我认为这是可能的4台是的,而不是三个。* *(大约)均匀的假设速度* *我们得到以下问题。假设v_p v_s美元和美元是常数之间的地震和你所有的接收器,收益率以下系统:\{对齐}开始d & = \√6{\δx ^ 2 + \δy ^ 2 + \δz ^ 2} ={3已知3未知数}\文本,文本\四\{为}\左\{\开始{数组}{rl} \δx & =间{地震}\文本间\文本{接收机}\ \ \δy & = y_ \{地震}- y_ \文本{接收机}\ \ \δz & = z_ \文本{地震}- z_ \文本{接收机}\{数组}{分裂}\ \开始结束结束{分裂}\。,\ \ \δt & = t_s - t_p ={测量}\文本,\ \ p & = \离开(\压裂{1}{v_s} - \压裂{1}{v_p} \右)= \文本未知{1}。结束\{对齐}重申问题的方程给出了提出问题:$ $ \δt = p d = p \√6{\δx ^ 2 + \δy ^ 2 + \δz ^ 2}。例如,$ $ $ v_p = 2000 m / s和v_s美元= 1200 $ m / s,我们发现p = 1/3000美元,可以计算如$ d = 600 $, $ \δt = 600/3000 = 0.2 $ s。* *,现在我们必须设置= * *逆问题。你通常至少需要尽可能多的方程未知数。在这种情况下,我们有4个未知数(速度和(x, y, z)美元地震位置)。这意味着我们需要4(独特的)方程。我认为最简单的方法获得4独特的方程结合测量:$ $ p \ {pmatrix} d_1开始\ \ d_2结束\ \ d_3 \ \ d_4 \ {pmatrix} = {pmatrix} \ \开始δδt_2 t_1 \ \ \ \ \ \δδt_4 t_3 \ \ \ \ {pmatrix}结束,表示美元和美元间的{1,2,3,4}$和$ y_{1, 2, 3, 4} $接收器的经度和纬度,并假设测量表面,我们得到:$ $ p {pmatrix} \ \开始√6{(间\文本{地震}-x_1) ^ 2 + (y_ \文本{地震}-y_1) ^ 2 + z_文本{地震}^ 2}\ \ \ \√6{(间\文本{地震}-x_2) ^ 2 + (y_ \文本{地震}-y_2) ^ 2 + z_文本{地震}^ 2}\ \ \ \√6{(间\文本{地震}-x_3) ^ 2 + (y_ \文本{地震}-y_3) ^ 2 + z_文本{地震}^ 2}\ \ \ \√6{(间\文本{地震}-x_4) ^ 2 + (y_ \文本{地震}-y_4) ^ 2 + z_文本{地震}^ 2}\ \结束{pmatrix} = {pmatrix} \ \开始δδt_2 t_1 \ \ \ \ \ \δδt_4 t_3 \ \ \ \ {pmatrix}结束。$ $ * *最后,我们必须解决逆问题* *。 Unfortunately, the system is non-linear (the unknowns are 'hidden' in the root of the square and can't be separated). You must therefore use some non-linear solver. For example, in MATLAB you can do this % True earthquake location x_earthquake=0; y_earthquake=0; z_earthquake=1000; % True station locations x_1 = 1000; y_1 = 1000; x_2 = 500; y_2 = -300; x_3 = -400; y_3 = -100; x_4 = -10; y_4 = 800; X = [x_1;x_2;x_3;x_4]; Y = [y_1;y_2;y_3;y_4]; % True velocity structure p0 = 1/3000; % Forward equation (F(1)=x_eartquake, F(2)=y_earthquake, F(3)=z_earthquake, F(4)=p. t =@(F) F(4) * sqrt( (F(1)-X).^2 + (F(2)-Y).^2 + F(3).^2); % True recordings measured_times = t([x_earthquake,y_earthquake,z_earthquake,p0]); % Misfit function (=0 at optimum) G =@(F) t(F) - measured_times; % Invert options = optimoptions('fsolve','FiniteDifferenceType','central'); F_inv = fsolve( G, [1000,1000,1000,0.1],options) x_earthquake_inv = F_inv(1) y_earthquake_inv = F_inv(2) z_earthquake_inv = F_inv(3) p_inv = F_inv(4) The results find an earthquake at (0,0,1000) and p=1/3000 (with small errors after 12 digits).