Homework 3

  1. Recall the Volterra operator:

    (Vf)(x):=0xf(y)dy
    on
    L2[0,1]
    .
    Compute the singular value decomposition of
    V
    . What is its norm? Is it trace-class? Does it have a bounded inverse?

  2. Show that a general second order ODE

    a2(x)u(x)+a1(x)u(x)+a0(x)u(x)=f(x)
    with
    a2(x)0,uC2[a,b]

    has the same solution space as one of the form
    (p(x)u(x))+q(x)u(x)=F(x)

    This is called Sturm-Liouville Form. How does this transformation affect the boundary conditions?

  3. Show that a general Sturm-Liouville eigenvalue equation

    (p(x)u(x))+q(x)u(x)=λu(x),
    p(x)0
    , can be converted to Schrodinger form (which we studied in the previous lecture)
    U(s)+Q(s)U(s)=λU(s)

    via the change of variables (i.e., diffeomorphism):
    dxds=p(x),f(s):=(p(x(s)))1/4,Q(s)=q(x(s))+f(s)f(s).

    How does this affect the boundary conditions and domain of definition of U?
    Put the Kimura equation
    x(1x)u(x)=λu(x)

    in Schrodinger form.

  4. Show that

    uC2[a,b] satisfies the ODE
    u(x)+q(x)u(x)=λu(x)
    if and only if there are functions
    ρ(x),θ(x)
    such that the vector
    (u(x),u(x))R2
    is equal to
    (ρ(x),θ(x))
    with
    u(x)=ρ(x)sin(θ(x))
    and
    u(x)=ρ(x)cos(θ(x))
    , and they satisfy the system of (nonlinear) ODE:
    θ(x)=cos(θ(x))2+(λq(x))sin(θ(x))2,ρ(x)=ρ(1+q(x)λ)sin(2θ(x))2

  5. Show that the subspace

    {fC2[a,b]:f(a)=f(b)=0} is dense in
    L2[a,b]
    .

  6. Show that if

    TL(H) is finite rank, then
    dimker(IT)=dimker(IT)<.

  7. Read Section 2 of http://lie.math.okstate.edu/~binegar/4233/4233-l17.pdf, which gives an example of how Sturm-Liouville problems arise naturally from physical PDE via separation of variables. Note that these are exampels of singular SL problems, which refers to any problem not satisfying all of the hypotheses: (1)

    [a,b] is a finite interval (2)
    p(x)
    is continuous and bounded on
    (a,b)
    (3)
    p(x)0
    on
    [a,b]
    . Similar conclusions can be reached for such problems, but it is much more complicated and will be discussed later.