scipy.linalg.solve_banded

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

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

The matrix a is stored in ab using the matrix diagonal ordered form:

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

Example of 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 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.

Examples

Solve the banded system a x = b, where:

    [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]

There is one nonzero diagonal below the main diagonal (l = 1), and two above (u = 2). The diagonal banded form of the matrix is:

     [*  * -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 ])