A first-order partial differential equation (PDE) is a relationship between an unknown function $u\left(x_1, x_2, \ldots, x_n\right)$ of $n$ independent variables and its first partial derivatives. The method of characteristics is a powerful technique for solving first-order PDEs. The core idea is to transform the PDE into a system of ordinary differential equations (ODEs) along certain curves, called characteristic curves. Along these curves, the PDE simplifies, making it easier to solve.

Example:

Determine and sketch the characteristics of the partial differential equation $\left(2 x_2-3 x_1\right) \partial_{x_1} u-x_2 \partial_{x_2} u=x_2^2\left(2 x_2-5 x_1\right)$ and determine the solution $u\left(x_1, x_2\right)$ for the initial value $u\left(x_1, 1\right)=x_1+1$.

Solution:

Step 1: Method of Characteristics

We write the PDE in the form:

$$ a\left(x_1, x_2\right) \frac{\partial u}{\partial x_1}+b\left(x_1, x_2\right) \frac{\partial u}{\partial x_2}=f\left(x_1, x_2\right) $$

Where:

The characteristic equations are:

$$ \frac{d x_1}{d t}=2 x_2-3 x_1, \quad \frac{d x_2}{d t}=-x_2, \quad \frac{d u}{d t}=x_2^2\left(2 x_2-5 x_1\right) $$

Step 2: Solve the characteristic system

  1. Solve $\frac{d x_2}{d t}=-x_2$

    This is separable:

    $\frac{d x_2}{x_2}=-d t \Rightarrow \ln \left|x_2\right|=-t+C \Rightarrow x_2(t)=C_1 e^{-t}$

    Let's set $x_2(0)=1 \Rightarrow C_1=1$, so:

    $x_2(t)=e^{-t}$

  2. Plug into $\frac{d x_1}{d t}=2 x_2-3 x_1$

    Now $x_2(t)=e^{-t}$, so:

    $$ ⁍ $$

    This is a linear ODE. Use integrating factor:

    $$ \frac{d x_1}{d t}+3 x_1=2 e^{-t} $$

    Integrating factor: $\mu(t)=e^{3 t}$

    $$ \frac{d}{d t}\left(x_1 e^{3 t}\right)=2 e^{2 t} \Rightarrow x_1 e^{3 t}=\int 2 e^{2 t} d t=\frac{2}{2} e^{2 t}+C=e^{2 t}+C \Rightarrow x_1(t)=e^{-t}+C e^{-3 t} $$

    From $x_2(0)=1$, let $x_1(0)=\xi$. Then:

    $$ \xi=x_1(0)=1+C \Rightarrow C=\xi-1 \Rightarrow x_1(t)=e^{-t}+(\xi-1) e^{-3 t} $$

    So, along characteristics:

  3. Now solve for $u(t)$ along characteristics:

    $$ \frac{d u}{d t}=x_2^2\left(2 x_2-5 x_1\right) $$

    Substitute:

    So:

    $$ \begin{gathered} \frac{d u}{d t}=e^{-2 t}\left(2 e^{-t}-5\left[e^{-t}+(\xi-1) e^{-3 t}\right]\right) \\ =e^{-2 t}\left(2 e^{-t}-5 e^{-t}-5(\xi-1) e^{-3 t}\right) \\ =e^{-2 t}\left(-3 e^{-t}-5(\xi-1) e^{-3 t}\right) \\ =-3 e^{-3 t}-5(\xi-1) e^{-5 t} \end{gathered} $$

    Now integrate:

    $$ \begin{gathered} u(t)=\int\left(-3 e^{-3 t}-5(\xi-1) e^{-5 t}\right) d t+D \\ =e^{-3 t}+(\xi-1) e^{-5 t}+D \\ u(t)=-e^{-3 t}-(\xi-1) e^{-5 t}+D \end{gathered} $$

    At $t=0, x_1=\xi, x_2=1 \Rightarrow u(0)=u(\xi, 1)=\xi+1$ So:

    $$ u(0)=-1-(\xi-1)+D=-\xi+D \Rightarrow \xi+1=-\xi+D \Rightarrow D=2 \xi+1 $$

    So the solution along characteristics is:

    $$ u(t)=-e^{-3 t}-(\xi-1) e^{-5 t}+2 \xi+1 $$

Step 3: Express $u$ in terms of $x_1, x_2$

We have: