solve_banded(l_and_u, ab, b, overwrite_ab=False, overwrite_b=False, debug=None, check_finite=True)

The matrix a is stored in :None:None:`ab` using the matrix diagonal ordered form:

ab[u + i - j, j] == a[i,j]

Example of :None:None:`ab` (shape of a is (6,6), u =1, l =2):

*    a01  a12  a23  a34  a45
a00  a11  a22  a33  a44  a55
a10  a21  a32  a43  a54   *
a20  a31  a42  a53   *    *

Parameters

(l, u) : (integer, integer)

Number of non-zero lower and upper diagonals

ab : (`l` + `u` + 1, M) array_like

Banded matrix

b : (M,) or (M, K) array_like

Right-hand side

overwrite_ab : bool, optional

Discard data in :None:None:`ab` (may enhance performance)

overwrite_b : bool, optional

Discard data in b (may enhance performance)

check_finite : bool, optional

Whether to check that the input matrices contain 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

x : (M,) or (M, K) ndarray

The solution to the system a x = b. Returned shape depends on the shape of b.

Solve the equation a x = b for x, assuming a is banded matrix.

Examples

[5 2 -1 0 0] [0] [1 4 2 -1 0] [1]

a = [0 1 3 2 -1] b = [2]

[0 0 1 2 2] [2] [0 0 0 1 1] [3]

[* * -1 -1 -1]

ab = [* 2 2 2 2]

[5 4 3 2 1] [1 1 1 1 *]

>>> from scipy.linalg import solve_banded
... ab = np.array([[0,  0, -1, -1, -1],
...                [0,  2,  2,  2,  2],
...                [5,  4,  3,  2,  1],
...                [1,  1,  1,  1,  0]])
... b = np.array([0, 1, 2, 2, 3])
... x = solve_banded((1, 2), ab, b)
... x
array([-2.37288136,  3.93220339, -4.        ,  4.3559322 , -1.3559322 ])

See :

Back References

The following pages refer to to this document either explicitly or contain code examples using this.

scipy.spatial.transform._rotation_spline._create_block_3_diagonal_matrix scipy.linalg._basic.solve_banded scipy.integrate._ivp.lsoda.LSODA scipy.integrate._ivp.ivp.solve_ivp

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GitHub : /scipy/linalg/_basic.py#367
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