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https://onlinelibrary.wiley.com/doi/full/10.1002/zamm.201700251
DOI: 10.1002/zamm.201700251
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DOI: 10.1002/zamm.201700251
O R I G I N A L PA P E R
Delta shock waves for a system of Keyfitz-Kranzer type
Richard De la cruz
1Marcelo Santos
21School of Mathematics and Statistics,
Universidad Pedagógica y Tecnológica de Colombia, Av. Central del Norte 39-115, Tunja, Boyacá, Colombia
2Institute of Mathematics, Statistics and
Scientific Computing – IMECC, State University of Campinas – UNICAMP, Rua Sérgio Buarque de Holanda 651, Campinas, SP, Brazil
Correspondence
Richard De la cruz, School of Mathematics and Statistics, Universidad Pedagógica y Tecnológica de Colombia, Av. Central del Norte 39-115, Tunja, Boyacá, Colombia. Email: richard.delacruz@uptc.edu.co Funding information
Fundação de Amparo à Pesquisa do Estado de São Paulo, Grant/Award Number: 2016/ 19502-4
This paper is concerned with a hyperbolic system of conservation laws of Keyfitz-Kranzer type. We show the existence of a delta shock wave solution using the vanishing viscosity method. The obtained solution satisfies the generalized Rankine-Hugoniot relation and the entropy condition. As a consequence, we have also a uniqueness result for the obtained solution.
K E Y W O R D S
delta shock wave, keyfitz–kranzer type, vanishing viscosity method
1
INTRODUCTION
Many natural phenomena are modeled by systems of Keyfitz-Kranzer type {
𝜌𝑡+ (𝜌𝜙(𝜌, 𝑢))𝑥= 0,
(𝜌𝑢)𝑡+ (𝜌𝑢𝜙(𝜌, 𝑢))𝑥= 0, (KK)
where𝜙(𝜌, 𝑢) = 𝑓(𝑢) − 𝑃 (𝜌). For example, when 𝑓(𝑢) = 𝑢 the system (KK) was introduced as a macroscopic model for traffic flow by Aw and Rascle[1]where𝜌 and 𝑢 are, respectively, the density and the velocity of cars on a roadway, and 𝑃 is a smooth and strictly increasing function satisfying
𝜌𝑃′′(𝜌) + 2𝑃′(𝜌) > 0 for 𝜌 > 0.
Another important study on the system (KK) is due to Blake Temple. In 10, he consider the case when𝜌𝜙(𝜌, 𝑢) is not a convex function (in this case, the system describes how the addition of a polymer affects the flow of water and oil in a reservoir) and prove the existence of a global weak solution to the Cauchy problem. Lu[8]showed the existence of a global weak solution of the Cauchy problem for (KK) when𝑓 is a non negative convex function and 𝑃 satisfies the following condition:
𝑃 (𝜌) ≤ 0 for 𝜌 > 0, 𝑃 (0) = 0, lim 𝜌→0𝜌𝑃
′(𝜌) = 0, lim
𝜌→∞𝑃 (𝜌) = ∞, 𝜌𝑃′′(𝜌) + 2𝑃′(𝜌) < 0 for 𝜌 > 0.
Using delta-shock waves, H. Cheng[3]solved the Riemann problem to the non symmetric system of Keyfitz-Kranzer type (KK) when 𝑃 (𝜌) is the function 𝑃 (𝜌) = −1𝜌,𝜌 > 0, and 𝑓(𝑢) is the indentity function, i.e. 𝑓(𝑢) = 𝑢 for all 𝑢 ∈ ℝ. In this case, 𝑃 satisfies the condition𝜌𝑃′′(𝜌) + 2𝑃′(𝜌) = 0 and it is not possible to solve the Riemann problem to (KK) using only classical
Z Angew Math Mech. 2019;99:e201700251. www.zamm-journal.org © 2018 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim 1 of 21 https://doi.org/10.1002/zamm.201700251
waves. In this paper, we generalize the result of 3 for an arbitrary strictly increasing function𝑓(𝑢). More precisely, we solve the Riemann problem for the following system,
⎧ ⎪ ⎨ ⎪ ⎩ 𝜌𝑡+ (𝜌𝑓(𝑢))𝑥= 0, (𝜌𝑢)𝑡+(𝜌𝑢(𝑓(𝑢) +1𝜌)) 𝑥= 0, (1)
where𝑓 ∈ 𝐶1(ℝ), with 𝑓′(𝑢) > 0 for all 𝑢 ∈ ℝ, and 𝜌 > 0.
The concept of delta shock wave is a generalization of a classical shock wave. This generalization was introdduced by Korchin-ski in the year of 1977 in his PhD thesis.[6]Motivated by some numerical results, he constructed the unique Riemann solution using generalized delta functions to obtain singular shocks in the sense of distributions. In 1994, Tan, Zhang and Zheng estab-lished in 9 the existence, uniqueness and stability of delta shock waves for a viscous perturbation of the system studied by Korchinski. Other works dealing with delta shock waves are due to Ercole[5]who in 2000 obtained a delta shock solution as
a limit of smooth solutions by the vanishing viscosity method. In 2005, Brenier[2] considered the Riemann problem for the Chaplygin gas system and he showed the existence of solutions with concentration. Some more recent works on delta shocks for general hyperbolic conservation laws are due to Danilov and Shelkovich,[4]where they described delta shock wave generation from continuous initial data by using smooth approximations in the weak sense.
Using the vanishing viscosity method, and following works by Tan, Zhang and Zheng,[9]Li, Zhang and Yang,[7]and Yang,[11]
we show the existence of a delta shock wave solution for the system (1). The main difficulty in applying the vanishing viscosity method is to choose a suitable Banach space and a bounded convex closed subset to use the Schauder fixed point theorem and, thus, obtain the solution of the viscous system (7). Passing to the limit when the viscosity tends to zero in solutions of the system (7), we obtain the existence of a delta shock wave for the system (1). Moreover, adding an entropy condition, we establish the uniqueness of our solution.
The remainder of this paper is organized as follows. In Section 2, we present the classical solutions and a short description of the delta shock wave solution for the Riemann problem for the system of Keyfitz-Kranzer type (1). In Section 3, we solve the viscous system (7). Also, in this Section, we obtain the existence and uniqueness of a solution for the system (1) involving delta shock waves.
2
RIEMANN PROBLEM
In this section, we consider the system (1) with an initial data (𝜌(𝑥, 0), 𝑢(𝑥, 0)) =
{
(𝜌−, 𝑢−), if 𝑥 < 0
(𝜌+, 𝑢+), if 𝑥 > 0 (2) for arbitrary constant states(𝜌±, 𝑢±).
The eigenvalues of the system (1), with respective right eigenvectors, are given by
𝜆1(𝜌, 𝑢) = 𝑓(𝑢), 𝐫1= (1, 0) and 𝜆2(𝜌, 𝑢) = 𝑓(𝑢) + 1𝜌, 𝐫2= (𝑓′(𝑢), 1∕𝜌2).
Notice that both characteristics are linearly degenerate and thus the system is in the Temple class. More specifically, shock wave curves coincide with rarefaction wave curves in the phase plane(𝜌, 𝑢), corresponding to the so called contact discontinuities in the(𝑥, 𝑡) plane. Since our system is a 2 × 2 system, we have two families of such curves in the (𝜌, 𝑢) plane. In our case, these families through a point(𝜌−, 𝑢−) are given by
Family 1: it is the vertical line𝑢 = 𝑢−,𝜌 > 0;
Family 2: it consists of all(𝜌, 𝑢) such that 𝑓(𝑢) +1𝜌 = 𝑓(𝑢+) +𝜌1
+,𝜌 > 0.
We shall denote these families, respectively, by𝐽1≡ 𝐽1(𝜌−, 𝑢−) and 𝐽2≡ 𝐽2(𝜌−, 𝑢−). It is a straightforward computation to verify that the shock and rarefactions wave curves for the system (1) through a point(𝜌−, 𝑢−), which coincide, are given by 𝐽1 and𝐽2, and so we omit this computation here.
F I G U R E 1 Phase plane𝜌 − 𝑢. Regions I, II, III, IV and V. 𝐽1, 𝐽2: contact discontinuites;𝐽2:𝑓(𝑢) +1 𝜌 = 𝑓(𝑢−) + 1 𝜌−.𝑆𝛿:𝑓(𝑢) + 1 𝜌 = 𝑓(𝑢−)
2.1
Classical Riemann problem
Consider the left constant state(𝜌−, 𝑢−). The phase plane is divided into five regions as follows (see Figure 1) • 𝐼 = {(𝜌, 𝑢) ∶ 𝑢−< 𝑢 < 𝑓−1(𝑓(𝑢−) +𝜌1 −), 1 𝑓(𝑢−)+𝜌−1 −𝑓(𝑢) < 𝜌 < ∞} • II= {(𝜌, 𝑢) ∶ 𝑢−< 𝑢 < 𝑓−1(𝑓(𝑢−) +𝜌1 −), 0 < 𝜌 < 1 𝑓(𝑢−)+𝜌−1−𝑓(𝑢)} ∪ {(𝜌, 𝑢) ∶ 𝑓 −1(𝑓(𝑢 −) +𝜌1−) ≤ 𝑢 < ∞, 0 < 𝜌 < ∞} • III= {(𝜌, 𝑢) ∶ −∞ < 𝑢 < 𝑢−, 0 < 𝜌 < 1 𝑓(𝑢−)+𝜌−1 −𝑓(𝑢)} • IV= {(𝜌, 𝑢) ∶ −∞ < 𝑢 < 𝑢−, 1 𝑓(𝑢−)+𝜌−1 −𝑓(𝑢) < 𝜌 < 1 𝑓(𝑢−)−𝑓(𝑢)} • V= {(𝜌, 𝑢) ∶ −∞ < 𝑢 < 𝑢−,𝑓(𝑢 1 −)−𝑓(𝑢) < 𝜌 < ∞}
For fixed(𝜌−, 𝑢−), we consider the family of curves
= {𝐽2(𝜌∗, 𝑢∗) ∶ (𝜌∗, 𝑢∗) ∈ 𝐽1(𝜌−, 𝑢−)}.
