Lines Matching refs:L
11 Consider the matrix problem $ A x = b$, where $A = L + U + D$.
38 x^{1/2} = \omega (L + D)^{-1} b
41 x = (1 - \omega) x^{1/2} + \omega (U + D)^{-1}(b - L x^{1/2}) = x^{1/2} + \omega (U+D)^{-1}(b - A x…
58 …b $ and is updated a column of the matrix at a time to contain the value of $ (b - L x^{1/2})$ that
86 x = \omega (L + D)^{-1}b
97 x = (L + D/\omega)^{-1}b
100 Rather than applying the left preconditioner obtained by apply the two step process $ (L + D/\omega…
103 (L + D/\omega)^{-1} A (U + D/\omega)^{-1} y = (L + D/\omega)^{-1} b.
108 …(L + D/\omega)^{-1} A (U + D/\omega)^{-1} & = & (L + D/\omega)^{-1} (L + D + U) (U + D/\omega)^{-…
109 & = & (L + D/\omega)^{-1} (L + D/\omega + U + D/\omega + D - 2D/\omega) (U + D/\omega)^{-1} \\
110 & = & (U + D/\omega)^{-1} + (L+D/\omega)^{-1}(I + \frac{\omega - 2}{\omega}D(U + D/\omega)^{-1}).