"""Frechet derivative of the matrix exponential.""" import numpy as np import scipy.linalg __all__ = ['expm_frechet', 'expm_cond'] def expm_frechet(A, E, method=None, compute_expm=True, check_finite=True): """ Frechet derivative of the matrix exponential of A in the direction E. Parameters ---------- A : (N, N) array_like Matrix of which to take the matrix exponential. E : (N, N) array_like Matrix direction in which to take the Frechet derivative. method : str, optional Choice of algorithm. Should be one of - `SPS` (default) - `blockEnlarge` compute_expm : bool, optional Whether to compute also `expm_A` in addition to `expm_frechet_AE`. Default is True. check_finite : bool, optional Whether to check that the input matrix contains only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs. Returns ------- expm_A : ndarray Matrix exponential of A. expm_frechet_AE : ndarray Frechet derivative of the matrix exponential of A in the direction E. For ``compute_expm = False``, only `expm_frechet_AE` is returned. See Also -------- expm : Compute the exponential of a matrix. Notes ----- This section describes the available implementations that can be selected by the `method` parameter. The default method is *SPS*. Method *blockEnlarge* is a naive algorithm. Method *SPS* is Scaling-Pade-Squaring [1]_. It is a sophisticated implementation which should take only about 3/8 as much time as the naive implementation. The asymptotics are the same. .. versionadded:: 0.13.0 References ---------- .. [1] Awad H. Al-Mohy and Nicholas J. Higham (2009) Computing the Frechet Derivative of the Matrix Exponential, with an application to Condition Number Estimation. SIAM Journal On Matrix Analysis and Applications., 30 (4). pp. 1639-1657. ISSN 1095-7162 Examples -------- >>> import numpy as np >>> from scipy import linalg >>> rng = np.random.default_rng() >>> A = rng.standard_normal((3, 3)) >>> E = rng.standard_normal((3, 3)) >>> expm_A, expm_frechet_AE = linalg.expm_frechet(A, E) >>> expm_A.shape, expm_frechet_AE.shape ((3, 3), (3, 3)) Create a 6x6 matrix containing [[A, E], [0, A]]: >>> M = np.zeros((6, 6)) >>> M[:3, :3] = A >>> M[:3, 3:] = E >>> M[3:, 3:] = A >>> expm_M = linalg.expm(M) >>> np.allclose(expm_A, expm_M[:3, :3]) True >>> np.allclose(expm_frechet_AE, expm_M[:3, 3:]) True """ if check_finite: A = np.asarray_chkfinite(A) E = np.asarray_chkfinite(E) else: A = np.asarray(A) E = np.asarray(E) if A.ndim != 2 or A.shape[0] != A.shape[1]: raise ValueError('expected A to be a square matrix') if E.ndim != 2 or E.shape[0] != E.shape[1]: raise ValueError('expected E to be a square matrix') if A.shape != E.shape: raise ValueError('expected A and E to be the same shape') if method is None: method = 'SPS' if method == 'SPS': expm_A, expm_frechet_AE = expm_frechet_algo_64(A, E) elif method == 'blockEnlarge': expm_A, expm_frechet_AE = expm_frechet_block_enlarge(A, E) else: raise ValueError('Unknown implementation %s' % method) if compute_expm: return expm_A, expm_frechet_AE else: return expm_frechet_AE def expm_frechet_block_enlarge(A, E): """ This is a helper function, mostly for testing and profiling. Return expm(A), frechet(A, E) """ n = A.shape[0] M = np.vstack([ np.hstack([A, E]), np.hstack([np.zeros_like(A), A])]) expm_M = scipy.linalg.expm(M) return expm_M[:n, :n], expm_M[:n, n:] """ Maximal values ell_m of ||2**-s A|| such that the