Then the solution to the Riemann problem is given when we connect(𝜌∗, 𝑢∗) to (𝜌−, 𝑢−) by a 1-contact discontinuity, denoted by
𝐽1, and we connect(𝜌+, 𝑢+) to (𝜌∗, 𝑢∗) by a 2-contact discontinuity, 𝐽2. When(𝜌+, 𝑢+) belongs to I,II, III or IV, one can get the solution consisting of two different (or just one) contact discontinuities (see Figure 2). The intermediate state(𝜌∗, 𝑢∗) connecting two contact discontinuities satisfies
𝑢∗= 𝑢− and 𝑓(𝑢∗) + 1
𝜌∗ = 𝑓(𝑢+) + 1𝜌+.
In this case we have that𝑓(𝑢−) < 𝑓(𝑢+) +𝜌1
+ and the solution to the Riemann problem (1)–(2) is given by
(𝜌(𝑥, 𝑡), 𝑢(𝑥, 𝑡)) = ⎧ ⎪ ⎨ ⎪ ⎩ (𝜌−, 𝑢−), if 𝑥 < 𝜆1(𝜌−, 𝑢−)𝑡, (𝜌∗, 𝑢∗), if 𝜆1(𝜌−, 𝑢−)𝑡 < 𝑥 < 𝜆2(𝜌+, 𝑢+)𝑡, (𝜌+, 𝑢+), if 𝑥 > 𝜆2(𝜌+, 𝑢+)𝑡, where 𝑢∗= 𝑢− and 𝑓(𝑢∗) + 1 𝜌∗ = 𝑓(𝑢+) + 1𝜌+.
We shall denote the common boundary between regions𝐼𝑉 and 𝑉 by 𝑆𝛿. The reason for this notation is that for right states (𝜌+, 𝑢+) in region 𝑉 we cannot connect (𝜌−, 𝑢−) and (𝜌+, 𝑢+) by the classical shock waves 𝐽1and𝐽2. Thus, it is for right states
F I G U R E 2 𝜆1(𝜌−, 𝑢−) < 𝜆2(𝜌+, 𝑢+) and Riemann solution involving two contact discontinuities
F I G U R E 3 Characteristic analysis of the delta shock wave
in region𝑉 that we must use delta shock waves to solve the Riemman problem for the system (1). Notice that the curve 𝑆𝛿 has the line𝑢 = 𝑢−as its asymptotic line. When (𝜌+, 𝑢+) ∈ 𝑉 , we have that 𝑓(𝑢+) +𝜌1
+ = 𝜆2(𝜌+, 𝑢+) ≤ 𝜆1(𝜌−, 𝑢−) = 𝑓(𝑢−).
The characteristic lines from initial data will overlap in a domainΩ = {(𝑥, 𝑡) ∶ 𝜆2(𝜌+, 𝑢+)𝑡 ≤ 𝑥 ≤ 𝜆2(𝜌+, 𝑢+)𝑡, 𝑡 > 0} shown in Figure 3. So singularity must happen inΩ.
Suppose(𝜌+, 𝑢+) ∈ 𝑉 . Let (𝜌∗, 𝑢∗) be any point on 𝐽1(𝜌−, 𝑢−). For the point (𝜌+, 𝑢′) on 𝐽2(𝜌∗, 𝑢∗), we have
𝑓(𝑢′) + 1
𝜌+ = 𝑓(𝑢∗) + 1𝜌∗ = 𝑓(𝑢−) + 1𝜌∗ ≥ 𝑓(𝑢+) + 1𝜌+ + 1𝜌∗.
Thus,
𝑓(𝑢′) ≥ 𝑓(𝑢+) + 1𝜌
∗, (𝜌∗> 0),
and(𝜌+, 𝑢+) cannot lie on any curve in , the singularity is impossible to be a jump with finite amplitude because the Rankine-Hugoniot relation is not satisfied on the bounded jump.
2.2
Delta shock solution
In this section, we discuss the case when𝜆1(𝜌−, 𝑢−) ≥ 𝜆2(𝜌+, 𝑢+). If 𝜆1(𝜌−, 𝑢−) ≥ 𝜆2(𝜌+, 𝑢+), then (𝜌+, 𝑢+) ∈ 𝑉 . The charac-teristic lines from the initial data will overlap in a domainΩ = {(𝑥, 𝑡) ∶ 𝑓(𝑢+) +𝜌1
+ ≤ 𝑥∕𝑡 ≤ 𝑓(𝑢−), 𝑡 > 0}. Hence, the
singu-larity will develop inΩ, while this singularity cannot be a jump with a finite amplitude. To analyze the singularity in Ω for the special case𝜆2(𝜌+, 𝑢+) = 𝜆1(𝜌−, 𝑢−), let us consider the limit of the solution 𝜌(𝜉) and 𝑢(𝜉) when 𝜌−,𝑢−and𝜌+are fixed,
𝑢+→ 𝑓−1(𝑓(𝑢
−) −𝜌1+) + 0 and the right state (𝜌+, 𝑢+) belongs to IV (in this situation the solution is given by 𝐽1+ 𝐽2). The
intermediate state(𝜌∗, 𝑢∗) determined by the intersection point of 𝐽1and𝐽2satisfies
𝑓(𝑢−) = 𝑓(𝑢∗) and 𝑓(𝑢+) + 1𝜌
+ = 𝑓(𝑢∗) + 1𝜌∗.
Notice that defining𝑎 = 𝑓−1(𝑓(𝑢−) −𝜌1
+), we have that
lim
𝑢+→𝑎+0𝜌∗= ∞ and lim𝑢+→𝑎+0𝑢∗= 𝑢−< ∞.
Now let us calculate the total quantities of𝜌 and 𝑢 between 𝐽1and𝐽2as𝜌−,𝑢−and𝜌+are fixed,𝑢+→ 𝑓−1(𝑓(𝑢−) −𝜌1
+) + 0.
From the first Equation of 1, with𝜉 = 𝑥∕𝑡, it follows that
0 = 𝜉=𝑓(𝑢+)+𝜌+1 +0 ∫ 𝜉=𝑓(𝑢−)−0 (−𝜉𝑑𝜌 + 𝑑 (𝜌𝑓(𝑢))) = (−𝜉𝜌)|𝜉=𝑓(𝑢+)+ 1 𝜌++0 𝜉=𝑓(𝑢−)−0 + 𝜉=𝑓(𝑢+)+𝜌+1 +0 ∫ 𝜉=𝑓(𝑢−)−0 𝜌𝑑𝜉 + (𝜌𝑓(𝑢))|𝜉=𝑓(𝑢+)+ 1 𝜌++0 𝜉=𝑓(𝑢−)−0 so lim 𝑢+→𝑎+0 𝜉=𝑓(𝑢+)+𝜌+1+0 ∫ 𝜉=𝑓(𝑢−)−0 𝜌(𝜉)𝑑𝜉 = 𝑓(𝑢−)+0 ∫ 𝑓(𝑢−)−0 𝜌(𝜉)𝑑𝜉 ≠ 0.
The last equality shows that𝜌(𝜉) has the same singularity as a weighted Dirac delta function at 𝜉 = 𝑓(𝑢−). We shall call this type of nonlinear hyperbolic waves a delta shock wave. For the case𝜆2(𝜌+, 𝑢+) < 𝜆1(𝜌−, 𝑢−), we suggest that the solution of the Riemann problem is also a delta shock wave defined by the speed𝜎 satisfying
𝜆2(𝜌+, 𝑢+) ≤ 𝜎 ≤ 𝜆1(𝜌−, 𝑢−). (3) Next, we recall the following definition:
Definition 1. A two-dimensional weighted delta function𝑤(𝑠)𝛿𝐿supported on a smooth curve𝐿 = {(𝑥(𝑠), 𝑡(𝑠)) ∶ 𝑎 < 𝑠 < 𝑏}, for𝑤 ∈ 𝐿1((𝑎, 𝑏)), is defined as ⟨𝑤(⋅)𝛿𝐿, 𝜙(⋅, ⋅)⟩ = 𝑏 ∫ 𝑎 𝑤(𝑠)𝜙(𝑥(𝑠), 𝑡(𝑠)) 𝑑𝑠, 𝜙 ∈ 𝐶∞ 0 (ℝ × [0, ∞)).
Now, we define a delta shock wave solution for the system (1) with initial data (2).
Definition 2. A distribution pair(𝜌, 𝑢) is a delta shock wave solution of (1) and (2) in the sense of distribution if there exist a smooth curve𝐿 and a function 𝑤 ∈ 𝐶1(𝐿) such that 𝜌 and 𝑢 are represented in the following form
𝜌 = ̃𝜌(𝑥, 𝑡) + 𝑤𝛿𝐿and𝑢 = ̃𝑢(𝑥, 𝑡),
̃𝜌, ̃𝑢 ∈ 𝐿∞(ℝ × (0, ∞); ℝ) and
{
⟨𝜌, 𝜑𝑡⟩ + ⟨𝜌𝑓(𝑢), 𝜑𝑥⟩ = 0,
⟨𝜌𝑢, 𝜑𝑡⟩ + ⟨𝜌𝑢𝑓(𝑢), 𝜑𝑥⟩ + ∫0∞∫ℝ𝑢𝜙𝑥𝑑𝑥𝑑𝑡 = 0, (4)
for all the test functions𝜑 ∈ 𝐶0∞(ℝ × (0, ∞)), where 𝑢|𝐿= 𝑢𝛿(𝑡) and
⟨𝜌, 𝜑⟩ = ∞ ∫ 0 ∫ℝ ̃𝜌𝜑 𝑑𝑥𝑑𝑡 + ⟨𝑤𝛿𝐿, 𝜑⟩, ⟨𝜌𝐺(𝑢), 𝜑⟩ = ∞ ∫ 0 ∫ℝ ̃𝜌𝐺(̃𝑢)𝜑 𝑑𝑥𝑑𝑡 + ⟨𝑤𝐺(𝑢𝛿)𝛿𝐿, 𝜑⟩.
