# An inequality in Evans' PDE

In Section $9.2$ Theorem $5$ of Lawrence Evans' Partial Differential Equations, First Edition the author proves that for a large enough $\lambda$, the equation

$$\begin{array}-\Delta u+b(\nabla u)+\lambda u=0\ &\mbox{ in } U\\ u=0&\mbox{on }\partial U\end{array}$$

has a solution in $H_0^1(U)$.

On page 507, the author writes

$$\int_UC(|\nabla u|+1)|u|dx\leq\frac{1}{2}\int_U|\nabla u|^2dx+C\int_U(|u|^2+1)dx\ \mbox{ for }u\in H_0^1(U).$$

Here $C$ is the Lipschitz constant for the Lipschitz function $b$.

My problem is that I cannot show this no matter how much I try. How is the gradient term becoming independent of $C$? Could someone please help!

• This is just Peter-Paul: the C is not really the same C as on the left. It is $C^2/2$. – Glen Wheeler Oct 13 '12 at 20:09
• Yes, I know but how does the constant for the first term become 1/2? The "1/2" is crucial in that proof. – user38404 Oct 14 '12 at 5:02
• It is just $ab \le \frac12a^2 + \frac12b^2$. – Glen Wheeler Oct 14 '12 at 16:18

Since the OP didn't seem to understand my comment, I'll make it into an answer. The Peter-Paul inequality (one guy big, the other one small) is the simple arithmetic estimate ($ab\in\mathbb{R}$, $\varepsilon>0$) $$ab \le \varepsilon a^2 + \frac1{4\varepsilon}b^2\,.$$ This simple little guy is all you need to establish the OP's desired estimate, once you accept the fact that the $C$ is different on each side.
In detail, we obtain the estimates $$C|u| \le \frac{C^2}2|u|^2 + \frac12\,,$$ and $$C|\nabla u|\,|u| \le \frac12|\nabla u|^2 + \frac{C^2}2|u|^2\,,$$ which we integrate and add together to conclude $$\int_UC(|\nabla u|+1)|u|dx \leq\frac{1}{2}\int_U|\nabla u|^2dx + C^2\int_U|u|^2dx + \frac12\int_Udx\,,$$ as desired.