📂Partial Differential Equations

Characteristics of Nonlinear First-Order Partial Differential Equations

• Regarding x and p, when emphasizing that they are variables of a partial differential equation, they are denoted in normal font as $x,p \in \mathbb{R}^{n}$, and when emphasizing that they are functions of $s$, they are denoted in bold font as $\mathbf{x}, \mathbf{p} \in \mathbb{R}^{n}$.

Characteristic Method¹

Let’s suppose that an open set $\Omega \subset \mathbb{R}^{n}$ is given. Assume that $u\in C^{2}(\Omega)$ is a solution to the following nonlinear first-order partial differential equation.

$$ F(Du,\ u,\ x)=0 $$

And let’s set $\mathbf{x}, z, \mathbf{p}$ as follows.

$$ \begin{align*} \mathbf{x}(s) &=(x^{1}(s), \dots, x^{n}(s)) \in C^1(I;\Omega)\quad (s\in I \subset \mathbb{R}) \\ z(s) &= u(\mathbf{x}(s)) \\ \mathbf{p}(s) &= Du(\mathbf{x}(s)) \end{align*} $$

Here, suppose $\mathbf{x}$ satisfies the following equation.

$$ \dot{\mathbf{x}}(s) = D_{p}F\big(\mathbf{p}(s), z(s), \mathbf{x}(s) \big) $$

Then, $\mathbf{p}(s)$ and $z(s)$ each become the solution of the following ordinary differential equations.

$$ \begin{cases} \dot{\mathbf{p}} (s) = -D_{x}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big)-D_{z}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big)\mathbf{p}(s) \\ \dot{z}(s) = D_{p}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big) \cdot \mathbf{p}(s) \\ \dot{\mathbf{x}}(s) = D_{p}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big) \end{cases} $$

Description

The method of characteristics is one of the methods to solve nonlinear first-order partial differential equations, which represents a single partial differential equation as a system of ODE and solves it.

Derivation

Let’s consider the following nonlinear first-order partial differential equation is given.

$$ \begin{equation} F(Du, u, x) = 0 \label{eq1} \end{equation} $$

In this case, $\Omega \subset \mathbb{R}^{n}$ is an open set and it is assumed to be $F\in C^{\infty}(\mathbb{R}^n \times \mathbb{R} \times \Omega)$. Let’s also assume that the following boundary conditions are given.

$$ \begin{equation} u=g \quad \text{ on } \Gamma \label{eq2} \end{equation} $$

The idea of the characteristic method is to find the line in $\Omega$ that connects fixed $x\in\Omega$ and $x^0 \in \Gamma$ using the given boundary conditions. In other words, when $u \in C^{2}$ is a solution that satisfies $\eqref{eq1}, \eqref{eq2}$, the goal is to obtain the value of $u$ along the line shown in the following figure.

Let’s assume the line in the figure is expressed by the following function.

$$ \mathbf{x}(s)=\big( x^1(s),\ \cdots,\ x^n(s) \big), \quad s\in I\subset \mathbb{R} $$

Set as $\mathbf{x}(0)=x^{0}$, and think of $\mathbf{x}$ moving along the line inside $\Omega$ as $s$ increases. Let’s represent the values of $u$ according to $s$ as a function $z$.

$$ \begin{equation} z(s):= u(\mathbf{x}(s)) \label{eq3} \end{equation} $$

Similarly, let’s say the values of $Du$ according to $s$ are $\mathbf{p}$.

$$ \mathbf{p}(s) := Du(\mathbf{x}(s)) $$

Then, when it is $\mathbf{p}(s)=(p^{1}(s), \dots, p^{n}(s))$, each component is as follows.

$$ \begin{equation} p^i(s)=u_{x_{i}}( \mathbf{x}(s)) \label{eq4} \end{equation} $$

Let’s denote the derivative with respect to $s$ simply as $\dfrac{d f}{ds} = \dot{f}$. Since $dp^i(s)=u_{x_{i}x_{1}}dx^1+\cdots u_{x_{i}x_{n}}dx^n$, $\dot{p}^i(s)$ is as follows.

