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Ok, so I got this equation:

[itex]y^2 \frac{∂^2 u}{∂x^2} + 2xy \frac{∂^2 u}{∂x∂y} + x^2 \frac{∂^2 u}{∂y^2} = 0 [/itex]

[itex]A = y^2[/itex]

[itex]B = xy[/itex]

[itex]C = x^2[/itex]

Now I want to see what type it is, so I compute [itex]B^2 - A C = 0[/itex] which by definition is parabolic. However, according to an earlier statement in my book a parabolic PDE is one that has a quadratic form (at some point) that consists of fewer than n squares, not necessarily all of the same sign. n signifies the space we're in (that would be n=2). However, to me it seems like there are two squares in the equation, so it can't be parabolic?

Are they "reducing" it by writing [itex] (y \frac{∂u}{∂x} + x \frac{∂u}{∂y} )^2 = 0 [/itex] and saying it contains less than n squares after dropping the square?

[itex]y^2 \frac{∂^2 u}{∂x^2} + 2xy \frac{∂^2 u}{∂x∂y} + x^2 \frac{∂^2 u}{∂y^2} = 0 [/itex]

[itex]A = y^2[/itex]

[itex]B = xy[/itex]

[itex]C = x^2[/itex]

Now I want to see what type it is, so I compute [itex]B^2 - A C = 0[/itex] which by definition is parabolic. However, according to an earlier statement in my book a parabolic PDE is one that has a quadratic form (at some point) that consists of fewer than n squares, not necessarily all of the same sign. n signifies the space we're in (that would be n=2). However, to me it seems like there are two squares in the equation, so it can't be parabolic?

Are they "reducing" it by writing [itex] (y \frac{∂u}{∂x} + x \frac{∂u}{∂y} )^2 = 0 [/itex] and saying it contains less than n squares after dropping the square?

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