1

Let $\Omega\subset\mathbb{R}^2$ be a bounded domain and $u\in C^2(\Omega)\cap C^1(\overline{\Omega})$. Show, that $$ (1+x^2)w_{xx}-2xw_{xy}+(1+u)w_{yy}-(1+u^2)w_x+(1+u_x)w_y-w=1 $$ is uniformly elliptic.

Hello! We defined uniformly elliptic as follows.

$$ \exists 0<\lambda\leq\Lambda~\forall~x\in\Omega~\forall\xi\in\mathbb{R}^n: \lambda\xi^2\leq\sum_{i,j=1}^{n}a_{ij}(x)\xi_i\xi_j\leq\Lambda\xi^2, \xi^2:=\sum_{i=1}^{n}\xi_i^2 $$

So here I have to show:

$$ \exists 0<\lambda\leq\Lambda~\forall~x\in\Omega\forall~\xi\in\mathbb{R}^2: \lambda(\xi_1^2+\xi_2^2)\leq (1+x^2)\xi_1^2-2x\xi_1\xi_2+(1+u)\xi_2^2\leq\Lambda(\xi_1^2+\xi_2^2) $$

Could you please explain me, how I can find $\lambda$ and $\Lambda$ here?

I do not know how to do it.

Sincerely yours,

math12

  • is that $u$ instead of $w$? –  Nov 11 '13 at 19:56
  • No, the original PDE was non-linear containing only u. And the w was introduced to get a linear one. –  Nov 11 '13 at 19:57
  • 1
    It is not uniformly elliptic without more information on $u$. Consider what happens if $1+u\le0$, for an extreme case. Or if $u$ is unbounded, in the opposite direction. – Harald Hanche-Olsen Nov 11 '13 at 20:01
  • I am sorry, I forgot to mention: u non-negative –  Nov 11 '13 at 20:01
  • 1
    That is better. (The boundedness of $u$ is OK, of course. I didn't notice that at first.) A hint: Consider the eigenvalues of the quadratic form here. – Harald Hanche-Olsen Nov 11 '13 at 20:04
  • How can I determine the eigenvalues of a quadratic form? Are this the eigenvalues of the matrix $\begin{pmatrix}1+x^2 & -x\-x & 1+u\end{pmatrix}$? If yes, the eigenvalues are $\lambda_1=\frac{1}{2}(2+x^2+u+\sqrt{u^2-2ux^2+x^4+4x^2})$ and $\lambda_2=\frac{1}{2}(2+x^2+u-\sqrt{u^2-2ux^2+x^4+4x^2})$. –  Nov 11 '13 at 20:09
  • How can I now use $\lambda_1$ and $\lambda_2$ to find $\lambda$ and $\Lambda$? –  Nov 11 '13 at 20:15
  • Why no reaction anymore? –  Nov 12 '13 at 12:59

0 Answers0