With the previous definitions, we are going to find a solution with discontinuity𝑥 = 𝑥(𝑡) for (1) of the form (𝜌(𝑥, 𝑡), 𝑢(𝑥, 𝑡)) = ⎧ ⎪ ⎨ ⎪ ⎩ (𝜌−(𝑥, 𝑡), 𝑢−(𝑥, 𝑡)), if 𝑥 < 𝑥(𝑡), (𝑤(𝑡)𝛿𝐿, 𝑢𝛿(𝑡)), if𝑥 = 𝑥(𝑡), (𝜌+(𝑥, 𝑡), 𝑢+(𝑥, 𝑡)), if 𝑥 > 𝑥(𝑡), (5)
where𝜌±(𝑥, 𝑡), 𝑢±(𝑥, 𝑡) are piecewise smooth solutions of system (1), 𝛿(⋅) is the Dirac measure supported on the curve 𝑥(𝑡) ∈ 𝐶1, and𝑥(𝑡), 𝑤(𝑡) and 𝑢𝛿(𝑡) are to be determined.
We define 1 𝜌∶= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1 𝜌−, if 𝑥 < 𝑥(𝑡) 0, if𝑥 = 𝑥(𝑡) 1 𝜌+, if 𝑥 > 𝑥(𝑡).
Now, we have the following theorem:
Theorem 1. If the curves𝑥(𝑡), 𝑤(𝑡) and 𝑢𝛿(𝑡) solve
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 𝑑𝑥(𝑡) 𝑑𝑡 = 𝑓(𝑢𝛿(𝑡)), 𝑑𝑤(𝑡) 𝑑𝑡 = −𝑓(𝑢𝛿(𝑡))[𝜌] + [𝜌𝑓(𝑢)], 𝑑(𝑤(𝑡)𝑢𝛿(𝑡)) 𝑑𝑡 = −𝑓(𝑢𝛿(𝑡))[𝜌𝑢] + [ 𝜌𝑢 ( 𝑓(𝑢) + 1 𝜌 )] (6)
then the solution(𝜌(𝑥, 𝑡), 𝑢(𝑥, 𝑡)) defined in (5) satisfies (1) in the sense of distributions.
The relations given by (6) are called the generalized Rankine-Hugoniot conditions.
Proof. If Equations 6 holds, then, for any test functions𝜑 ∈ 𝐶0∞(ℝ × (0, ∞)), we obtain
⟨𝜌𝑢, 𝜑𝑡⟩ + ⟨𝜌𝑢𝑓(𝑢), 𝜑𝑥⟩ + ∞ ∫ 0 ∫ℝ 𝑢𝜙𝑥𝑑𝑥𝑑𝑡 = ∞ ∫ 0 ∫ℝ (𝜌𝑢𝜑𝑡+ 𝜌𝑢 ( 𝑓(𝑢) + 1 𝜌 ) 𝜑𝑥)𝑑𝑥𝑑𝑡 + ∞ ∫ 0 (𝑤(𝑡)𝑢𝛿(𝑡)𝜑𝑡+ 𝑤(𝑡)𝑢𝛿(𝑡)𝑓(𝑢𝛿(𝑡))𝜑𝑥)𝑑𝑡 = ∞ ∫ 0 𝑥(𝑡) ∫ −∞ (𝜌−𝑢−𝜑𝑡+ 𝜌−𝑢− ( 𝑓(𝑢−) + 1𝜌 − ) 𝜑𝑥)𝑑𝑥𝑑𝑡 + ∞ ∫ 0 ∞ ∫ 𝑥(𝑡) (𝜌+𝑢+𝜑𝑡+ 𝜌+𝑢+ ( 𝑓(𝑢+) + 1𝜌 + ) 𝜑𝑥)𝑑𝑥𝑑𝑡 + ∞ ∫ 0 𝑤(𝑡)𝑢𝛿(𝑡)(𝜑𝑡+ 𝑓(𝑢𝛿(𝑡))𝜑𝑥)𝑑𝑡 = − ∮ − ( 𝜌−𝑢− ( 𝑓(𝑢−) + 1 𝜌− ) 𝜑 ) 𝑑𝑡 + (𝜌−𝑢−𝜑)𝑑𝑥
+ ∮ − ( 𝜌+𝑢+ ( 𝑓(𝑢+) + 1 𝜌+ ) 𝜑 ) 𝑑𝑡 + (𝜌+𝑢+𝜑)𝑑𝑥 − ∞ ∫ 0 𝜑𝑑(𝑤(𝑡)𝑢𝛿(𝑡)) 𝑑𝑡 𝑑𝑡 = ∞ ∫ 0 𝜑 ( −𝑓(𝑢𝛿)[𝜌𝑢] + [𝜌𝑢𝜙(𝜌, 𝑢)] −𝑑(𝑤(𝑡)𝑢𝛿(𝑡)) 𝑑𝑡 ) 𝑑𝑡 = 0
which implies the second Equation of 4. A completely similar argument leads to the first Equation of 4. □ Thereby, the Riemann problem is reduced to find𝑥(𝑡), 𝑤(𝑡) and 𝑢𝛿(𝑡) such that they solve (6) with the initial data (at 𝑡 = 0):
𝑥(0) = 0, 𝑤(0) = 0, 𝑢𝛿(0) = 0.
Observe that if𝑢𝛿(𝑡) is a constant, say 𝑢𝛿, then by (6) we have that𝑥(𝑡) and 𝑤(𝑡) are linear functions of 𝑡. In the next section, we will utilize the vanishing viscosity method to show that𝑢𝛿(𝑡) is a constant and therefore we obtain existence of delta shock waves. Moreover, we show uniqueness of delta shock wave under the entropy condition (3).
3
VANISHING VISCOSITY METHOD AND DELTA SHOCK WAVE
In this section we only focus on the study of the case𝑓(𝑢−) > 𝑓(𝑢+) + 1∕𝜌+. We consider the following problem {𝜌
𝑡+ (𝜌𝑓(𝑢))𝑥= 0,
(𝜌𝑢)𝑡+ (𝜌𝑢(𝑓(𝑢) +1𝜌)𝑥= 𝜀𝑡𝑢𝑥𝑥, (7)
with initial data
(𝜌(𝑥, 0), 𝑢(𝑥, 0)) = {
(𝜌−, 𝑢−), if 𝑥 < 0,
(𝜌+, 𝑢+), if 𝑥 > 0. (8) By the self-similar transformation𝜉 = 𝑥∕𝑡, we have
{
−𝜉𝜌𝜉+ (𝜌𝑓(𝑢))𝜉= 0,
−𝜉(𝜌𝑢)𝜉+ (𝜌𝑢(𝑓(𝑢) +1𝜌))𝜉= 𝜀𝑢𝜉𝜉 (9) with data at infinity given by
(𝜌(±∞), 𝑢(±∞)) = (𝜌±, 𝑢±). (10)
3.1
Existence of solutions to the viscous system (9)–(10)
Let𝑅1and𝑅2be positive numbers such that𝑅1≥ |𝑓(𝑢−)| and 𝑅2≥ |𝑓(𝑢+) +𝜌1
+|. Then, for 𝑅 ≥ max{𝑅1, 𝑅2, |𝑢−|, |𝑢+|} we
consider the Banach space𝐶([−𝑅, 𝑅]), endowed with the supremum norm, and we take the set 𝐾 given by
𝐾 = {𝑈 ∈ 𝐶([−𝑅, 𝑅]) | 𝑈 is monotone decreasing with 𝑈(−𝑅) = 𝑢−and𝑈(𝑅) = 𝑢+} which is bounded and a convex closed set in𝐶([−𝑅, 𝑅]).
Lemma 1. Suppose𝑈 ∈ 𝐾 ∩ 𝐶1([−𝑅, 𝑅]). Let
𝜌(𝜉) =
{
𝜌1(𝜉), if − 𝑅 ≤ 𝜉 < 𝜉𝜎,
where𝜉𝜎 is the unique solution of the equation 𝑓(𝑈(𝜉𝜎)) = 𝜉𝜎 (which solution exists because𝑓(𝑢−) > 𝑓(𝑢+) and 𝑅 is big enough), 𝜌1(𝜉) ∶= 𝜌−𝑓(𝑈(𝜉)) − 𝜉𝑓(𝑢−) + 𝑅 exp⎛⎜⎜ ⎝ − 𝜉 ∫ −𝑅 𝑑𝑠 𝑓(𝑈(𝑠)) − 𝑠 ⎞ ⎟ ⎟ ⎠ (12) and 𝜌2(𝜉) ∶= 𝜌+𝜉 − 𝑓(𝑈(𝜉))𝑅 − 𝑓(𝑢+) exp⎛⎜⎜ ⎝ 𝑅 ∫ 𝜉 𝑑𝑠 𝑓(𝑈(𝑠)) − 𝑠 ⎞ ⎟ ⎟ ⎠ . (13)
Then𝜌 ∈ 𝐿1([−𝑅, 𝑅]), 𝜌 is continuous in [−𝑅, 𝜉𝜎) ∪ (𝜉𝜎, 𝑅] and it is a weak solution for
−𝜉𝜌𝜉+ (𝜌𝑓(𝑈))𝜉= 0, (14)
and𝜌(±𝑅) = 𝜌±.
