Problem: Determine if the following PDEs are parabolic for all $x,y\in \mathbb{R} $?
-
$x^2 \frac{\partial ^2 u}{\partial x \partial y} - 2xy \frac{\partial u}{\partial y}+ y^2 = 0$
-
$x^2 \frac{\partial ^2 u}{\partial x^2}-2xy \frac{\partial ^2 u}{\partial x\partial y}+ y^2 \frac{\partial ^2 u}{\partial y^2} = 0$
-
$x^2 \frac{\partial ^2 u}{\partial x^2} + 2xy \frac{\partial ^2 u}{\partial x\partial y}+ y^2 \frac{\partial ^2 u}{\partial y^2} = 0$
-
$x^2 \frac{\partial ^2 u}{\partial x^2} - 2xy \frac{\partial ^2 u}{\partial x\partial y}+ y^2 \frac{\partial ^2 u}{\partial y^2} + x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}= 0$
Solution: We recall that a second order PDE of the form
\[
a(x,y) \frac{\partial ^2 u}{\partial x^2} + b(x,y) \frac{\partial ^2 u}{\partial x\partial y} + c(x,y) \frac{\partial ^2 u}{\partial y^2} = f\left( x,y,u,\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} \right)
\]
is
-
elliptic if $b^2 - 4ac \lt 0$;
-
parabolic if $b^2 - 4ac = 0$;
-
hyperbolic if $b^2 - 4ac \gt 0$.
We will examine each of the given PDEs.
-
Here $a(x,y) = 0,~ b(x,y) = x^2$ and $c(x,y) = 0$. So,
\[
b^2 - 4ac = x^2,
\]
and hence the given PDE is not parabolic.
-
Here $a(x,y) = x^2,~ b(x,y) = -2xy$ and $c(x,y) = y^2$. So,
\[
b^2 - 4ac = 4x^2 y^2 - 4x^2 y^2 - 0,
\]
and hence, the given PDE is parabolic.
-
We have $a(x,y) = x^2, ~b(x,y) = 2xy$ and $c(x,y) = y^2$. Since,
\[
b^2 - 4ac = 4x^2 - 4x^2 y^2 = 0,
\]
the given PDE is parabolic.
-
Finally, $a(x,y) = x^2,~ b(x,y) = -2xy$ and $c(x,y) = y^2$ which gives $b^2 - 4ac = 0$, and hence the given PDE is parabolic.