[Math] Dirac delta convolution with function

convolutiondirac deltaintegration

I've come into a bit of a snag, and thought some more talented mathematicians could maybe help.
I am trying to do the following integral:

$$S(x,t) = \int I(z)\delta(x-G(z,t)) \mathrm{d}z,$$

where $G(z,t)$ is a function which 'pushes' the original function $I(z)$ into $S(x,t)$ at some later time.
I've tried using some Dirac delta identities but have not had much success.
Any help would be very much appreciated. Thank you.

Best Answer

Have you tried to use the decomposition of the "composite" delta function $\delta(f(x))$,

http://en.wikipedia.org/wiki/Dirac_delta_function#Composition_with_a_function

In your case you have $$\delta(x - G(z,t)) = \sum_{i} \frac{\delta(z-z_i)}{|\partial_z G(z,t)|_{z=z_i}|}$$ where the sum goes over the solutions $z_i(x,t)$ of the equation $G(z,t) = x$, so that $$S(x,t) = \sum_{i} \frac{I(z_i)}{|\partial_z G(z,t)|_{z=z_i}|}$$ Potential problems might arise at points where $\partial_z G(z,t)|_{z=z_i} = 0$.

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Set $z=\partial_x y$. Then $z=z(x)$ satisfies the equation $$ \partial_x z+\delta(x)\,z=0, $$ with solution $$ z(x)=z(0)\,\mathrm{e}^{-\int_0^x\delta(\xi)\,d\xi}. $$ Hence, in order to obtain $y=y(x)$ you should integrate $z$ once.

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I have a Gaussian noise source $V(x,t)=\int d y\, e^{-A (x-y)^2} \xi(x,t)$, where $\xi$ is a Brownian noise, possessing an average of zero $\langle \langle \xi(x,t)\rangle \rangle=0$, and an autocorrelation function $\langle \langle \xi(x_1,t),\xi(x_2,t_2)\rangle \rangle=\gamma\, \delta(x_1-x_2)\delta(t_1-t_2)$.

You must be mistaken, still.

You can rearrange that integral as $V(x, t) = \xi(x, t)\int dy~e^{-A (x - y)^2};$ it cannot fundamentally be a different sort of noise than $\xi$ is, except that it might have a different mean/variance.

Brownian noise does not have a $\delta$ autocorrelation; the (discrete) definition of Brownian noise is (in JavaScript ES6; Python is similar)

function* white_noise() {
    var r, a;
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        a = 2 * Math.PI * Math.random();
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function* brown_noise(r) { 
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}

and as you can see, it explicitly takes an autocorrelation argument and produces a stream of values which depend on prior history.

Best Answer

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