Proof. The Equation 14 can be rewritten as
(𝑓(𝑈(𝜉)) − 𝜉)𝜌′+ 𝜌(𝑓(𝑈(𝜉)))′= 0. (15)
Integrating (15) on[−𝑅, 𝜉] for −𝑅 < 𝜉 < 𝜉𝜎, we get
(𝑓(𝑈(𝜉)) − 𝜉)𝜌1(𝜉) − (𝑓(𝑢−) + 𝑅)𝜌−+ 𝜉 ∫ −𝑅 𝜌1(𝑠)𝑑𝑠 = 0. (16) Let 𝑝(𝜉) = 𝜉 ∫ −𝑅 𝜌1(𝑠)𝑑𝑠, 𝐴1= (𝑓(𝑢−) + 𝑅)𝜌− and𝑎(𝜉) = (𝑓(𝑈(𝜉)) − 𝜉). Then (16) can be written as
{ 𝑎(𝜉)𝑝′(𝜉) + 𝑝(𝜉) = 𝐴 1, 𝑝(−𝑅) = 0. It follows that 𝑝(𝜉) = 𝐴1 ⎧ ⎪ ⎨ ⎪ ⎩ 1 − exp⎛⎜⎜ ⎝ − 𝜉 ∫ −𝑅 𝑑𝑠 𝑎(𝑠) ⎞ ⎟ ⎟ ⎠ ⎫ ⎪ ⎬ ⎪ ⎭ Noting that𝑎(𝜉) > 0 and 𝑎(𝜉) = 𝑂(|𝜉 − 𝜉𝜎|) as 𝜉 → 𝜉𝜎−, we obtain
lim 𝜉→𝜉𝜎− 𝜉 ∫ −𝑅 𝜌1(𝑠)𝑑𝑠 = lim 𝜉→𝜉𝜎−𝑝(𝜉) = 𝐴1. (17) Hence lim 𝜉→𝜉𝜎−(𝑓(𝑈(𝜉)) − 𝜉)𝜌1(𝜉) = 0. (18)
Similarly, one can get lim 𝜉→𝜉𝜎+ 𝑅 ∫ 𝜉 𝜌2(𝑠)𝑑𝑠 = 𝐴2, (19) lim 𝜉→𝜉𝜎+(𝑓(𝑈(𝜉)) − 𝜉)𝜌2(𝜉) = 0,
where𝐴2= (𝑓(𝑢+) − 𝑅)𝜌+. The equalities (17) and (19) imply that𝜌(𝜉) ∈ 𝐿1([−𝑅, 𝑅]). Now, for arbitrary𝜙 ∈ 𝐶0∞([−𝑅, 𝑅]), we verify that
𝐼 ≡ − 𝑅 ∫ −𝑅 (𝑓(𝑈(𝜉)) − 𝜉)𝜌(𝜉)𝜙′(𝜉)𝑑𝜉 + 𝑅 ∫ −𝑅 𝜌(𝜉)𝜙(𝜉)𝑑𝜉 = 0.
For any𝜉1, 𝜉2, such that−𝑅 < 𝜉1< 𝜉𝜎< 𝜉2< 𝑅 we can write 𝐼 = 𝐼1+ 𝐼2+ 𝐼3, where
𝐼1= 𝜉1 ∫ −𝑅 (−(𝑓(𝑈(𝜉)) − 𝜉)𝜌(𝜉)𝜙′(𝜉) + 𝜌(𝜉)𝜙(𝜉))𝑑𝜉, 𝐼2= 𝜉2 ∫ 𝜉1 (−(𝑓(𝑈(𝜉)) − 𝜉)𝜌(𝜉)𝜙′(𝜉) + 𝜌(𝜉)𝜙(𝜉))𝑑𝜉 and 𝐼3= 𝑅 ∫ 𝜉2 (−(𝑓(𝑈(𝜉)) − 𝜉)𝜌(𝜉)𝜙′(𝜉) + 𝜌(𝜉)𝜙(𝜉))𝑑𝜉. Observe that |𝐼1| =|||| || | −(𝑓(𝑈(𝜉1)) − 𝜉1)𝜌1(𝜉1)𝜙(𝜉1) + 𝜉1 ∫ −𝑅 (((𝑓(𝑈(𝜉)) − 𝜉)𝜌(𝜉))′𝜙(𝜉) + 𝜌(𝜉)𝜙(𝜉)))𝑑𝜉|||| || | = ||(𝑓(𝑈(𝜉1)) − 𝜉1)𝜌1(𝜉1)𝜙(𝜉1)|| By (18), we have that lim 𝜉1→𝜉𝜎−|𝐼1| = lim𝜉1→𝜉𝜎−||(𝑓(𝑈(𝜉1)) − 𝜉1)𝜌1(𝜉1)𝜙(𝜉1)|| = 0.
In similar way, we show that
lim 𝜉2→𝜉𝜎+|𝐼3| = lim𝜉2→𝜉𝜎+||(𝑓(𝑈(𝜉2)) − 𝜉2)𝜌2(𝜉2)𝜙(𝜉2)|| = 0. Since𝜌 ∈ 𝐿1([−𝑅, 𝑅]), |𝐼2| ≤ 𝜉2 ∫ 𝜉1 | − (𝑓(𝑈(𝜉)) − 𝜉)𝜙′(𝜉) + 𝜙(𝜉)||𝜌(𝜉)|𝑑𝜉 → 0, as𝜉1→ 𝜉𝜎−, 𝜉2→ 𝜉𝜎+ .
Define an operator𝑇 ∶ 𝐾 → 𝐶2([−𝑅, 𝑅]) as follows: for any 𝑈 ∈ 𝐾, 𝑢 = 𝑇 𝑈 is the unique solution of the boundary value problem
{
𝜀𝑢′′= (𝜌(𝑈, 𝜉)(𝑓(𝑈(𝜉)) − 𝜉) + 1) 𝑢′
𝑢(±𝑅) = 𝑢± (20)
where𝜌(𝑈, 𝜉) ≡ 𝜌(𝜉) is defined in (12) or (13). In fact, the solution to this problem can be found explicitly and it is given by
𝑢(𝜉) = 𝑢−+ (𝑢+− 𝑢−) ∫ 𝜉 −𝑅exp ( ∫−𝑅𝑟 𝜌(𝑈,𝑠)(𝑓(𝑈(𝑠))−𝑠)+1𝜀 𝑑𝑠)𝑑𝑟 ∫−𝑅𝑅 exp(∫−𝑅𝑟 𝜌(𝑈,𝑠)(𝑓(𝑈(𝑠))−𝑠)+1𝜀 𝑑𝑠)𝑑𝑟 . (21)
Lemma 2. 𝑇 ∶ 𝐾 → 𝐾 is a continuous operator.
Proof. Choose{𝑈𝑛} in 𝐾 such that 𝑈𝑛→ 𝑈. As 𝑈 belongs to 𝐾, then each 𝑢𝑛= 𝑇 𝑈𝑛and𝑢 = 𝑇 𝑈 satisfy the problem (20). Now, we have the following problem
{
𝜀(𝑢𝑛− 𝑢)′′= (𝜌(𝑈
𝑛, 𝜉)(𝑓(𝑈𝑛(𝜉)) − 𝜉) + 1)(𝑢𝑛− 𝑢)′+ (𝜌(𝑈𝑛, 𝜉)(𝑓(𝑈𝑛(𝜉)) − 𝜉) − 𝜌(𝑈, 𝜉)(𝑓(𝑈(𝜉)) − 𝜉))𝑢′
(𝑢𝑛− 𝑢)(±𝑅) = 0. (22)
Setting𝑝𝑛(𝜉) = 𝜌(𝑈𝑛, 𝜉)(𝑓(𝑈𝑛(𝜉)) − 𝜉) and 𝑞𝑛(𝜉) = (𝜌(𝑈𝑛, 𝜉)(𝑓(𝑈𝑛(𝜉)) − 𝜉) − 𝜌(𝑈, 𝜉)(𝑓(𝑈(𝜉)) − 𝜉))𝑢′, from problem (22) we have (𝑢𝑛− 𝑢)′(𝜉) = −∫ 𝑅 −𝑅∫−𝑅𝑦 𝑞𝑛𝜀(𝑟)exp ( ∫𝑟𝑦𝑝𝑛(𝑠)+1 𝜀 𝑑𝑠 ) 𝑑𝑟𝑑𝑦 ∫−𝑅𝑅 exp(∫−𝑅𝑟 𝑝𝑛(𝑠)+1 𝜀 𝑑𝑠 ) 𝑑𝑟 exp ⎛ ⎜ ⎜ ⎝ 𝜉 ∫ −𝑅 𝑝𝑛(𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎠ + 𝜉 ∫ −𝑅 𝑞𝑛(𝑟) 𝜀 exp ⎛ ⎜ ⎜ ⎝ 𝜉 ∫ −𝑟 𝑝𝑛(𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎠ 𝑑𝑟 (23) (𝑢𝑛− 𝑢)(𝜉) = −∫ 𝑅 −𝑅∫−𝑅𝑦 𝑞𝑛𝜀(𝑟)exp ( ∫𝑟𝑦𝑝𝑛(𝑠)+1 𝜀 𝑑𝑠 ) 𝑑𝑟𝑑𝑦 ∫−𝑅𝑅 exp(∫−𝑅𝑟 𝑝𝑛(𝑠)+1 𝜀 𝑑𝑠 ) 𝑑𝑟 𝜉 ∫ −𝑅 exp⎛⎜⎜ ⎝ 𝑟 ∫ −𝑅 𝑝𝑛(𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎠ 𝑑𝑟 + 𝜉 ∫ −𝑅 𝑦 ∫ −𝑅 𝑞𝑛(𝑟) 𝜀 exp ⎛ ⎜ ⎜ ⎝ 𝑦 ∫ −𝑟 𝑝𝑛(𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎠ 𝑑𝑟𝑑𝑦 (24) From (14), we have ((𝑓(𝑈(𝜉)) − 𝜉)𝜌(𝜉))′= −𝜌(𝜉) < 0, ((𝑓(𝑈𝑛(𝜉)) − 𝜉)𝜌𝑛(𝜉))′= −𝜌 𝑛(𝜉) < 0 for𝑛 = 1, 2, … .
