Partial Differential Equations III/IV

[6pt] Exercise Sheet 3

Lecturer: Alpár R. Mészáros


  1. 1.

    Existence of classical solutions. Consider the conservation law

    ut+u3ux=0 for (x,t)×(0,T),
    u(x,0)=u0(x) for x,

    where

    u0(x)={1x<0,(1-x2)20x1,0x>1.

    Find the largest value of T for which this conservation law has a classical solution u:×[0,T).

  2. 2.

    Breakdown of classical solutions. Consider the conservation law

    ut+uux=0 for (x,t)×(0,),
    u(x,0)=11+x2 for x.
    • (i)

      Show that ut and ux blow up at (xc,tc) where tc=8/(33) and xc is to be determined.

    • (ii)

      Sketch the characteristics up until time tc.

    • (iii)

      Find an implicit equation for u that does not contain partial derivatives.

  3. 3.

    Geometric interpretation of tc: crossing characteristics. Consider the conservation law

    ut+c(u)ux=0 for (x,t)×(0,),
    u(x,0)=u0(x) for x,

    where c, u0C1(), u0 and u0 are bounded. Assume that

    I={s:c(u0(s))s<0}

    is nonempty and suppose that there exists scI such that

    tc=minsI-1c(u0(s))s=minsI-1c(u0(s))u0(s)=-1c(u0(sc))u0(sc).
    • (i)

      Let t>tc and x=sc+c(u0(sc))t. Show that there exists s, ssc, such that x=s+c(u0(s))t. Therefore the point (x,t) lies on (at least) two characteristics. Hint: see the proof of Theorem 3.5. Let h(s,t)=s+c(u0(s))t. Show that hs(sc,t)<0. Recall that h(s,t)± as s±.

    • (ii)

      Suppose that there are two characteristics given by x1(t)=s1+tc(u0(s1)) and x2(t)=s2+tc(u0(s2)) with s1,s2I, s1s2, that are crossing each other at point (x,t). Show that ttc, that is the time coordinate of the crossing point must be greater than the critical time. Hint: use the fact that the function sc(u0(s)) is Lipschitz continuous.

  4. 4.

    The Fundamental Lemma of the Calculus of Variations. Let g:Ω be a continuous function on an open set Ωn. Prove that if

    Ωg(𝒙)ψ(𝒙)𝑑𝒙=0for all ψCc(Ω),

    then g=0. Hint: Assume for contradiction that g0.

  5. 5.

    Weak formulation of scalar conservation laws. State the weak formulation of the conservation law

    ut+(ln|u|+1)ux=0 for (x,t)×(0,),
    u(x,0)=u0(x) for x,

    where

    u0(x)={1+cosxif |x|π,0if |x|>π.

    Hint:

    0u(ln|s|+1)𝑑s=0u(dsdsln|s|+1)𝑑s.
  6. 6.

    Weak formulation of scalar conservation laws in several spatial variables. Scalar conservation laws in several spatial variables take the form

    ut+div𝒇(u)=0 for (𝒙,t)n×(0,), (1)
    u(𝒙,0)=u0(𝒙) for 𝒙n,

    where 𝒇:n and u0:n are prescribed, and u:n×[0,) is the unknown. We assume that 𝒇C1(;n), u0L1(n). The weak formulation of (1) is the following: Find uL(n×(0,)) such that

    0n[u(𝒙,t)φt(𝒙,t)+𝒇(u(𝒙,t))φ(𝒙,t)]𝑑𝒙𝑑t+nu0(𝒙)φ(𝒙,0)𝑑𝒙=0 (2)

    for all φCc(n×[0,)).

    • (i)

      Let uC1(n×[0,)) be a bounded, classical solution, i.e., let u satisfy (1). Show that u is also a weak solution, i.e., that it satisfies (2).

    • (ii)

      Let uL(n×[0,)) be a weak solution. In addition, assume that u is continuously differentiable, uC1(n×[0,)). Show that u is a classical solution.

  7. 7.

    The Rankine-Hugoniot condition. Consider the initial value problem

    ut+uux=0 for (x,t)×(0,),
    u(x,0)=u0(x) for x,

    where

    u0(x)={2if x<0,1if 0<x<1,0if x>1.
    • (i)

      Use the Rankine-Hugoniot condition to find the shock that originates from x=0 and the shock that originates from x=1. Show that these shocks meet at (x,t)=(3/2,1).

    • (ii)

      Write down the weak solution for 0<t<1.

    • (iii)

      Find a weak solution for all t>0 by introducing a new shock at time t=1. Sketch the characteristics and the shocks.

    • (iv)

      Is the weak solution found in (iii) satisfying Lax’s entropy condition? Is it an entropy solution? Justify your answers!

  8. 8.

