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