Partial Differential Equations III & V

[6pt] Exercise Sheet 1

Lecturer: Alpár R. Mészáros


  1. 1.

    Examples of PDEs. State the order and type (linear, semilinear, quasilinear, fully nonlinear) of the following PDEs:

    • (i)

      The Schrödinger equation:

      iΨt=-22mΔΨ+VΨ.

      This is the Schrödinger equation for a single particle moving in an electric field. The function Ψ(𝒙,t) is the wave function, is the reduced Planck constant, m is the reduced mass of the particle, and V(𝒙,t) is the potential of the electric field. The squared modulus of the wave function |Ψ(𝒙,t)|2 is the probability density of finding the particle at position 𝒙 at time t.

    • (ii)

      The p–Laplace equation:

      div(|u|p-2u)=0,

      where p[2,). Observe that we recover Laplace’s equation when p=2.

    • (iii)

      The Cahn-Hilliard equation:

      ut=kΔ(u3-u-ε2Δu),

      where k>0, ε>0 are constants. This PDE describes a mixture of two fluids. It was originally introduced to model a molten mixture of iron and nickel (occurring in the manufacturing process of an iron-nickel alloy). The function u(x,t) represents the amount of each fluid present at position x and time t. If u=1, then only fluid A is present; if u=-1, then only fluid B is present; if u(-1,1), then both fluids are present, e.g., if u=0 then there is an equal quantity of fluid A and B. For more information see the PDE Coffee Table book by L.N. Trefethen & K. Embree (editors): https://people.maths.ox.ac.uk/trefethen/pdectb.html

    • (iv)

      The nonlinear wave equation:

      ρ𝒓tt=x(N(|𝒓x|)𝒓x|𝒓x|)+𝒇.

      This system of PDEs describes large deformations of a uniform elastic string (unlike the wave equation, which only describes small deformations). The variable x[0,L] represents points of the string, and the unknown 𝒓(x,t)3 is the position of point x at time t. The constant ρ>0 is the density of the string and 𝒇(x,t) is the applied force on the string. The function N:[0,) describes the stress-strain behaviour of the string; N>0 if the string is under tension (|𝒓x|>1), N<0 if the string is under compression (|𝒓x|<1), and N=0 if the string is unstressed (|𝒓x|=1).

  2. 2.

    Characterisation of PDEs. Write down the general form of a

    • (i)

      second-order scalar PDE in two independent variables;

    • (ii)

      linear, second-order scalar PDE in two independent variables;

    • (iii)

      semilinear, second-order scalar PDE in two independent variables;

    • (iv)

      quasilinear, second-order scalar PDE in two independent variables.

    Remark: You can write your answers in a more succinct form by using vector notation. In particular, the following notation is useful: Let D2u denote the matrix of second-order partial derivatives of u (also known as the Hessian matrix of u),

    D2u=(uxxuxyuxyuyy),

    and let A:B denote the inner product of two n-by-n matrices A and B, which is defined by

    A:B=i,j=1naijbij.
  3. 3.

    The transport equation: Derivation of the travelling wave solution. In this exercise we derive the travelling wave solution u(x,t)=g(x-ct) of the transport equation

    ut+cux=0 for (x,t)×(0,),u(x,0)=g(x) for x,

    where c and g:.

    • (i)

      Show that the transport equation can be written as

      u(c1)=0

      where u=(ux,ut)T.

    • (ii)

      It follows that u is constant in the direction (c,1)T. In particular, for each x0, u is constant on the line that passes through (x,t)=(x0,0) and has direction (c,1)T. By finding the equation of this line and using the fact that u(x,t)=u(x0,0) for all points (x,t) on this line, show that

      u(x,t)=g(x-ct).

      We will see a more systematic way of solving the transport equation in Chapter 2.

  4. 4.

    The transport equation on n. Find the travelling wave solution of the initial value problem

    ut+𝒄u=0 for (𝒙,t)n×(0,),
    u(𝒙,0)=g(𝒙) for x,

    where 𝒄n is a constant vector and u:n×[0,), (𝒙,t)u(𝒙,t). Hint: Guess the solution by comparing with the transport equation in one spatial variable, ut+cux=0.

  5. 5.

    The transport equation with boundary conditions. This exercise illustrates why care must be taken when prescribing boundary conditions for the transport equation. It is adapted from M. Shearer & R. Levy (2015) Partial Differential Equations, Princeton University Press.

