Evaluation of earthquake mechanism through waveform inversion

1. Evaluation of earthquake mechanism through waveform inversion S.N. Bhattacharya Dept. of Geology and Geophysics IIT, Kharagpur 721302 email <sn_bhattacharya@hotmail.com

2. Contents • Shear dislocation & synthetic seismogram • Moment tensor and its decomposition • Synthetic seismogram for a source represented by a moment tensor • Procedure for moment tensor evaluation • Application

3. Up N  Strike dir.   V Receiver R T  Fault plane D A  Epc.

4. Ground motion at an epicentral distance  Fourier transform We represent U(,) = M() E(,) Where M() is depends on moment time function m(t) = SA(t)D(t) And E(,) depends on (1) Medium (2) Focal depth (3) Focal mechanism (4) Epicentral Distance (5) Azimuth of receiver from strike dir ()

5. Source time function Moment time function m(t) = M0 s(t), where s(t) source time function and M0 = s A D is seismic moment. s(t) = 0 for t  0 = g(t) for 0 < t  = 1 for t >  Fourier transformation gives M () = M0 S()

6. Expression of g(t) (i) Ramp fn. g(t) = t/ (ii) Brustle and Muller (1983) g (t) = (9/16){- cos (t/) +(1/9) cos 3(t/)} + 1/2 s(t) s(t) t/ s(t) s(t) t/

7. Expression of g(t) (continued) Some time g’(t) is chosen directly. While choosing g’(t), we note that   g(t) dt = 1 0 (iii) Wang and Herrmann (1980) suggested 2 T g (t) = 0.5 (t/T)2, 0 < t  T/4 = - 0.5 (t/T)2 + 2 (t/T) - 1, T < t  3 T = 0.5 (t/T)2 - 4 (t/T) + 8, 3 T < t  4 T where  = 4 T. s(t) t/

8. Expression of g(t) (continued) (iv) g(t) is a trapezoidal function g (t) 0  (v) g(t) is a simple triangular function g (t) 2/  0 

9. Generating E(,) Z Ground level Receiver * Source

10. Reference synthetic seismograms for V U(,) = M() E(,) where M() = M0 S() and E(,) depends on (1) Medium (2) Focal depth * (3) Focal mechanism (, ) (4) Epc. Distance * (5) Az. of receiver from strike dir () For vertical component, we obtain EVDD(,), EVDS(,), EVSS(,) which depends on (1), (2) and (4). Each of three E’s are for a basic focal mechanism DD, DS, SS and at a fixed . So that UVDD(,) = S() EVDD(,), we generate seismogram for unit  moment e.g. 1024 dyne.cm uVDD(t, ) = (1/ 2) UVDD(,) exp(it) d - 

11. 3 Reference synthetic seismograms each for Z and R For vertical component, in addition to uVDD(,t) we generate 2 more reference synthetic seismograms uVDS(,t) and uVSS(,t). For radial components we similarly generate 3 reference synthetic seismograms uRDD(,t), uRDS(,t) and uRSS(,t) While we generate these reference seismograms we take a unit seismic moment, say 1024 dyne.cm

12. Projection of upper hemisphere and the point where the 3 reference synthetic seismograms are generated DD DS SS D C D C D D C C D  = 450 ,  = 900 = 900,  = - 900 = 900,  = 00 at  = 450 at  = 900 at  = 450 uVDD(,t) uVDS(,t) uVSS(,t) uRDD(,t) uRDS(,t) uRSS(,t)

13. Vertical & Radial seismogram for any (, ,  ) uV(,t) = M0 [ a uVDD(,t) + b uVDS(,t) +c uVSS(,t) ] uR(,t) = M0 [ a uRDD(,t) + b uRDS(,t) + c uRSS(,t) ] where M0 = Seismic moment in the unit under consideration a = sin  sin 2 b = - cos  cos  cos  + sin  cos 2 sin  c = 0.5 sin  sin 2 cos 2 + cos  sin  sin 2 We shall choose , ,  in such a way that uZ(,t) and uR(,t) match with the shape of corrsponding observed ground motions and absolute amplitude is matched with selection of M0 . Thus we evaluate focal mechanism.

