Method of Characteristics

1Introduction

This lecture explains how a first-order PDE can be reduced to ODEs along characteristic curves and how a local classical solution can be reconstructed from those ODEs. Existence, uniqueness, or regularity may fail when the data curve is tangent to the characteristic direction or when distinct characteristics intersect.

2Terminology and definition

A {characteristic curve} follows the differential direction specified by a PDE. Along such a curve, the PDE reduces to an ODE.

3General form and characteristic ODEs

Let $$D\subset R^3$$D\subset\mathbb R^3 be open in the variables $$(x,t,z)$$(x,t,z), and suppose that $$a,b,F\in C^1\left(D\right)$$a,b,F\in C^1(D). A first-order PDE is {quasilinear} if its highest-order derivatives occur linearly, while their coefficients may depend on the independent variables and on the unknown function. Consider

Set $$z\left(s\right)=u\left(x\right(s),t(s\left)\right)$$z(s)=u(x(s),t(s)). The chain rule gives

so the characteristic equations are

Because the coefficients are $$C^1$$C^1, this ODE system has a unique local solution near each initial point. If $$F\equiv 0$$F\equiv0, then $$z$$z, and hence $$u$$u, is constant along every characteristic. In general, $$z$$z evolves according to its own ODE.

4The curve carrying the data

Let $$x_0,t_0,h\in C^1\left(I\right)$$x_0,t_0,h\in C^1(I) on an interval $$I$$I, with $$\left(x_0\right(r),t_0(r),h(r\left)\right)\in D$$(x_0(r),t_0(r),h(r))\in D, and prescribe

The characteristic direction on this curve is $$(a,b)$$(a,b). The initial curve is {non-characteristic} when it is transverse to this direction, namely when

This determinant is precisely the condition that the characteristic coordinate map from $$(r,s)$$(r,s) to $$(x,t)$$(x,t) have a local inverse near the initial curve. The function $$z(r,s)$$z(r,s) obtained from the characteristic ODEs can therefore be reconstructed as $$u(x,t)$$u(x,t).

Indeed, write the local inverse as $$(r,s)=\left(r\right(x,t),s(x,t\left)\right)$$(r,s)=(r(x,t),s(x,t)) and define $$u(x,t)=z\left(r\right(x,t),s(x,t\left)\right)$$u(x,t)=z(r(x,t),s(x,t)). The characteristic ODEs and the chain rule give

At $$s=0$$s=0, one has $$(x,t,z)=\left(x_0\right(r),t_0(r),h(r\left)\right)$$(x,t,z)=(x_0(r),t_0(r),h(r)), so $$u\left(x_0\right(r),t_0(r\left)\right)=h\left(r\right)$$u(x_0(r),t_0(r))=h(r). Thus the reconstructed function satisfies both the PDE and the initial condition.

For the normal form

take $$x_0\left(r\right)=r$$x_0(r)=r and $$t_0\left(r\right)=0$$t_0(r)=0. The determinant above is then 1. Thus, provided $$h\in C^1\left(I\right)$$h\in C^1(I) and the lifted curve remains in $$D$$D, the standard initial line $$t=0$$t=0 is non-characteristic and supports data for constructing a local classical solution.

The construction remains valid only while the lifted characteristic $$(x,t,z)$$(x,t,z) stays in $$D$$D, the ODE solution exists, and the Jacobian of the characteristic coordinate map is nonzero. The domain of the PDE solution is the image of this map; it does not automatically extend to all space and all time. A global solution additionally requires global existence of the ODEs and global injectivity and coverage of the coordinate map.

If the determinant vanishes, arbitrary data on the characteristic curve may be inconsistent along a single characteristic, so no solution exists, or may fail to determine values away from that characteristic, so uniqueness is lost. The suitability of the data is determined not merely by the number of curves but by their position relative to the differential direction of the PDE.

