Partial Differential Equations III/IV

[6pt] Exercise Sheet 4


  1. 1.

    Green’s functions. Consider Poisson’s equation in one dimension with mixed Dirichlet-Neumann boundary conditions:

    -u′′(x)=f(x),x(0,1),
    u(0)=0,u(1)=0,

    where fC([0,1]). Find the unique solution of this equation. Write your solution in the form

    u(x)=01G(x,y)f(y)𝑑y.
  2. 2.

    Homogenization. Let α>0 be constant and a:[α,) be 1–periodic, i.e., a(x+1)=a(x) for all x. For ε>0, define the ε–periodic function aε(x)=a(xε). Consider the steady diffusion equation

    -(aε(x)uε(x))=f(x),x(0,1), (1)
    uε(0)=uε(1)=0, (2)

    where f:[0,1] is continuous. This models the equilibrium temperature distribution in a metal bar with heat source f, where the bar is made of a nonhomogeneous material. The thermal conductivity aε depends on position and represents a metal bar composed of repeating segments of length ε. We will discover that, for small ε, the bar behaves as if it were made of a homogenous material with constant thermal conductivity a0. You might guess that a0 is some sort of average of aε, but what is the correct notion of average? This is called a homogenization problem. Homogenization is an active research area.

    • (i)

      Derive the following solution to (1), (2):

      uε(x)=011aε(z)0zf(y)𝑑y𝑑z0x1aε(z)𝑑z(011aε(z)𝑑z)-1-0x1aε(z)0zf(y)𝑑y𝑑z.
    • (ii)

      Let εn=1n. Prove that limnuεn(x)=u0(x), where

      u0(x)=x(1a)¯010zf(y)𝑑y𝑑z-(1a)¯0x0zf(y)𝑑y𝑑z

      and where

      (1a)¯=011a(y)𝑑y

      is the average of 1/a over one period.

      Hint: Use the following deep result, which you do not need to prove: Let gL() be 1–periodic. For any interval [c,d],

      limncdg(nz)h(z)𝑑z=cdg¯h(z)𝑑z  hL1()C1() (3)

      where g¯ is the average of g over one period, i.e., g¯=01g(y)𝑑y. This also holds under the weaker assumption that hL1(). Define gn(y):=g(ny). We say that gn converges weak-* in L() to g¯ as n, but that’s another story for a functional analysis course.

    • (iii)

      Write u0 in the form

      u0(x)=01G(x,y)f(y)𝑑y

      for some symmetric Green’s function G, which you should determine.

    • (iv)

      Use part (ii) to show that u0 satisfies the steady diffusion equation

      -a0u0′′(x)=f(x),x(0,1),
      u0(0)=u0(1)=0,

      where the thermal conductivity a0 is the constant

      a0=1(1a)¯.
    • (v)

      Observe that a0 is the reciprocal of the average of the reciprocal of a. In general this is not the same as the average of a, as we now illustrate. Let

      a(x)={12x(0,12),1x(12,1),

      and extend a by periodicity to the real line. This represents a composite metal bar composed of segments of two homogeneous materials with different thermal conductivities. Compute a0 and a¯. Verify that a0a¯.

    • (vi)

      Bonus, optional question (hard): Prove (3). This is a form of the Riemann-Lebesgue Lemma. It is a generalisation of the Riemann-Lebesgue Lemma for Fourier series and the Fourier transform.

      Hint: Start by writing

      cdg(nz)h(z)𝑑z=cd(1n0nzg(y)𝑑y)zh(z)𝑑z.
  3. 3.

    Radial symmetry of Laplace’s equation on n. Let

    O(n,)={Mn×n:MMT=MTM=I}

    be the set of real, n–by–n orthogonal matrices, which represent rotations and reflections of n. Let v:n be a harmonic function and let RO(n,). Define w:n by w(𝒙):=v(R𝒙). Prove that w is also harmonic.

  4. 4.

    Fundamental solution of Poisson’s equation in 3D. Let Φ be the fundamental solution of Poisson’s equation in 3:

    Φ(𝒙)=14π1|𝒙|.
    • (i)

      Let R>0. Compute ΦL1(BR(𝟎)).

      Hint: Use spherical polar coordinates or, simpler, the following formula:

      BR(𝟎)f(𝒙)𝑑𝒙=0R(Br(𝟎)f(𝒚)𝑑S(𝒚))𝑑r.
    • (ii)

      Prove that ΦLloc1(3).

