The Buckley-Leverett equation models how oil and water move in a reservoir. The unknownu is the saturation of water,0 ≤ u ≤ 1. The equation is
ut+ f (u)x = 0 where
f (u) = u2
u2+ a(1 − u2)2. Unlike previous examples, the fluxf here is a non-convex function.
5.2 Basic theory
Let consider scalar conservation law
ut+ f (u)x= 0. (2.5)
The equation can be viewed as a directional derivative∂t+ f′(u)∂x ofu is zero. That implies u is constant along the characteristic curve
dx
dt = f′(u(x, t)).
This yields that the characteristic curve is indeed a straight line. Using this we can solve the Cauchy problem of (2.5) with initial datau0implicitly:
u = u0(x − ut).
For instance, for inviscid Burgers’ equation withu0(x) = x, the solution u is given by u = x − ut, oru = x/(1 + t).
Homeworks.
1. If f is convex and u0 is increasing, then the Cauchy problem for equation (2.5) has global solution.
2. Iff is convex and u′0< 0 at some point, then ux→ −∞ at finite time.
The solution may blow up (i.e.|ux| → ∞) in finite time due to the intersection of characteristic curves. A shock wave (discontinuity) is formed. We have to extend our solution class to to include these discontinuous solutions. We can view (2.5) in “weak sense.” That is, for every smooth test functionφ with compact support in R × [0, ∞),
Z ∞
0
Z ∞
−∞
φ[ut+ f (u)x] dx dt = 0
Integrate by part, we obtain Z ∞
0
Z ∞
−∞
[φtu + φxf (u)] dx dt + Z ∞
−∞
φ(x, 0)u(x, 0) dx = 0, (2.6) In this formulation, it allowsu to be discontinuous.
Definition 2.12. A functionu is called a weak solution of (2.5) if it satisfies (2.6) for all smooth test functionφ with compact support in R × [0, ∞).
Lemma 5.1. Suppose u is a weak solution with discontinuity across a curve x(t). Suppose u is smooth on the two sides ofx(t). Then u satisfies the following jump condition across x(t):
dx
dt[u] = [f (u)] (2.7)
where[u] := u(x(t)+, t) − u(x(t)−, t).
Homeworks.Work out this by yourself.
5.2.1 Riemann problem
The Riemann problem is a Cauchy problem of (2.5) with the following initial data u(x, 0) =
uℓ forx < 0
ur forx > 0 (2.8)
The reasons why Riemann problem is important are:
(i) Discontinuities are generic, therefore Riemann problem is generic locally.
(ii) In physical problems, the far field states are usually two constant states. Because of the hyperbolicity, at large time, we expect the solution is a perturbation of solution to the Riemann problem. Therefore, Riemann problem is also generic globally.
(iii) Both the equation (2.5) and the Riemann data (2.8) are invariant under the Galilean transform:
x → λx, t → λt for all λ > 0. If the uniqueness is true, the solution to the Riemann problem is self-similar. That is,u = u(x/t). The PDE problem is then reduced to an ODE problem.
Whenf′′6= 0, say, f′′> 0, here are two important solutions.
1. shock wave:uℓ≥ ur
u(x, t) =
uℓ forx < σt
ur forx > σt (2.9)
whereσ = (f (ur) − f (uℓ))/(ur− uℓ).
5.2. BASIC THEORY 71 2. rarefaction wave: uℓ < ur
u(x, t) =
uℓ forx < λℓt
u forλℓ< λ(u) = xt < λr ur forx > λrt
(2.10)
whereλ(u) = f′(u) is an increasing function.
These two solution are of fundamental importance. We shall denote them by(uℓ, ur).
The weak solution is not unique. For instance, in the case of uℓ < ur, both (2.10) and (2.9) are weak solutions. Indeed, there are infinite many weak solutions to such a Riemann problem.
Therefore, additional condition is needed to guarantee uniqueness. Such a condition is called an entropy condition.
