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Let $u$ be a smooth solution of the uniformly elliptic equation $Lu=-\sum_{i,j=1}^n a^{ij}(x)u_{x_i x_j}$ in $U$. Assume that the coefficients have bounded derivatives.

Set $v:=|Du|^2+\lambda u^2$ and show that $$Lv \le 0 \quad \text{in }U$$ if $\lambda$ is large enough. Deduce $$\|Du\|_{L^\infty(U)}\le C(\|Du\|_{L^\infty(\partial U)}+\|u\|_{L^\infty(\partial U)}).$$

This is PDE Evans, 2nd edition: Chapter 6, Exercise 8.

I believe I was able to show $Lv \le 0$ for large enough $\lambda$ already. Specifically, given $v$, we obtain $$v_{x_i x_j} = 2[(Du_{x_j}\cdot Du_{x_i}+Du \cdot Du_{x_i x_j})+(\lambda u_{x_j}u_{x_i}+\lambda uu_{x_i x_j})],$$ which means, for large enough $\lambda$, $$Lv=-2\sum_{i,j=1}^n a^{ij}(x) Du_{x_j}\cdot Du_{x_i}-2\lambda\sum_{i,j=1}^n a^{ij}(x) u_{x_j} u_{x_i} \le 0.$$

But how can I get started on deriving the estimate $$\|Du\|_{L^\infty(U)}\le C(\|Du\|_{L^\infty(\partial U)}+\|u\|_{L^\infty(\partial U)})$$ Any hints on this part would be helpful.

Community
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Cookie
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  • Apply the maximum principle to $v$ and translate that information into a bound for $u$. – Jose27 Feb 02 '15 at 02:54
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    The calculation in the line for getting $Lv\leq 0$ is wrong. –  Dec 06 '16 at 03:36
  • @Jack and Cookie. The expression of $v_{x_ix_j}$ has four terms. The first and the third terms yield the terms in the line of $Lv$ (which is wrong according to Jack). The last term yields zero because $Lu=0$. Could you explain how the second term is estimated (and what is wrong in the said line)? – Pedro May 02 '17 at 19:26
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    Well, this post answers my question. – Pedro May 02 '17 at 20:22

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I believe this can indeed be handled by the maximum principle. Since $Lu,Lv\leq0$, it tells us that $$\max_{\partial U}u=\max_{\overline{U}}u\quad\text{and}\quad\max_{\partial U}v=\max_{\overline{U}}v.$$ It helps considerably to rewrite this in terms of the infinity norm and the definition of $v$, which reads $$\left\|u^2\right\|_{L^\infty(\partial U)}=\left\|u^2\right\|_{L^\infty(U)}\quad\text{and}\quad\left\||Du|^2+\lambda u^2\right\|_{L^\infty(\partial U)}=\left\||Du|^2+\lambda u^2\right\|_{L^\infty(U)}.$$ Applying the triangle inequality to both sides of the right equation gives $$\left\||Du|^2\right\|_{L^\infty(\partial U)}+\left\|\lambda u^2\right\|_{L^\infty(\partial U)}\geq\left\||Du|^2\right\|_{L^\infty(U)}-\left\|\lambda u^2\right\|_{L^\infty(U)}$$ So we can then add $\lambda$ times the left equation to this, which gives

$$\left\||Du|^2\right\|_{L^\infty(\partial U)}+2\left\|\lambda u^2\right\|_{L^\infty(\partial U)}\geq\left\||Du|^2\right\|_{L^\infty(U)}$$ We can then add the appropriate terms to the left side (which are all positive, of course) to allow us to complete the square. Taking square roots then gives exactly the desired estimate.

MathAnon
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  • When I apply the triangle inequality on the left side and the reverse triangle inequality on the right side of your second equation, then I get $$|Du|^2_{L^\infty(\partial U)} + |\lambda| |u|^2_{L^\infty(\partial U)}\ge |Du|^2_{L^\infty(U)} - |\lambda| |u|^2_{L^\infty(U)},$$ and after employing the first equality, I obtain actually $$|Du|{L^\infty(\partial U)} \ge |Du|^2{L^\infty(U)} - 2|\lambda| |u|^2_{L^\infty(\partial U)}.$$ This result is different from what you deduced, but the desired result follows anyway (if I'm not mistaken). – Cookie Feb 13 '15 at 22:48