    • (iii)

      Prove that ΦL1(3).

    • (iv)

      Prove that ΦLloc1(3).

  5. 5.

    Fundamental solution of Poisson’s equation in 1D. Let fCc2() be twice continuously differentiable with compact support. Define

    Φ(x):={xifx0,0ifx0.

    We call Φ the fundamental solution of Poisson’s equation in . Define u:=Φ*f. Prove that u satisfies

    -u′′(x)=f(x),x.

    Hint: Write

    u′′(x)=(Φ*f)′′(x)=(f*Φ)′′(x)=-f′′(x-y)Φ(y)𝑑y.

    Now integrate by parts. Unlike for the case of Poisson’s equation in n, n2, you do not need to remove a ball of radius ε around the origin since Φ does not have a singularity at the origin in 1D. In fact, unlike in higher dimensions, Φ is continuous in 1D.

  6. 6.

    The function spaces L1 and Lloc1. Let f:, f(x)=|x|k. Let R>0. Find all the values of k for which

    • (i)

      fL1((-R,R)),

    • (ii)

      fL1((R,)),

    • (iii)

      fLloc1(),

    • (iv)

      fL1().

  7. 7.

    Properties of the convolution. Let φLloc1(), fCc(). Prove

    • (i)

      |(φ*f)(x)|< for all x;

    • (ii)

      if φL1(), then φ*fL();

    • (iii)

      the convolution is commutative: φ*f=f*φ.

    Remark: It can be shown that φ*fL1() if φ,fL1(). Consequently (L1(),*) is an algebra.

  8. 8.

    The Poincaré inequality for functions that vanish on the boundary. Prove that there exists a constant C>0 such that

    ab|f(x)|2𝑑xCab|f(x)|2𝑑x

    for all fC1([a,b]) satisfying f(a)=f(b)=0.
    Hint: The proof is similar to, and simpler than, the version we proved in Section 4.3.

  9. 9.

    The Poincaré inequality on unbounded domains.

    • (i)

      Construct a sequence fnC1()L2() such that fnL2() and

      fnL2()=constant,fnL2()n.

      This means that the Poincaré inequality does not hold on , i.e, there does not exists any C>0 such that

      -|f(x)|2𝑑xC-|f(x)|2𝑑x

      for all fC1()L2() with fL2().

    • (ii)

      Let Ω be the unbounded domain Ω=(a,b)×(-,). Prove that there exists C>0 such that

      Ω|f(𝒙)|2𝑑𝒙CΩ|f(𝒙)|2𝑑𝒙

      for all fC1(Ω¯)L2(Ω) with fL2(Ω) and with f(a,y)=f(b,y)=0 for all y. More generally, the Poincaré inequality is true if Ω is bounded between two parallel hyperplanes (lines in 2D, planes in 3D, etc). In this example Ω is bounded between the lines x=a and x=b.

  10. 10.

    The Poincaré constant depends on the domain. Let C1>0 satisfy

    01|f(x)|2𝑑xC101|f(x)|2𝑑x (4)

    for all fC1([0,1]) with f(0)=f(1)=0. By using a change of variables, use (4) to prove that

    0L|g(x)|2𝑑xCL0L|g(x)|2𝑑x (5)

    for all gC1([0,L]) with g(0)=g(L)=0, where

    CL=L2C1.

    Remark: Those with a good physical intuition will see that CL must have units of length squared, otherwise the units in equation (5) do not match: if g is dimensionless, then 0L|g|2𝑑x has units of length whereas 0L|g|2𝑑x has units of 1/length.

  11. 11.

    Eigenvalues of -Δ: Can you hear the shape of a drum? Let Ω2 be open and bounded with smooth boundary. Let u:Ω¯ be a smooth eigenfunction of -Δ that vanishes on Ω, which means u0 and

    -Δu=λu in Ω,
    u=0 on Ω,

    for some λ. We say that λ is the eigenvalue associated to the eigenfunction u. Use the energy method to prove that λ is real and that λ>0.

    Hint: Start by multiplying the PDE by u*, the complex conjugate of u.

