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Suppose $P=\sum_i a_i \partial/\partial x^i$. It seems for me that there always exists a nonzero $\xi=\{\xi_1, \cdots, \xi_n\}\in \mathbb R^n\setminus 0$ such that the principle symbol $\sigma(P)(x, \xi)=\sum_i a_i\xi_i=0$. Hence by definition it cannot be elliptic. Did I miss anything?

So for differential operators acting on functions defined over $U\subset\mathbb R^n$, The only elliptic operator must be of second order?

How to define the symbol of $P$ for function $f:\mathbb R^n\to\mathbb R^n$? Can you show me an example of first order ellliptic operator?

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A differential operator of first order with real coefficients cannot be elliptic (unless $n=1$ which we should and do exclude). But if complex coefficients are allowed, then in dimension $2$ there are elliptic first order operators, such as $\frac{\partial}{\partial x_1}+i\frac{\partial}{\partial x_2}$ which you may recognize as a form of the Cauchy-Riemann operator. This is why we can have nice things in complex analysis.

In dimensions higher than $2$ even complex coefficients do not help, since any $\mathbb R$-linear map from $\mathbb R^n$ to $\mathbb C$ has nontrivial kernel. Suppressing the temptation to dust off quaternions (hm, quaternion-valued elliptic PDE?) we admit that any elliptic operator in dimensions above $2$ must have order at least $2$. Like the Laplacian $\Delta$.

But the order can be higher; for example the bi-Laplacian $\Delta^2$ is elliptic and has order $4$. The powers of $\Delta$ provide easy examples of elliptic operators of any even order (and there are many more). But there are no elliptic operators of odd orders $3,5,7,\dots$ (except in dimension $2$, where we can take powers of the Cauchy-Riemann operator).

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The Euclidean Dirac operator $\sum_{i=1}^n X_i \nabla_{X_i}$ is an example of a first order differential operator that is elliptic on any $\mathbb{R}^n$. –  fuzzytron Mar 17 '13 at 2:49
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