$$ \begin{equation} \dot{p}^i(s)=\sum \limits_{j=1}^n u_{x_{i}x_{j}}( \mathbf{x}(s))\dot{x}^j(s) \label{eq5} \end{equation} $$

At this time, since $dF=F_{p_{1}}dp_{1}+\cdots F_{p_{n}}dp_{n}+F_{z}dz+F_{x_{1}}dx_{} + \cdots + F_{x_{n}}dx_{n}$, differentiating $\eqref{eq1}$ with respect to $x_{i}$ yields the following.

$$ \sum \limits_{j=1}^n F_{p_{j}}\big( \mathbf{p}(s), z(s), \mathbf{x}(s) \big)p^j_{x_{i}}(s)+F_{z} \big( \mathbf{p}(s), z(s), \mathbf{x}(s) \big)z_{x_{i}}(s)+F_{x_{i}} \big( \mathbf{p}(s), z(s), \mathbf{x}(s) \big) = 0 $$

At this time, since $\eqref{eq3}, \eqref{eq4}$ implies $z_{x_{i}}=u_{x_{i}}=p^{i}$ and $\eqref{eq4}$ implies $p^{j}_{x_{i}}=u_{x_{j}x_{i}}$, substituting them into the above equation yields the following.

$$ \begin{equation} \sum \limits_{j=1}^{n} F_{p_{j}}\big( \mathbf{p}(s), z(s), \mathbf{x}(s) \big)u_{x_{j}x_{i}}(\mathbf{x}(s))+F_{z}\big( \mathbf{p}(s), z(s), \mathbf{x}(s) \big) p^{i}(s)+F_{x_{i}}\big( \mathbf{p}(s), z(s), \mathbf{x}(s) \big) = 0 \label{eq6} \end{equation} $$

However, looking at the above equation, we can see that a second-order derivative term $u_{x_{i}x_{j}}$ emerges due to $\eqref{eq5}$. Since it is not correct to solve a second-order derivative to solve a first-order partial differential equation, it is necessary to eliminate this term. For this purpose, assume that each component of $\dot{\mathbf{x}}$ satisfies the following.

$$ \begin{equation} \dot{x}^{j}(s)=F_{p_{j}}(\mathbf{p}(s), z(s), \mathbf{x}(s)), \quad \forall\ s\in I \subset \mathbb{R} \label{eq7} \end{equation} $$

Substituting $\eqref{eq6}$ for $\eqref{eq7}$ and using $\eqref{eq5}$ leads to the following equation.

$$ \begin{equation} \dot{p}^{i} (s) = -F_{z} \big( \mathbf{p}(s), z(s), \mathbf{x}(s) \big)p^i(s)-F_{x_{i}} \big( \mathbf{p}(s), z(s), \mathbf{x}(s) \big), \quad \forall\ s\in I(i=1,\dots,n) \label{eq8} \end{equation} $$

Also, differentiating $\eqref{eq3}$ with respect to $s$ yields the following.

$$ \dot{z}(s)=\sum \limits_{j=1}^n u_{x_{j}}(\mathbf{x}(s))\dot{x}^j(s) $$

Substituting $\eqref{eq4} p^{i}=u_{x_{i}}$ and $\eqref{eq7} \dot{x}^{j} = F_{p_{j}}$ into this yields the following.

$$ \begin{equation} \dot{z}(s)=\sum \limits_{j=1}^n p^j(s)F_{p_{j}}\big( \mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big), \quad s\in I \subset \mathbb{R}, \label{eq9} \end{equation} $$

Arranging and bundling the obtained $\eqref{eq7}, \eqref{eq8}, \eqref{eq9}$ in the following manner into $2n+1$ simultaneous equations is called the characteristic equations. Also, each unknown $\mathbf{p}(s), z(s), \mathbf{x}(s)$ is called the characteristics of $\eqref{eq1}$.

$$ \begin{cases} \dot{\mathbf{p}} (s) = -D_{x}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big)-D_{z}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big)\mathbf{p}(s) \\ \dot{z}(s) = D_{p}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big) \cdot \mathbf{p}(s) \\ \dot{\mathbf{x}}(s) = D_{p}F\big(\mathbf{p}(s),\ z(s),\ \mathbf{x}(s) \big) \end{cases} $$

■