A seond order PDE for two independent variables \( x \) and \( y \) can be written as: $$ \begin{equation} A \frac{\partial^2\phi}{\partial x^2}+B \frac{\partial^2\phi}{\partial x \partial y}+C \frac{\partial^2\phi}{\partial y^2}+f=0 \tag{5.19} \end{equation} $$ If \( A, B, C \) and \( f \) are functions of \( x, y,\phi,\frac{\partial\phi}{\partial x} \) and \( \frac{\partial\phi}{\partial y} \), (5.19) is said to be quasi-linear. If \( A, B \) and \( C \) are only functions of \( x \) and \( y \), (5.19) is semi-linear. If \( f \) is only function of \( x \) and \( y \), (5.19) is linear.
A PDE that can be written in the following form: $$ \begin{equation} A \frac{\partial^2\phi}{\partial x^2}+B \frac{\partial^2\phi}{\partial x \partial y}+C \frac{\partial^2\phi}{\partial x^2}+D \frac{\partial\phi}{\partial x}+E \frac{\partial\phi}{\partial y}+F\phi+G=0 \tag{5.23} \end{equation} $$ where \( A, B, C, D, E, F \), and \( G \) are only functions of \( x \) and \( y \), is a 2nd order linear PDE. (5.23) is consequently a special case of (5.19). Notice that usually we will use the term non-linear as opposed to linear. Let us now investigate whether (5.19) has characteristics or not.
Consider a function \( \phi \) such that \( u=\frac{\partial\phi}{\partial x} \) and \( v=\frac{\partial\phi}{\partial y} \), which in turn implies that \( \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} \).
(5.19) can then be written as a system of two 1st order PDEs: $$ \begin{align} A \frac{\partial u}{\partial x}+B \frac{\partial u}{\partial y}+C \frac{\partial v}{\partial y}+f &=0 \tag{5.24} \\ \frac{\partial u}{\partial y}-\frac{\partial v}{\partial x} & =0 \tag{5.25} \end{align} $$
It can be shown that a higher order (quasi-linear) PDE can always be written as a system of 1st order PDEs. Moreover, the resulting first order system can have many forms. If, for example, (5.19) is the potential equation in gas dynamics, we can denote \( \phi \) as the potential speed. (5.25) is the condition of irrotational flow.
We will now try to write (5.24) and (5.25) as a total differential (see (5.13) – (5.15)). Multiply (5.25) with any scalar \( \sigma \) and add to (5.24): $$ \begin{equation} A\left[\frac{\partial u}{\partial x}+\left(\frac{B+\sigma}{A}\right)\frac{\partial u}{\partial y}\right]-\sigma\left[\frac{\partial v}{\partial x}-\frac{C}{\sigma}\frac{\partial v}{\partial y}\right]+f=0 \tag{5.26} \end{equation} $$ Now we have: $$ \begin{equation*} \frac{du}{dx}=\frac{\partial u}{\partial x}+\frac{dy}{dx}\frac{\partial u}{\partial y},\ \frac{dv}{dx}=\frac{\partial v}{\partial x}+\frac{dy}{dx} \frac{\partial v}{\partial y} \end{equation*} $$
Hence: $$ \begin{equation} \frac{du}{dx}=\frac{\partial u}{\partial x}+\lambda\frac{\partial u}{\partial y},\ \frac{dv}{dx}=\frac{\partial v}{\partial x}+\lambda \frac{\partial v}{\partial y} \tag{5.27} \end{equation} $$ where we have defined \( \lambda \) as: $$ \begin{equation} \lambda=\frac{dy}{dx} \tag{5.28} \end{equation} $$
By comparing (5.26) and (5.27): $$ \begin{equation} \frac{dy}{dx}=\lambda=\frac{B+\sigma}{A}=-\frac{C}{\sigma} \tag{5.29} \end{equation} $$
(5.29) inserted in (5.26) gives the compatibility equation $$ \begin{equation} A \frac{du}{dx}-\sigma \frac{dv}{dx}+f=0 \tag{5.30} \end{equation} $$ If the characteristics are real, i.e. \( \lambda \) is real, we have transformed the original PDE into an ODE given by (5.30) along the directions defined by (5.28).
From (5.29): $$ \begin{equation} \sigma=-\frac{C}{\lambda} \tag{5.31} \end{equation} $$ which also gives: $$ \begin{equation} \lambda=\frac{B-\frac{C}{\lambda}}{A} \tag{5.32} \end{equation} $$
The following 2nd order equation can be obtained from (5.32) to determine \( \lambda \): $$ \begin{equation} A\lambda^2-B\lambda+C=0 \tag{5.33} \end{equation} $$ or by using (5.28): $$ \begin{equation} A\cdot(dy)^2-B\cdot dy\cdot dx+C\cdot(dx)^2=0 \tag{5.34} \end{equation} $$
After \( \lambda \) is found from (5.33) and (5.34), \( \sigma \) can be found from (5.31) and (5.32) so that the compatibility equation in (5.30) can be determined. Instead of using (5.30), we can insert \( \sigma \) from (5.29) in (5.30) so that we get the following compatibility equation: $$ \begin{equation} A \frac{du}{dx}+\frac{C}{\lambda}\frac{dv}{dx}+f=0 \tag{5.35} \end{equation} $$ 2nd degree equations (5.33) and (5.34) have the roots \( \lambda_1 \) and \( \lambda_2 \) $$ \begin{equation} \lambda_{1,2}=\frac{B\pm \sqrt{B^2-4AC}}{2A} \tag{5.36} \end{equation} $$
We have three possibilities for the roots in (5.36):
The roots \( \lambda_1 \) and \( \lambda_2 \) are characteristics. Hence:
The wave equation \( \dfrac{\partial^2u}{\partial x^2}-\dfrac{\partial^2u}{\partial y^2}=0 \) is hyperbolic since \( B^2-4AC=4 > 0 \) Characteristics are given by \( \lambda^2=1\to \dfrac{dy}{dx}=\pm 1 \)
Laplace equation \( \dfrac{\partial^2u}{\partial x^2}+\dfrac{\partial^2u}{\partial y^2}=0 \) is elliptic since \( B^2-4AC=-4 < 0 \)
The diffusion equation \( \dfrac{\partial u}{\partial y}=\dfrac{\partial^2u}{\partial x^2} \) is parabolic since \( B^2-4AC=0 \)
Linearisert potensial-ligning for kompressibel strømning: $$ \begin{equation} (1-M^2)\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}=0,\ M=\text{Mach-number.} \tag{5.37} \end{equation} $$