14. 2 Reference synthetic seismograms for T U(,) = M() E(,) where M() = M0 S() and E(,) depends on (1) Medium (2) Focal depth * (3) Focal mechanism(, ) (4) Epc. Distance * (5) Az. of receiver from strike dir () For Transverse component, we obtain ETDS(,), ETSS(,) which depends on (1), (2) and (4). Each of the two E’s are for a basic focal mechanism DS, SS and at a fixed . So that UTDS(,) = S() ETDS(,), we generate seismogram for unit  moment e.g. 1024 dyne.cm uTDS(t, ) = (1/ 2) UTDS(,) exp(it) d - 

15. Projection of upper hemisphere and the point where the 2 reference synthetic seismograms are generated DS SS D C C D C D  = 900,  = -900 = 900,  = 00 at  = 00 at  = 00 uTDS(,t) uTSS(,t)

16. Transverse seismogram for any (, , ) uT(,t) = M0 [ a uTDS(,t) + b uTSS(,t) ] with M0 = Seismic moment in the unit under consideration a = sin  cos 2 cos  + cos  cos  sin  b = cos  sin  cos 2 - 0.5 sin  sin 2 sin 2 We shall choose , , in such a way that not only uZ(,t) and uR(,t) but alsouT(,t) match with the shape of the corresponding observed ground motions and absolute amplitude is matched with selection of M0 . Thus we evaluate focal mechanism.

17. Distance = 45 km

18. Distance = 132 km

19. PUNE 18030 N Study area Mumbai Goa Chennai 18000 Koyna Reservoir 17030 KARD 1997 Apr 25 Warna Reservoir 17000 N 74000 E 73030 E

20. 45 km 132 km Vertical Radial Transverse

21. Vertical Vertical

22. Double couple and principle axes X3 Auxiliary plane X2 Fault plane X1 M13 X3 T M31 P X1 M13 = M31 = M0 =  At Da

23. Moment tensor We define a force couple Mij in a Cartesian coordinate system as pair of opposite forces pointing in the i direction separated in the j direction. The magnitude of Mij = f., when  0 and f. X3 X3 X3 X1 X1 X1 M13 & M31 M11 M13 Force couple (a & b) are opposing forces separated by small distance. A double couple (c ) is pair of complimentary couples with no torque

24. Moment tensors Z Z Z MXX Y MXY Y MXZ Y X X X Z Z Z MYX Y MYY Y MYZ Y X X X Z Z Z MZX Y MZY Y MZZ Y X X X

25. Moment tensor • There can be 9 different force couples. The condition that angular moment be conserved requires that Mij = Mji . Thus we define a moment tensor as • [M] = = • Thus moment tensor [M] has only 6 independent elements and is symmetric matrix.

26. Evaluation of moment tensor • We write moment tensor as • m1 Mxx • m2 MYY • [m] = m3 = MZZ • m4 MXY • m5 MXZ • m6 MYZ • We define Gij(t) as a seismogram (or Green’s function) at the ith component or seismometer due to moment tensor component mj . The Gij(t) takes care of source time function, epicentral distance, focal depth, azimuth of the station from epicentre, medium (including attenuation). So the ith seismogram is the sum of Green’s functions weighted by the moment tensor components, • ui (t) = Gij(t) mj

27. Evaluation of moment tensor ui (t) = Gij(t) mj Because we have many seismograms, we can write [u] = [G] [m] where [u] is vector composed of the seismogram at N stations and [G] is Green’s functions matrix. [G] has N rows and 6 columns. Then we can find the moment tensor that matches the observed seismogram in a least square sense use generalized inverse [m] =  [G]T[G] ] -1 [G]T [u]

28. Decomposition of Moment tensor [M] = = Q + Where Q = (1/3) tr[M] and tr [M[ = MXX + MYY + MXY In above first term is isotropic part and second term is deviatoric part.

