Question on Evans' treatment of elliptic 2n order equations I am working through L.C.Evans' Partial Differential Equations -- the chapter on second-order elliptic equations.
I have got a general question on symmetric vs. non-symmetric elliptic operators. Consider an operator of the form
$\displaystyle Lu = \sum_{i,j}a^{ij} u_{,ij} + \sum_i b^i u_{,i} + cu$
In his book, Evans mainly treats the case of a symmetric highest order part (i.e. the coefficients for the second order derivatives form a symmetric matrix, $a^{ij}=a^{ji}$). The author frequently tells this restriction is 'without loss of generality'. However, I don't see what he means by this. Hence I wonder


*

*You can you generalize to the case of $(a^{ij})_{ij}$ being non-symmetric?

*How does this relate with the operator as a whole being symmetric?


It would be great help if you could clarify this picture.
 A: The reason for this generalization is actually simple. If you assume your function is sufficiently smooth, then you have $\frac{\partial^2 u}{\partial x_i \partial x_j} = \frac{\partial^2 u}{\partial x_j \partial x_i}$.
For instance if $n=2$, the highest order term in the PDE is of the form
$$a_{11} \frac{\partial^2 u}{\partial x_1^2} + a_{12} \frac{\partial^2 u}{\partial x_1 x_2} + a_{21} \frac{\partial^2 u}{\partial x_2 x_1} + a_{22} \frac{\partial^2 u}{\partial x_2^2}$$.
Note that as said before is $u$ is smooth enough, then $$\frac{\partial^2 u}{\partial x_1 x_2} = \frac{\partial^2 u}{\partial x_2 x_1}$$
Hence, the leading order term in the PDE becomes
$$a_{11} \frac{\partial^2 u}{\partial x_1^2} + (a_{12} + a_{21}) \frac{\partial^2 u}{\partial x_1 x_2} + a_{22} \frac{\partial^2 u}{\partial x_2^2}$$
which could be re-written as
$$a_{11} \frac{\partial^2 u}{\partial x_1^2} + \frac{(a_{12} + a_{21})}{2} \frac{\partial^2 u}{\partial x_1 x_2} + \frac{(a_{21} + a_{12})}{2} \frac{\partial^2 u}{\partial x_2 x_1} + a_{22} \frac{\partial^2 u}{\partial x_2^2}$$
Hence, the leading order term in the PDE becomes
$$a'_{11} \frac{\partial^2 u}{\partial x_1^2} + a'_{12} \frac{\partial^2 u}{\partial x_1 x_2} + a'_{21} \frac{\partial^2 u}{\partial x_2 x_1} + a'_{22} \frac{\partial^2 u}{\partial x_2^2}$$
where
$$\displaystyle
\begin{align*}
a'_{11} & = & a_{11}\\
a'_{22} & = & a_{22}\\
a'_{12} & = \frac{(a'_{12} + a'_{21})}{2} = \frac{(a'_{21} + a'_{12})}{2} = & a'_{21}
\end{align*}
$$
Hence, you can always write the leading order term of the PDE as a symmetric part.
