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Math Help - Linear Map Question

  1. #1
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    Linear Map Question

    Let V denote the vector space of twice differentiable functions on R. Define a linear map L on V by the formula

    L(u)=a\frac{d^2}{dx^2}u+b\frac{d}{dx}u+cu

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

    This is my working out so far:

    Since u_{1},u_{2} is a basis for the solution space of L(u)=0, then u_{1},u_{2} is a basis for the kernal of L.

    Therefore we can write

    u=\lambda u_{1}+\mu u_{2} ,\ \ \ \ \ \ \ \forall u\ \epsilon \ Ker(L), \ \lambda,\mu \ \epsilon \ R

    So Nullity(L)=2 since Ker(L) is spanned by two linearly independent vectors.

    Now, consider L(L(u))=0. Since every solution of L(u)=0 can be represented as  \lambda u_{1}+\mu u_{2},
    the equation L(L(u))=0 becomes

     L(u)=\lambda u_{1}+\mu u_{2}, which then becomes

    a\frac{d^2}{dx^2}u+b\frac{d}{dx}u+cu=\lambda u_{1}+\mu u_{2}.

    Let Y_{H }, Y_{P } denote the homogenous solution and the particular solution respectively.

    Obviously Y_{H }=\lambda u_{1}+\mu u_{2} as u_{1},u_{2} is a basis for the solution space of L(u)=0.

    Now this is where I got stuck. I did’t know what the particular solution is. I let Y_{P}=mxu_{1}+nxu_{2},\ \ \ \ \ \ m,n\ \epsilon \ R

    and tried to solve for the constants m,n. It was a mess and I don’t think I was on the right track.

    I was thinking that if I can find Y_{P}, then Y_{H }, Y_{P } would be the basis of the fourth order differential equation L(L(u))=0. Am I correct?

    Any help will be appreciated.

    Thanks.
    Last edited by sakodo; January 16th 2011 at 03:32 AM.
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  2. #2
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    I think you are "overworking" this. For any linear transformation, L, L(0)= 0 so that, if u is in the kernel of L, Lu= 0, the L^2u= L(Lu)= L(0)= 0. That is, for any linear transformation, L, the kernel of L is a subspace of the kernel of L^2.
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  3. #3
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    Thanks. Yep I understand what you mean by 'Ker(L) is a subspace of Ker(L^2)''.

    I think Nullity(L^2)=4 because I have read somewhere that the solution space of an nth order homogeneous differential equation forms an nth dimensional vector space. Is L(L(u))=0 a fourth order homogeneous equation? How would you find the basis?
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  4. #4
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    Bump. Anyone know how to find the basis for L(L(u))? Many thanks.
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