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Is it possible to assign a meaning to $$ \int_{-\infty}^\infty dt \; \delta(a-x(t)) \delta(b-y(t)) $$

Is it a distribution? Since the usual definition of the Dirac delta involves a single integral for a single Dirac delta $$ \int_{-\infty}^\infty dt \; \delta(t-a) f(t) = f(a) $$

using this to do the integral over one of the delta we are left with the other Dirac delta.

For a single Dirac delta it is straightforward to show by variable substitution that $$ \int_{-\infty}^\infty dt \; \delta(a-x(t))= \left ( \frac{dx}{dt} \Bigg |_{x(t)=a} \right )^{-1} $$

(assuming that $x(t)$ is injective such that there is only one $t$ where $x(t)=a$).

By some intuitive argument I want this to be the same as: if $x(t)$ and $y(t)$ are both injective and $x(t_1)=a$ and $y(t_2)=b$ then it is $1/(\text{some derivative...})$ if $t_1=t_2$ and $0$ otherwise - but this is not correct, right?

Thanks!

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  • $\begingroup$ If $a-x(t)$ and $b-y(t)$ are not zero for the same $t$ then it should be possible. $\endgroup$
    – md2perpe
    Commented Aug 21, 2017 at 21:09
  • $\begingroup$ If they are not zero for the same $t$? Can you elaborate - is it a distribution? $\endgroup$
    – smörkex
    Commented Aug 21, 2017 at 21:14
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    $\begingroup$ No $\delta(t)^2$ doesn't mean anything. But $\delta \ast \delta(t)$ is well-defined as a distribution as well as $T(x,y)=\int_a^b \delta(x-\gamma(t),y-\gamma(t))dt$ where $\delta(x,y)=\delta(x)\delta(y)$ is the Dirac delta in $\mathbb{R}^2$ $\endgroup$
    – reuns
    Commented Aug 21, 2017 at 21:20
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    $\begingroup$ If $a-x(t)$ and $b-y(t)$ are not zero for the samt $t$ then the mapping $\phi \mapsto \int_{-\infty}^{\infty} \delta(a-x(t)) \, \delta(b-y(t)) \, \phi(t) \, dt$ can be given a meaning as a distribution; it will however be the zero distribution. $\endgroup$
    – md2perpe
    Commented Aug 21, 2017 at 21:27
  • $\begingroup$ Two quite relevant links: math.stackexchange.com/questions/2389100/… math.stackexchange.com/questions/2336866/… $\endgroup$
    – md2perpe
    Commented Aug 21, 2017 at 21:28

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