# 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

1. You can you generalize to the case of $(a^{ij})_{ij}$ being non-symmetric?
2. How does this relate with the operator as a whole being symmetric?

It would be great help if you could clarify this picture.

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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.

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what about if it is not smooth enough? –  Moritzplatz Jul 4 '13 at 15:40