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Dec 8, 2015 at 20:54 comment added username Now done, as announced by @FanZheng earlier, $H^1$ is sufficient.
Dec 8, 2015 at 20:52 history edited username CC BY-SA 3.0
made the argument more precise
Dec 8, 2015 at 9:50 vote accept A random mathematician
Dec 8, 2015 at 8:22 comment added username @MathStudent I don't see what you mean. But 2.b is not optimal: if we allow $f\in H^1$ (as suggested) then it can be improved (by bootstrap) to get to $a$ in a slightly bigger space (I think). I'll edit it.
Dec 8, 2015 at 7:20 comment added A random mathematician This leads to the estimate ||a(x)u||H1(Ω)≤(1+C||a||W1,pn) ∥|f||L2(Ω), and not ||a(x)u||H1(Ω)≤C||a|| ∥|f||L2(Ω). The second inequality gives a better estimate for small a
Dec 7, 2015 at 23:04 history answered username CC BY-SA 3.0