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Homework Help: Second order differential equation

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Let V denote a vector space of twice differentiable functions on R. Define a linear map L on V by the formula:

    [tex]L(u)=au''+bu'+cu[/tex]

    Suppose that [tex]u_{1},u_{2}[/tex] is a basis for the solution space of L(u)=0. Find a basis for the solution space of the fourth order equation L(L(u))=0. What can you say about the kernels of L and [tex]L^{2}[/tex]?

    2. Relevant equations



    3. The attempt at a solution
    Since u1,u2 is a basis for the solution space of L(u)=0, then the kernel of L is a plane spanned by [tex]u_{1},u_{2}[/tex].

    Now, we want to find the basis for L(L(u))=0. Expanding L(L(u)), we get:

    [tex]a(au''+bu'+c)''+b(au''+bu'+c)'+c(au''+bu'+c)=0[/tex], simplifying we get:

    [tex]a^{2}u''''+2abu'''+(2ac+b^{2})u''+2bau'+c^{2}u=0[/tex]

    So the characteristic equation is:

    [tex]a^{2}\lambda^{4}+2ab\lambda^{3}+(2ac+b^{2})\lambda^{2}+2bc\lambda+c^{2}=0[/tex]

    [tex]a^{2}\lambda^{4}+2ab\lambda^{3}+b^{2}\lambda^{2}+2ac\lambda^{2}+2bc\lambda+c^{2}=0[/tex]

    [tex]\lambda^{2}(a\lambda+b)^{2}+2c\lambda(a\lambda+b)+c^{2}=0[/tex]

    [tex](\lambda(a\lambda+b)+c)^{2}=0[/tex]

    [tex]a\lambda^{2}+b\lambda+c=0[/tex]

    Now, since [tex]u_{1},u_{2}[/tex] is a basis for L(u)=0, then [tex]u_{1},u_{2}[/tex] must satisfies the characteristic equation [tex]a\lambda^{2}+b\lambda+c=0[/tex].

    Thus, a basis for L(L(u))=0 is [tex]u_{1},u_{2}[/tex]. The kernel of [tex]L^{2}[/tex] is also a plane spanned by [tex]u_{1},u_{2}[/tex].

    If this is true, then the solution space of L(L(u))=0 is exactly the same as L(u)=0. I don't really understand the meaning of this, as in why is it the same? I think something is wrong but I am not sure.

    Any help would be appreciated.
     
  2. jcsd
  3. Sep 22, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    No, the solution space to L(L(u))= 0 is not the same as the solution space to L(u)= 0. If L(u)= 0, then certainly L(L(u))= L(0)= 0 so the solution set for L(u)= 0 is a subspace of the solution set for L(u).

    But further, you have already that L(u)= 0 only for vectors of the form [itex]au_1+ bu_2[/itex] for numbers a and b, so that if L(L(u))= 0 then you must have [itex]L(u)= au_1+ bu_2[/itex]. Think of that as a non-homogeneous second order differential equation. Of course, [itex]u_1[/itex] and [itex]u_2[/itex] are already solutions to the homogeneous equation so you should try solutions like [itex]u= xu_1(x)[/itex] and [itex]u= xu_2(x)[/itex].
     
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