Then,𝜌(𝑓(𝑈) − 𝜉) and 𝜌𝑛(𝑓(𝑈𝑛) − 𝜉), 𝑛 = 1, 2, …, are monotone decreasing and continuous functions. Because the sequence of monotone functions which converges to a continuous function must converge uniformly, we get that𝑞𝑛(𝜉) converges to zero uniformly. Then, from (22), (23) and (24) it follows that
𝑢𝑛→ 𝑢 in 𝐶2([−𝑅, 𝑅]), as 𝑛 → ∞.
Therefore𝑇 ∶ 𝐾 → 𝐶2([−𝑅, 𝑅]) is continuous. In addition, from (21), we have
𝑢′(𝜉) = (𝑢+− 𝑢−) exp
(
∫−𝑅𝜉 𝜌(𝑈,𝑠)(𝑓(𝑈(𝑠))−𝑠)+1𝜀 𝑑𝑠)
∫−𝑅𝑅 exp(∫−𝑅𝑟 𝜌(𝑈,𝑠)(𝑓(𝑈(𝑠))−𝑠)+1𝜀 𝑑𝑠)𝑑𝑟
Lemma 3. 𝑇 𝐾 is a bounded set in 𝐶2([−𝑅, 𝑅]).
Proof. For any𝑈 ∈ 𝐾, if 𝑠 < 𝜉𝜎, we have
0 < 𝜌(𝑈, 𝑠)(𝑓(𝑈(𝑠)) − 𝑠) = 𝜌−(𝑓(𝑢−) + 𝑅) − 𝑠 ∫ −𝑅 𝜌(𝑟)𝑑𝑟 < 𝜌−(𝑓(𝑢−) + 𝑅) (25) and if𝑠 > 𝜉𝜎, 0 > 𝜌(𝑈, 𝑠)(𝑓(𝑈(𝑠)) − 𝑠) = 𝜌+(𝑓(𝑢+) − 𝑅) + 𝑅 ∫ 𝑠 𝜌(𝑟)𝑑𝑟 > 𝜌+(𝑓(𝑢+) − 𝑅). (26)
From (20), we can deduce that
𝑢′′(𝜉) < 0, 𝜉 ∈ [−𝑅, 𝜉 𝜎). Then,𝑢′(𝜉) ≤ 𝑢′(−𝑅) < 0, 𝜉 ∈ [−𝑅, 𝜉𝜎), and 𝑢−− 𝑢+> 𝑢(−𝑅) − 𝑢(𝜉𝜎) = 𝑢′(𝜁)(−𝑅 − 𝜉𝜎) > 𝑢′(𝜁)(−𝑅 − 𝑓(𝑢+)), 𝜁 ∈ (−𝑅, 𝜉𝜎). Thus, 0 > 𝑢′(−𝑅) > 𝑢′(𝜁) > − 𝑢−− 𝑢+ 𝑅 + 𝑓(𝑢+). Also, from (20) we have
𝑢′(𝜉) = 𝑢′(−𝑅) exp⎛⎜ ⎜ ⎝ 𝜉 ∫ −𝑅 𝜌(𝑓(𝑈) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎠
and by (25) and (26), we conclude that𝑢′is uniformly bounded. Consequently,𝑢′′is also uniformly bounded.
So,𝑇 𝐾 is a bounded set in 𝐶2([−𝑅, 𝑅]). □
Lemma 4. 𝑇 𝐾 is precompact in 𝐶([−𝑅, 𝑅]).
Proof. This is a consequence of the compact embedding𝐶2([−𝑅, 𝑅]) → 𝐶([−𝑅, 𝑅]). □
From the above lemmas, by virtue of Schauder fixed point theorem, we get the following result.
Theorem 2. For each𝑅 ≥ max{𝑅1, 𝑅2, |𝑢−|, |𝑢+|}, there exists a weak solution
(𝜌𝑅, 𝑢𝑅) ∈ 𝐿1([−𝑅, 𝑅]) × 𝐶2([−𝑅, 𝑅])
for the system (9) with boundary value(𝜌𝑅(±𝑅), 𝑢𝑅(±𝑅)) = (𝜌±, 𝑢±), and, in addition, being 𝑢𝑅a decreasing function.
The next step is to obtain from this family of solutions a sequence𝑅𝑘→ ∞ such that (𝜌𝑅𝑘, 𝑢𝑅𝑘) converges to a weak solution of (9)–(10). To this end, we need the following lemma.
Lemma 5.
1. 𝑢𝑅(𝜉), 𝑢′𝑅(𝜉) and 𝑢′′𝑅(𝜉) are uniformly bounded, with respect to 𝑅 and 𝜉 ∈ [−𝑅, 𝑅]. 2. There exist a sequence𝑅𝑘→ ∞ and a decreasing function 𝑢 ∈ 𝐶1(ℝ) such that 𝑢𝑅
𝑘 converges to𝑢 in 𝐶1([−𝑀, 𝑀]), for
each positive number𝑀 (i.e. 𝑢𝑅
𝑘,𝑢′𝑅𝑘converge uniformly in compact sets ofℝ to 𝑢, 𝑢
′, respectively).
3. 𝜌𝑅
𝑘(𝑢𝑅𝑘, 𝜉) converges to 𝜌(𝑢, 𝜉), as 𝑅𝑘→ ∞, for each 𝜉 ∈ ℝ ⧵ {𝜉𝜎}, where 𝜌𝑅𝑘(𝑢𝑅𝑘, 𝜉), 𝜌(𝑢, 𝜉) are defined accordingly with
Proof.
1. To simplify the notation in this proof, we shall use𝑢, 𝑢′and𝑢′′instead of𝑢𝑅,𝑢′𝑅and𝑢′′𝑅. Observe that𝑓(𝑢+) +𝜌1
+ < 𝜉𝜎 < 𝑓(𝑢−). We choose 𝜉1such that−𝑅 < 𝜉1< 𝑓(𝑢+) +
1
𝜌+. From (9) it follows that
𝑢′(𝜉) = 𝑢′(𝜉 1) exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝜉 ∫ 𝜉1 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ .
As𝑢′′(𝜉) < 0 for 𝜉 ∈ (−𝑅, 𝜉𝜎), then we have that 𝑢′(𝜉) < 𝑢′(𝜉1) < 0, 𝜉 ∈ (𝜉1, 𝜉𝜎). Since
𝑢−− 𝑢+> 𝑢(𝜉1) − 𝑢(𝜉𝜎) = 𝑢′(𝜁)(𝜉1− 𝜉𝜎) > 𝑢′(𝜁)(𝜉1− 𝑓(𝑢+)), where𝜁 ∈ (𝜉1, 𝜉𝜎), we get 𝑢′(𝜁) > 𝑢−− 𝑢+ 𝜉1− 𝑓(𝑢+), 𝜁 ∈ (𝜉1, 𝜉𝜎). It follows that 0 > 𝑢′(𝜉1) > 𝜉𝑢−− 𝑢+ 1− 𝑓(𝑢+). When𝜉 < 𝜉1, exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝜉 ∫ 𝜉1 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ < 1.
When𝜉1< 𝜉 < 𝜉𝜎, observe that
𝜌1(𝜉1) = 𝜌− 𝑓(𝑢−) + 𝑅 𝑓(𝑢(𝜉1)) − 𝜉1exp ⎛ ⎜ ⎜ ⎜ ⎝ − 𝜉1 ∫ −𝑅 𝑑𝑠 𝑓(𝑢(𝑠)) − 𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ ≤ 𝜌−𝑓(𝑢(𝜉𝑓(𝑢−) + 𝑅 1)) − 𝜉1exp ⎛ ⎜ ⎜ ⎜ ⎝ − 𝜉1 ∫ −𝑅 𝑑𝑠 𝑓(𝑢−) − 𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ = 𝜌−𝑓(𝑢(𝜉𝑓(𝑢−) − 𝜉1 1)) − 𝜉1 and 𝜌(𝜉)(𝑓(𝑢(𝜉)) − 𝜉) + 1 = 𝜌(𝜉1)(𝑓(𝑢(𝜉1)) − 𝜉1) + 1 − 𝜉 ∫ 𝜉1 𝜌(𝑠)𝑑𝑠 ≤ 𝜌(𝜉1)(𝑓(𝑢(𝜉1)) − 𝜉1) + 1 ≤ 𝜌−(𝑓(𝑢−) − 𝜉1) + 1, and we obtain exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝜉 ∫ 𝜉1 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ ≤ exp ((𝜌 −(𝑓(𝑢−) − 𝜉1) + 1)(𝑓(𝑢−) − 𝜉1) 𝜀 ) .
When𝜉 > 𝜉𝜎, we have 𝜉 ∫ 𝜉1 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 = 𝜉𝜎 ∫ 𝜉1 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 + 𝜉 ∫ 𝜉𝜎 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 < 𝜉𝜎 ∫ 𝜉1 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 or exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝜉 ∫ 𝜉1 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ < exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝜉𝜎 ∫ 𝜉1 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ .
Therefore,𝑢′(𝜉) and 𝑢(𝜉) are uniformly bounded. From (20) it follows that 𝑢′′𝑅(𝜉) is also uniformly bounded, with respect to
𝑅 and 𝜉 ∈ [−𝑅, 𝑅].
2. Fixing 𝑀 > 0, we consider 𝑅 >> 𝑀 and apply the Arzelá-Ascoli theorem to obtain a sequence (𝑢𝑅𝑘) converging in
𝐶1([−𝑀, 𝑀]) to a decreasing function 𝑢. Then, by a diagonalization process, we obtain a sequence (𝑢
𝑅𝑘), which we do
not relabel, such that(𝑢𝑅
𝑘) converges to a decreasing function 𝑢 ∈ 𝐶1(ℝ), uniformly in compact sets in ℝ, (𝑢′𝑅𝑘) also
con-verges uniformly in compact sets inℝ to 𝑢′, and𝑢(−∞) = 𝑢−,𝑢(∞) = 𝑢+.
