Question

If the partial differential equation \[ (x + 2) \frac{\partial^2 u}{\partial x^2} + 2(x + y) \frac{\partial^2 u}{\partial x \partial y} + 2(y - 1) \frac{\partial^2 u}{\partial y^2} - 3y^2 \frac{\partial u}{\partial y} = 0 \] is parabolic on the circle \( (x - a)^2 + (y - b)^2 = r^2 \), then the values of \( a \), \( b \), and \( r \) are given by:

• A. ( a = 1, b = 2, r = 1 \)
• B. ( a = -1, b = 2, r = 1 \)
• C. ( a = 1, b = -2, r = 1 \)
• D. ( a = -1, b = -2, r = 1 \)

Hint:

For a PDE to be parabolic, its discriminant (the quantity involving the coefficients of the second derivatives) must be zero. This condition can be used to identify the geometry of the solution.

Answer 1

The Correct Option is B

The given equation is a second-order partial differential equation in two variables, \( x \) and \( y \). We are tasked with finding the values of \( a \), \( b \), and \( r \) such that the equation is parabolic on the circle \( (x - a)^2 + (y - b)^2 = r^2 \). This can be done by comparing the given equation with the standard form of a parabolic equation. 1. For the equation to be parabolic, the discriminant of the quadratic form must be zero. The discriminant for a second-order PDE is given by: \[ \Delta = B^2 - 4AC \] where \( A \), \( B \), and \( C \) are the coefficients of the second derivatives in the equation. For this problem: \[ A = (x + 2), \quad B = 2(x + y), \quad C = 2(y - 1). \] We now calculate the discriminant: \[ \Delta = [2(x + y)]^2 - 4(x + 2)(2(y - 1)). \] By simplifying and substituting \( x = -1 \) and \( y = 2 \), we get: \[ \Delta = 0. \] Thus, the equation becomes parabolic for \( a = -1 \), \( b = 2 \), and \( r = 1 \). Therefore, the correct answer is (B): \( a = -1, b = 2, r = 1 \).