Stability of finite difference schemes for hyperbolic initial - HAL
... can be dissipative only if both A1 and A2 do not have 0 as an eigenvalue, which is far more restrictive than assuming that the boundary of the half-space {x2 > 0} is non-characteristic. It is also rather restrictive from the point of view of applications (one can think of the linearized gas dynamics ...
... can be dissipative only if both A1 and A2 do not have 0 as an eigenvalue, which is far more restrictive than assuming that the boundary of the half-space {x2 > 0} is non-characteristic. It is also rather restrictive from the point of view of applications (one can think of the linearized gas dynamics ...
Square Deal: Lower Bounds and Improved Relaxations for Tensor
... ranktc (X 0 )} is empty.2 The recovery performance of (2.1) depends heavily on the properties of G. Suppose (2.1) fails to recover X 0 ∈ Tr . Then there exists another X 0 ∈ Tr such that G[X 0 ] = G[X 0 ]. To guarantee that (2.1) recovers any X 0 ∈ Tr , a necessary and sufficient condition is that G ...
... ranktc (X 0 )} is empty.2 The recovery performance of (2.1) depends heavily on the properties of G. Suppose (2.1) fails to recover X 0 ∈ Tr . Then there exists another X 0 ∈ Tr such that G[X 0 ] = G[X 0 ]. To guarantee that (2.1) recovers any X 0 ∈ Tr , a necessary and sufficient condition is that G ...
Group Theory in Solid State Physics I
... one chosen in the figure in such that along the principal diagonal one finds the operation e. Exercise: Check the result of the operation σu C4 . Exercise: Prove that each element of the group occurs once and only once in each column or in each row. Exercise: prove that a cyclic group (cyclic means th ...
... one chosen in the figure in such that along the principal diagonal one finds the operation e. Exercise: Check the result of the operation σu C4 . Exercise: Prove that each element of the group occurs once and only once in each column or in each row. Exercise: prove that a cyclic group (cyclic means th ...
3. Linear function
... (d) Redoing part (a) using the formula, we have L([2, 3] ) = [2, 0] . For part (b), we seek the set of all x for which L(x) = 0, that is, [x1 , 0]T = [0, 0]T . This last equation forces x1 = 0 but places no restriction on x2 , so ker L = {[0, x2 ]T | x2 ∈ R} (which is the same as the set obtained in ...
... (d) Redoing part (a) using the formula, we have L([2, 3] ) = [2, 0] . For part (b), we seek the set of all x for which L(x) = 0, that is, [x1 , 0]T = [0, 0]T . This last equation forces x1 = 0 but places no restriction on x2 , so ker L = {[0, x2 ]T | x2 ∈ R} (which is the same as the set obtained in ...