3 Conservation Laws3.1 MotivationExample 1. (Burgers’ Equation) Consider the initial-value problem for Burgers’ equation,a first-order quasilinear equation of the form(ut+ uux= 0u(x, 0) = φ(x).This equation models wave motion, where u(x, t) is the height of the wave at point x,time t. As described earlier, if φ0(x) < 0, we may have projected characteristic curvesintersecting, resulting in a difficulty in defining the solution beyond the point of intersection.This phenomenon occurs as a result of wave breaking. In order to define solutions which“solve” this problem, we need to determine how to define solutions b eyond this point wherethe projected characteristic curves intersect. In particular, we need to allow for solutionswhich may not even be continuous! In what sense can a function which is not even continuous,let alone differentiable, satisfy a differential equation? We will discuss that momentarily.First, we present another example.¦Example 2. (Traffic Flow Problem) Consider a street starting at point x1and ending atpoint x2. Let u(x, t) be the density of cars at point x, time t. Therefore, the total numberof cars between point x1and x2at time t can be represented byZx2x1u(x, t) dx.Now the rate of change in the number of cars between points x1and x2at time t is given byddtZx2x1u(x, t) dx = f(u(x1, t)) − f(u(x2, t))where f represents the flow rate onto and off the street. Assuming u and f are continuouslydifferentiable functions, we see thatZx2x1ut(x, t) dx = f(u(x1, t)) − f(u(x2, t)),and, therefore,1x2− x1Zx2x1ut(x, t) dx =f(u(x1, t)) − f(u(x2, t))x2− x1.Taking the limit as x2→ x1, we getut= −[f(u)]x.Therefore, we can say that the density of cars at point x at time t satisfies the PDE:ut+ [f(u)]x= 01for some smooth function f. However, this is assuming the density of cars is a continuousfunction. We would like to derive some sort of notion to say that a function u which is noteven differentiable will “solve” the PDE.¦3.2 Weak SolutionsIn this section we introduce the notion of a solution u to a partial differential equation whereu may not even be a differentiable function. This type of solution will be known as a weaksolution of a PDE. First, however, we define the notion of a strong solution of a partialdifferential equation. Consider the initial-value problem for the k-th order PDE,(F (~x, u, Du, . . . , Dku) = 0 x ∈ Rn, t > 0u(~x, 0) = φ(~x)(3.1)We say u is a strong or classical solution of (3.1) if u is k times continuously differentiableand u satisfies (3.1). This is probably the idea you have had in mind when you think ofsolving a PDE. However, as described in the examples above, sometimes a classical solutiondoes not always exist, or you may want to allow for “solutions” which are not differentiable,or even continuous. What do we mean by this?In this section we will define the notion of a weak solution of a first-order, quasilinearinitial-value problem of the form(ut+ [f(u)]x= 0 x ∈ R, t ≥ 0u(x, 0) = φ(x).(3.2)Before doing so, however, we give some preliminary definitions. We say a subset of Rniscompact if it is closed and bounded. We say a function v : Rn→ R has compact supportif v ≡ 0 outside some compact set. Now, we say that u is a weak solution of (3.2) ifZ∞0Z∞−∞[uvt+ f(u)vx] dx dt +Z∞−∞φ(x)v(x, 0) dx = 0 (3.3)for all smooth functions v ∈ C∞(R × [0, ∞)) with compact support.Notice that a function u need not be differentiable or even continuous for it to satisfy(3.3). Functions u which satisfy (3.3) may not be classical solutions of (3.2). However, theyshould satisfy (3.2) in some sense. Where did (3.3) come from? So far, we have just made adefinition. We will now prove that if u is a strong solution of (3.2), then u is a weak solutionof (3.2); that is, u will satisfy (3.3). In this sense, condition (3.3) is a natural extension ofthe notion of a “solution” to (3.2).Theorem 3. If u is a strong solution of (3.2), then u is a weak solution of (3.2).Proof. If u is a classical solution of (3.2), then u is continuously differentiable and(ut+ [f(u)]x= 0, t ≥ 0u(x, 0) = φ(x).2In addition, for any smooth function v : R × [0, ∞) → R with compact support,Z∞0Z∞−∞[ut+ [f(u)]x]v dx dt = 0. (3.4)Integrating (3.4) by parts and using the fact that v vanishes at infinity, we see thatZ∞0Z∞∞[uvt+ f(u)vx] dx dt +Z∞−∞φ(x)v(x, 0) dx = 0.But this is true for all functions v ∈ C∞(R ×[0, ∞)) with compact support. Therefore, u isa weak solution of (3.2).As mentioned above, the notion of weak solution allows for solutions u which need noteven be continuous. However, weak solutions u have some restrictions on types of disconti-nuities, etc. For example, suppose u is a weak solution of (3.2) such that u is discontinuousacross some curve x = ξ(t), but u is smooth on either side of the curve. Let u−(x, t) be thelimit of u approaching (x, t) from the left and let u+(x, t) be the limit of u approaching (x, t)from the right. We claim that the curve x = ξ(t) cannot be arbitrary, but rather there is arelation between x = ξ(t), u−and u+.xtx=ξ(t)u=uu=u+-Theorem 4. If u is a weak solution of (3.2) such that u is discontinuous across the curvex = ξ(t) but u is smooth on either side of x = ξ(t), then u must satisfy the conditionf(u−) − f(u+)u−− u+= ξ0(t) (3.5)across the curve of discontinuity, where u−(x, t) is the limit of u approaching (x, t) from theleft and u+(x, t) is the limit of u approaching (x, t) from the right.Proof. If u is a weak solution of (3.2), thenZ∞0Z∞−∞[uvt+ f(u)vx] dx dt +Z∞−∞φ(x)v(x, 0) dx = 0for all smo oth functions v ∈ C∞(R × [0, ∞)) with compact support. Let v be a smoothfunction such that v(x, 0) = 0, and break up the first integral into the regions Ω−, Ω+whereΩ−≡ {(x, t) : 0 < t < ∞, −∞ < x < ξ(t)}Ω+≡ {(x, t) : 0 < t < ∞, ξ(t) < x < +∞}.3Therefore,0 =Z∞0Z∞−∞[uvt+ f(u)vx] dx dt +Z∞−∞φ(x)v(x, 0) dx=ZZΩ−[uvt+ f(u)vx] dx dt +ZZΩ+[uvt+ f(u)vx] dx dt.(3.6)Combining the Divergence Theorem with the fact that v has compact support and v(x, 0) =0, we haveZZΩ−[uvt+f(u)vx]dx dt=−ZZΩ−[ut+ (f(u))x]v dx dt+Zx=ξ(t)[u−vν2+f(u−)vν1]ds(3.7)where ν = (ν1, ν2) is the outward unit normal to Ω−.xtx=ξ(t)u=uu=u+-ΩΩ−+νν12( , )Similarly, we see thatZZΩ+[uvt+ f(u)vx] dx dt = −ZZΩ+[ut+ (f(u))x]v dx dt −Zx=ξ(t)[u+vν2+ f(u+)vν1] ds (3.8)By assumption, u is a weak