Question

For the following partial differential equation, \[ x \frac{\partial^2 f}{\partial x^2} + y \frac{\partial^2 f}{\partial y^2} = \frac{x^2 + y^2}{2} \] which of the following option(s) is/are CORRECT?

• A. elliptic for \( x>0 \) and \( y>0 \)
• B. parabolic for \( x>0 \) and \( y>0 \)
• C. elliptic for \( x = 0 \) and \( y>0 \)
• D. hyperbolic for \( x<0 \) and \( y>0 \)

Hint:

To classify second-order PDEs, examine the coefficients of the highest-order terms. Elliptic equations have coefficients of the same sign, while hyperbolic equations have coefficients of opposite signs.

Answer 1

The Correct Option is A

Step 1: The given partial differential equation can be rewritten in the general form: \[ a \frac{\partial^2 f}{\partial x^2} + b \frac{\partial^2 f}{\partial y^2} = f(x, y) \] where: \[ a = x, \quad b = y \] Step 2: The classification of the PDE depends on the sign of the coefficients: If \( a>0 \) and \( b>0 \), the equation is elliptic. If \( a<0 \) and \( b>0 \), the equation is hyperbolic. If either \( a \) or \( b \) equals zero, the equation may degenerate to a different type. Step 3: Evaluating the given options: For \( x>0 \) and \( y>0 \), both coefficients are positive, so the equation is elliptic. Option (A) is correct. For \( x<0 \) and \( y>0 \), the coefficients have opposite signs, making the equation hyperbolic. Option (D) is correct. Conclusion: The correct answers are (A) and (D), confirming the classification of the given differential equation.