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Unknown integral kernel

  1. Sep 18, 2012 #1

    vmw

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    Consider,

    [tex] f(\mathbf{w}) = \int K(\mathbf{w,\mathbf{v}}) g(\mathbf{v}) d\mathbf{v} [/tex]

    where [itex]\mathbf{v},\mathbf{w} \in \mathbb{R^3}[/itex].

    Is it possible to solve for the integral kernel, [itex] K(\mathbf{w,\mathbf{v}}) [/itex], if [itex] f(\mathbf{w}) [/itex] and [itex] g(\mathbf{v}) [/itex], are known scalar functions and we require [itex] \int K(\mathbf{w,\mathbf{v}}) d\mathbf{v} = 1 [/itex]? These are definite integrals: [itex]\int \rightarrow \int_{a1}^{b1}\int_{a2}^{b2}\int_{a3}^{b3}[/itex]

    Thank you for any solution/advice/insight!
     
    Last edited: Sep 18, 2012
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  3. Sep 25, 2012 #2

    haruspex

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    Hi vmw,
    Suppose you have a K'(w,v) which satisfies the =1 integral constraint. Consider any function h(v), and define h'(v) = h(v)-∫h(v).dv/∫dv. So ∫h'(v).dv = 0.
    If it happens that ∫h'(v)g(v).dv is nonzero then set K = K' + h'(v).j(w) where
    j(w) = (f(w) - ∫K'(w,v)g(v).dv)/∫h'(v)g(v).dv.
     
    Last edited: Sep 25, 2012
  4. Sep 25, 2012 #3

    vmw

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    It seems like you are making an argument that there are infinitely many [itex]K(\mathbf{w},\mathbf{v}) [/itex] so one cannot solve for a unique kernel. This is a conclusion I have already come to. However, I do not understand your argument. How can you say [itex]h'(\mathbf{v}) = h(\mathbf{v}) - \frac{\int h(\mathbf{v}) d\mathbf{v}}{\int d\mathbf{v}}[/itex] implies [itex]\int h'(\mathbf{v}) d\mathbf{v}= 0[/itex]? Certainly this is true when [itex]h(\mathbf{v})[/itex] is a constant but how can you say this is so when [itex]h(\mathbf{v})[/itex] is any function of [itex]\mathbf{v}[/itex]?
     
  5. Sep 25, 2012 #4

    haruspex

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    Maybe it would have been clearer if I'd written
    [itex]h'(\mathbf{v}) = h(\mathbf{v}) - \frac{\int h(\mathbf{x}) d\mathbf{x}}{\int d\mathbf{x}}[/itex]
     
  6. Sep 25, 2012 #5

    vmw

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    I still can't see why defining [itex] h'(\mathbf{v}) = h(\mathbf{v}) + \frac{\int h(\mathbf{x}) d \mathbf{x}}{\int d \mathbf{x}}[/itex] implies [itex] \int h'(\mathbf{v}) d \mathbf{v} = 0 [/itex]. I am probably just being dense.

    Instead of getting caught up in the details... Is it accurate to say you are showing me that given a kernel K' that satisfies my condition, you can construct a different kernel K from K' that still satisfies my conditions. Thus K' is not unique?
     
  7. Sep 26, 2012 #6

    haruspex

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    It's -, not +:
    Write [itex]c = \frac{\int h(\mathbf{x}) d \mathbf{x}}{\int d \mathbf{x}}= \frac{\int h(\mathbf{v}) d \mathbf{v}}{\int d \mathbf{v}}[/itex]
    [itex] \int h'(\mathbf{v}) d \mathbf{v} = \int h(\mathbf{v}) d \mathbf{v} - \int c.d \mathbf{v} = c \int d \mathbf{v} - \int c.d \mathbf{v}= 0[/itex].
    Not quite. I'm saying that given a K' that satisfies one criterion (the = 1 criterion), it's not hard to construct one that satisfies both. There is the degenrate case where g is a constant. In this case there is no solution unless f happens to be the same (the integral of h'g will always be 0). It would be interesting to prove that an h' can be found which makes h'g integral nonzero when g is not (almost everywhere) constant.
     
    Last edited: Sep 26, 2012
  8. Sep 26, 2012 #7

    vmw

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    I see. I was, indeed, being dense. You make an interesting point. Thank you.

    Ultimately, I am now uninterested in the problem since I realized I cannot solve for a unique K. If you consider the discrete case of my problem where f and g are length-N vectors, there are infinitely many matrices, K, which satisfy the two relationships I have proposed. The continuous cause can be (not-so-rigorously) thought of as the case where N->infinity.
     
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