PDE notes: Method of Characteristic

1. Introduction

The method of characteristics is a general technique for solving first-order equations. The basic idea is to reduce the determination of explicit solutions to solving ODE.

Initial value problem in one space variable ${x}$ and time ${t}$ take the form

$\displaystyle u_t+cu_x=0, t>0 \ \ \ \ \ (1)$

$\displaystyle u(x,0)=f(x) \ \ \ \ \ (2)$

We solve (1) using the method of characteristics, which reduces the initial value problem to an initial value problem for a system of ODE. In this method, we depend on the observation that if ${\{(x(t),t):t\geq 0\}}$ is smooth curve, then along the curve, ${u(x(t),t)}$ has rate of change

$\displaystyle \frac{d}{dt} u(x(t),t) = u_t+\frac{dx(t)}{dt}u_x \ \ \ \ \ (3)$

given by the chain rule. Comparing (3) with (1), we see that the expressions are identical if we set

$\displaystyle \frac{dx(t)}{dt}=c \ \ \ \ \ (4)$

and interpret ${c}$ as speed. The left-hand side of the PDE can also be interpreted as the derivative of ${u(x,t)}$ , in the direction ${(c,1)}$ in ${x-t}$ space.

Now the PDE in (1) can be replaced by the ODE system,

$\displaystyle \frac{dx(t)}{dt}=c \ \ \ \ \ (5)$

$\displaystyle \frac{d}{dt}[u(x(t),t)]=0 \ \ \ \ \ (6)$

These ODEs are called the characteristic equations. Take note that ${u(x(t),t)}$ is constant along the curve ${x=x(t)}$ . We solve (5) to get

$\displaystyle x(t)=ct+a \ \ \ \ \ (7)$

where ${a}$ is an unknown constant. Equation (7) is the equation for the family of parallel characteristic curves of (1). We observe that at ${t=0}$ , we have ${x(0)=a}$ , and hence this constant ${a}$ is the point where the characteristic curve starts. We call it the anchor point. If a point ${(x,t)}$ is given, we can always find the corresponding anchor point as

$\displaystyle a=x-ct \ \ \ \ \ (8)$

Now we solve the second characteristic ODE to get

$\displaystyle u(x(t),t)=k \ \ \ \ \ (9)$

where ${k}$ is an arbitrary constant.

Initial conditions for the ODE system are derived from the initial condition ${u(x,0)=f(x)}$ for the PDE problem (1). We already know that the initial condition for ODE (5) is ${x(0)=a}$ . We write ${u(t)}$ in place of ${u(x(t),t)}$ , then the initial condition for ODE (6) is

$\displaystyle u(0)=f(x(0))=f(a)=f(x(t)-ct) \ \ \ \ \ (10)$

Since u is constant along the characteristic curve,

$\displaystyle u(x(t),t)=f(x(0),0)=f(a) \ \ \ \ \ (11)$

Thus, given a point ${(x,t)}$ , there is a unique characteristic curve passing through ${(x,t)}$ , with anchor point at ${a=x-ct}$ , and the general solution at ${(x,t)}$ is

$\displaystyle u(x,t)=f(x-ct) \ \ \ \ \ (12)$

Next we extend the method to allow for linear forcing terms. We try to solve the following PDE for ${u(x,t)}$ on ${-\infty < x < \infty}$

$\displaystyle u_t+\alpha u_x+\gamma u=0 \ \ \ \ \ (13)$

$\displaystyle u(x,0)=f(x) \ \ \ \ \ (14)$

Solution From the chain rule we have

$\displaystyle \frac{du}{dt}= u_t+\frac{du}{dx}u_x \ \ \ \ \ (15)$

Hence the characteristic equation becomes

$\displaystyle \frac{dx}{dt}=\alpha \ \ \ \ \ (16)$

$\displaystyle \frac{du}{dt}=-\gamma u \ \ \ \ \ (17)$

The initial conditions are

$\displaystyle x(0)=a \ \ \ \ \ (18)$

$\displaystyle u(0)=f(a) \ \ \ \ \ (19)$

The solution for the first characteristic equation is

$\displaystyle x=\alpha t + a \ \ \ \ \ (20)$

Thus, the anchor point for the characteristic passing through the point ${(x,t)}$ is

$\displaystyle a=x-\alpha t \ \ \ \ \ (21)$

The solution for the second characteristic equation is

$\displaystyle u= u(x(0),0)e^{-\gamma t} \ \ \ \ \ (22)$

and the solution along the characteristic ${x=x(t)}$ is

$\displaystyle u=u(a,0)=f(x-\alpha t)e^{-\gamma t} \ \ \ \ \ (23)$

${\square}$

For this kind of initial value problem, the method of characteristic is summarized as:

Rewrite the initial value problem (13) as a system of ODE consisting of the characteristic equations (16) and (17) with initial conditions (18) and (19).
Solve the ODE and initial condition for ${x(t)}$ and ${u(t)}$ , with the anchor point ${x(0)=a}$ to get the solution along the characteristic.
Solve for ${a}$ as a function of ${x,t}$ . This effectively changes variables from ${t,a}$ to ${x,t}$ .
Write the solution ${u=u(x,t)}$ .

