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SVD of a differential operator

  1. Nov 10, 2011 #1
    I'm just checking my work; please correct me if I'm wrong. Also, if someone has a suggestion for an intro book that covers SVD of operators, please let me know! I'm not fully confident with my procedure for general problems yet, so the more info, the better. Many thanks.

    1. The problem statement, all variables and given/known data

    Given

    [tex]\mathscr{D}\, f = \frac{df}{dx} + cf [/tex]

    defined in [itex]\mathbb{L}_2([0,1])[/itex] with [itex]f(0)=f(1)=0[/itex], obtain the eigenfunction for the operator and its singular value decomposition.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    1) Test this function:
    [tex]f(x)=A\sin (Bx)[/tex]
    If [itex]f(0)=A\sin (B0) = 0[/itex] OK
    If [itex]f(1)=A\sin (B) = 0 \therefore B = m\pi[/itex]
    Then [itex]f(x)=A\sin (m\pi \, x)[/itex] satisfies the boundary conditions and is an eigenfunction

    2) Find [itex]A[/itex] by taking the inner product of the eigenfunction [itex]f(x)=A\sin (m\pi \, x)[/itex] with itself in [itex]\mathbb{L}_2([0,1])[/itex]:

    [tex]\int _0 ^1 \left[ A \sin (m\pi \, x) \right] \left[ A \sin (m\pi \, x) \right] \, dx = 1[/tex]
    [tex]\frac{A^2}{2(m\pi \, x)}\left[ m\pi \, x - \cos (m\pi \, x) \sin (m\pi \, x) \right] = 1[/tex]
    [tex]\frac{A^2}{2} = 1 \therefore A = \pm \sqrt{2}[/tex]

    Arbitrarily choose the positive root, so [itex]f(x)=\sqrt{2}\sin (m\pi \, x)[/itex].

    3) Use the eigenvalue equation for the [itex]\mathscr{D} ^{\dagger} \mathscr{D}[/itex]:

    [tex]\mathscr{D} ^{\dagger} \mathscr{D} \, \vec{u}_n = \mu_n \vec{u}_n[/tex]

    That is, if

    [tex]\mathscr{D}\, f = \frac{df}{dx} + cf = \sqrt{2} (m\pi) \cos (m\pi x) + c \sqrt{2} \sin (m\pi x) = f^{\prime} [/tex]

    then because in this case [itex]\mathscr{D} ^{\dagger} = \mathscr{D}[/itex]

    [tex] \mathscr{D}^{\dagger}\, \mathscr{D}\, f = \mathscr{D}\, f^{\prime} = \frac{df^{\prime}}{dx} + cf^{\prime} = 2\sqrt{2} c(m\pi) \cos (m\pi x) + (c^2 - m^2 \pi ^2) \sqrt{2} \sin (m\pi x) [/tex]

    But if [itex]u_n = f(x) =\sqrt{2} \sin (m\pi x) [/itex] is one of the two singular vectors, then the cos(.) term must be set to zero (VALID STEP?), and we have [itex]\mu_n = c^2 - m^2 \pi ^2[/itex] as the singular value.

    To find the other singular vector, use the relationship:

    [tex]v_n = \frac{1}{\sqrt{\mu_n}} \mathscr{D} u_n[/tex]

    so


    [tex]v_n = \frac{1}{\sqrt{c^2 - m^2 \pi ^2}} \sqrt{2}\left[ (m\pi) \cos (m\pi x) + c \sin (m\pi x) \right][/tex]

    and this completes the SVD.
     
    Last edited: Nov 10, 2011
  2. jcsd
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