# General Method of Characteristics

In the general discussion about PDEs we have seen the second order equation

\begin{align} au_{xx}+2bu_{xy}+cu_{yy}+du_{x}+eu_y+fu=g \tag{1} \end{align}

which can be written in compact form as

\begin{align} \sum_{i,j=1}^n a_{ij} u_{x_ix_j} + \sum_{i=1}^n b_iu_{xi} + cu = d \end{align}

We already introduced the concept of characteristics at the example of the first order transport equation. Now we generalize the concept for second order equation. The idea behind the method is to reduce the complexity of the equation by the right coordinate transformation. \begin{align} A:(x,y) & \rightarrow B:( \xi(x,y), \eta(x,y))\\ u(x,y) & \rightarrow u(\xi, \eta) \end{align}

The result will be, that we simplify the priciple part of the PDE. We assume that the determinant of the Jacobian \begin{align} \Delta L=det \begin{pmatrix} \xi_x & \xi_y \\ \eta_x & \eta_y \end{pmatrix} \not = 0 \end{align} at each point $(x_0,y_0) \in U$. We can use the chain rule, to determine how $u$ changes with $x,y$ in the coordinate system $B$. \begin{align} \frac{\partial u(\xi,\eta)}{\partial x}=\frac{\partial u(\xi,\eta)}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u(\xi,\eta)}{\partial \eta}\frac{\partial \eta}{\partial y} \end{align}

$x,y$ in the coordinate system B

In short notation this is \begin{align} u_x=u_{\xi}\xi_x+u_{\eta} \eta_x \end{align} and similarly \begin{align} u_y=u_{\xi}\xi_y+u_{\eta} \eta_y \end{align} Now, derive the following second derivatives on your own, be watchful, this is a good thinking task \begin{align} \frac{\partial^2 u(\xi,\eta)}{\partial x^2}&=\frac{\partial }{\partial x} \left( \frac{\partial u(\xi,\eta)}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u(\xi,\eta)}{\partial \eta}\frac{\partial \eta}{\partial x} \right)\\ u_{xx}&=u_{\xi \xi} \xi_{x}^2+u_{\xi \eta} \eta_x \xi_{x}+u_{\xi} \xi_{xx}+u_{\eta \xi} \xi_x \eta_x+u_{\eta \eta} \eta_{x}^2+u_{\eta} \eta_{xx}\\ &=u_{\xi \xi} \xi_{x}^2+ 2 \eta_x \xi_{x} u_{\xi \eta} +u_{\eta \eta} \eta_{x}^2+u_{\xi} \xi_{xx}+u_{\eta} \eta_{xx}\\ u_{yy}&=u_{\xi \xi} \xi_{y}^2+ 2 \eta_y \xi_{y} u_{\xi \eta} +u_{\eta \eta} \eta_{y}^2+u_{\xi} \xi_{xx}+u_{\eta} \eta_{yy}\\ \end{align} \begin{align} \frac{\partial^2 u(\xi,\eta)}{\partial x \partial y}&=\frac{\partial }{\partial y} \left( \frac{\partial u(\xi,\eta)}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u(\xi,\eta)}{\partial \eta}\frac{\partial \eta}{\partial x} \right)\\ u_{xy}&=u_{\xi \xi} \xi_x \xi_y + u_{\xi \eta} \eta_y \xi_x + u_{\xi} \xi_{xy} + u_{\eta \eta} \eta_y^2 + u_{\eta \xi} \xi_y \eta_{x} +u_{\eta} \eta_{xy} \end{align} If we plug this results into (1) and simplify, we get the equation \begin{align} Au_{\xi \xi} + 2B u_{\xi \eta} + C u_{\eta \eta} + F(\xi, \eta, u, u_{\xi}, u_{\eta})=0 \end{align} with \begin{align} A&=a \xi_x^2 + 2b \xi_x \xi_y + c \xi_y^2 \\ B&=a \xi_x \xi_y + b(\xi_x \eta_y+ \xi_y \eta_x) + c \xi_y \eta_y\\ C&=a \eta_x^2 + 2b \eta_x \eta_y + c \eta_y^2 \\ \end{align} which is in a the more compact matrix notation \begin{align} \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 & \eta_y \\ \xi_x & \eta_y \end{pmatrix} \end{align} The determinant of a product of matrices is the product of the determinants of each matrix \begin{align} det \begin{pmatrix} A & B \\ C & D \end{pmatrix} = AC-B^2 = (ac-b^2)(\xi_x \eta_y - \eta_x \xi_y)^2 \end{align} This means that no matter what coordinate transform we perform, as long as $\xi_x \eta_y - \eta_x \xi_y$ does not vanish, it holds that \begin{align} AC-B^2=(ac-b^2) \Longleftrightarrow \xi_x \eta_y - \eta_x \xi_y \not = 0 \end{align} so the coordinate transform does not change the PDE type like we classifyed them in the general discussion about PDEs.

## Method

To solve a PDE with the method of characteristics we choose a coordinate transformation where $A$ and $C$ are zero. So $\xi$ and $\eta$ have to satisfy the equations \begin{align} A&=a \xi_x^2 + 2b \xi_x \xi_y + c \xi_y^2 =0 \\ C&=a \eta_x^2 + 2b \eta_x \eta_y + c \eta_y^2 =0\\ \end{align} respectively. If we rewrite this equations as \begin{align} A&=a \frac{\xi_x^2}{\xi_y^2} + 2b \frac{\xi_x}{\xi_y} + c =0 \\ C&=a \frac{\eta_x^2}{\eta_y^2} + 2b \frac{\eta_x}{\eta_y} + c =0\\ \end{align} we see with \begin{align} \frac{\xi_x}{\xi_y} = \frac{dy}{dx}\\ \frac{\eta_x}{\eta_y} = \frac{dy}{dx}\\ \end{align} that $dy/dx$ must satisfy the the quadratic equation \begin{align} a \left( \frac{dy}{dx} \right)^2 + 2b \left( \frac{dy}{dx} \right) + c =0 \\ \end{align} with the solutions \begin{align} \frac{dy}{dx}= \frac{b \pm \sqrt{b^2-ac}}{a} \end{align} Integration of this two solutions gives the family of characteristic curves \begin{align} \xi=\phi(x,y)=const.\\ \eta = \psi(x,y)=const. \end{align}

Example

The wave equation \begin{align} a^2u_{xx}-u_{tt}=0 \end{align} has the characteristic equation \begin{align} a^2 \left( \frac{dy}{dx} \right)^2 -1 =0 \\ \end{align} with the solutions \begin{align} \frac{dy}{dx}= \frac{\pm \sqrt{-a^2(-1)}}{a^2}=\pm \frac{1}{a} \end{align} Integration gives \begin{align} \frac{x}{a}+t=const.\\ \frac{x}{a}-t=const. \end{align} or equivalently \begin{align} \xi=x+at=const.\\ \eta=x-at=const. \end{align}

The resulting equation \begin{align} Au_{\xi \xi} + 2B u_{\xi \eta} + C u_{\eta \eta} + F(\xi, \eta, u, u_{\xi}, u_{\eta})=0 \end{align} has $A=C=F=0$ so we are left with \begin{align} u_{\xi \eta}=0 \end{align} Integration gives us the, from the derivation of d'Alemberts formula familiar solution \begin{align} u(x,t)=f(\xi)+g(\eta)=f(x+ct)+g(x-ct) \end{align}