5.2.2 Entropy conditions
To find a suitable entropy condition for general hyperbolic conservation laws, let us go back to study the gas dynamic problems. The hyperbolic conservation laws are simplified equations. The original physical equations usually contain a viscous termνuxx, as that in the Navier-Stokes equa-tion. We assume the viscous equation has uniqueness property. Therefore let us make the following definition.
Definition 2.13. A weak solution is called admissible if it is the limit of
uǫt+ f (uǫ)x= ǫuǫxx, (2.11) asǫ → 0+.
We shall label this condition by (A). In gas dynamics, the viscosity causes the physical entropy increases as gas particles passing through a shock front. One can show that such a condition is equivalent to the admissibility condition. Notice that this entropy increasing condition does not involve viscosity explicitly. Rather, it is a limiting condition asǫ → 0+. This kind of conditions is what we are looking for. For general hyperbolic conservation laws, there are many of them. We list some of them below.
(L) Lax’s entropy condition: across a shock(uℓ, ur) with speed σ, the Lax’s entropy condition is
λℓ> σ > λr (2.12)
whereλℓ(λr) is the left (right) characteristic speed of the shock.
The meaning of this condition is that the information can only enter into a shock, then disap-pear. It is not allowed to have information coming out of a shock. Thus, if we draw character-istic curve from any point(x, t) backward in time, we can always meet the initial axis. It can not stop at a shock in the middle of time because it would violate the entropy condition. In other words, all information can be traced back to initial time. This is a causality property. It is also time irreversible, which is consistent to the second law of thermodynamics. However, Lax’s entropy is only suitable for fluxf with f′′6= 0.
(OL) Oleinik-Liu’s entropy condition: Let
σ(u, v) := f (u) − f (v) u − v . The Oleinik-Liu’s entropy condition is that, across a shock
σ(uℓ, v) ≥ σ(uℓ, ur) (2.13)
for allv between uℓandur. This condition is applicable to nonconvex fluxes.
(GL) The above two conditions are conditions across a shock. Lax proposed another global entropy condition. First, he define entropy-entropy flux: a pair of function(η(u), q(u)) is called an entropy-entropy flux for equation (2.5) A weak solution u(x, t) is said to satisfy entropy condition if for any entropy-entropy flux pair(η, q), u(x, t) satisfies
η(u(x, t))t+ q(u(x, t))x ≤ 0 (2.14) in weak sense.
(K) Another global entropy proposed by Kruzkov is for any constantc, Z ∞
0
Z ∞
−∞[|u − c|φt+ sign(u − c)(f (u) − f (c))φx] dx ≥ 0 (2.15) for all positive smoothφ with compact support inR × (0, ∞). (GL) ⇒ (K):
For anyc, we choose η(u) = |u − c|, which is a convex function. One can check the cor-respondingq(u) = sign(u − c)(f (u) − f (c)). Thus, (K) is a special case of (GL). We may remark here that we can choose even simplier entropy-entropy flux:
η(u) = u ∨ c, q(u) = f (u ∨ c), whereu ∨ c := max{u, c}.
When the flux is convex, each of the above conditions is equivalent to the admissibility condi-tion. Whenf is not convex, each but the Lax’s entropy condition is equivalent to the admissibility condition.
We shall not provide general proof here. Rather, we study special case: the weak solution is only a single shock(uℓ, ur) with speed σ.
Theorem 5.1. Consider the scalar conservation law (2.5) with convex flux f . Let (uℓ, ur) be its shock with speedσ. Then the above entropy conditions are all equivalent.
Proof. (L)⇔ (OL);
We need to assumef to be convex. This part is easy. It follows from the convexity of f . We leave the proof to the reader.
5.2. BASIC THEORY 73 (A)⇔ (OL):
We also need to assumef to be convex. Suppose (uℓ, ur) is a shock. Its speed σ = f (ur) − f (uℓ)
ur− uℓ
.
