Consider the second order linear PDE in two independent variables, \begin{equation} \label{2nd} a(x,y) u_{xx} + 2b(x,y) u_{xy} + c(x,y) u_{yy} + f(x,y,u, u_x, u_y) = 0. \end{equation} In order to seek a classification of such equations, we can search for properties of the PDE \eqref{2nd} that are unchanged by our chosen coordinate system. We introduce the change in coordinates using \[ (x,y) \mapsto (\xi(x,y), \eta(x,y)), \] where in order to be a well-defined invertible mapping, we require that the Jacobian determinant is not equal to zero, \[ \pd{(\xi, \eta)}{(x,y)} = \begin{vmatrix} \xi_x & \xi_y \\ \eta_x & \eta_y \end{vmatrix} \neq 0. \] Let the solution in this coordinate system be written as \[ u(x,y) = \phi(\xi, \eta). \] Then we have for the derivatives \begin{align*} u_x &= \xi_x \phi_\xi + \eta_x \phi_\eta \\ u_{xx} &= \xi_x^2 \phi_{\xi\xi} + 2 \xi_x \eta_x \phi_{\xi\eta} + \eta_x^2 \phi_{\eta\eta} + \xi_{xx} \phi_\xi + \eta_{xx} \phi_\eta. \end{align*} and similarly for the other derivatives. The equation \eqref{2nd} can now be placed in the form of \begin{equation} A\phi_{\xi\xi} + 2B \phi_{\xi\eta} + C \phi_{\eta\eta} + F(\xi, \eta, \phi, \phi_x, \phi_y) = 0, \end{equation} where \begin{gather*} A = a\xi_x^2 + 2b \xi_x \xi_y + c\xi_y^2 \\ B = a \xi_x \eta_x + b(\eta_x \xi_y + \xi_x \eta_y) + c \xi_y \eta_y \\ C = a \eta_x^2 + 2b \eta_x \eta_y + c \eta_y^2. \end{gather*}

This three-by-three system of equations can be placed in the form of \[ \begin{pmatrix} A & B \\ C & D \end{pmatrix} = \begin{pmatrix} \xi_x & \xi_y \\ \eta_x & \eta_y \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \xi_x & \xi_y \\ \eta_x & \eta_y \end{pmatrix}, \] and by taking the determinant of both sides, we find \[ \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \text{det} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \left[ \pd{(\xi, \eta)}{(x,y)}\right]^2, \]

The key, then, is that the sign of $\Delta = a^2 - bc$ is an invariant of the PDE, and does not change for different choices of the coordinate system. This leads to the following classification:

- A
*hyperbolic*equation has $\Delta > 0$ - A
*parabolic*equation has $\Delta = 0$ - An
*elliptic*equation has $\Delta < 0$.

Before we go further, note that we can write \begin{align*} A &= \xi_y^2 \left( a (\xi_x/\xi_y)^2 + 2b(\xi_x/\xi_y) + c\right) \\ &= \xi_y^2 \left((\xi_x/\xi_y) - \lambda_1\right) \left((\xi_x/\xi_y) - \lambda_2\right) \\ &= \left(\xi_x + \lambda_1\xi_y \right) \left(\xi_x + \lambda_2\xi_y \right) \end{align*} where the two roots are given by solving \[ a \lambda^2 - 2b \lambda + c = 0, \] so \[ \lambda_{1, 2} = b \pm \sqrt{\Delta} = b \pm \sqrt{b^2 - ac}. \]

*Notice the change in the sign of* $b$! *I've done this to be consistent with your notes.*

Similarly, $C$ can be placed in the form \[ C = \left(\eta_x + \lambda_1\eta_y \right) \left(\eta_x + \lambda_2\eta_y \right). \]

The goal of the reduction is to choose the coordinates $\xi$ and $\eta$ such that the PDE reduces to a simpler form, where we can get rid of as many second derivatives as possible.

For hyperbolic equations, $\Delta > 0$ and so the two roots $\lambda_{1, 2}$ are real. We can try to choose $\xi$ and $\eta$ so as to set both $A = C = 0$. For example, we can choose to solve the pair of equations \begin{gather*} \left(\xi_x + \lambda_1\xi_y \right) = 0 \\ \left(\eta_x + \lambda_2\eta_y \right) = 0. \end{gather*}

However, you already know how to solve such equations! These are simply first-order linear equations for which you can apply the method of characteristics. We know then that the (first-order) characteristics are given by the set of curves $y = y(x)$ such that \[ \dd{y}{x} = \lambda_{1,2}(x,y) \] and along these curves, both $\xi$ and $\eta$ are constant. These are precisely the constant coordinate lines we seek. For example, in the simple case that $\lambda_{1,2} = c_{1,2}$ and is constant, then we have \begin{gather*} \phi(x,y) = f(y - c_1 x) \\ \xi(x,y) = g(y - c_2 x). \end{gather*} Any choice of $\phi$ and $\xi$ according to this form will suffice, and typically we would chose the simplest option, \begin{gather*} \phi(x,y) = y - c_1 x \\ \xi(x,y) = y - c_2 x. \end{gather*} In cases where the characteristics are not straight lines, the solution would be similar.

Once a choice of $\phi$ and $\xi$ is chosen so that $A = C = 0$, then we are left with the reduced PDE \begin{equation*} \phi_{\xi\eta} = \tilde{F}(x, y, \phi, \phi_\xi, \phi_\eta). \end{equation*} This is the canonical form for second-order linear hyperbolic PDEs.

For parabolic equations, we have \begin{gather*} A = (\xi_x + \lambda \xi_y)^2 \\ C = (\eta_x + \lambda \eta_y)^2 \\ \end{gather*} where the roots from before are equal $\lambda_1 = \lambda_2 = \lambda$. We can choose $\xi$ to be constant along lines with $dy/dx = \lambda$, and this guarantees that $A = 0$. Since $AC = B^2$, then this also implies that $B = 0$. The choice of $\eta$ is arbitrary so long as lines with $\eta$ constant is never parallel to the characteristics, and $C$ is never zero. This yields the canonical form \begin{equation} \phi_{\eta\eta} = \tilde{G}(\xi, \eta, \phi, \phi_\xi, \phi_\eta). \end{equation}

For elliptic equations, neither $A$ nor $C$ can be chosen to be zero, since there are no real characteristic curves. Instead, set $A = C$ and $B = 0$. We will not go through this case, but it effectively involves defining $\chi = \xi + i\eta$ and then solving the equation \[ a\chi_x^2 + 2b\chi_x \chi_y + c\chi_y^2 = 0. \]