Question

The partial differential equation \[ (1 + x^2) \frac{\partial^2 u}{\partial x^2} + 2x(1 - y^2) \frac{\partial^2 u}{\partial x \partial y} + (1 - y^2) \frac{\partial^2 u}{\partial y^2} + x \frac{\partial u}{\partial x} + (1 - y^2) \frac{\partial u}{\partial y} = 0 \] is:

• A. elliptic in the region \( \{(x, y) \in \mathbb{R}^2 : |y| \leq 1 \} \)
• B. hyperbolic in the region \( \{(x, y) \in \mathbb{R}^2 : |y|>1 \} \)
• C. elliptic in the region \( \{(x, y) \in \mathbb{R}^2 : |y|>1 \} \)
• D. hyperbolic in the region \( \{(x, y) \in \mathbb{R}^2 : |y| \leq 1 \} \)

Hint:

To classify a second-order PDE, calculate the discriminant \( \Delta = B^2 - AC \). Based on its value, classify the equation as hyperbolic (\( \Delta>0 \)), elliptic (\( \Delta<0 \)), or parabolic (\( \Delta = 0 \)).

Answer 1

The Correct Option is B

To classify the given partial differential equation (PDE), we need to check the discriminant of the corresponding second-order terms. 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 \( A \), \( B \), and \( C \) are the coefficients of the second-order derivatives. The classification of the PDE is based on the discriminant:

\[ \Delta = B^2 - AC \]

• If \( \Delta > 0 \), the PDE is hyperbolic.
• If \( \Delta = 0 \), the PDE is parabolic.
• If \( \Delta < 0 \), the PDE is elliptic.

Now, for the given equation:
\[ A = 1 + x^2,\quad B = x(1 - y^2),\quad C = 1 - y^2 \]
We compute the discriminant:
\[ \Delta = B^2 - AC = \left[x(1 - y^2)\right]^2 - (1 + x^2)(1 - y^2) \]
Expanding both terms:
\[ \Delta = x^2(1 - y^2)^2 - (1 + x^2)(1 - y^2) \]
Factor \( (1 - y^2) \) from both terms:
\[ \Delta = (1 - y^2)\left[x^2(1 - y^2) - (1 + x^2)\right] \]
Now simplify the expression in the parentheses:
\[ x^2(1 - y^2) - (1 + x^2) = x^2 - x^2 y^2 - 1 - x^2 = -1 - x^2 y^2 \]
Thus:
\[ \Delta = (1 - y^2)(-1 - x^2 y^2) \]
We now examine the sign of this expression. Note that:

• If \( |y| > 1 \), then \( 1 - y^2 < 0 \)
• And \( -1 - x^2 y^2 < 0 \) for all \( x, y \in \mathbb{R} \)

So the product of two negative quantities is positive:
\[ \Delta > 0 \text{ when } |y| > 1 \]
Therefore, the PDE is hyperbolic in the region \( \{(x, y) \in \mathbb{R}^2 : |y| > 1 \} \).

Final Answer
\[ \boxed{(B) \quad \text{hyperbolic in the region } \{(x, y) \in \mathbb{R}^2 : |y| > 1\}} \]