3. Claim 3 is obtained from (12), (13) by passing to the limit as𝑅𝑘→ ∞, for each fixed 𝜉 ≠ 𝜉𝜎, noting that, up to a subsequence, we can assume that𝜉𝜎𝑅𝑘, defined by𝑓(𝑢𝑅𝑘(𝜉𝜎𝑅𝑘)) = 𝜉𝜎𝑅𝑘, converges to𝜉𝜎(where𝜉𝜎is defined by𝑓(𝑢(𝜉𝜎)) = 𝜉𝜎). □
Theorem 3. Let𝑢 be the function obtained in Lemma 5. Then, for each 𝜀 > 0, 𝑢 satisfies
{ 𝜀𝑢′′= (𝜌(𝑢, 𝜉)(𝑓(𝑢) − 𝜉) + 1)𝑢′, 𝑢(±∞) = 𝑢±, and 𝜌(𝜉) = { 𝜌1(𝜉), if − ∞ < 𝜉 < 𝜉𝜎, 𝜌2(𝜉), if 𝜉𝜎< 𝜉 < ∞, where𝜉𝜎satisfies𝑓(𝑢(𝜉𝜎)) = 𝜉𝜎, 𝜌1(𝜉) = 𝜌−exp⎛⎜⎜ ⎝ − 𝜉 ∫ −∞ (𝑓(𝑢(𝑠)))′ 𝑓(𝑢(𝑠)) − 𝑠𝑑𝑠 ⎞ ⎟ ⎟ ⎠ and 𝜌2(𝜉) = 𝜌+exp ⎛ ⎜ ⎜ ⎝ +∞ ∫ 𝜉 (𝑓(𝑢(𝑠)))′ 𝑓(𝑢(𝑠)) − 𝑠𝑑𝑠 ⎞ ⎟ ⎟ ⎠ .
Proof. Denote by(𝜌𝑅(𝜉), 𝑢𝑅(𝜉)) the solution of the problem (9) with boundary value (𝜌(±𝑅), 𝑢(±𝑅)) = (𝜌±, 𝑢±). Fixing 𝜉2and integrating (20) from𝜉2to𝜉, we obtain
𝜀(𝑢′𝑅(𝜉) − 𝑢′𝑅(𝜉2)) = (𝜌𝑅(𝜉)(𝑓(𝑢𝑅(𝜉)) − 𝜉) + 1)𝑢𝑅(𝜉) − (𝜌𝑅(𝜉2)(𝑓(𝑢𝑅(𝜉2)) − 𝜉2) + 1)𝑢𝑅(𝜉2) +
𝜉
∫
𝜉2
𝜌𝑅(𝑠)𝑢𝑅(𝑠)𝑑𝑠 (independently of whether𝜉𝜎is between𝜉2and𝜉). Letting 𝑅 → +∞, by the Lebesgue Convergence Theorem it follows that
𝜀(𝑢′(𝜉) − 𝑢′(𝜉 2)) = (𝜌(𝜉)(𝑓(𝑢(𝜉)) − 𝜉) + 1)𝑢(𝜉) − (𝜌(𝜉2)(𝑓(𝑢(𝜉2)) − 𝜉2) + 1)𝑢(𝜉2) + 𝜉 ∫ 𝜉2 𝜌(𝑠)𝑢(𝑠)𝑑𝑠. (27)
Differentiating (27) with respect to𝜉, we obtain
𝜀𝑢′′= (𝜌(𝑓(𝑢) − 𝜉) + 1)𝑢′,
and from (21) we have𝑢(±∞) = 𝑢±. □
Theorem 4. There exists a weak solution(𝜌, 𝑢) ∈ 𝐿1𝑙𝑜𝑐((−∞, +∞)) × 𝐶2((−∞, +∞)) for the boundary value problem (9)–(10).
Proof. Let(𝜌, 𝑢) be defined in Theorem 3. By Lemma 5 we know that 𝑢 is decreasing and of class 𝐶1inℝ. Then 𝜌 is of class
𝐶1in(−∞, 𝜉
𝜎) ∪ (𝜉𝜎, ∞). In addition, it is also bounded, hence, locally integrable. From (27) it follows that 𝑢 is of class 𝐶2.
The first equation in (9) comes by differentiating𝜌1and𝜌2, and the second is equivalent to the first and the equation stated in
Theorem 3. □
3.2
The limit solutions of (7)–(8) as viscosity vanishes
We continue this section studying the case when𝑓(𝑢−) > 𝑓(𝑢+) +𝜌1
+ and we are interested in analyzing the behavior of the
solutions(𝜌𝜀, 𝑢𝜀) of (9)–(10) as 𝜀 → 0+.
Lemma 6. Let𝜉𝜎𝜀be the unique point satisfying𝑓(𝑢𝜀(𝜉𝜀𝜎)) = 𝜉𝜎𝜀, and let𝜉𝜎be the limit 𝜉𝜎 = lim
𝜀→0+𝜉 𝜀 𝜎 (passing to a subsequence if necessary). Then for any𝜂 > 0,
lim 𝜀→0+𝑢 𝜀 𝜉(𝜉) = 0, for|𝜉 − 𝜉𝜎| ≥ 𝜂, lim 𝜀→0+𝑢 𝜀(𝜉) = { 𝑢−, if 𝜉 ≤ 𝜉𝜎− 𝜂, 𝑢+, if 𝜉 ≥ 𝜉𝜎+ 𝜂,
uniformly in the above intervals.
Proof. To simplify the notation in this proof, we shall use𝜌, 𝑢 instead of 𝜌𝜀,𝑢𝜀. Take𝜉3= 𝜉𝜎+ 𝜂∕2, and let 𝜀 be so small such that 𝜉𝜎𝜀< 𝜉3− 𝜂∕4. For 𝜉 > 𝜉𝜎,
𝜌(𝜉) = 𝜌+exp⎛⎜⎜ ⎝ +∞ ∫ 𝜉 (𝑓(𝑢(𝑠)))′ 𝑓(𝑢(𝑠)) − 𝑠𝑑𝑠 ⎞ ⎟ ⎟ ⎠ = lim 𝑅→+∞𝜌+ 𝑅 − 𝑓(𝑢+) 𝜉 − 𝑓(𝑢(𝜉))exp ⎛ ⎜ ⎜ ⎝ 𝑅 ∫ 𝜉 𝑑𝑠 𝑓(𝑢(𝑠)) − 𝑠 ⎞ ⎟ ⎟ ⎠ ≤ lim 𝑅→+∞𝜌+ 𝑅 − 𝑓(𝑢+) 𝜉 − 𝑓(𝑢(𝜉))exp ⎛ ⎜ ⎜ ⎝ 𝑅 ∫ 𝜉 𝑑𝑠 𝑓(𝑢+) − 𝑠 ⎞ ⎟ ⎟ ⎠ = lim 𝑅→+∞𝜌+ 𝑅 − 𝑓(𝑢+) 𝜉 − 𝑓(𝑢(𝜉)) 𝜉 − 𝑓(𝑢+) 𝑅 − 𝑓(𝑢+) = 𝜌+ 𝜉 − 𝑓(𝑢+) 𝜉 − 𝑓(𝑢(𝜉)), and we have 𝜌(𝜉)(𝑓(𝑢(𝜉)) − 𝜉) + 1 ≥ 𝜌+(𝑓(𝑢+) − 𝜉) + 1, 𝜉 ∈ (𝜉𝜎, +∞).
Now, integrating the second Equation of 9 twice on[𝜉3, 𝜉], we get 𝑢(𝜉3) − 𝑢(𝜉) = −𝑢′(𝜉 3) 𝜉 ∫ 𝜉3 exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝑟 ∫ 𝜉3 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ 𝑑𝑟 ≥ −𝑢′(𝜉 3) 𝜉 ∫ 𝜉3 exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝑟 ∫ 𝜉3 𝜌+(𝑓(𝑢+) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ 𝑑𝑟 = −𝑢′(𝜉3) 𝜉 ∫ 𝜉3 exp(𝜌𝜀+ (( 𝑓(𝑢+) + 1𝜌 + − 𝜉3 ) (𝑟 − 𝜉3) − 12(𝑟 − 𝜉3)2 )) 𝑑𝑟 = −𝑢′(𝜉 3) 𝜉−𝜉3 ∫ 0 exp(𝜌+ 𝜀 (( 𝑓(𝑢+) + 1 𝜌+− 𝜉3 ) 𝑟 − 1 2𝑟2 )) 𝑑𝑟. Letting𝜉 → +∞, we get 𝑢−− 𝑢+≥ −𝑢′(𝜉 3) +∞ ∫ 0 exp (𝜌 + 𝜀 (( 𝑓(𝑢+) + 1 𝜌+ − 𝜉3 ) 𝑟 − 1 2𝑟2 )) 𝑑𝑟 ≥ −𝑢′(𝜉3) 2𝜀 ∫ 0 exp(𝜌𝜀+ (( 𝑓(𝑢+) + 1𝜌 +− 𝜉3 ) 𝑟 − 12𝑟2 )) 𝑑𝑟 ≥ −𝑢′(𝜉 3) √ 𝜀𝐴3
for0 ≤ 𝜀 ≤ 1, where 𝐴3is a constant independent of𝜀. Thus |𝑢′(𝜉 3)| ≤ 𝑢√−− 𝑢+ 𝜀𝐴3 . So |𝑢′(𝜉)| ≤ 𝑢√−− 𝑢+ 𝜀𝐴3 exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝜉 ∫ 𝜉3 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ . (28) For𝜉 > 𝜉3, 𝜌(𝜉) = lim 𝑅→+∞𝜌+ 𝑅 − 𝑓(𝑢+) 𝜉 − 𝑓(𝑢(𝜉))exp ⎛ ⎜ ⎜ ⎝ 𝑅 ∫ 𝜉 𝑑𝑠 𝑓(𝑢(𝑠)) − 𝑠 ⎞ ⎟ ⎟ ⎠ ≥ lim 𝑅→+∞𝜌+ 𝑅 − 𝑓(𝑢+) 𝜉 − 𝑓(𝑢(𝜉))exp ⎛ ⎜ ⎜ ⎝ 𝑅 ∫ 𝜉 𝑑𝑠 𝑓(𝑢(𝜉3)) − 𝑠 ⎞ ⎟ ⎟ ⎠ = 𝜌+𝜉 − 𝑓(𝑢(𝜉3)) 𝜉 − 𝑓(𝑢(𝜉)) 𝑅→+∞lim 𝑅 − 𝑓(𝑢+) 𝑅 − 𝑓(𝑢(𝜉3)) = 𝜌+𝜉 − 𝑓(𝑢(𝜉𝜉 − 𝑓(𝑢(𝜉))3))
and we have
𝜌(𝜉)(𝑓(𝑢(𝜉)) − 𝜉) + 1 ≤ 𝜌+(𝑓(𝑢(𝜉3)) − 𝜉) + 1, 𝜉 > 𝜉3. (29)
From (28) and (29) we have
|𝑢′(𝜉)| ≤ 𝑢√−− 𝑢+ 𝜀𝐴3 exp ⎛ ⎜ ⎜ ⎜ ⎝ −𝜌𝜀+ 𝜉 ∫ 𝜉3 ( 𝑠 − ( 𝑓(𝑢(𝜉3)) + 1𝜌 + )) 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ which implies that
lim
𝜀→0+𝑢 𝜀
𝜉(𝜉) = 0, uniformly for𝜉 ≥ 𝜉𝜎+ 𝜂.
