Method of Characteristics for First-Order Partial Differential Equations

General Steps

Consider the partial differential equation: \boldsymbol{\alpha}(\boldsymbol{x},u) \cdot \frac{\partial}{\partial \boldsymbol{x}} u = \beta(\boldsymbol{x},u) where \boldsymbol{\alpha} is an n-dimensional vector function, \beta is a scalar function, \cdot denotes the dot product of vectors, u \equiv u(\boldsymbol{x}) is a function of n variables, and \boldsymbol{x} represents its independent variables.

The idea behind the method of characteristics is to imagine \boldsymbol{x} as a function of some parameter s. In this case, \boldsymbol{x}(s) is actually the parametric equation of a high-dimensional curve, known as the characteristic line. Consequently, u also becomes a function of the parameter s. We then have: \frac{du}{ds} = \frac{\partial u}{\partial \boldsymbol{x}} \cdot \frac{d\boldsymbol{x}}{ds} Comparing this with the original partial differential equation (1), we find that we can set: \frac{d\boldsymbol{x}}{ds} = \boldsymbol{\alpha}(\boldsymbol{x},u) Then it follows that: \frac{du}{ds} = \beta(\boldsymbol{x},u) Combining (3) and (4), we obtain a system of ordinary differential equations consisting of n+1 equations: \left\{\begin{aligned} &\frac{d\boldsymbol{x}}{ds} = \boldsymbol{\alpha}(\boldsymbol{x},u) \\ &\frac{du}{ds} = \beta(\boldsymbol{x},u) \end{aligned}\right. Since s is merely an auxiliary variable introduced for the derivation, we can, in principle, solve for results independent of s: \boldsymbol{c} = \boldsymbol{f}(\boldsymbol{x},u) where \boldsymbol{c} is an n-dimensional vector representing the integration constants of the system of ODEs, and \boldsymbol{f} is an n-dimensional vector function. The remaining task is to determine the relationship between these integration constants based on the initial conditions. If a general solution expression is required, it takes the form: G(\boldsymbol{f}(\boldsymbol{x},u)) = 0 where G is an arbitrary function of n variables. In principle, we can solve this to obtain the expression for u in terms of \boldsymbol{x}.

Simple Example

The steps above are quite theoretical. In practical problem-solving, one can be more flexible. Let us solve: \frac{\partial u}{\partial t} + x \frac{\partial u}{\partial x} = u^2, \quad u(x,0) = f(x) We obtain the characteristic equations: dt = \frac{dx}{x} = \frac{du}{u^2} Solving these yields: x = C_1 e^t, \quad u = \frac{1}{C_2 - t} When t=0, we have x = C_1 and u = \frac{1}{C_2} = f(C_1), which leads to C_2 = \frac{1}{f(C_1)}. Since we have C_1 = xe^{-t} and C_2 = u^{-1} + t, substituting these gives: u^{-1} + t = \frac{1}{f(xe^{-t})} That is: u = \frac{f(xe^{-t})}{1 - t \cdot f(xe^{-t})}

A Brief Discussion

What is the actual meaning of a characteristic line? For beginners, the process above might seem like a magic trick—solving for constants and then eliminating them—without grasping the core logic. This was my own confusion when I first started learning the method of characteristics.

In fact, we can think of it this way: the characteristic line is already a solution to the partial differential equation, but it is only a single line on the solution surface. In its entirety, the solution should be a high-dimensional surface. Obviously, points move to form lines, and lines move to form surfaces. By finding a way to make these lines “move,” we can obtain the equation of this surface. That is to say, the various integration constants \boldsymbol{c} need to “move.”

Of course, they cannot move randomly without constraints; if they moved arbitrarily, they might cover the entire space. How they move depends on the initial conditions. Therefore, we use the initial conditions to determine the constraint relationships between the integration constants. Once determined, we have obtained the parametric equation of this surface. From a solving perspective, it is not strictly necessary to eliminate the constants, but since we often prefer explicit solutions, we find ways to eliminate them.

This is the general logic of the process.