    • (i)

      Solve the initial-boundary value problem

      ut+4ux=0 for (x,t)(0,)×(0,),
      u(x,0)=0 for x(0,),
      u(0,t)=t2e-t for t[0,).

      Hint: Either guess the solution using the physical intuition that this equation transports information from the left to the right with speed 4, in particular that the information on the boundary x=0 will be transported to the right with speed 4. Or, easier, seek a travelling wave solution of the form u(x,t)=f(x-4t) for some function f: to be determined.

    • (ii)

      Show that if the PDE is changed to ut-4ux=0 (with the same initial and boundary conditions), then there is no travelling wave solution of the form u(x,t)=f(x+4t). Can you also give a physical explanation for this?

  6. 6.

    The heat equation on the real line. Let k>0 and gC() be bounded. Define u:×(0,) by

    u(x,t)=14πkt-e-(x-y)24ktg(y)𝑑y.
    • (i)

      Prove that u satisfies ut=kuxx for x, t>0.
      Hint: Compute ut,ux,uxx by bringing the partial derivatives inside the integral. You do not need to prove that you can interchange the order of differentiation and integration.
      In Chapter 5 we will prove that limt0u(x,t)=g(x).

    • (ii)

      Assume that g(x)=u0 is a constant function, u0. Find an explicit expression for u(x,t).
      Hint: See Appendix A.7 of the lecture notes.

    • (iii)

      Let L1() be the set of functions defined by

      L1()={φ::-|φ(x)|𝑑x<}.

      Assume additionally that gL1(). Prove that

      limtu(x,t)=0

      for all x. Explain why this does not contradict your answer to (ii).
      Hints: Prove an upper bound on |u(x,t)| using inequalities (i) and (ii) from Appendix A.5. Then use your estimate to show that |u(x,t)|0 as t.

    • (iv)

      Let 0gL1(). Show that the total heat energy of the system is conserved, i.e., show that

      -u(x,t)𝑑x=-g(x)𝑑xfor all t>0.

      Hint: Interchange the order of integration. You do not need to prove that the order of integration can be interchanged.

  7. 7.

    Revision of vector calculus and integration by parts in many variables.

    • (i)

      Let 𝒇:nn, φ:n, n2. Prove the generalised product rule

      div(φ𝒇)=φ𝒇+φdiv𝒇.
    • (ii)

      Let Ωn be an open and bounded set with smooth boundary, which we denote by Ω. Let 𝒏 denote the outward-pointing unit normal vector field to Ω. Use part (i) to prove the following generalised integration by parts formula:

      Ωφdiv𝒇𝑑𝒙=Ωφ𝒇𝒏𝑑S-Ωφ𝒇d𝒙.

      Hints: Recall the proof of the integration by parts formula for functions of one variable. You will also need to use the Divergence Theorem:

      Ωdiv𝒇𝑑𝒙=Ω𝒇𝒏𝑑S.
    • (iii)

      Prove that Δu=divu.

    • (iv)

      Put everything together to conclude that

      ΩφΔu𝑑𝒙=Ωφu𝒏dS-Ωuφd𝒙.
  8. 8.

    Poisson’s equation with Neumann boundary conditions. Consider Poisson’s equation with Neumann boundary conditions:

    -Δu=f in Ω, (1)
    u𝒏=g on Ω,

    where Ωn is an open and bounded set with smooth boundary, n2, and 𝒏 is the outward-pointing unit normal vector field to Ω. The given data for the problem are f:Ω and g:Ω, and the unknown is u:Ω¯. This equation describes the equilibrium temperature u(𝒙) of a body Ω subject to heat source f in Ω and a heat flux g through Ω.

    • (i)

      Prove that a necessary condition for the existence of a solution to (1) is

      Ωf𝑑𝒙+Ωg𝑑S=0.

      Hint: See Example 1.15 in the lecture notes.

    • (ii)

      Prove that (1) has either no solution or infinitely many solutions.
      Hint: If u is a solution, how could you modify it to obtain another solution?

  9. 9.

    Implicit form of Burger’s equation. Let u0: be a continuously differentiable function with bounded derivative. Suppose that u:×[0,tc) is a continuously differentiable function satisfying

    u(x,t)=u0(x-u(x,t)t)

    where tc=inf{-1u0(s):s,u0(s)<0}. Show that u satisfies Burger’s equation:

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