    Remark: This eigenvalue problem arises if you seek a solution of the form v(𝒙,t)=u(𝒙)eiωt of the wave equation vtt=cΔv with clamped boundary conditions, which models small vibrations of a drum of shape Ω. The eigenvalues λ are related to the principal frequencies ω of the drum by λ=ω2/c. It can be shown that there are countably-many eigenvalues {λi}i. Moreover, H. Weyl showed that the eigenvalues determine the area of Ω; you can hear the area of a drum. In 1966 M. Kac asked whether the eigenvalues determine the shape of Ω; can you hear the shape of a drum? This was disproved in 1992 by Gordon, Webb and Wolpert, who constructed two distinct, non-convex polygons with the same principal frequencies. As a final twist, it is possible to hear the shape of a convex drum; two distinct convex sets have different principal frequencies.

  12. 12.

    The optimal Poincaré constant and eigenvalues of -Δ.

    • (i)

      Use the energy method to show that if the pair (λ,u)×C2(Ω¯), u0, satisfies

      -Δu =λuin Ω, (6)
      u =0   on Ω,

      then

      λ=Ω|u|2𝑑𝒙Ω|u|2𝑑𝒙. (7)

      This is the Rayleigh quotient formula for the eigenvalue λ in terms of the eigenfunction u. It can be shown that there are countably many eigenvalues and that 0<λ1<λ2λ3.
      Remark: The Rayleigh quotient formula for the matrix eigenvalue problem A𝒙=λ𝒙 is

      λ=𝒙TA𝒙𝒙T𝒙.

      This has the same form of (7) but with the L2–inner product in (7) replaced by the dot product.

    • (ii)

      Define

      V={φC1(Ω¯):φ=0 on Ω,φ0}

      and define the functional E:V by

      E[v]=Ω|v|2𝑑𝒙Ω|v|2𝑑𝒙.

      Suppose that uC2(Ω¯)V minimises E, i.e.,

      E[u]=minvVE[v].

      Prove that

      -Δu =λ1uin Ω,
      u =0   on Ω,

      where λ1 is the smallest eigenvalue of (6), and that

      E[u]=λ1.

      Remark: We have shown that the minimum value of the Rayleigh quotient E is the smallest eigenvalue of the operator -Δ on V, and that E is minimised by the corresponding eigenfunction. This is analogous to the result that if A is a symmetric positive definite matrix, then the minimum value of the Rayleigh quotient 𝒙TA𝒙/𝒙T𝒙 is the smallest eigenvalue of A, and it is minimised by the corresponding eigenvector. (Recall also that the maximum value of the Rayleigh quotient is the largest eigenvalue of A). Equivalently, the minimum value of the quadratic form 𝒙TA𝒙 over the sphere |𝒙|=1 is the smallest eigenvalue of A.

    • (iii)

      The optimal Poincaré constant is the smallest value of C>0 such that

      fL2(Ω)CfL2(Ω) (8)

      for all fC1(Ω¯) with f=0 on Ω. Let us denote this value of C by CP. Show that

      1CP=inffVfL2(Ω)fL2(Ω).
    • (iv)

      Combine parts (ii) and (iii) to conclude that the optimal Poincaré constant is

      CP=1λ1.
    • (v)

      Use part (iv) to show that the optimal Poincaré constant for the domain Ω=(0,2π) is CP=2. Compare this with the constant you obtained in Q8 with a=0, b=2π.
      Remark: The optimal constant CP=2 can also be obtained using Fourier series.

  13. 13.

    Uniqueness for Poisson’s equation with Robin boundary conditions. Let Ωn be open and bounded with smooth boundary. Let α>0. Use the energy method to show that there is at most one smooth solution of

    -Δu=f in Ω,
    u𝒏+αu=g on Ω,

    where 𝒏 denotes the outward-pointing unit normal vector field to Ω. This type of boundary condition is called a Robin boundary condition.

  14. 14.

    Uniqueness and stability for a more general elliptic problem. Consider the linear, second-order, elliptic PDE

    -div(Au)+𝒃u+cu=f in Ω, (9)
    u=g on Ω,

    where Ωn is open and bounded with smooth boundary, AC1(Ω¯;n×n), 𝒃C(Ω¯;n), and c,f,gC(Ω¯). Assume that c is nonnegative, div𝒃=0, and A is uniformly positive definite, i.e., there exists a constant α>0 such that 𝒚TA(𝒙)𝒚α|𝒚|2 for all 𝒚n, 𝒙Ω.

    • (i)

      Prove that (9) has at most one solution uC2(Ω¯).

    • (ii)

      Let (An)n be a sequence of matrix valued functions satisfying the previous conditions (in particular they are all uniformly positive definite with the same α>0). The rest of the data is fixed and satisfy all the previous assumptions. Suppose that un is the unique solutions to

      -div(Anun)+𝒃un+cun=f in Ω,
      un=g on Ω,

    Show that if AnA uniformly in Ω¯ as n+ and if the problem with the limit matrix A has a unique solution u, then unu in L2(Ω), as n+.

