Question

Let \[ u(x, y) = (x^2 - y^2)v(x, y) \] be such that both \( u(x, y) \) and \( v(x, y) \) satisfy the Laplace equation in a domain \( \Omega \) of the xy-plane. Then, which one of the following is TRUE in \( \Omega \)?

• A. \( x \frac{\partial v}{\partial x} - y \frac{\partial v}{\partial y} = 0 \)
• B. \( x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = 0 \)
• C. \( x \frac{\partial v}{\partial y} - y \frac{\partial v}{\partial x} = 0 \)
• D. \( x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial x} = 0 \)

Hint:

When both functions satisfy the Laplace equation, carefully differentiate and apply the conditions to find the relationship between them.

Answer 1

The Correct Option is A

Given that both \( u(x, y) = (x^2 - y^2)v(x, y) \) and \( v(x, y) \) satisfy the Laplace equation, we know: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0, \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0. \] To find the relationship, we first differentiate \( u(x, y) \) with respect to \( x \) and \( y \) and then use the given conditions for the Laplace equation. The first partial derivatives of \( u(x, y) \) are: \[ \frac{\partial u}{\partial x} = 2xv(x, y) + (x^2 - y^2)\frac{\partial v}{\partial x}, \] \[ \frac{\partial u}{\partial y} = -2yv(x, y) + (x^2 - y^2)\frac{\partial v}{\partial y}. \] To satisfy the Laplace equation, we apply these derivatives to the given equation. After simplifying, we obtain the condition: \[ x \frac{\partial v}{\partial x} - y \frac{\partial v}{\partial y} = 0. \] Final Answer: (A) \( x \frac{\partial v}{\partial x} - y \frac{\partial v}{\partial y} = 0 \)