Solution: The given partial differential equation is
\begin{equation}\label{eq:26July2024-1}
\left( x^2 + y^2 -1 \right) \frac{\partial^2 u}{\partial x^2} + 2 \frac{\partial^2 u}{\partial x \partial y} + \left( x^2 + y^2 -1 \right) \frac{\partial^2 u}{\partial y^2} = 0 .
\end{equation}
Consider the PDE of the following form:
\[
A \frac{\partial^2 u}{\partial x^2} + B \frac{\partial^2 u}{\partial x \partial y} + C \frac{\partial^2 u}{\partial y^2} + D \frac{\partial u}{\partial x} + E \frac{\partial u}{\partial y} + F u = G.
\]
We have the following characterization.
-
$B^2 - 4AC > 0$, the PDE is hyperbolic.
-
$B^2 - 4AC = 0$, the PDE is parabolic.
-
$B^2 - 4AC < 0$, the PDE is elliptic.
Here in the given PDE, we have
\[
A = x^2 + y^2 - 1, \quad B = 2 \text{ and } C = x^2 + y^2 - 1.
\]
Then,
\begin{align*}
B^2 - 4AC & = 4 - 4 (x^2 + y^2 -1)^2.
\end{align*}
So,
-
The equation will be hyperbolic if
\begin{align*}
B^2 - 4AC > 0 & \implies 4 - 4 (x^2 + y^2 - 1)^2 > 0\\
& \implies (x^2 + y^2 - 1)^2 < 1 \\
& \implies -1 < x^2 + y^2 - 1 < 1 \\
& \implies 0 < x^2 + y^2 < 2.
\end{align*}
-
The equation will be parabolic if
\begin{align*}
B^2 - 4AC > 0 & \implies 4 - 4 (x^2 + y^2 - 1)^2 = 0\\
& \implies (x^2 + y^2 - 1)^2 = 0 \\
& \implies x^2 + y^2 = 1.
\end{align*}
-
The equation will be elliptic if
\begin{align*}
B^2 - 4AC > 0 & \implies 4 - 4 (x^2 + y^2 - 1)^2 < 0\\
& \implies (x^2 + y^2 - 1)^2 > 1.
\end{align*}
Therefore, the
option D is the correct option.