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Jul 5, 2013 at 11:46 comment added Willie Wong Let $B(t)u = b(t,x,hD)u$. The operator $B(0)u = u + O(h^\infty)$. We write that as $(B(0) - \mathrm{Id}) = O(h^\infty)$. Now, if I read correctly: note that we already have the propagator $F(t)$ which acts well on "error terms". So $$ U(t) = B(t) - F(t)(B(0) - \mathrm{Id}) $$ should solve (10.2.2)
Jul 5, 2013 at 5:21 review First posts
Jul 5, 2013 at 5:35
Jul 5, 2013 at 5:03 history asked Sean Gomes CC BY-SA 3.0