# Bounding $H_{0}^{1}$ norm

I'm currently reading "Partial differential equation" by L.C. Evans (1st ed). On page 382, formula (25), Evans claimed that $$||u_m||_{H_{0}^{1}}^{2}\leq A [u_m,u_m;t]$$ by applying the uniform hyperbolicity inequality, i.e., $\int|Du|^{2}dx\leq A[u,u;t]$ where $A [u,v;t]=\int\sum_{{i}{j}}a^{ij}u_{x_i}v_{x_j} dx$.

I don't understand how he drived the above inequality, since $||u_m||_{H_{0}^{1}}^{2}$ involves also $L^2$ norm of $u$, and I can't see why the $L^2$ norm is also bounded by $A[u,u;t]$.

Any explainations?

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If you have parabolic instead of hyperbolic then think of when $a^{ij}$ is the identity and remember Poincaré's inequality. – Jose27 Nov 19 '12 at 2:57
@Jose27 I don't think Poincare's inequality would work since that requires $1\leq p < n$. In this case, $p=2$, but we don't know anything about $n$. – CC_Azusa Nov 19 '12 at 3:44
The Poincaré inequality for $H_0^1$ doesn't have that restriction since you can prove it without the Sobolev embedding. By the way could you write out the conditions on the $a^{ij}$? – Jose27 Nov 19 '12 at 4:12
@Jose27 I see. I believe $a^{ij}$ should be symmetric, and belong to $L^{\infty}$. – CC_Azusa Nov 19 '12 at 5:20

## 1 Answer

Okay, here's an argument: By Poincaré's inequality we have that there exists $C>0$ such that for every $u\in H_0^1$

$$C^{-1}\int_{} |Du|^2 \leq \| u\|_{H^1} \leq C \int_{} |Du|^2$$

so we define $\| u\|_{H_0^1} = \int_{\Omega} |Du|^2$. The rest follows from the hyperbolicity condition since

$$a^{ij}u_{x_i}u_{x_j} \geq \theta |Du|^2$$

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