29. Principal axes of moment tensor Let us consider that elements of moment tensor have evaluated through waveform inversion. We may decompose it to evaluate the double couple part and the other part. We may write [M] as   a1x a2x a3x1 0 0 a1x a1y a1z [M] = a1y a2y a3y 0 2 0 a2x a2y a2z a1z a2z a3z 0 0 3 a3x a3y a3z where m1, m2 and m3 are eigenvalues of [M] and [aix aiy aiz]T are the eigenvector corresponding to mi. M0 = 0.5 [( 12 + 22 + 32)]. We shall assume | 1|  | 2|  | 3| ,

30. Moment tensor of a double couple   a1x a2x a3xM0 0 0 a1x a1y a1z [M] = a1y a2y a3y 0 -M00 a2x a2y a2z a1z a2z a3z 0 0 0a3x a3y a3z In case of double couple m1 = - m2 = M0 and m3 = 0 (a1x , a1y , a1z) are components of unit vector along T-axis (a2x , a2y , a2z) are components of unit vector along P-axis (a3x , a3y , a3z ) are components of unit vector along P-axis

31. Moment tensor for double couple For the double couple shown earlier we have Mxz = Mzx = M0 0 0 Mxz 0 0 M0 [M] = 0 0 0 = 0 0 0 P Z T Mzx 0 0 M0 0 0 X =

32. Decomposition of deviatoric Moment tensor to major & minor double couple Since MXX + MYY + MXY = 0 therefore 1 + 2 + 3 = 0, i.e. 2 = - 1 - 3 1 0 0 1 0 0 0 0 0 0 2 0 = 0 - 1 0 + 0 - 3 0 0 0 3 0 0 0 0 0 3 The first term gives major double couple and second term gives minor double couple.

33. Decomposition of deviatoric moment tensorto best double couple and CLVD A compensated linear vector dipole (CLVD) are set of 3 force dipoles that are compensated with one dipole – 2 times the magnitude of the others. Consider F = - 3 / 1 (note 0 < F < 0.5). Further since 2 = - 1 - 3 , we can write 1 0 0 1 0 0 0 2 0 = 1 0 2/ 1 0 0 0 3 0 0 3 / 1 = 1 (1-2F) + 1 F Percentage of Double Couple is (1-2F)*100

34. There are 3 type of pure mechanisms represented by the seismic moment tensor, isotropic, double couple, and compensated linear vector dipole.

35. Deviatoric Moment Tensor • Here Mxx + Myy + Mzz = 0 • m1 = Mxx, m2 = Myy, m3 = Mxy , m4 = Mxz , m5 = Myz • Gi1 = Gxxi , Gi2 = Gyyi , Gi3 = Gxyi , Gi4 = Gxzi , Gi5 = Gyzi • i = V, R, T

36. Green’s functions (Vertical & Radial) are obtained from above Expressions for by replacing V with R

37. Green’s functions (Transverse) Transverse

39. Formation of equation • Minimize d(t) = u(t) [d] = [G] [m]

41. [d] = [G1 G2 G3 G4 G5]

42. Inversion • [m] = [G]-g [d] where [G]-g = (GT G)-1 GT Or, better [G]-g = [V] []-1 [UT] where singular value decomposition of [G] is [G] = [U] [] [VT] Synthetic seismogram corresponding to the solution is

45. 2000 September 05 Harvard This study USGS Strike 1830 Dip 480 Rake -720 Strike 1810 Dip 480 Rake -920 Strike 1780 Dip 690 Rake -890  : T-axis,  : P-axis

46. Study area 18030N PUNE Mumbai Goa Chennai 18000 Koyna Reservoir 1 3 17030 6 8 KARD 2 Warna Reservoir 9 5 7 4 17000N 73o30E 74000E