The first-order equations with real coefficients are particularly simple tohandle. The method of characteristics reduces the givenfirst-order partial differential equation (PDE) to a system of first-orderordinary differential equations (ODE) along some special curves called thecharacteristics of the given PDE. This will, in turn,help us to prove the existence of a solution to the Cauchyproblem or initial value problem (IVP)associated with the PDE. Complications do arise in case of quasilinear ornon-linear equations resulting only in local existence; thegeometry of the characteristics also becomes more involved and nonuniquenessof (smooth) solutions may also result. To motivate the ideas we begin by asimple example.

Example 3.1. Consider the transport equation in two independentvariables t and x, namely

for t 0, x ∈ ℝ, wherec > 0 is a given constant. This is a linear,first-order PDE. Consider the curve x =x(t) in the (x, t)plane given by the slope condition. These are straight lines with slope1/c and are represented by the equationx − ct =x0, where x0 is the pointat which the curve meets the line t = 0 (see Figure3.1(a)). These curves, straight lines in this case, are called thecharacteristic curves or simply thecharacteristics of (3.1). When c is afunction of t and x, the characteristiccurves need not be straight lines.

Now restrict the solution u(x, t) to acharacteristic x(t) = ct+ x0, that is, consider the function ofone variable U(t) =u(x(t,t). By the chain rule, it is easy to see that

using (3.1). Therefore, U ≡ constant and thusu is constant along the characteristicx − ct =x0. This observation can be used to solvethe IVP for the PDE (3.1) as follows:

Suppose the initial values of u are given on the linet = 0, that is, u(x,0) = u0(x) is given andu0 is a C1function. Now, for any point (x, t), t> 0 in the upper half plane, draw the characteristicpassing through the point (x, t). It is easy to see thatthis is given by the line with slope 1/c meeting theline t = 0 at the point (x –ct, 0). As shown above, we haveu(x, t) =u(x – ct, 0) =u0(x –ct).