    Remarks: Uniqueness may fail if c is negative; see the PDEs exam from May 2017, Q4(b). Another obstacle to uniqueness is unbounded domains; see Exercise Sheet 5. We say that the PDE (9) has divergence form, which is the most convenient form for energy methods (compare (9) with the general form of elliptic PDEs given in Definition 4.1).

  15. 15.

    Uniqueness for a degenerate diffusion equation. Let m>1 be a constant. Show that the following steady degenerate diffusion equation has a unique positive solution:

    Δum=0 in Ω,
    u=π on Ω.

    Remark: Observe that Δum=div(um)=div(mum-1u)=div(a(u)u) with a(u)=mum-1. We call the equation Δum=0 the degenerate diffusion equation since the diffusion coefficient a(u)=mum-1 vanishes when u=0.

  16. 16.

    The H01 and H1 norms. Let

    V={fC1([a,b]):f(a)=f(b)=0}.
    • (i)

      Prove that L2([a,b]) is a norm on C([a,b]).

      Hint: The only difficulty is proving the triangle inequality. Write

      f+gL2([a,b])2=fL2([a,b])2+gL2([a,b])2+2abf(x)g(x)𝑑x

      and use the Cauchy-Schwarz inequality.

    • (ii)

      Prove that H1([a,b]) is a norm on C1([a,b]).

    • (iii)

      Prove that H01([a,b]) is a norm on V. Is it a norm on C1([a,b])?

    • (iv)

      Prove that the norms H1([a,b]) and H01([a,b]) are equivalent on V, which means that there exist constants c,C>0 such that

      cfH01([a,b])fH1([a,b])CfH01([a,b])fV.

      Hint: Use the Poincaré inequality to find C.

      Remark: If two norms are equivalent, then a sequence converges in one norm if and only if it converges in the other.

  17. 17.

    Continuous dependence. Let Ωn be open and bounded with smooth boundary. Let uC2(Ω¯) satisfy

    -div(Au)+cu=f in Ω,
    u=0 on Ω,

    where fC(Ω¯), AC1(Ω¯;n×n) is uniformly elliptic (see Q9), and c>0 is a constant. Prove that there exists a constant C>0 such that

    uH1(Ω)CfL2(Ω).
  18. 18.

    Continuous dependence with a first-order term. Let Ωn be open and bounded with smooth boundary. Let k>0, c>0 be constants and let f:Ω¯, 𝒃:Ω¯n be continuous. Let uC2(Ω¯) satisfy

    -kΔu+𝒃u+cu=f in Ω, (10)
    u=0 on Ω.
    • (a)

      Prove that

      kuL2(Ω)2+Ω(𝒃u)u𝑑𝒙+cuL2(Ω)2fL2(Ω)uL2(Ω).
    • (b)

      Prove that for all ε>0

      |Ω(𝒃u)u𝑑𝒙|𝒃L(Ω)(εuL2(Ω)2+14εuL2(Ω)2).

      Hint: You may use the Young inequality, which states that

      αβ12α2+12β2α,β>0.
    • (c)

      Prove that for all ε>0

      (k-ε𝒃L(Ω))uL2(Ω)2+(c-𝒃L(Ω)4ε)uL2(Ω)2fL2(Ω)uL2(Ω).
    • (d)

      Find a constant c0>0 such that if c>c0, then

      uH1(Ω)MfL2(Ω)

      for some constant M>0.

    • (e)

      Show that if c>c0, then u is the only solution of (10).

  19. 19.

    Neumann boundary conditions for variational problems. Let Ωn be open and bounded with smooth boundary. Define E:C1(Ω¯) by

    E[v]=12Ω|v|2𝑑𝒙-Ωfv𝑑𝒙.

    Suppose that uC1(Ω¯) minimises E:

    E[u]=minvC1(Ω¯)E[v].
    • (i)

      Show that

      Ωuφd𝒙=Ωfφ𝑑𝒙for all φC1(Ω¯).

      This is the weak formulation of Poisson’s equation with zero Neumann boundary conditions, as the following part demonstrates:

    • (ii)

      Show that, if in addition uC2(Ω¯), then

      -Δu=f in Ω,
      u𝒏=0 on Ω,

      where 𝒏 denotes the outward-pointing unit normal vector field to Ω.

