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320 Evan PDE 2nd edition. I have trouble verifying (ii) of theorem 2, i.e. ellipticity for the following elliptic operator $$ Lu=-D_j(a^{ij}D_i u)+cu$$ Multiply by test function then do integration by parts, I found $$B[u,v]=\int a^{ij} D_i u D_j v+cuv dx$$

I understand boundedness of $B$ is bounded follows from Holder inequality. However, I am stuck on the 2nd part, ellipticity. Nevertheless, theorem 2 is below. Hope someone help me with checking (ii). I have tried Poincare inequality...

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  • $\begingroup$ That follows from the fact that $a^{ij}$ satisfies $ c|v|^2 \le a^{ij} v_iv_j \le C |v|^2$ and plugging in $v = Du$. $\endgroup$ – user99914 Jun 5 '16 at 4:00
  • $\begingroup$ But how about the $cu^2$ term? $\endgroup$ – math101 Jun 5 '16 at 4:25
  • $\begingroup$ Are you trying to prove it yourself or understand Evans' proof? $\endgroup$ – Chee Han Jun 5 '16 at 6:37
  • $\begingroup$ @JohnMa. is $c=1/C$? $\endgroup$ – math101 Jun 5 '16 at 7:06
  • $\begingroup$ @math101 In evan's book such an upper bounded is not assumed. But you really need the lower bound only. $\endgroup$ – user99914 Jun 5 '16 at 7:09
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Since $\lambda |v|^2 \le a^{ij} v_i v_j$ for some positive $\lambda$,

$$\begin{split} \lambda \|u\|_{ H^1(U)}^2 &= \lambda \| \nabla u\|_{L^2(U)}^2 + \lambda \| u\|^2_{L^2(U)}\\ & \le \int_{U}a^{ij} D_i u D_jv + \lambda\|u\|_{L^2(U)}^2 \\ & = B[u,u] - \int_U c u^2+ \lambda\|u\|_{L^2(U)}^2 \\ &\le B[u,u] + (\|c\|_{\infty} + \lambda) \|u\|_{L^2(U)}^2 \end{split}$$

Now set $\beta= \lambda$ and $\gamma = \|c\|_\infty + \lambda$.

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