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19 Jan 2018: Minmod(a, b) =. 0 if ab < 0. a if ab > 0 and |a| < |b|b if ab > 0 and |b| < |a|. ... a if ab > 0 and(|a| < 12|b| or |b| < |a| < 2|b|. ),.

4 Feb 2018: Minmod(a, b) =. 0 if ab < 0. a if ab > 0 and |a| < |b|b if ab > 0 and |b| < |a|. ... a if ab > 0 and(|a| < 12|b| or |b| < |a| < 2|b|. ),.

31 Jan 2018: Details in 1D. d2f. dx2= ρ in a < x < b, with f (a) = A and f (b) = B,. ... where ρ(x),A and B given. Divide [a,b] into N equal segments h = (b a)/N.Use linear finite elements with basis functions.

13 Dec 2018: erf. [. A. B. AX. B. )]. B. AX. (. erf. ). A. AX. (. erf. [. A. B. AX. B. )]. ... 1. (. t. y. D. x. Q. -. =. where A=a/(2((Kt)), B = b//(2((Kt)) and X = x/a.

13 Dec 2018: there. exists p, say, such that p _-> V {P2,Pt} and b o t h ( b , , - b ) p and (be - b)p are. ... a b o v e we can suppose, for each po => pl, that (b - b l ) p l = 0, b y replacing each b.

13 Dec 2018: If AZ issufficiently small so that B is almost constant in Z over thatdistance, say B - B(X), then fz 8z B dZ - AZ B = B', say.Consider the error ... u) = [exp(A' B') - exp(A') exp(B')]ubetween relations (3) and (4).

13 Dec 2018: and B’ = B’({ti}t {pi}) = -c 2 [p(zi - zj) p(yi - yj) - 2p(li - yj)]. ... It. follows that [A, B] is small on the solution space, as required.

4 Feb 2018: Minmod(a, b) =. 0 if ab < 0. a if ab > 0 and |a| < |b|b if ab > 0 and |b| < |a|. ... a if ab > 0 and(|a| < 12|b| or |b| < |a| < 2|b|. ),.

19 Jan 2018: Details in 1D. d2f. dx2= ρ in a < x < b, with f (a) = A and f (b) = B,. ... d2f. dx2= ρ in a < x < b, with f (a) = A and f (b) = B,.

25 Oct 2018: Minmod(a, b) =. 0 if ab < 0. a if ab > 0 and |a| < |b|b if ab > 0 and |b| < |a|. ... a if ab > 0 and(|a| < 12|b| or |b| < |a| < 2|b|. ),.