By substituting $z = h(t)$ show that $\delta(h(t))=\sum\limits_{i}\frac{\delta(t−t_i)}{\mid h^{\prime}(t_i)\mid}$

Question

Firstly, apologies in advance for using an abuse of notation by placing the Dirac-delta inside an integral. But for my level of understanding, I have no choice.

This question involves one of the Dirac-delta properties: $$\delta(bt)=\frac{1}{\mid b \mid}\delta(t)$$ where $b \gt 0$ and $b \in \mathbb{R^+}$; which was proven in this previous question asked by myself.

This is essentially a word by word copy of a derivation given on page 470:

Given $$\delta(bt)=\frac{1}{\mid b \mid}\delta(t)$$ and by considering an integral of the form $$\int f(t)\delta(h(t))\mathrm{d}t$$ then making a change of variables to $z = h(t)$, we may show that $$\displaystyle\delta(h(t)) = \operatorname*{\Large\sum}\limits_{i}\cfrac{\delta(t − t_i)}{\mid h^{\prime}(t_i)\mid}$$ where the $t_i$ are those values of $t$ for which $h(t) = 0$ and $h^{\prime}(t)$ stands for $\cfrac{\mathrm{d}h}{\mathrm{d}t}$.

Could someone please show the steps in this derivation as I have no idea how they reached $$\color{#180}{\displaystyle\delta(h(t)) = \operatorname*{\Large\sum}\limits_{i}\cfrac{\delta(t − t_i)}{\mid h^{\prime}(t_i)\mid}}$$

Many thanks.

Answer 1

I think the way to think about this is as a local phenomenon which can be extended easily. If $h$ is differentiable and $h'$ is nowhere zero, then the inverse function theorem says that $h$ is locally invertible. (The true inverse function theorem is a little bit stronger than this, but I'm just going with this version.) Therefore if we consider the interval $[a,b]$ (which is small enough so that $h$ is invertible on it), then by doing a change of variable we get

$$\int_a^b \delta(h(t)) f(t)\,dt = \int_{h(a)}^{h(b)} \delta(z) f(h^{-1}(z))\frac{1}{h'(h^{-1}(z))}\,dz.$$

If $0\in [h(a),h(b)]$, then this becomes $\frac{f(h^{-1}(0))}{h'(h^{-1}(0))}.$ If you sum all of the local bits (the integrals over $[a,b]$), you have your result.

Side note: This approach really bothers me because you have to first give meaning to $\delta\circ h$ in the first place. As a distribution, this isn't defined a priori, but the integral approach makes it easy to argue why this should be the definition of $\delta\circ h$.

Answer 2

The definition of integration against a dirac delta is,

$$\int_{-\infty}^\infty f(t) \delta(t) \ \mathrm d t = \lim_{n\rightarrow \infty} \int_{-\infty}^\infty f(t) \frac{1}{\sqrt{2\pi n^2}}e^{-t^2/2n^2} \ \mathrm d t, $$

we will show that the identity is vaid by showing that both expressions will have the same result when this limiting process is applied.

To do this I am going to have to make some assumptions about $h$. These might not be the most narrow assumptions possible, but they will work in most applications of this identity.

• I am going to assume that $h$ is differentiable and that the first derivative is continuous.

• I will also assume that $h$ only has simple roots, i.e., the first deriviative is nonzero at each root.

• I am also going to assume that none of the roots cluster, i.e., every root has a small enough open neighborhood containing it and no other roots.

The integral we are trying to compute is,

$$\int_{-\infty}^\infty f(t) \delta(h(t)) \ \mathrm d t = \lim_{n\rightarrow \infty} \int_{-\infty}^\infty f(t) \frac{1}{\sqrt{2\pi n^2}}e^{-h^2(t)/2n^2} \ \mathrm d t. $$

We would like to change variables, but $h$ is almost certainly not monotonic.

