Discussion
Consider the second order Partial Differential Equation (PDE)
\[ \begin{aligned} &4x^{2}\frac{\partial^{2}u}{\partial x^{2}} + 4(x + y)\frac{\partial^{2}u}{\partial x \partial y} \\ &+ (x^{2} + y^{2})\frac{\partial^{2}u}{\partial y^{2}} - u = 0. \end{aligned} \]
Then which one of the following statements is correct ?
Answer: A
The correct statement is The PDE is hyperbolic in the region -1 < x < 0, y < 0.
For \(Au_{xx} + Bu_{xy} + Cu_{yy} + \dots = 0\) the type is set by the discriminant \(B^{2} - 4AC\). Here A = 4x2, B = 4(x + y), C = x2 + y2:
\[ B^{2}-4AC = 16\left[(x+y)^{2} - x^{2}(x^{2}+y^{2})\right] \]
In the region -1 < x < 0, y < 0, both x and y are negative, so 2xy > 0 and \((x+y)^{2} \gt x^{2}+y^{2}\), while \(x^{2} \lt 1\) gives \(x^{2}+y^{2} \ge x^{2}(x^{2}+y^{2})\). The discriminant is positive everywhere there: hyperbolic.
For x > 1 and y < 0 the bracket is negative at, for example, x = 2, y = -0.1 (3.61 - 16.04), so the PDE is not hyperbolic throughout that region, and it is not parabolic there either, which would need the discriminant to be zero. For 0 < x < 1, y > 0 the discriminant is positive, so the PDE is hyperbolic there, not elliptic.
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