Derivation

To this end, we first denote: \boldsymbol{p} = \frac{\partial u}{\partial\boldsymbol{x}} Then, differentiating F\left(\boldsymbol{x}, u, \boldsymbol{p}\right) = 0 with respect to s, we get: \begin{aligned} 0 &= \frac{\partial F}{\partial\boldsymbol{x}} \cdot \frac{d\boldsymbol{x}}{ds} + \frac{\partial F}{\partial u} \frac{\partial u}{\partial \boldsymbol{x}} \cdot \frac{d\boldsymbol{x}}{ds} + \frac{\partial F}{\partial\boldsymbol{p}} \cdot \frac{d\boldsymbol{p}}{ds} \\ &= \left(\frac{\partial F}{\partial\boldsymbol{x}} + \frac{\partial F}{\partial u}\boldsymbol{p}\right) \cdot \frac{d\boldsymbol{x}}{ds} + \frac{\partial F}{\partial\boldsymbol{p}} \cdot \frac{d\boldsymbol{p}}{ds} \end{aligned} We can see that the above is the sum of two dot products of vectors equaling zero. An interesting way to solve this is to let: \frac{d\boldsymbol{x}}{ds} = \frac{\partial F}{\partial\boldsymbol{p}}, \quad \frac{d\boldsymbol{p}}{ds} = -\frac{\partial F}{\partial\boldsymbol{x}} - \frac{\partial F}{\partial u}\boldsymbol{p} Furthermore, we have: \frac{du}{ds} = \frac{\partial u}{\partial\boldsymbol{x}} \cdot \frac{d\boldsymbol{x}}{ds} = \boldsymbol{p} \cdot \frac{\partial F}{\partial\boldsymbol{p}} Combining these, we obtain the system of ordinary differential equations: \left\{\begin{aligned} &\frac{d\boldsymbol{x}}{ds} = \frac{\partial F}{\partial\boldsymbol{p}} \\ &\frac{d\boldsymbol{p}}{ds} = -\frac{\partial F}{\partial\boldsymbol{x}} - \frac{\partial F}{\partial u}\boldsymbol{p} \\ &\frac{du}{ds} = \boldsymbol{p} \cdot \frac{\partial F}{\partial\boldsymbol{p}} \\ &F\left(\boldsymbol{x}, u, \boldsymbol{p}\right) = 0 \end{aligned}\right. The subsequent steps are no different from the quasilinear case, except that n additional variables \boldsymbol{p} are introduced. Solving this system yields 2n integration constants (independent of s), and the relationships between these constants are found using the initial conditions. The difference is that because of the extra n variables \boldsymbol{p}, partial derivatives must also be considered for the initial conditions, making the solving process more complex. Please see the following example.

Another Example

\frac{\partial u}{\partial t} = \left(\frac{\partial u}{\partial x}\right)^2, \quad u(x,0) = f(x) This is 0 = F(\boldsymbol{p}) = p_t - p_x^2. According to equation (18), we obtain the characteristic equations: \left\{\begin{aligned} &\frac{dt}{ds} = 1, \quad \frac{dx}{ds} = -2p_x \\ &\frac{dp_t}{ds} = 0, \quad \frac{dp_x}{ds} = 0 \\ &\frac{du}{ds} = p_t - 2p_x^2 = -p_x^2 \end{aligned}\right. Setting s=t, we find that p_t = C_1 and p_x = C_2 are constants, which leads to x = -2C_2 t + C_3 and u = -C_2^2 t + C_4. According to the initial conditions, when t=0, we have x = C_3 and u = C_4, which implies C_4 = f(C_3). Substituting this gives: u = -C_2^2 t + f(x + 2C_2 t) Next, taking the partial derivative of the initial condition with respect to x, at t=0 we have: C_2 = p_x = f'(x) = f'(C_3) Note that f is a pre-specified initial function, so the above is actually an algebraic equation. Thus we have: \begin{aligned} u &= -t \cdot f'(C_3)^2 + f\Big(x + 2t \cdot f'(C_3)\Big) \\ &= -t \cdot f'(C_3)^2 + f(C_3) \end{aligned} And: x = -2C_2 t + C_3 = -2t \cdot f'(C_3) + C_3 By solving for C_3 from this equation and substituting it into the previous one, we obtain the complete solution. For example, when f(x) = x^2, we solve to find: C_3 = \frac{x}{1-4t} Substituting this yields: u = \frac{x^2}{1-4t}

Top-down Process

Formally, the method of characteristics for the general case looks quite different from the quasilinear case. Can the general method be reduced to the quasilinear case?

In fact, by substituting F = \boldsymbol{\alpha} \cdot \boldsymbol{p} - \beta into equation (18), we obtain: \left\{\begin{aligned} &\frac{d\boldsymbol{x}}{ds} = \boldsymbol{\alpha} \\ &\frac{du}{ds} = \boldsymbol{p} \cdot \boldsymbol{\alpha} \end{aligned}\right. The equation for \boldsymbol{p} need not be written out because we have \boldsymbol{p} \cdot \boldsymbol{\alpha} = \beta. Substituting this into the above yields a closed system of equations, which reduces to equation (5).

Additionally, a particularly simple case is when F is a function of \boldsymbol{p} only. In this case, \frac{\partial F}{\partial\boldsymbol{x}} and \frac{\partial F}{\partial u} are both 0. Consequently, \boldsymbol{p} is constant in the characteristic equations, making \boldsymbol{x} and u linear functions of s. The entire equation becomes completely solvable, turning into a purely algebraic problem.