Now, we choose𝜉 and 𝜉4such that𝜉 > 𝜉4≥ 𝜉𝛽+ 𝜂. From
𝑢(𝜉4) − 𝑢(𝜉) = −𝑢′(𝜉 4) 𝜉 ∫ 𝜉4 exp ⎛ ⎜ ⎜ ⎜ ⎝ 𝑟 ∫ 𝜉4 𝜌(𝑠)(𝑓(𝑢(𝑠)) − 𝑠) + 1 𝜀 𝑑𝑠 ⎞ ⎟ ⎟ ⎟ ⎠ 𝑑𝑟, we get |𝑢(𝜉4) − 𝑢(𝜉)| ≤ |𝑢′(𝜉 4)| 𝜉 ∫ 𝜉4 exp ( −𝐴4 𝜀 (𝑟 − 𝜉4) ) 𝑑𝑟 ≤ 𝜀 𝐴4|𝑢′(𝜉4)| ( 1 − exp ( 𝐴4 𝜀 (𝜉4− 𝜉) )) , where𝐴4= 𝜌+(𝜉4− (𝑓(𝑢(𝜉4)) +𝜌1 +)). When 𝜉 → +∞, we obtain |𝑢(𝜉4) − 𝑢+| ≤ 𝜀 𝐴4|𝑢′(𝜉4)|,
which implies that
lim
𝜀→0+𝑢 𝜀(𝜉) = 𝑢
+, uniformly for𝜉 ≥ 𝜉𝜎+ 𝜂.
The results for𝜉 < 𝜉𝜎− 𝜂 can be obtained analogously. □
Lemma 7. For any𝜂 > 0,
lim 𝜀→0+𝜌 𝜀(𝜉) = { 𝜌−, if 𝜉 < 𝜉𝜎− 𝜂, 𝜌+, if 𝜉 > 𝜉𝜎+ 𝜂,
uniformly, with respect to𝜉.
Proof. Take𝜀0> 0 so small such that |𝜉𝜎𝜀− 𝜉𝜎| < 2𝜂 whenever0 < 𝜀 < 𝜀0. For any𝜉 > 𝜉𝜎+ 𝜂 and 𝜀 < 𝜀0, we have
𝜉 > 𝜉𝜀 𝜎+𝜂2 and 𝜌𝜀(𝜉) = 𝜌+exp⎛⎜⎜ ⎝ ∞ ∫ 𝜉 (𝑓(𝑢𝜀(𝑠)))′ 𝑓(𝑢𝜀(𝑠)) − 𝑠𝑑𝑠 ⎞ ⎟ ⎟ ⎠ .
For any𝑠 ∈ [𝜉, +∞), we have
𝑓(𝑢𝜀(𝑠)) − 𝑠 < 𝑓(𝑢𝜀(𝜉)) − 𝜉 = (1 − (𝑓(𝑢𝜀(𝜁)))′)(𝜉𝜀 𝜎− 𝜉)
≤ −𝜂 2.
As𝑓′> 0 and 𝑢 is decreasing, we have that (𝑓(𝑢(𝜉)))′= 𝑓′(𝑢(𝜉))𝑢′(𝜉) < 0, and (𝑓(𝑢(𝑠)))′
𝑓(𝑢𝜀(𝑠)) − 𝑠 < −2𝜂(𝑓(𝑢(𝑠)))′, for any𝑠 ∈ [𝜉, +∞).
Now, in the last inequality, integrating on[𝜉, +∞) we have
0 ≤ ∞ ∫ 𝜉 (𝑓(𝑢𝜀(𝑠)))′ 𝑓(𝑢𝜀(𝑠)) − 𝑠𝑑𝑠 ≤ −2𝜂 ∞ ∫ 𝜉 (𝑓(𝑢𝜀(𝑠)))′𝑑𝑠 = −2𝜂(𝑓(𝑢+) − 𝑓(𝑢𝜀(𝜉))), so 1 ≤ exp⎛⎜⎜ ⎝ ∞ ∫ 𝜉 (𝑓(𝑢𝜀(𝑠)))′ 𝑓(𝑢𝜀(𝑠)) − 𝑠𝑑𝑠 ⎞ ⎟ ⎟ ⎠ ≤ exp ( −2 𝜂(𝑓(𝑢+) − 𝑓(𝑢𝜀(𝜉))) ) . (30)
By Lemma 6 we have that lim
𝜀→0+𝑢𝜀(𝜉) = 𝑢+, and from (30) we have
lim 𝜀→0+exp ⎛ ⎜ ⎜ ⎝ ∞ ∫ 𝜉 (𝑓(𝑢𝜀(𝑠)))′ 𝑓(𝑢𝜀(𝑠)) − 𝑠𝑑𝑠 ⎞ ⎟ ⎟ ⎠ = 1 and lim 𝜀→0+𝜌 𝜀(𝜉) = lim 𝜀→0+𝜌+exp ⎛ ⎜ ⎜ ⎝ ∞ ∫ 𝜉 (𝑓(𝑢𝜀(𝑠)))′ 𝑓(𝑢𝜀(𝑠)) − 𝑠𝑑𝑠 ⎞ ⎟ ⎟ ⎠ = 𝜌+, uniformly for 𝜉 > 𝜉𝜎+ 𝜂. Similarly, we obtain alsolim
𝜀→0𝜌𝜀(𝜉) = 𝜌−, uniformly for𝜉 < 𝜉𝜎− 𝜂. □
Now, we study the limit behavior of(𝜌𝜀, 𝑢𝜀) in the neighborhood of 𝜉𝜎 as𝜀 → 0+.
Theorem 5. Denote 𝜎 = 𝜉𝜎 = lim 𝜀→0+𝜉 𝜀 𝜎 = lim𝜀→0+𝑓(𝑢𝜀(𝜉𝜎𝜀)) = 𝑓(𝑢(𝜎)). (31) Then lim 𝜀→0+(𝜌 𝜀(𝜉), 𝑢𝜀(𝜉)) = ⎧ ⎪ ⎨ ⎪ ⎩ (𝜌−, 𝑢−), if𝜉 < 𝜎, (𝑤0⋅ 𝛿, 𝑢𝛿), if 𝜉 = 𝜎, (𝜌+, 𝑢+), if𝜉 > 𝜎,
where𝜌𝜀(𝜉) converges in the sense of the distributions to the sum of a step function and a Dirac measure 𝛿 with weight 𝑤0=
−𝜎(𝜌−− 𝜌+) + (𝜌−𝑓(𝑢−) − 𝜌+𝑓(𝑢+)).
Proof. As𝜎 = 𝜉𝜎 = lim
𝜀→0+𝑓(𝑢𝜀(𝜉𝜎𝜀)) = 𝑓(𝑢(𝜎)), then we have 𝑓(𝑢+) + 1
Let𝜉1and𝜉2be real numbers such that𝜉1< 𝜎 < 𝜉2and𝜙 ∈ 𝐶0∞([𝜉1, 𝜉2]) such that 𝜙(𝜉) ≡ 𝜙(𝜎) for 𝜉 in a neighborhood Ω of
𝜎, Ω ⊂ (𝜉1, 𝜉2)1. Then𝜉𝜀
𝜎∈ Ω whenever 0 < 𝜀 < 𝜀0. From (9) we have
− 𝜉2 ∫ 𝜉1 𝜌𝜀(𝑓(𝑢𝜀) − 𝜉)𝜙′𝑑𝜉 + 𝜉2 ∫ 𝜉1 𝜌𝜀𝜙𝑑𝜉 = 0. (33)
For𝛼1, 𝛼2∈ Ω, 𝛼1, 𝛼2near𝜎 such that 𝛼1< 𝜎 < 𝛼2, we write
𝜉2 ∫ 𝜉1 𝜌𝜀(𝑓(𝑢𝜀) − 𝜉)𝜙′𝑑𝜉 = 𝛼1 ∫ 𝜉1 𝜌𝜀(𝑓(𝑢𝜀) − 𝜉)𝜙′𝑑𝜉 + 𝜉2 ∫ 𝛼2 𝜌𝜀(𝑓(𝑢𝜀) − 𝜉)𝜙′𝑑𝜉,
and from Lemmas 6 and 7, we obtain
lim 𝜀→0+ 𝜉2 ∫ 𝜉1 𝜌𝜀(𝑓(𝑢𝜀) − 𝜉)𝜙′𝑑𝜉 = 𝛼1 ∫ 𝜉1 𝜌−(𝑓(𝑢−) − 𝜉)𝜙′𝑑𝜉 + 𝜉2 ∫ 𝛼2 𝜌+(𝑓(𝑢+) − 𝜉)𝜙′𝑑𝜉 =(𝜌−𝑓(𝑢−) − 𝜌+𝑓(𝑢+) − 𝜌−𝛼1+ 𝜌+𝛼2)𝜙(𝜎) + 𝛼1 ∫ 𝜉1 𝜌−𝜙(𝜉)𝑑𝜉 + 𝜉2 ∫ 𝛼2 𝜌+𝜙(𝜉)𝑑𝜉
Then taking𝛼1→ 𝜎−, 𝛼2→ 𝜎+, we arrive at
lim 𝜀→0+ 𝜉2 ∫ 𝜉1 𝜌𝜀(𝑓(𝑢𝜀) − 𝜉)𝜙′𝑑𝜉 = (−[𝜌]𝜎 + [𝜌𝑓(𝑢)]) 𝜙(𝜎) + 𝜉2 ∫ 𝜉1 𝐽(𝜉 − 𝜎)𝜙(𝜉)𝑑𝜉 (34) where[𝑞] = 𝑞−− 𝑞+and 𝐽(𝑥) = { 𝜌−, if 𝑥 < 0, 𝜌+, if 𝑥 > 0.