We can extend our concepts to first-order equations with non-constant coefficients in the form

$\displaystyle A(x,y)u_x+B(x,y)u_y=C(x,y,u) \ \ \ \ \ (24)$

In applications, this PDE is usually accompanied by a side condition of the form

$\displaystyle u(x,y)=f(x,y) \ \ \ \ \ (25)$

for ${(x,y) \in \Gamma_a}$ , where ${\Gamma_a}$ is a curve of anchor points and ${f}$ is a given function. Suppose that ${u(x,y)}$ is the solution to (24) subject to the side condition (25). We can think of ${z=u(x,y)}$ as a two-dimensional surface in the ${x-y-z}$ space. Denote by ${\Gamma}$ the curve on the surface ${z=u(x,y)}$ whose projection onto the ${xy-}$ plane is ${\Gamma_a}$ . The curve ${\Gamma}$ is called the initial curve. We parameterize the initial curve ${\Gamma}$ using the anchor points to get

$\displaystyle \Gamma: \left\{ \begin{array}{lr} x=x_0(a) \\ y=y_0(a) \\ z=z_0(a)=f(x_0(z),y_0(a)) \end{array} \right. \ \ \ \ \ (26)$

Equation (24) states that the vector field ${\textbf{F}=(A(x,y),B(x,y),C(x,y,u))}$ is tangent to the solution surface ${z=u(x,y)}$ , since the solution surface has normal

$\displaystyle \nabla(u(x,y)-z)=(u_x,u_y,-1) \ \ \ \ \ (27)$

The solution surface can therefore be generated by integrating along the vector field, starting at each point of the curve ${\Gamma}$ .

Integral curves

$\displaystyle \textbf{r}(s)=(x(s),y(s),z(s)) \ \ \ \ \ (28)$

of the vector field ${\textbf{F}}$ which starts from the initial curve ${\Gamma}$ satisfy the following vector ODE:

$\displaystyle \frac{d\textbf{r}}{ds}=\textbf{F} \ \ \ \ \ (29)$

$\displaystyle \textbf{r}(0)=(x_0(a),y_0(a),z_0(a)) \ \ \ \ \ (30)$

or, in component form, the system of ODEs:

$\displaystyle \left\{ \begin{array}{lr} \frac{dx}{ds}=A(x,y) \\ x(0)=x_0(a) \\ \end{array} \right. \ \ \ \ \ (31)$

$\displaystyle \left\{ \begin{array}{lr} \frac{dy}{ds}=B(x,y) \\ y(0)=y_0(a) \\ \end{array} \right. \ \ \ \ \ (32)$

$\displaystyle \left\{ \begin{array}{lr} \frac{dz}{ds}=Z(x,y) \\ z(0)=z_0(a) \\ \end{array} \right. \ \ \ \ \ (33)$

The system of ODEs (31), (32), (33) are called the characteristic equations for the PDE (24). The solutions to the characteristic equations are called the characteristic curves for the PDE.

The procedure to solve the PDE (24) and (25) is as follows:

Solve the first two characteristic equations (31) and (32)to get ${x}$ and ${y}$ in terms of the characteristic variable ${s}$ and the anchor point ${a}$ :
$\displaystyle x=X(s,a), y=Y(s,a) \ \ \ \ \ (34)$
Insert the solution from the previous step into equation (33) and solve the resulting equation for ${z}$ :
$\displaystyle z=Z(s,a) \ \ \ \ \ (35)$
Apply the Inverse Function Theorem. In this step we solve the equations

$\displaystyle x=X(s,a), y=Y(s,a) \ \ \ \ \ (36)$

for

$\displaystyle s=S(x,y), a=\wedge(x,y) \ \ \ \ \ (37)$

The solution is guaranteed by the Inverse function theorem.
Write the solution for ${z}$ in terms of ${x}$ and ${y}$ to get the solution to the original PDE:
$\displaystyle u(x,y)=Z(S(x,y),\wedge(x,y)) \ \ \ \ \ (38)$

This procedure will work as long as the transformation in (36) and (37) is invertible. We can guarantee this locally by appealing to the Inverse Function Theorem.