Discussion
The partial differential equation
\[ \begin{aligned} &(1 + x^{2})\frac{\partial^{2} u}{\partial x^{2}} + 2x(1 - y^{2})\frac{\partial^{2} u}{\partial y \partial x} \\ &+ (1 - y^{2})\frac{\partial^{2} u}{\partial y^{2}} + x\frac{\partial u}{\partial x} + (1 - y^{2})\frac{\partial u}{\partial y} = 0 \end{aligned} \]
is
Answer: B
The equation is hyperbolic in the region |y| > 1.
For \(Au_{xx} + 2Bu_{xy} + Cu_{yy} + \ldots = 0\), the type is fixed by the sign of \(B^{2} - AC\). Here \(A = 1 + x^{2}\), \(2B = 2x(1 - y^{2})\), \(C = 1 - y^{2}\):
\[ \begin{aligned} B^{2} - AC &= x^{2}(1 - y^{2})^{2} - (1 + x^{2})(1 - y^{2}) \\ &= (1 - y^{2})(x^{2} - x^{2}y^{2} - 1 - x^{2}) \\ &= (y^{2} - 1)(1 + x^{2}y^{2}) \end{aligned} \]
Since \(1 + x^{2}y^{2} \gt 0\), the sign is that of \(y^{2} - 1\): positive (hyperbolic) for |y| > 1, negative (elliptic) for |y| < 1, and zero (parabolic) on |y| = 1. So 'elliptic for |y| ≤ 1' wrongly includes the parabolic lines, and the other two options reverse the types.
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