Question

The partial differential equation \[ (1 + x^2) \frac{\partial^2 u}{\partial x^2} + 2x(1 - y^2) \frac{\partial^2 u}{\partial x \partial y} + (1 - y^2) \frac{\partial^2 u}{\partial y^2} + x \frac{\partial u}{\partial x} + (1 - y^2) \frac{\partial u}{\partial y} = 0 \] is:

• A. elliptic in the region \( \{(x, y) \in \mathbb{R}^2 : |y| \leq 1 \} \)
• B. hyperbolic in the region \( \{(x, y) \in \mathbb{R}^2 : |y|>1 \} \)
• C. elliptic in the region \( \{(x, y) \in \mathbb{R}^2 : |y|>1 \} \)
• D. hyperbolic in the region \( \{(x, y) \in \mathbb{R}^2 : |y| \leq 1 \} \)

Hint:

To classify a second-order PDE, calculate the discriminant \( \Delta = B^2 - AC \). Based on its value, classify the equation as hyperbolic (\( \Delta>0 \)), elliptic (\( \Delta<0 \)), or parabolic (\( \Delta = 0 \)).

Answer 1

The Correct Option is B

To determine the nature of the given partial differential equation (PDE), we analyze the general form of a second-order PDE:

\(A \frac{\partial^2 u}{\partial x^2} + 2B \frac{\partial^2 u}{\partial x \partial y} + C \frac{\partial^2 u}{\partial y^2} + \text{lower-order terms} = 0\) 

where \(A\), \(B\), and \(C\) are coefficients that may be functions of \(x\) and \(y\). The nature of the PDE is determined by the discriminant \(D = B^2 - AC\):

• If \(D > 0\), the PDE is hyperbolic.
• If \(D = 0\), the PDE is parabolic.
• If \(D < 0\), the PDE is elliptic.

For the given PDE:

\(A = 1 + x^2\)

\(B = x(1 - y^2)\)

\(C = 1 - y^2\)

Substitute these into the formula for \(D\):

\(D = B^2 - AC = (x(1 - y^2))^2 - (1 + x^2)(1 - y^2)\)

Expanding and simplifying:

\(B^2 = x^2(1 - y^2)^2\)

\(AC = (1 + x^2)(1 - y^2) = (1 - y^2) + x^2(1 - y^2)\)

Thus:

\(D = x^2(1 - 2y^2 + y^4) - (1 - y^2) - x^2(1 - y^2)\)

Simplifying further:

\(D = x^2y^4 - (1 - y^2)\)

\(D = y^2(x^2y^2 - 1) + y^2\)

Analyzing \(D\) based on the region:

• For \(|y| \gt 1\), \(D > 0\), indicating the PDE is hyperbolic.
• For \(|y| \le 1\), \(D < 0\), indicating the PDE could be elliptic or parabolic depending on further evaluation.

Therefore, the equation is hyperbolic in the region \(\{(x, y) \in \mathbb{R}^2 : |y| > 1 \}\).