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Proving properties of the Dirac delta function

  1. Apr 12, 2015 #1
    I've been thinking about the properties of the Dirac delta function recently, and having been trying to prove them. I'm not a pure mathematician but come from a physics background, so the following aren't rigorous to the extent of a full proof, but are they correct enough?

    First I aim to prove that [itex] x\delta (x) =0[/itex]. Let [itex] f[/itex] be an arbitrary (integrable) function. Then, [tex] \int_{-\infty} ^{\infty} (xf(x)) \delta (x) \; dx=0f(0)=0[/tex] where we have used the filtering property of [itex] \delta[/itex]-function. As [itex] f[/itex] was chosen arbitrarily we must conclude that [itex] x\delta (x) [/itex].

    Next, I aim to prove that [itex] \delta (x^{2} - a^{2}) =\frac{1}{2 \vert a\vert} (\delta (x - a)+ \delta (x+a)) [/itex]. Consider the following, [tex] \int_{-\infty} ^{\infty}\delta (x^{2} - a^{2})f(x)\;dx[/tex] Now, let [itex] x=\sqrt{u} [/itex] then [itex] dx=\frac{1}{2\sqrt{u}} du[/itex] and so [tex]\int_{0} ^{\infty}\delta (x^{2} - a^{2})f(x)\;dx=\int_{0} ^{\infty}\delta (u - a^{2})\frac{f(\sqrt{u})} {2 \sqrt{u}} \;dx =\frac{f(\vert a\vert)} {2 \vert a\vert} [/tex] Similarly, let [itex] x=-\sqrt{u} [/itex] then [itex] dx=-\frac{1}{2\sqrt{u}} du[/itex] and so [tex]\int_{-\infty} ^{0}\delta (x^{2} - a^{2})f(x)\;dx=\int_{0} ^{\infty}\delta (u - a^{2})\frac{f(-\sqrt{u})} {2 \sqrt{u}} \;dx =\frac{f(-\vert a\vert)} {2 \vert a\vert} [/tex] and hence [tex]\int_{-\infty} ^{\infty}\delta (x^{2} - a^{2})f(x)\;dx=\int_{0} ^{\infty}\delta (x^{2} - a^{2})f(x)\;dx+\int_{-\infty} ^{0}\delta (x^{2} - a^{2})f(x)\;dx=\frac{f(\vert a\vert)} {2 \vert a\vert}+\frac{f(-\vert a\vert)} {2 \vert a\vert}\\ \qquad\qquad\qquad\qquad\quad\;\;\;=\frac{1} {2\vert a\vert}(f(a)+f(-a))=\int_{-\infty} ^{\infty}\frac{1}{2 \vert a\vert} (\delta (x - a)+ \delta (x+a))f(x) \;dx[/tex] Thus implying that [itex] \delta (x^{2} - a^{2}) =\frac{1}{2 \vert a\vert} (\delta (x - a)+ \delta (x+a)) [/itex].

    Would this be correct at all?
     
  2. jcsd
  3. Apr 15, 2015 #2

    wabbit

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    Gold Member

    Yes, the statements about these integrals are in a sense what's rigorous about the delta function. To make them complete you would need to be precise about what test functions are used and whether the integrals written do converge, but as a handwavy proof this is fine I think : the calculations would be the same in a rigorous proof, with just extra care added to make sure no infinity is lurking.
     
  4. Apr 15, 2015 #3
    Ok cool. As much as I would love to be able to prove them rigorously I don't think I have quite the level of mathematical training to do so, so I guess I'll have to settle for the formal (but non-rigorous) proof in my first post.
     
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