    The Lax entropy condition. Find all values of the positive constants a, b, c, d for which the function

    u(x,t)={-a(t+bx+ct2)if x>-dt2,0if x<-dt2,

    is an admissible weak solution of Burger’s equations for x, t[0,), where admissible means that it satisfies the Lax entropy condition. For these values of the parameters, is the weak solution an entropy solution? Justify your answer!

    Hint: The PDE and the Rankine-Hugoniot condition give three equations for a, b, c, d. This question is reproduced from Shearer & Levy (2015, page 212).

  9. 9.

    The Lax entropy condition again. Consider the PDE

    ut+13(u3)x=0.
    • (i)

      State the Rankine-Hugoniot condition for a shock wave of the form

      u(x,t)={u-if x<st,u+if x>st, (3)

      where u- and u+ are constants.

    • (ii)

      Find all shock waves of the form (3) with u-=1 that satisfy the Lax entropy condition.

    • (iii)

      Is the solution found by the arguments in (ii) an entropy solution? Justify your answer!

    This question is reproduced from Shearer & Levy (2015, page 212).

  10. 10.

    Rarefaction waves and shocks. Consider the initial value problem

    ut+u4ux=0 for (x,t)×(0,),
    u(x,0)=u0(x) for x,

    where

    u0(x)={0if x<0,1if 0<x<3,0if x>3.
    • (i)

      Sketch the characteristics. Explain why the problem has no classical solution for t>0.

    • (ii)

      Find a flux function f(u) such that the PDE can be written in the form of a conservation law. Hence write down the Rankine-Hugoniot jump condition for a shock of the form x=σ(t).

    • (iii)

      Derive the following weak solution for 0<t<15/4:

      u(x,t)={0if x<0,(x/t)1/4if 0<x<t,1if t<x<t/5+3,0if x>t/5+3.
    • (iv)

      Explain what happens at time t=15/4. Use the Rankine-Hugoniot condition to find a weak solution for all t>0. Sketch the characteristics and the shocks. Hint: The shape of the shock changes after time t=15/4; it is no longer a straight line.

  11. 11.

    Watching paint dry. Consider the initial value problem

    ut+u2ux=0 for (x,t)×(0,),
    u(x,0)=u0(x) for x,

    where

    u0(x)={0if x<0,1if 0<x<1,0if x>1.

    This PDE models paint flowing down a wall, where u(x,t) is the thickness of the paint at height x at time t. The initial condition corresponds to applying a strip of paint at time t=0. See J. Ockendon, S. Howison, A. Lacey & A. Movchan (2003) Applied Partial Differential Equations, Oxford, page 6.

    • (i)

      Derive the following weak solution for 0<t<3/2:

      u(x,t)={0if x<0,(x/t)1/2if 0<x<t,1if t<x<σ(t),0if x>σ(t),

      where σ(t)=1+t/3.

    • (ii)

      Explain what happens at time t=3/2. Use the Rankine-Hugoniot condition to find a weak solution for all t>0. Sketch the characteristics and the shocks. Hint: The shape of the shock changes after time t=3/2; it is no longer a straight line.

  12. 12.

    Traffic flow. Consider the traffic flow model

    ρt+vm(1-2ρ)ρx=0

    where ρ(x,t)[0,1] is the traffic density and vm is the speed limit. We saw this model in Example 2.1.

    • (i)

      Show that this PDE can be written in the form of a conservation law with flux f(ρ)=ρv(ρ), where v(ρ)=vm(1-ρ) is the velocity of the traffic.

    • (ii)

      Suppose that the initial density is constant except for a block of higher density, slower moving traffic:

      ρ(x,0)={1/4if x<0,1/3if 0<x<1,1/4if x>1.

      Find the weak solution ρ(x,t) that satisfies the entropy condition.

    • (iii)

      Sketch the solution ρ(x,t) as a function of x at times t=0, 6/vm, 24/vm. Describe the effect as t of the initial block of slower moving traffic.

    This question is based on a traffic model proposed in the paper M.J. Lighthill & G.B. Whitham (1955) On Kinematic Waves. II. A Theory of Traffic Flow on Long Crowded Roads, Proc. R. Soc. Lond. A, 229, 317–345.

  13. 13.

    Strong solutions. Show that if u is continuous and piecewise smooth and satisfies a scalar conservation law piecewise, then u is a weak solution of the conservation law. To be concrete, let g:[0,) be continuously differentiable and let u:×[0,) be bounded and continuous, and be continuously differentiable on the sets Vl={(x,t):t>0,x<g(t)}, Vr={(x,t):t>0,x>g(t)}. Assume that u satisfies ut+f(u)x=0 in Vl and Vr, in the classical sense, and satisfies u(x,0)=u0(x), where u0C() and fC1() are prescribed. Show that u is a weak solution of the scalar conservation law

    ut+f(u)x=0 for (x,t)×(0,),
    u(x,0)=u0(x) for x.

    Solutions of this form are called strong solutions. For those familiar with measure theory, u satisfies the classical form of the PDE almost everywhere.