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I have seen that there is a classification of elliptic pde's. It says the pde $$au_{xx} + 2bu_{xy} + cu_{yy} + du_x + eu_y+f =0$$ is elliptic if $b^2 - ac < 0$.
There is another definition that is used : The operator $$ Lu = \sum_{i,j} a^{ij}(x)D_{ij}(x) + \sum_{i}b^i(x)D_i(x) + cu$$ is elliptic if there is a constant $\lambda >0$ such that $\sum_{ij}a^{ij}(x)\xi_i\xi_j \geq \lambda|\xi|^2$, for all $x, \xi$

So consider $u_{xx} + y^2 u_{yy} = 0$. According to the first definition, this is elliptic since $0-1y^2 \leq 0$, (I guess it's parabolic for $y=0$?)

According to the second definition, this is not elliptic. If so, there would be a constant $\lambda$ such that $\xi_1^2 + y^2\xi_2^2 \geq \lambda (\xi_1^2 + \xi_2^2)$ for all $(x,y), (\xi_1,\xi_2)$. But now, take $\xi_1 = 0, \xi_2 \not= 0$. Then we get $y^2 \geq \lambda$, which cannot be satisfied, since we needed $\lambda > 0$. So what is going on here?

Euler....IS_ALIVE
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    Nothing much, just the second definition is slightly stronger than the first. The opening paragraphs of http://en.wikipedia.org/wiki/Elliptic_operator refer to the second criterion as "uniform ellipticity". – Erick Wong Mar 12 '13 at 06:39
  • The second definition you have written is wrong. The first term on the RHS should be negative (refer Evans for the detailed definition). Thus, the equation is uniformly elliptic for all $y>0$. For any $0 < \delta < 1$, the operator $L$ is uniformly elliptic in the strip ${(x,y) : \delta< y < 1}$, with $\lambda =\delta$ , but it is not uniformly elliptic in ${(x,y) : 0 < y < 1}$. – Spoilt Milk Jun 04 '19 at 12:33

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