Question

Consider the second-order partial differential equation (PDE) \[ \frac{\partial^2 u}{\partial x^2} + 4 \frac{\partial^2 u}{\partial x \partial y} + (x^2 + 4y^2) \frac{\partial^2 u}{\partial y^2} = \sin(x + y) \] Consider the following statements: P: The PDE is parabolic on the ellipse \( \frac{x^2}{4} + y^2 = 1 \).
Q: The PDE is hyperbolic inside the ellipse \( \frac{x^2}{4} + y^2 = 1 \). Then:

• A. both P and Q are TRUE
• B. P is TRUE and Q is FALSE
• C. P is FALSE and Q is TRUE
• D. both P and Q are FALSE

Hint:

The classification of a PDE depends on the discriminant of the coefficient matrix. Parabolic, hyperbolic, and elliptic are determined by the value of \( \Delta \).

Answer 1

The Correct Option is A

To determine the nature of the PDE, we look at its coefficient matrix. The general form of a second-order PDE is: \[ A \frac{\partial^2 u}{\partial x^2} + 2B \frac{\partial^2 u}{\partial x \partial y} + C \frac{\partial^2 u}{\partial y^2} = 0 \] where the discriminant \( \Delta = B^2 - AC \) helps classify the equation as parabolic, hyperbolic, or elliptic: - If \( \Delta = 0 \), the equation is parabolic.
- If \( \Delta > 0 \), the equation is hyperbolic.
- If \( \Delta < 0 \), the equation is elliptic.
For the given PDE, we have: \[ A = 1, B = 2, C = x^2 + 4y^2 \] The discriminant is: \[ \Delta = B^2 - AC = 2^2 - 1(x^2 + 4y^2) = 4 - (x^2 + 4y^2) \] On the ellipse \( \frac{x^2}{4} + y^2 = 1 \), we find that \( \Delta = 0 \), which means the PDE is parabolic along the ellipse. Inside the ellipse, \( \Delta > 0 \), meaning the PDE is hyperbolic inside the ellipse. Thus, both statements P and Q are TRUE. Final Answer: (A) both P and Q are TRUE