U ] 1 = [ ] T = / 1/ = = = First, smmetricall stretch the unit circle using [U] [ U ] = [ U ][ X] [ X] R = [ R] [ U ][ X] [ U ][ X] Second, rotate the ellipse (not the reference frame) using [R] 10/3/13 GG303 44

23 [F] = [V] [R] F = F = [ F] [ X] [ X] [ X ] = [ F] [ X]; [ F] = [ V ][ R] [ F] = ; F [ V ] = [ F] [ F] T [ R] = [ V ] 1 [ F] = [ ] T = / 1/ 8 3 = = = First, rotate the unit circle using [R] R = [ R] X [ V ] R = = [ ] [ X] R V [ ][ R] [ X] [ ][ X] Second, stretch the rotated unit circle smmetricall using [V] 10/3/13 GG VIII RotaNons in homogeneous deformanon DecomposiNon of F = VR b method of Ramsa and Huber (for D). Consider the effect of an irrotanonal strain V that follows a pure rotanon (R*) of the object (not a rigid rotanon of the reference frame) F = a c b d = A B B D cosω sinω sinω = VR cosω 10/3/13 GG

24 VIII RotaNons in homogeneous deformanon Ke fact about rotanon matrices: [R] - 1 = [R] T ( ) = R ω R 1 = R ω R T = cosω sinω ( ) = cosω sinω sinω cosω cosω sinω sinω cosω sinω cosω [ X ] = [ R] [ X] [ X] = [ R] 1 [ X ] 10/3/13 GG VIII RotaNons in homogeneous deformanon Ke fact about rotanon matrices: [R] - 1 = [R] T 3D treatment: rotanng a reference frame does not change the length of a vector [ X ] = [ R] [ X] X X = X X X X = [ X] T [ X] = X T X X = [ R] [ X] R [ ] T [ I ][ X] [ ][ X] = [ X] T [ R] T [ R] [ X] [ R] T [ R] = [ I ], but [ R] 1 [ R] = [ I ] [ R] T = [ R] 1 10/3/13 GG

25 VIII RotaNons in homogeneous deformanon 1 a b A F = = c d B a b = c d B cosω D sinω sinω = VR cosω Acosω + Bsinω Asinω + Bcosω Bcosω + Dsinω Bsinω + D cosω c- b = (A+D)sinω, and a+d = (A+D)cosω c b a + d = tanω 3 If c=b (F is smmetric), then ω=0! From 3 one can obtain ω and hence R. [ R] = cosω sinω sinω cosω 10/3/13 GG VIII RotaNons in homogeneous deformanon Post- mulnpling both sides of (1) b [R] - 1 = R T ields V, the smmetric part of F. F = VR F[R] - 1 = VR[R] - 1 = VR[R] T = V a c b d cosω sinω sinω cosω 1 = a c b d cosω sinω sinω cosω = A B B D = V 10/3/13 GG

26 IX Closing comments 1 Our solunons so far depend on knowing the displacement field. With satellite imaging we can get an approimate value for the displacement field at the surface of the Earth for current deformanons 3 EvaluaNng strains for past deformanons require certain assumpnons about ininal sizes and shapes of bodies, the original locanons of point, and/or the displacement field. 4 AlternaNve approach: formulanon and solunon of boundar value problems to solve for the displacement and strain fields. 5 The deformanon gradient matri F has strain and rotanon intertwined; the two can be separated using matri mulnplicanon. In the infinitesimal strain matri [ε], the rotanon is alread separated. 6 References a Ramsa, J.G., and Huber, M.I., 1983, The techniques of modern structural geolog, volume 1: strain analsis: Academic Press, London, 307 p. (See equanons of secnon 5, p. 91). b Ramsa, J.G., and Lisle, M.I., 1983, The techniques of modern structural geolog, volume 3: applicanons of connnuum mechanics in structural geolog: Academic Press, London, 307 p. (See especiall sessions 33 and 36). c Malvern, L.E., 1969, IntroducNon to the mechanics of a connnuous medium: PrenNce- Hall, Englewood Cliffs, New Jerse, 713 p. (See equanons 4.6.1, a, 4.6.3b on p ).) 10/3/13 GG