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Adjoint of a Differential Operator

  1. Dec 9, 2009 #1
    If the inner product is defined on V with dimension less than or equal to 3 as [tex]\left\langle f,g \right\rangle = \int_{0}^{1}f(x)g(x)dx[/tex], I'm trying to find D* such that [tex]\left\langle Df,g \right\rangle[/tex] = [tex]\left\langle f,D*g \right\rangle[/tex], and I thought I had a closed form of D*. If {1, x, (x^2)/2, (x^3)/3}. Call each element e_i, i = 0,...,3 form a basis for V, then [tex]\left\langle Df,g \right\rangle[/tex] = [tex]\left\langle f,D*g \right\rangle[/tex] holds given D* is defined as i times the (derivative of g(x)) if g(x) has degree greater than 0, but I can't think of a closed form so that it will work for all elements. Does anyone have any suggestions?
     
  2. jcsd
  3. Dec 10, 2009 #2
    Would [itex]\delta(x-1)-\delta(x)-D[/itex] work?

    [tex]\int_0^1 Df(x) g(x)dx = \int_0^1 (df(x)/dx) g(x)dx = f(x) g(x) |^1_0 - \int^1_0 f(x) Dg(x) dx = \int^1_0 f(x) [(\delta(x-1)-\delta(x)) - D]g(x) dx[/tex]
     
  4. Dec 10, 2009 #3

    HallsofIvy

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    ??? Please be careful to tell us exactly what you are doing! You start by talking about a vector space "V with dimension less than or equal to 3" but then start integrating? A vector space, of any dimension, does not necessarily contain functions at all. "A vector space with dimension less than or equal to 3" is NOT the same as "polynomials of degree 3 or less". (In fact, the space of polynomials of degree 3 or less has dimension four, not three, as you make clear by giving a basis consisting of four functions.)

    In any case, what D* is such that <Du, v>= <u, D*v> depends upon what D is! Do you mean that D is the derivative operator?
     
  5. Dec 10, 2009 #4
    Exactly: D is the derivative operator and I'm trying to find D*, which is simply the adjoint operator. I apologize for writing it so messily. I'm a first year graduate student still working on my writing skills, so if you have any suggestions, either for finding D* or improving writing skills, it would be greatly appreciated.
     
  6. Dec 11, 2009 #5
    Figured this out. Since every operator takes one some matrix representation, and we consider the orthonormal basis with respect to the differentiation operator, we can extend that basis to matrix form. But theres a theorem that states the matrix representation of D* is simply the transpose of the matrix represented by D. Therefore, I'm all set. Thanks for the suggestions!
     
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