backward error bound does not exceed 2**-53. """ ell_table_61 = ( None, # 1 2.11e-8, 3.56e-4, 1.08e-2, 6.49e-2, 2.00e-1, 4.37e-1, 7.83e-1, 1.23e0, 1.78e0, 2.42e0, # 11 3.13e0, 3.90e0, 4.74e0, 5.63e0, 6.56e0, 7.52e0, 8.53e0, 9.56e0, 1.06e1, 1.17e1, ) # The b vectors and U and V are copypasted # from scipy.sparse.linalg.matfuncs.py. # M, Lu, Lv follow (6.11), (6.12), (6.13), (3.3) def _diff_pade3(A, E, ident): b = (120., 60., 12., 1.) A2 = A.dot(A) M2 = np.dot(A, E) + np.dot(E, A) U = A.dot(b[3]*A2 + b[1]*ident) V = b[2]*A2 + b[0]*ident Lu = A.dot(b[3]*M2) + E.dot(b[3]*A2 + b[1]*ident) Lv = b[2]*M2 return U, V, Lu, Lv def _diff_pade5(A, E, ident): b = (30240., 15120., 3360., 420., 30., 1.) A2 = A.dot(A) M2 = np.dot(A, E) + np.dot(E, A) A4 = np.dot(A2, A2) M4 = np.dot(A2, M2) + np.dot(M2, A2) U = A.dot(b[5]*A4 + b[3]*A2 + b[1]*ident) V = b[4]*A4 + b[2]*A2 + b[0]*ident Lu = (A.dot(b[5]*M4 + b[3]*M2) + E.dot(b[5]*A4 + b[3]*A2 + b[1]*ident)) Lv = b[4]*M4 + b[2]*M2 return U, V, Lu, Lv def _diff_pade7(A, E, ident): b = (17297280., 8648640., 1995840., 277200., 25200., 1512., 56., 1.) A2 = A.dot(A) M2 = np.dot(A, E) + np.dot(E, A) A4 = np.dot(A2, A2) M4 = np.dot(A2, M2) + np.dot(M2, A2) A6 = np.dot(A2, A4) M6 = np.dot(A4, M2) + np.dot(M4, A2) U = A.dot(b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident) V = b[6]*A6 + b[4]*A4 + b[2]*A2 + b[0]*ident Lu = (A.dot(b[7]*M6 + b[5]*M4 + b[3]*M2) + E.dot(b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident)) Lv = b[6]*M6 + b[4]*M4 + b[2]*M2 return U, V, Lu, Lv def _diff_pade9(A, E, ident): b = (17643225600., 8821612800., 2075673600., 302702400., 30270240., 2162160., 110880., 3960., 90., 1.) A2 = A.dot(A) M2 = np.dot(A, E) + np.dot(E, A) A4 = np.dot(A2, A2) M4 = np.dot(A2, M2) + np.dot(M2, A2) A6 = np.dot(A2, A4) M6 = np.dot(A4, M2) + np.dot(M4, A2) A8 = np.dot(A4, A4) M8 = np.dot(A4, M4) + np.dot(M4, A4) U = A.dot(b[9]*A8 + b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident) V = b[8]*A8 + b[6]*A6 + b[4]*A4 + b[2]*A2 + b[0]*ident Lu = (A.dot(b[9]*M8 + b[7]*M6 + b[5]*M4 + b[3]*M2) + E.dot(b[9]*A8 + b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident)) Lv = b[8]*M8 + b[6]*M6 + b[4]*M4 + b[2]*M2 return U, V, Lu, Lv def expm_frechet_algo_64(A, E): n = A.shape[0] s = None ident = np.identity(n) A_norm_1 = scipy.linalg.norm(A, 1) m_pade_pairs = ( (3, _diff_pade3), (5, _diff_pade5), (7, _diff_pade7), (9, _diff_pade9)) for m, pade in m_pade_pairs: if A_norm_1 <= ell_table_61[m]: U, V, Lu, Lv = pade(A, E, ident) s = 0 break if s is None: # scaling s = max(0, int(np.ceil(np.log2(A_norm_1 / ell_table_61[13])))) A = A * 2.0**-s E = E * 2.0**-s # pade order 13 A2 = np.dot(A, A) M2 = np.dot(A, E) + np.dot(E, A) A4 = np.dot(A2, A2) M4 = np.dot(A2, M2) + np.dot(M2, A2) A6 = np.dot(A2, A4) M6 = np.dot(A4, M2) + np.dot(M4, A2) b = (64764752532480000., 32382376266240000., 7771770303897600., 1187353796428800., 129060195264000., 10559470521600., 670442572800., 33522128640., 1323241920., 40840800., 960960., 16380., 182., 1.) W1 = b[13]*A6 + b[11]*A4 + b[9]*A2 W2 = b[7]*A6 + b[5]*A4 + b[3]*A2 + b[1]*ident Z1 = b[12]*A6 + b[10]*A4 + b[8]*A2 Z2 = b[6]*A6 + b[4]*A4 + b[2]*A2 + b[0]*ident W = np.dot(A6, W1) + W2 U = np.dot(A, W) V = np.dot(A6, Z1) + Z2 Lw1 = b[13]*M6 + b[11]*M4 + b[9]*M2 Lw2 = b[7]*M6 + b[5]*M4 + b[3]*M2 Lz1 = b[12]*M6 + b[10]*M4 + b[8]*M2 Lz2 = b[6]*M6 + b[4]*M4 + b[2]*M2 Lw = np.dot(A6, Lw1) + np.dot(M6, W1) + Lw2 Lu = np.dot(A, Lw) + np.dot(E, W) Lv = np.dot(A6, Lz1) + np.dot(M6, Z1) + Lz2 # factor once and solve twice lu_piv = scipy.linalg.lu_factor(-U + V) R = scipy.linalg.lu_solve(lu_piv, U + V) L = scipy.linalg.lu_solve(lu_piv, Lu + Lv + np.dot((Lu - Lv), R)) # squaring for k in range(s): L = np.dot(R, L) + np.dot(L, R) R = np.dot(R, R) return R, L def vec(M): """ Stack columns of M to construct a single vector. This is somewhat standard notation in linear algebra. Parameters ---------- M : 2-D array_like Input matrix Returns ------- v : 1-D ndarray Output vector """ return M.T.ravel() def expm_frechet_kronform(A, method=None, check_finite=True): """ Construct the Kronecker form of the Frechet derivative of expm. Parameters ---------- A : array_like with shape (N, N) Matrix to be expm'd. method : str, optional Extra keyword to be passed to expm_frechet. check_finite : bool, optional Whether to check that the input matrix contains only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs. Returns ------- K : 2-D ndarray with shape (N*N, N*N) Kronecker form of the Frechet derivative of the matrix exponential. Notes ----- This function is used to help compute the condition number of the matrix exponential. See Also -------- expm : Compute a matrix exponential. expm_frechet : Compute the Frechet derivative of the matrix exponential. expm_cond : Compute the relative condition number of the matrix exponential in the Frobenius norm. """ if check_finite: A = np.asarray_chkfinite(A) else: A = np.asarray(A) if len(A.shape) != 2 or A.shape[0] != A.shape[1]: raise ValueError('expected a square matrix') n = A.shape[0] ident = np.identity(n) cols = [] for i in range(n): for j in range(n): E = np.outer(ident[i], ident[j]) F = expm_frechet(A, E, method=method, compute_expm=False, check_finite=False) cols.append(vec(F)) return np.vstack(cols).T def expm_cond(A, check_finite=True): """ Relative condition number of the matrix exponential in the Frobenius norm. Parameters ---------- A : 2-D array_like Square input matrix with shape (N, N). check_finite : bool, optional Whether to check that the input matrix contains only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs. Returns ------- kappa : float The relative condition number of the matrix exponential in the Frobenius norm See Also -------- expm : Compute the exponential of a matrix. expm_frechet : Compute the Frechet derivative of the matrix exponential. Notes ----- A faster estimate for the condition number in the 1-norm has been published but is not yet implemented in SciPy. .. versionadded:: 0.14.0 Examples -------- >>> import numpy as np >>> from scipy.linalg import expm_cond >>> A = np.array([[-0.3, 0.2, 0.6], [0.6, 0.3, -0.1], [-0.7, 1.2, 0.9]]) >>> k = expm_cond(A) >>> k 1.7787805864469866 """ if check_finite: A = np.asarray_chkfinite(A) else: A = np.asarray(A) if len(A.shape) != 2 or A.shape[0] != A.shape[1]: raise ValueError('expected a square matrix') X = scipy.linalg.expm(A) K = expm_frechet_kronform(A, check_finite=False) # The following norm choices are deliberate. # The norms of A and X are Frobenius norms, # and the norm of K is the induced 2-norm. A_norm = scipy.linalg.norm(A, 'fro') X_norm = scipy.linalg.norm(X, 'fro') K_norm = scipy.linalg.norm(K, 2) kappa = (K_norm * A_norm) / X_norm return kappa