Let the roots of $h$ be $\lbrace r_1,r_2,\dots\rbrace$. Rolles theorem tells us that between every pair of roots there is a at least one point where the derivative changes sign. This means around each root we can identify a interval for which $h$ is monotonic, let the interval assosciated with $r_i$ be $I(r_i)$; furthermore lets require that each interval contains only one of the $r_i$. Define $I_T = U_i I(r_i)$.

We can break up the integral according to the following sum,

$$ \color{red}{\int_{\mathbb{R}-I}f(t) \frac{1}{\sqrt{2\pi n^2}}e^{-h^2(t)/2n^2} \ \mathrm d t} + \sum_i \int_{I(r_i)}f(t) \frac{1}{\sqrt{2\pi n^2}}e^{-h^2(t)/2n^2} \ \mathrm d t. $$

The integral on the $\color{red}{\text{left}}$ goes to zero in the limit because the domain of integration, by definition, doesn't contain any roots of $h$. We will neglect this integral from now on.

Now consider what we have left. Each integral in the sum $$\sum_i \int_{I(r_i)}f(t) \frac{1}{\sqrt{2\pi n^2}}e^{-h^2(t)/2n^2} \ \mathrm d t $$

Now near one of its roots its possible to approximate $h$ by,

$$ h(t) = h'(r_i) (t-r_i) + o((t-r_i)^2), $$

squaring this we find,

$$ h(t)^2 = h'(r_i)^2 (t-r_i)^2 + o((t-r_i)^3). $$

Now consider the fact that in the limit we have,

$$ \int \frac{1}{n} e^{-t^3/n^2} \ \mathrm dt = \frac{\text{const.}}{n^{1/3}} \rightarrow 0,$$

this means that our approximation of $h^2$ commits less and less error in evaluating the integral as $n$ increases. This is a symptom of the fact that the Gaussian is becomming narrower and taller causing our intervals to effectively shrink.

So in the limit we have,

$$\int_{-\infty}^\infty f(t) \delta(h(t)) \ \mathrm d t = \lim_{n\rightarrow \infty} \sum_i \int_{I(r_i)}f(t) \frac{1}{\sqrt{2\pi n^2}}e^{-h^2(t)/2n^2} \ \mathrm d t $$ $$=\lim_{n\rightarrow \infty} \sum_i \int_{I(r_i)}f(t) \frac{1}{\sqrt{2\pi n^2}}e^{-h'(r_i)^2(t-r_i)^2/2n^2} \ \mathrm d t $$ $$= \lim_{n\rightarrow \infty} \sum_i \int_{I(r_i)}f(t) \frac{1}{\sqrt{2\pi n^2}}e^{-(t-r_i)^2/2n^2} \frac{\mathrm d t}{|h'(r_i)|} $$ $$ = \sum_i \int_{I(r_i)}f(t) \frac{\delta(t-r_i)}{|h'(r_i)|}$$ $$ = \sum_i \int_{-\infty}^\infty f(t) \frac{\delta(t-r_i)}{|h'(r_i)|}$$ $$ = \int_{-\infty}^\infty f(t) \sum_i \frac{\delta(t-r_i)}{|h'(r_i)|}$$

Note that the last two equalities arrose from the fact that their result is indistinguishable from the previous expression.

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Remember, by our definition the delta function doesn't have a valid existence outside of an integral. Two expressions involving delta functions are defined to be equal if integration against them gives the same answer for every function. This condition is satisfied by the equality we established becuase $f(t)$ was arbitrary (up to a point).

$$ \text{This } \boxed{ \int_{-\infty}^\infty f(t) \delta(h(t)) \ \mathrm d t = \int_{-\infty}^\infty f(t) \sum_i \frac{\delta(t-r_i)}{|h'(r_i)| } } $$ $$ \text{is the same as this } \boxed{\delta(h(t)) = \sum_i \frac{\delta(h(t-r_i))}{|h'(r_i)|}}.$$