5Constant-coefficient transport

Let $$g\in C^1\left(\mathbb{R}\right)$$g\in C^1(\mathbb R) and consider the whole-space Cauchy problem

The characteristics are $$x-2t=\xi$$x-2t=\xi, and $$du/dt=u_t+2u_x=0$$du/dt=u_t+2u_x=0 along them. Hence

so the initial profile travels to the right with speed 2.

6A source or decay term

Let $$c,\lambda \in \mathbb{R}$$c,\lambda\in\mathbb R and $$g\in C^1\left(\mathbb{R}\right)$$g\in C^1(\mathbb R). For

the characteristic is $$x=\xi +ct$$x=\xi+ct, and

Therefore

The profile travels with speed $$c$$c while its amplitude grows or decays according to $$\lambda$$\lambda. Constancy along characteristics applies only when the right-hand side vanishes; the method itself also applies when the value evolves by an ODE.

7A variable-coefficient example

Let $$g\in C^1\left(\mathbb{R}\right)$$g\in C^1(\mathbb R) and consider $$u_t+x{u}_x=0$$u_t+xu_x=0. The characteristic equations are $$x^{\prime }=x$$x'=x and $$t^{\prime }=1$$t'=1, hence $$x\left(t\right)=C{e}^t$$x(t)=Ce^t and $$x{e}^{-t}$$xe^{-t} is constant. With $$u(x,0)=g\left(x\right)$$u(x,0)=g(x),

Writing the initial point as $$\xi$$\xi gives $$x=\xi e^t$$x=\xi e^t. Since the inverse $$\xi =x{e}^{-t}$$\xi=xe^{-t} exists, values on characteristics define $$u(x,t)$$u(x,t). In general, solving the characteristic ODEs is insufficient by itself: one must also identify the region on which the map from the characteristic label and time to the spatial point is injective. That image is the domain of the classical solution.

8Burgers' equation and crossing characteristics

Assume $$g\in C^1\left(\mathbb{R}\right)$$g\in C^1(\mathbb R) and consider

Along the characteristic issued from $$\xi$$\xi, the value $$u=g\left(\xi \right)$$u=g(\xi) is constant, so

The derivative of the map from $$\xi$$\xi to $$x$$x is

While this quantity is nonzero, $$\xi$$\xi can be recovered locally from $$(x,t)$$(x,t) and a classical solution can be constructed. If it vanishes, the characteristic coordinate map loses local invertibility, and this construction cannot continue as a single-valued smooth solution. In a compressive region, distinct characteristics subsequently reach the same point and cross.

Implicit differentiation gives, within the classical regime,

Thus, as the coordinate map degenerates, the denominator can approach zero and the gradient can diverge. In particular, if

satisfies $$-\mathrm{\infty }<m<0$$-\infty<m<0, then the candidate time of the first coordinate degeneration and gradient breakdown is

To identify this expression with the exact breakdown time, one must additionally verify the regularity of $$g$$g and whether the infimum is attained. If $$g^{\prime }\left(\xi \right)\text{[PARSE ERROR: Undefined("Command("ge")")]}0$$g'(\xi)\ge0 for every $$\xi$$\xi, this mechanism produces no characteristic crossing at positive times. After crossing, one needs weak solutions based on the integral form of the conservation law and an entropy condition that selects the admissible solution. These notions are defined in the subsequent lecture on transport and conservation laws.

9An example outside the method's scope

For second-order PDEs, the reduction of a first directional derivative to an ODE does not apply directly. For example, the heat equation $$u_t=\kappa u_{xx}$$u_t=\kappa u_{xx} contains diffusion and is not a problem in which values are transported unchanged along a single curve.

Second-order PDEs also have characteristic directions, but their theory differs from reconstruction by the three characteristic ODEs used here. The next classification lecture treats the characteristics and types determined by the second-order principal part.

10Scope of validity

Constructing a classical solution by characteristics requires local existence and uniqueness for the characteristic ODEs, a non-characteristic initial curve, and local invertibility of the coordinate map. The procedure can fail if the coefficients lack sufficient regularity, if the ODE itself is nonunique, or if characteristics cross.

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