Question

Let $D = \{(x, y) \in \mathbb{R}^2 : x > 0 \text{ and } y > 0\}$. If the following second-order linear partial differential equation 
$y^2 \frac{\partial^2 u}{\partial x^2} - x^2 \frac{\partial^2 u}{\partial y^2} + y \frac{\partial u}{\partial y} = 0$ on $D$ 
is transformed to 
$\left( \frac{\partial^2 u}{\partial \eta^2} - \frac{\partial^2 u}{\partial \xi^2} \right) + \left( \frac{\partial u}{\partial \eta} + \frac{\partial u}{\partial \xi} \right) \frac{1}{2\eta} + \left( \frac{\partial u}{\partial \eta} - \frac{\partial u}{\partial \xi} \right) \frac{1}{2\xi} = 0$ on $D$, 
for some $a, b \in \mathbb{R}$, via the coordinate transform $\eta = \frac{x^2}{2}$ and $\xi = \frac{y^2}{2}$, then which one of the following is correct?

• A. \( a = 2, b = 0 \)
• B. \( a = 0, b = -1 \)
• C. \( a = 1, b = -1 \)
• D. \( a = 1, b = 0 \)

Hint:

For transformations in partial differential equations, carefully compute the derivatives and simplify to match the target form.

Answer 1

The Correct Option is B

Step 1: Transformation of variables. The coordinate transform \( \eta = \frac{x^2}{2} \) and \( \xi = \frac{y^2}{2} \) changes the differential terms accordingly. Step 2: Substituting into the equation. Using the chain rule, we rewrite the terms of the partial derivatives under the new variables \( \eta \) and \( \xi \). After simplifications, the transformed equation matches the given form if \( a = 0 \) and \( b = -1 \). Step 3: Conclusion. The correct values are \( {(B)} a = 0, b = -1 \).