1.3 Types of Second-Order Equations

Transformation Theorems

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

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

For a second-order PDE in standard form

a (x, y) u_{x x} + 2 b (x, y) u_{x y} + c (x, y) u_{y y} = 0

Δ = b^{2} - a c

If $Δ > 0$ : hyperbolic
If $Δ$ =0: parabolic
If $Δ < 0$ : elliptic

Linear Algebra Perspective

Now we have $n$ variables denoted $x_{1}, x_{2}, \dots, x_{n}$

\sum_{i . j = 1}^{n} a_{i j} u_{x_{i}, x_{j}} + \sum_{i = 1}^{n} a_{i} u_{x_{i}} + a_{0} u = 0

Assume $a_{i j} = a_{j i} .$ Let $x = (x_{1}, x_{2}, \dots, x_{n})$ . Considering any linear change of independent variables:

(ξ_{1}, \dots, ξ_{n}) = ξ = B x

We $B$ is an $n \times n$ matrix $ξ_{k} = \sum_{m} b_{k m} x_{m}$

Convert to the new variables using chain rule:

\frac{\partial}{\partial x_{i}} = \sum_{k}^{} \frac{\partial ξ_{k}}{\partial x_{i}} \frac{\partial}{\partial ξ_{k}}

note: \frac{\partial ξ_{k}}{\partial x_{m}} = \frac{\partial}{\partial x_{m}} (\sum_{r} b_{k r} x_{r}) = b_{k m} .

⟹ u_{x_{i}, x_{j}} = (\sum_{k}^{} b_{k i} \frac{\partial}{\partial ξ_{k}}) (\sum_{l}^{} b_{l j} \frac{\partial}{\partial ξ_{l}}) u

So the PDE is converted to

\sum_{i, j}^{} a_{i j} u_{x_{i}, x_{j}} = \sum_{k, l} (\sum_{i, j}^{} b_{i k} a_{i j} b_{l j}) u_{ξ_{k} ξ_{l}}

Which is a new second order equation and the new coefficient matrix
Since $A$ is a symmetric matrix there exists a rotational matrix $B$ with determinant 1

B A^{⊤} B = D = (\begin{matrix} d_{1} \\ d_{2} \\ ⋱ \\ d_{n} \end{matrix})

If all eigenvalues are negative or positive it is elliptic.
If none of the $d_{1}, \dots, d_{n}$ vanish and one has the opposite as the $(n - 1)$ others it is hyperbolic or if at least two are positive and at least two are negative it is ultrahyperbolic. If exactly one of the eigenvalues is zero and the others have the same sign it is parabolic.

Example (hyperbolic)

Start with

u_{x x} - 4 u_{x y} + u_{y y} = 0 \Rightarrow a = 1, 2 b = - 4 \Rightarrow b = - 2, c = 1.

Discriminant: $b^{2} - a c = 4 - 1 = 3 > 0 \Rightarrow$ hyperbolic, so we expect $u_{ξ ξ} - u_{η η}$ .

1) Package the second-order part as a matrix

A = (\begin{matrix} a & b \\ b & c \end{matrix}) = (\begin{matrix} 1 & - 2 \\ - 2 & 1 \end{matrix}) .

2) Find a linear change that diagonalizes $A$
Eigenpairs:
$λ_{1} = 3, v_{1} = (1, - 1)^{T}$ ,
$λ_{2} = - 1, v_{2} = (1, 1)^{T}$ .

Orthonormal basis:

{\hat{v}}_{1} = \frac{1}{\sqrt{2}} (1, - 1), {\hat{v}}_{2} = \frac{1}{\sqrt{2}} (1, 1) .

New variables:

(\begin{matrix} ξ \\ η \end{matrix}) \underset{B}{\underset{⏟}{(\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix})}} (\begin{matrix} x \\ y \end{matrix})

ξ = \frac{x - y}{\sqrt{2}}, η = \frac{x + y}{\sqrt{2}} .

Jacobian (rows are $\nabla ξ^{T}, \nabla η^{T}$ ):

B = (\begin{matrix} \partial ξ / \partial x & \partial ξ / \partial y \\ \partial η / \partial x & \partial η / \partial y \end{matrix}) = (\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix}) .

3) Transform the coefficients
For a linear change, the principal matrix transforms by congruence:

A^{'} = B A B^{T} = (\begin{matrix} 3 & 0 \\ 0 & - 1 \end{matrix}) .

Thus the PDE becomes

3 u_{ξ ξ} - u_{η η} = 0.

4) Normalize (optional)
Scale $\tilde{ξ} = \sqrt{3} ξ$ so that

u_{\tilde{ξ} \tilde{ξ}} - u_{η η} = 0,

the canonical hyperbolic form.

Chain rule: $u_{x_{i} x_{j}} = \sum_{k, l} b_{k i} b_{l j} u_{ξ_{k} ξ_{l}}$ .
So $\sum_{i, j} a_{i j} u_{x_{i} x_{j}} = \sum_{k, l} (B A B^{T})_{k l} u_{ξ_{k} ξ_{l}}$ .
Choosing $ξ, η$ along eigenvectors of $A$ makes $A^{'}$ diagonal.
Signs of eigenvalues (invariant by Sylvester’s law) classify PDE: elliptic $(+, +)$ , hyperbolic $(+, -)$ , parabolic (one $0$ ).

We are simply diagonalizing the weights of our hessian