      Hint: First choose test functions φC1(Ω¯) such that φ=0 on Ω. Use this to establish that -Δu=f. Then choose any test function φC1(Ω¯) and show that the boundary condition holds.

      Remark: Observe that the Neumann boundary condition arises naturally without including it in the domain of E (cf. the case of Dirichlet boundary conditions in Section 4.5, where the boundary condition is included in the domain of the energy functional). Consequently Neumann boundary conditions are sometimes referred to as natural boundary conditions.

  20. 20.

    The p–Laplacian operator. Let Ωn be open and bounded with smooth boundary. Let

    V={φC1(Ω¯):φ=0 on Ω}.

    For 1p<, define Ep:V by

    Ep[v]=1pΩ|v|p𝑑𝒙-Ωfv𝑑𝒙.

    We met the case p=2 in Section 4.5 and the previous question. Suppose that uC2(Ω¯)V minimises Ep:

    Ep[u]=minvVEp[v].
    • (i)

      Prove that u satisfies the PDE

      -Δpu=f in Ω,
      u=0 on Ω,

      where Δp is the p–Laplacian operator, which is defined by Δpv=div(|v|p-2v). By taking p=2 we recover the regular Laplacian operator: Δ2=Δ.

    • (ii)

      Show that

      Ep[u]=1-ppΩ|u|p𝑑𝒙=1-ppΩfu𝑑𝒙.
  21. 21.

    The minimal surface equation: PDEs and soap films. This question is adapted from the PDEs exam, May 2017. Let Ω2 be open and bounded with smooth boundary. Let g:Ω be a given smooth function and let

    V={φC1(Ω¯):φ=g on Ω}.

    Define A:V by

    A[v]=Ω1+|v|2𝑑𝒙.

    Observe that A[v] is the area of the surface {(x,y,v(x,y)):(x,y)Ω}, i.e., A[v] is the surface area of the graph of v. Suppose that the graph of uC2(Ω¯)V has minimal surface area amongst all graphs with given boundary g:

    A[u]=minvVA[v].

    Show that u satisfies the minimal surface equation

    div(u1+|u|2)=0in Ω.
  22. 22.

    Homogenization and the calculus of variations. In this question we revisit the homogenization problem from Q2 from the viewpoint of the calculus of variations. Let l,r be constants. Define

    V={φC1([0,1]):φ(0)=l,φ(1)=r}.

    Let α>0 be constant and a:[α,) be continuously differentiable. Let f:[0,1] be continuous. Define the energy functional E:V by

    E[v]=1201a(x)|v(x)|2𝑑x-01f(x)v(x)𝑑x.

    Observe that we recover the Dirichlet energy when a=1.

    • (i)

      Suppose that uC2([0,1])V minimises E:

      E[u]=minvVE[v].

      Show that u satisfies

      -(a(x)u(x))=f(x),x(0,1),
      u(0)=l,u(1)=r.
    • (ii)

      Now assume that a is 1–periodic and consider the energy

      En[v]=1201a(nx)|v(x)|2𝑑x-01f(x)v(x)𝑑x.

      Show that

      limnEn[v]=E[v]:=1201a¯|v(x)|2𝑑x-01f(x)v(x)𝑑x

      where

      a¯=01a(x)𝑑x.

      This means the nonhomogeneous energy En converges pointwise to the homogeneous energy E.

    • (iii)

      Let l=r=0. By part (i), if unC2([0,1])V is the minimiser of En, then un satisfies (1), (2) for ε=1/n. Let uC2([0,1])V be the minimiser of E. Show that

      limnun(x)u(x).

      Interpretation: We have shown that En converges pointwise to E, but the minimiser of En does not converge to the minimiser of E. The moral of the story is that pointwise convergence is not the ‘correct’ notion of convergence when considering energy functionals. The correct notion is something called Γ–convergence. It can be shown that En Γ–converges in a suitable sense to

      E0[v]=1201a0|v(x)|2𝑑x-01f(x)v(x)𝑑x

      where a0 was defined in Q2(iv). It is easy to check that the minimiser of En converges to the minimiser of E0. The subject of Γ–convergence goes beyond the scope of this course, but the important property of Γ–convergence is that minimisers converge to minimisers. This means that if you are modelling a system with an energy functional and you want to simplify the functional by sending a large parameter n or a small parameter ε0, then you should compute the Γ–limit, not the pointwise limit.