We shall find a solution of (2.11) such that its zero viscosity limit is(uℓ, ur). Consider a solution haing the formφ((x − σt)/ǫ). In order to have φ → (uℓ, ur), we need to require far field condition:
φ(ξ) →
uℓ ξ → −∞
ur ξ → ∞ (2.16)
Plugφ((x − σt)/ǫ) into (2.11), integrate in ξ once, we obtain
φ′ = F (φ). (2.17)
whereF (u) = f (u) − f (uℓ) − σ(u − uℓ). We find F (uℓ) = F (ur) = 0. This equation with far-field condition (2.16) if and only if, for allu between uℓ andur, (i)F′(u) > 0 when uℓ < ur, or (ii)F′(u) < 0 when uℓ> ur. One can check that (i) or (ii) is equivalent to (OL).
Next, we study global entropy conditions.
(A)⇒ (GL)
Ifu is an admissible solution. This means that it is the limit of uǫ which satisfy the viscous con-servation law (2.11). Let (η, q) be a pair of entropy-entropy flux. Multiply (2.11) by η′(uǫ), we obtain
η(uǫ)t+ q(uǫ)x = ǫη′(uǫ)uǫxx
= ǫη(uǫ)xx− ǫη′′(uǫx)2
≤ ǫη(uǫ)xx
We multiply this equation by any positive smooth test function φ with compact support in R × (0, ∞), then integrate by part, and take ǫ → 0, we obtain
Z ∞
0
Z ∞
−∞
[η(u)φt+ q(u)φx] dx dt ≥ 0 This means thatη(u)t+ q(u)x ≤ 0 in weak sense.
(K)⇒ (OL) for single shock:
Suppose(uℓ, ur) is a shock. Suppose it satisfies (K). We want to show it satisfies (OL). The condi-tion (GL), as applied to a single shock(uℓ, ur), is read as
−σ[η] + [q] ≤ 0.
Here, we chooseη = |u − c|. The condition becomes
−σ(|ur− c| − |uℓ− c|) + sign(ur− c)(f (ur) − f (c)) − sign(uℓ− c)(f (uℓ) − f (c)) ≤ 0 Or
−σ(uℓ, ur)(|ur− c| − |uℓ− c|) + |ur− c|σ(ur, c) − |uℓ− c|σ(uℓ, c) ≤ 0 (2.18) We claim that this condition is equivalent to (OL). First, ifc lies outside of uℓ and ur, then the left-hand side of (2.18) is zero. So (2.18) is always true in this case. Next, ifc lies betrween uℓand ur, one can easily check it is equivalent to (OL).
5.2.3 Rieman problem for nonconvex fluxes
The Oleinik-Liu’s entropy condition can be interpreted as the follows graphically. Suppose(uℓ, ur) is a shock, then the condition (OL) is equivalent to one of the follows. Eitheruℓ > urand the graph off between uℓ, ur lies below the secant(ur, f (ur)), (uℓ, f (uℓ)). Or uℓ < ur and the graph of f between uℓ, ur lies above the secant((uℓ, f (uℓ)), (ur, f (ur))). With this, we can construct the solution to the Riemann problem for nonconvex flux as the follows.
Ifuℓ > ur, then we connect (uℓ, f (uℓ)) and (ur, f (ur)) by a convex envelope of f (i.e. the largest convex function belowf ). The straight line of this envelope corresponds to an entropy shock.
In curved part,f′(u) increases, and this portion corresponds to a centered rarefaction wave. Thus, the solution is a composition of rarefaction waves and shocks. It is called a copmposite wave.
Ifuℓ < ur, we simply replace convex envelope by concave envelope.
Example. Consider the cubic flux:f (u) = 13u3. Ifuℓ< 0, ur > 0 From uℓ, we can draw a line tangent to the graph off at u∗ℓ = −uℓ/2. If ur > u∗ℓ, then the wave structure is a shock(uℓ, u∗ℓ) follows by a rarefaction wave(u∗ℓ, ur). If ur ≤ u∗ℓ, then the wave is a single shock. Notice that in a composite wave, the shock may contact to a rarefaction wave. Such a shock is called a contact shock.
Homeworks.
1. For the fluxf (u) = u3/3, construct the general solution to the Riemann problem for general left/right statesuℓandur.