1.6 Types of Second-Order Equations

Consider the general PDE:

a_{11} u_{x x} + 2 a_{12} u_{x y} + a_{22} u_{y y} + a_{1} u_{x} + a_{2} u_{y} + a_{0} u = 0

Transformation Theorems

After some linear transformation of variables

$x^{'} = b_{1} x + b_{2} y$ and $y^{'} = b_{3} x + b_{4} y$

i) Elliptic Case:

If $a_{12}^{2} < a_{11} a_{22}$ then it is reducible to

u_{x^{'} x^{'}} + u_{y^{'} y^{'}} + \dots = 0

ii) Hyperbolic Case:

If $a_{12}^{2} > a_{11} a_{22}$ then it is reducible to

u_{x^{'} x^{'}} - u_{y^{'} y^{'}} + \dots = 0

iii) Parabolic case:

If $a_{12}^{2} = a_{11} a_{22}$ then it is reducible to

u_{x^{'} x^{'}} + \dots = 0

(unless $a_{11} = a_{12} = a_{22} = 0$ )

Example

u_{x x} - 5 u_{x y} = 0

$a_{11} = 1, a_{12} = \frac{5}{2}, a_{22} = 0$

{(\frac{5}{2})}^{2} > a_{11} a_{22}

Hyperbolic

\begin{aligned} \frac{\partial}{\partial x} \frac{\partial}{\partial x} u - 5 \frac{\partial}{\partial x} \frac{\partial}{\partial y} u & = 0 \\ u ({(\frac{\partial}{\partial x})}^{2} - 5 (\frac{\partial}{\partial x} \frac{\partial}{\partial y}) + {(\frac{5}{2} \frac{\partial}{\partial y})}^{2} - {(\frac{5}{2} \frac{\partial}{\partial y})}^{2}) & = 0 \\ u ((\frac{\partial}{\partial x} - (\frac{5}{2}) (\frac{\partial}{\partial y} \frac{\partial}{\partial y})) - {(\frac{5}{2} \frac{\partial}{\partial y})}^{2}) & = 0 \\ u (\frac{\partial}{\partial x} - (\frac{5}{2}) (\frac{\partial}{\partial y} \frac{\partial}{\partial y})) - u {(\frac{5}{2} \frac{\partial}{\partial y})}^{2} & = 0 \end{aligned}

\begin{array}{r} (1) & want {\begin{cases} \frac{\partial}{\partial x^{'}} = \frac{\partial}{\partial x} - \frac{5}{2} \frac{\partial}{\partial y} \\ \frac{\partial}{\partial y^{'}} = \frac{5}{2} \frac{\partial}{\partial y} \end{cases} {\begin{cases} \frac{\partial}{\partial x} = \frac{\partial}{\partial x^{'}} + \frac{\partial}{\partial y^{'}} \\ \frac{\partial}{\partial y} = \frac{2}{5} \frac{\partial}{\partial y^{'}} \end{cases} \end{array}

$x^{'} = b_{1} x + b_{2} y, y^{'} = b_{3} x + b_{4} y$ whose inverse $(x, y) = (x (x^{'}, y^{'}), y (x^{'}, y^{'}))$

\begin{aligned} \partial_{x^{'}} & = (\partial x / \partial x^{'}) \partial_{x} + (\partial y / \partial x^{'}) \partial_{y} = \partial_{x} - \frac{5}{2} \partial_{y} \\ \partial_{y^{'}} & = (\partial x / \partial y^{'}) \partial_{x} + (\partial y / \partial y^{'}) \partial_{y} = \frac{5}{2} \partial_{y} . \end{aligned}

So looking at $(1) :$

$\frac{\partial x}{\partial x^{'}} = 1, \frac{\partial y}{\partial x^{'}} = - \frac{5}{2}; \frac{\partial x}{\partial y^{'}} = 0, \frac{\partial y}{\partial y^{'}} = \frac{5}{2}$

Integrating we get:
$x = x^{'}, y = \frac{5}{2} (y^{'} - x^{'})$

$⟹ x^{'} = x, y^{'} = x + \frac{2}{5} y$

$b_{1} = 1, b_{2} = 0, b_{3} = 1, b_{4} = \frac{2}{5}$

$u_{x^{'}} = \partial_{x^{'}} u = (\partial_{x} - \frac{5}{2} \partial_{y}) u = u_{x} - \frac{5}{2} u_{y}$

$u_{y^{'}} = \partial_{y^{'}} u = \frac{5}{2} \partial_{y} u = \frac{5}{2} u_{y}$

\begin{aligned} u_{x^{'} x^{'}} = \partial_{x^{'}} (u_{x^{'}}) & = (\partial_{x} - \frac{5}{2} \partial_{y}) (u_{x} - \frac{5}{2} u_{y}) \\ u_{x^{'} x^{'}} & = u_{x x} - 5 u_{x y} + \frac{25}{4} u_{y y} . \\ u_{y^{'} y^{'}} = \partial_{y^{'}} (u_{y^{'}}) & = \partial_{y^{'}} u_{y^{'}} = \partial_{y^{'}} (\frac{5}{2} u_{y}) = \frac{5}{2} \partial_{y^{'}} u_{y} \\ u_{y^{'} y^{'}} & = \frac{5}{2} \frac{5}{2} u_{y y} = \frac{25}{4} u_{y y} . \end{aligned}

\begin{aligned} u_{x^{'} x^{'}} - u_{y^{'} y^{'}} & = [u_{x x} - 5 u_{x y} + \frac{25}{4} u_{y y}] - \frac{25}{4} u_{y y} \\ ⟹ u_{x^{'} x^{'}} - u_{y^{'} y^{'}} = u_{x x} - 5 u_{x y} = 0 \end{aligned}