Scalar Conservation Laws and First Order Equations Introduction. Consider equations of the form (1) u t + q(u) x =, x R, t >. In general, u = u(x, t) represents the density or the concentration of a physical quantity Q and q(u) its flux function. Equation (1) constitutes a link between density and flux and expresses a (scalar) conservation law for the following reason. If we consider a control interval [x 1,x 2 ], the integral x2 x 1 u(x, t)dt gives the amount of Q between x 1 and x 2 at time t. A conservation law states that, without sources or sinks, the rate of changes of Q in the interior of [x 1,x 2 ] is determined by the net flux through the end points of the interval. If the flux is modelled by a function q = q(u), the law translates into the equation (2) d x2 u(x, t)dx = q(u(x 2,t)) + q(u(x 1,t)), dt x 1 where we assume that q>(q<) for a flux along the positive (negative) direction of the x axes. Ifu are q are smooth functions, equation (2) can be written in the form x2 x 1 [u t (x, t)+q(u(x, t)) x ]dx =, which implies (1), due to the arbitrariness of the interval [x 1,x 2 ]. In the next section we go back to the model of pollution in a channel, neglecting the diffusion and choosing for q a linear function of u, namely, q(u) = vu, where v is constant. The result is a pure transport model, in which the vector vi is the advection speed. In the sequel, to introduce and motivate some important concepts and results, we shall use a nonlinear model from traffic dynamics, with speed v depending on u. Typeset by AMS-TEX 1
2 Linear Transport Equation 1. Pollution In a Channel Go back to the simple model for the evolution of a pollutant in a narrow channel. When diffusion and transport are both relevant we have the equation c t = Dc xx vc x where c is the concentration of the pollutant and vi is the stream speed (c >, constant). We discuss here the case of the pure transport equation (4) c t + vc x =, i.e. D =. Introducing the vector v = vi + j equation (4) can be written in the form vc x + c t = c v =, pointing out the orthogonality of c and v. But c is orthogonal to the level lines of c, along which c is constant. Therefore the level lines of c are the straight lines parallel to v, of equation x = v t + x. These straight lines are called characteristics. Let us compute c along the characteristic x = vt + x, letting w(t) =c(x + vt, t). Since ẇ(t) =vc x (x + vt, t)+c t (x + vt, t), equation (4) becomes an o.d.e. ẇ(t) =, which implies that c is constant along the characteristic. We want to determine the evolution of the concentration c, by knowing its initial profile c(x, ) = g(x). The method to compute the solution at a point (x, t), t>, is very simple. Let x = vt + x be the equation of the characteristic passing through (x, t). Go back in time along this characteristic from (x, t) until the point (x, ), of intersection with the x-axis. Since c is constant along the characteristic and c(x, ) = g(x ), we have c(x, t) =g(x )=g(x vt). Thus, if g C 1 (R), the solution of the initial value problem (4), (5) is given by (6) c(x, t) =g(x vt). The solution (6) represents a travelling wave, moving with speed v in the positive x-direction.
3 2. Distributed Source The equation (7) c t + vc x = f(x, t), (8) c(x, ) = g(x), describes the effect of an external distributed source along the channel. The function f represents the intensity of the source, measured in concentration per unit time. Again, to compute the value of the solution u at a point (x, t) is not difficult. Let x = x + vt be the characteristic passing through (x, t) and compute u along this characteristic, setting From (7), w satisfies the o.d.e. Thus w(t) =c(x + vt, t). ẇ(t) =vc x (x + vt, t)+c t (x + vt, t) =f(x + vt, t) w(t) =g(x )+ w() = g(x ). f(x + vs, s)ds. Letting t = t and recalling that x = x vt, we get c(x, t) =w(t) =g(x vt)+ f(x v(t s),s)ds. Proposition 1. Let g C 1 (R) and f, f x C(R R + ). The solution of the initial value problem is given by the formula c t + vc x = f(x, t), x R, t >, c(x, ) = g(x) x R. c(x, t) =g(x vt)+ f(x v(t s),s)ds.