From (33) and (34), we get
lim 𝜀→0+ 𝜉2 ∫ 𝜉1 (𝜌𝜀− 𝐽(𝜉 − 𝜎))𝜙(𝜉)𝑑𝜉 = (−[𝜌]𝜎 + [𝜌𝑓(𝑢)]) 𝜙(𝜎). for all sloping test functions𝜙 ∈ 𝐶0∞([𝜉1, 𝜉2]).
For an arbitrary𝜓 ∈ 𝐶0∞([𝜉1, 𝜉2]), we take a sloping test function 𝜙, such that 𝜙(𝜎) = 𝜓(𝜎) and max
[𝜉1,𝜉2]|𝜓 − 𝜙| < 𝜇,
for a sufficiently small𝜇 > 0. As 𝜌𝜀∈ 𝐿1([𝜉1, 𝜉2) uniformly, we obtain
lim 𝜀→0+ 𝜉2 ∫ 𝜉1 (𝜌𝜀− 𝐽(𝜉 − 𝜎))𝜓(𝜉)𝑑𝜉 = lim 𝜀→0+ 𝜉2 ∫ 𝜉1 (𝜌𝜀− 𝐽(𝜉 − 𝜎))𝜙(𝜉)𝑑𝜉 + 𝑂(𝜇) = (−[𝜌]𝜎 + [𝜌𝑓(𝑢)]) 𝜙(𝜎) + 𝑂(𝜇) = (−[𝜌]𝜎 + [𝜌𝑓(𝑢)]) 𝜓(𝜎) + 𝑂(𝜇).
Then, when𝜇 → 0+, we find that lim 𝜀→0+ 𝜉2 ∫ 𝜉1 (𝜌𝜀− 𝐽(𝜉 − 𝜎))𝜓(𝜉)𝑑𝜉 = (−[𝜌]𝜎 + [𝜌𝑓(𝑢)]) 𝜓(𝜎) (35)
holds for all test functions𝜓 ∈ 𝐶0∞([𝜉1, 𝜉2]). Thus, 𝜌𝜀converges in the sense of the distributions to the sum of a step function and a Dirac delta function with strength−[𝜌]𝜎 + [𝜌𝑓(𝑢)]. In a similar way, from
− 𝜉2 ∫ 𝜉1 (𝜌𝜀(𝑓(𝑢𝜀) − 𝜉) + 1) 𝑢𝜀𝜙′𝑑𝜉 + 𝜉2 ∫ 𝜉1 𝜌𝜀𝑢𝜀𝜙𝑑𝜉 = 𝜀 𝜉2 ∫ 𝜉1 (𝑢𝜀)′′𝜙𝑑𝜉 (36) we can obtain lim 𝜀→0+ 𝜉2 ∫ 𝜉1 (𝜌𝜀𝑢𝜀− ̃𝐽(𝜉 − 𝜎))𝜙(𝜉)𝑑𝜉 =(−[𝜌𝑢]𝜎 +[𝜌𝑢(𝑓(𝑢) + 1 𝜌 )]) 𝜙(𝜎)
for all test functions𝜙 ∈ 𝐶0∞([𝜉1, 𝜉2]), where
̃ 𝐽(𝑥) =
{
𝜌−𝑢−, if 𝑥 < 0, 𝜌+𝑢+, if 𝑥 > 0.
Thus𝜌𝑢 also converges in the sense of the distributions to the sum of a step function and a Dirac delta function with strength −[𝜌𝑢]𝜎 + [𝜌𝑢(𝑓(𝑢) +1𝜌)].
If we take the test function in (36) as̃𝑢𝜀𝜓+𝜈,𝜈 > 0, where ̃𝑢𝜀is a modified function satisfying𝑢𝜀(𝜎) in Ω and 𝑢𝜀outsideΩ, and let𝜈 → 0+, we find lim 𝜀→0+ 𝜉2 ∫ 𝜉1 (𝜌𝜀− 𝐽(𝜉 − 𝜎))𝜓𝑑𝜉 ⋅ 𝑢(𝜎) =(−[𝜌𝑢]𝜎 +[𝜌𝑢(𝑓(𝑢) + 1 𝜌 )]) 𝜓(𝜎) (37)
for all test functions𝜓 ∈ 𝐶0∞([𝜉1, 𝜉2]). Let 𝑤0be the strength of the Dirac delta function in𝜌, and denote
𝑢𝛿= lim
𝜀→0+𝑢 𝜀(𝜉𝜀
𝜎) = 𝑢(𝜎).
From (31), (35) and (37) it follows that ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝜎 = 𝑓(𝑢𝛿), 𝑤0= −𝜎[𝜌] + [𝜌𝑓(𝑢)], 𝑤0𝑢𝛿= −𝜎[𝜌𝑢] +[𝜌𝑢(𝑓(𝑢) +1𝜌)]. (38)
Under the entropy condition (32) the system (38) admits a unique solution(𝜎, 𝑤0, 𝑢𝛿). □ Then we get the following theorem.
Theorem 6. Suppose𝑓(𝑢−) > 𝑓(𝑢+) + 1
𝜌+. Let(𝜌
𝜀(𝑥, 𝑡), 𝑢𝜀(𝑥, 𝑡)) be the self-similar solution of (7)–(8). Then the limit
lim
𝜀→0+(𝜌
exists in the measure sense and(𝜌, 𝑢) solves (1)–(2). Moreover, (𝜌(𝑥, 𝑡), 𝑢(𝑥, 𝑡)) = ⎧ ⎪ ⎨ ⎪ ⎩ (𝜌−, 𝑢−), if𝑥 < 𝜎𝑡, (𝑤0𝑡𝛿(𝑥 − 𝜎𝑡), 𝑢𝛿), if 𝑥 = 𝜎𝑡, (𝜌+, 𝑢+), if𝑥 > 𝜎𝑡, where𝜎, 𝑤0, and𝑢𝛿are determined uniquely by the entropy condition𝑓(𝑢+) + 1
𝜌+ < 𝜎 < 𝑓(𝑢−) and ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝜎 = 𝑓(𝑢𝛿), 𝑤0= −𝜎[𝜌] + [𝜌𝑓(𝑢)], 𝑤0𝑢𝛿= −𝜎[𝜌𝑢] +[𝜌𝑢(𝑓(𝑢) +1𝜌)].
Finally, from the previous theorem we can conclude that the function𝑢𝛿(𝑡) is a constant which we denote by 𝑢𝛿,𝑥(𝑡) = 𝑓(𝑢𝛿)𝑡 and𝑤(𝑡) = (−𝑓(𝑢𝛿)[𝜌] + [𝜌𝑓(𝑢)])𝑡. Moreover by the above results, the uniqueness of delta shock wave is guaranteed by entropy condition (3) (see also (32)).
Theorem 7. Assume that𝑓(𝑢) is a smooth and strictly monotone function and 𝑓(𝑢−) > 𝑓(𝑢+) + 1
𝜌+. Then the Riemann problem
(1)–(2) admits one and only one measure solution of the form
(𝜌(𝑥, 𝑡), 𝑢(𝑥, 𝑡)) = ⎧ ⎪ ⎨ ⎪ ⎩ (𝜌−, 𝑢−), if𝑥 < 𝜎𝑡, (𝑤0𝑡𝛿(𝑥 − 𝜎𝑡), 𝑢𝛿), if 𝑥 = 𝜎𝑡, (𝜌+, 𝑢+), if𝑥 > 𝜎𝑡, where𝜎, 𝑤0, and𝑢𝛿are determined uniquely by the entropy condition𝑓(𝑢+) + 1
𝜌+ < 𝜎 < 𝑓(𝑢−) and ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 𝜎 = 𝑓(𝑢𝛿), 𝑤0= −𝜎[𝜌] + [𝜌𝑓(𝑢)], 𝑤0𝑢𝛿= −𝜎[𝜌𝑢] +[𝜌𝑢(𝑓(𝑢) +1𝜌)]. ACKNOWLEDGMENTS
This research was supported by FAPESP (Fundação de Amparo à Pesquisa do Estado de São Paulo, Brazil) under the Grant #2016/19502-4.
ORCID
Richard De la cruz https://orcid.org/0000-0003-1342-7946
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How to cite this article: De la cruz R, Santos M. Delta shock waves for a system of Keyfitz-Kranzer type. Z Angew Math
