expm_multiply(A, B, start=None, stop=None, num=None, endpoint=None)
The optional arguments defining the sequence of evenly spaced time points are compatible with the arguments of numpy.linspace
.
The output ndarray shape is somewhat complicated so I explain it here. The ndim of the output could be either 1, 2, or 3. It would be 1 if you are computing the expm action on a single vector at a single time point. It would be 2 if you are computing the expm action on a vector at multiple time points, or if you are computing the expm action on a matrix at a single time point. It would be 3 if you want the action on a matrix with multiple columns at multiple time points. If multiple time points are requested, expm_A_B[0] will always be the action of the expm at the first time point, regardless of whether the action is on a vector or a matrix.
The operator whose exponential is of interest.
The matrix or vector to be multiplied by the matrix exponential of A.
The starting time point of the sequence.
The end time point of the sequence, unless :None:None:`endpoint` is set to False. In that case, the sequence consists of all but the last of num + 1
evenly spaced time points, so that :None:None:`stop` is excluded. Note that the step size changes when :None:None:`endpoint` is False.
Number of time points to use.
If True, :None:None:`stop` is the last time point. Otherwise, it is not included.
The result of the action $e^{t_k A} B$ .
Compute the action of the matrix exponential of A on B.
>>> from scipy.sparse import csc_matrix
... from scipy.sparse.linalg import expm, expm_multiply
... A = csc_matrix([[1, 0], [0, 1]])
... A.toarray() array([[1, 0], [0, 1]], dtype=int64)
>>> B = np.array([np.exp(-1.), np.exp(-2.)])
... B array([ 0.36787944, 0.13533528])
>>> expm_multiply(A, B, start=1, stop=2, num=3, endpoint=True) array([[ 1. , 0.36787944], [ 1.64872127, 0.60653066], [ 2.71828183, 1. ]])
>>> expm(A).dot(B) # Verify 1st timestep array([ 1. , 0.36787944])
>>> expm(1.5*A).dot(B) # Verify 2nd timestep array([ 1.64872127, 0.60653066])
>>> expm(2*A).dot(B) # Verify 3rd timestep array([ 2.71828183, 1. ])See :
The following pages refer to to this document either explicitly or contain code examples using this.
scipy.sparse.linalg._expm_multiply.expm_multiply
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