4 3. Decay And Localized Source Suppose that, due to biological decomposition, the pollutant decays at the rate r(x, t) = γc(x, t), γ >. Without external sources and diffusion, the mathematical model is c t + vc x = γc, Setting c(x, ) = g(x). (9) u(x, t) =c(x, t)e γ v x we have u x = (c x + γv c ) e γ v x and u t = c t e γ v x and therefore the equation for u is From Proposition 1, we obtain and from (9) u t + vu x = u(x, ) = g(x)e γ v x. u(x, t) =g(x vt)e γ v (x vt) c(x, t) =g(x vt)e γt, which is a damped travelling wave. We now examine the effect of a source of pollutant placed at a certain point of the channel, e.g. at x =. Typically, one can think of waste material from industrial machineries. Before the machine start working, for instance before time t =, we assume that the channel is clean. We want to determine the pollutant concentration, supposing that at x = it is kept at a constant level β>, for t>. To model this source we introduce the Heaviside function { 1, t H(t) =, t <, with the boundary condition (1) c(,t)=βh(t)
5 and the initial condition (11) c(x, )=, for x>. As before, let u(x, t) =c(x, t)e γ v x, which is a solution of Then u t + vu x =. u(x, ) =c(x, )e γ v x =, x > u(,t)=c(,t)=βh(t). Since u is constant along the characteristics, it must be of the form (12) u(x, t) =u (x vt) where u is to be determined from the boundary condition (1) and the initial condition (11). To compute u for x<vt, observe that a characteristic leaving the t-axis from a point (x, t) carries the data βh(t). Therefore, we must have Letting s = vt, we obtain u ( vt) =βh(t). u (s) =βh( s v ) and from (12) we obtain u(x, t) =βh(t x v ). This formula gives the solution also in the sector x>vt, t> since the characteristics leaving the x-axis carry zero data and hence we deduce u = c =. This means that the pollutant has not yet reached the point x at time t, ifx>vt. Finally, recalling (9), we find c(x, t) =βh(t x v )e γ v x. Observe that in (, ) there is a jump discontinuity which is transported along the characteristic x = vt.
6 4. Inflow And Outflow Characteristics. A Stability Estimate. The domain in the localized source problem is the quadrant x>, t>. To uniquely determine the solution we have used the initial data on the x-axis, x> and the boundary data on the t-axis, t>. The solution to this problem is uniquely determined. This is due to the fact that, since v>, when time increases, all the characteristics carry the information (the data) towards the interior of the quadrant x>, t>. In other words, the characteristics are inflow characteristics. More generally, consider the equation u t + au x = f(x, t) in the domain x>, t>, where a is constant (a ). The characteristics are the lines x at = constant. (i) If a>, we are in the case of the pollutant model: all the characteristics are inflow and the data must be assigned on both semi-axes. (ii) If a<, the characteristics leaving the x-axis are inflow, while those leaving the t-axis are outflow. In these case the initial data alone are sufficient to uniquely determine the solution, while no data has to be assigned on the semi-axis x =, t>. Coherently, a problem in the half-strip <x<r, t>, besides the initial data, requires a data assigning on the inflow boundary, namely, { u(,t)=h (t) if a>, u(r, t) =h R (t) if a<. The solution to the resulting initial-boundary-value problem is uniquely determined at every point in the strip by its values along the characteristics. Moreover a stability estimate can be proved as follows. Consider, for instance, the case a> and the problem u t + au x = <x<r, t>, (13) u(,t)=h(t) t>, u(x, ) = g(x) <x<r. Multiply the differential equation by u and write =uu t + auu x = 1 d 2 dt u2 + a d 2 dx u2 =. Integrating in x over (,R) we obtain d dt u 2 (x, t)dx + a[u 2 (R, t) u 2 (,t)] =.
7 Now use the data u(,t)=h(t) and the positivity of a to obtain d dt u 2 (x, t)dx ah 2 (t). Integrating in t we have, using the initial condition u(x, ) = g(x), (14) u 2 (x, t)dx g 2 (x, t)dx + a h 2 (s)ds. Now, let u 1 and u 2 be solutions of problem (13) with initial data g 1, g 2 and boundary data h 1, h 2 on x =. Then, by linearity, w = u 1 u 2 is a solution of the problem (13), with initial data h 1 h 2 on x =. Applying the inequality (14) to w, we have [u 1 (x, t) u 2 (x, t)] 2 dx [g 1 (x) g 2 (x)] 2 dx + a [h 1 (s) h 2 (s)] 2 ds. Thus, a least-square approximation of the data controls a least-square approximation of the corresponding solutions. In this sense, the solution of the problem depends continuously on the initial data and on the boundary data on x =. Remark. The values of u on x = R do not appear on (14).