Radon measure determined by the intersection of half lines in the plane

Question

Consider a vector $r$ in the euclidian plane $\mathbb R^2$ and two unit vectors $u,v\in\mathbb U$ ($\mathbb U$ is the unit circle). Let $s>0$ be a real number. I am looking for an expression of the following distribution on $\mathbb R^2\times\mathbb R^+\times\mathbb U\times\mathbb U$ $$\mu(r,s,u,v)=\int_0^s \delta^{(2)}(r-tu+(s-t)v)\mathrm dt\tag1$$ where $\delta^{(2)}$ is a 2-dimensional Dirac distribution, which is defined by $\delta^{(2)}((x,y))=\delta(x)\delta(y)$. I exclude the cases $u=v$ and $u=-v$.

Here is what I have done so far. It's detailed and hence rather long...

I have considered the following problem : there are two points $A$ and $B$ from where start two half lines with unit vectors $u$ and $v$ (see the figure below) and such that $\overrightarrow{AB}=r$.

The half-lines can cross, but under a condition that is to be determined. A point on the half-line $(A,u)$ is of the form $A+xu$ with $x\geq0$. Similarly a point on $(B,v)$ is of the form $B+yv$, with $y\geq0$. The intersection is given by the equation $A+xu=B+yv$, $$xu=r+yv\tag2.$$ If we write (2) as $r-xu+yv=0$, we see that problem (1) is related to the intersection equation (2) with the extra condition $x+y=s$.

As the vectors $u$ and $v$ are not colinear, it is possible to solve (2) by taking the scalar products with both $u$ and $v$ : $$\left\{\begin{array}{ccccc} x&=&r\cdot u&+&y\,u\cdot v\\ x\,u\cdot v&=&r\cdot v&+&y \end{array}\right.$$ We get $$x=\frac{r\cdot u-(r\cdot v)(u\cdot v)}{1-(u\cdot v)^2}\qquad y=\frac{r\cdot v-(r\cdot u)(u\cdot v)}{1-(u\cdot v)^2}.\tag3$$ and we conclude that the half-lines intersect if $r\cdot u\geq(r\cdot v)(u\cdot v)$ and $r\cdot v\geq(r\cdot u)(u\cdot v)$. If these conditions are not fulfilled, the integral (1) vanishes.

Supposing these conditions are satisfied, we can see that we have to impose $x+y=s$ which can simplified into $$r\cdot u+r\cdot v-(1+u\cdot v)s=0.\tag4$$ Interestingly the conditions $x\geq0$ and $y\geq0$ expressed with (3) and combined with the condition (4) give $s-r\cdot u\geq0$ and $s-r\cdot v\geq 0$. From there I am not so sure: I have tried to write $\mu$ as $$\mu(r,s,u,v)=f(r,s,u,v)\delta\big(r\cdot u+r\cdot v-(1+u\cdot v)s\big)\Theta(s-r\cdot u)\Theta(s-r\cdot v)\tag5$$ where $f$ is a regular function ($\Theta$ is the Heaviside step function). I have integrated this expression with respect to $v$ and obtained $$\int\mu(r,s,u,v)\mathrm dv=2f(r,s,u,v_1)\frac{\Theta(s-|r|)}{\sqrt{r^2-(r\cdot u)^2}}$$ where $v_1$ is such that $(r-su)\cdot v_1=s-r\cdot u$.
I have also integrated (1) with respect to $v$ and got $$\int\mu(r,s,u,v)\mathrm dv=\frac{\Theta(s-|r|)}{s-r\cdot u}.$$ I am wondering if I can deduce $f(r,s,u,v)$ from these computations. Can anyone (who went throught all of this) give me a little hint ? Thanks.

Answer 1

Let $r = (x_1,x_2)$, $u=(u_1,u_2)$, $v=(v_1,v_2)$ be the vectors once analyzed in the canonical basis of the plane $\mathbb R^2$. We set $u+v=w=(w_1,w_2)$ and $r+sv=y=(y_1,y_2)$.

$\mu(r,s,u,v)=\int_{0}^s \delta^{(2)}(r-tu+(s-t)v)dt=\int_{0}^s \delta^{(2)}(y-tw)dt$

$\mu(r,s,u,v)=\int_{0}^s \delta(y_1-tw_1) \delta(y_2-tw_2) dt$

$\mu(r,s,u,v)=\int_{0}^s \delta(y_1-tw_1) \delta(y_2-tw_2) dt$

We set $t_1 = y_1/w_1$ and $t_2=y_2/w_2$

Using the property of Dirac measure $\delta(ax)=\frac{1}{|a|}\delta(x)$ and symetry $\delta(-x)=\delta(x)$.

$\mu(r,s,u,v)=\frac{1}{|w_1w_2|}\int_{0}^s \delta(t-t_1) \delta(t-t_2) dt$

Consider now an expression of the dirac distribution as the limit of, let say, a family of gaussian distributions ("generalized functions")

$\delta(x) =\lim_{\sigma\rightarrow 0+} \frac{1}{\sigma\sqrt{2\pi}}\exp{-\frac{x^2}{2\sigma^2}}$

$\mu(r,s,u,v)=\frac{1}{|w_1w_2|}\lim_{\sigma_1\rightarrow 0+}\frac{1}{\sigma_1\sqrt{2\pi}} \lim_{\sigma_1\rightarrow 0+}\frac{1}{\sigma_1\sqrt{2\pi}} \int_{0}^s \exp{-\frac{(t-t_1)^2}{2\sigma_1^2}} \exp{-\frac{(t-t_2)^2}{2\sigma_2^2}} dt$

$\mu(r,s,u,v)=\frac{1}{|w_1w_2|}\lim_{\sigma_1\rightarrow 0+}\frac{1}{\sigma_2\sqrt{2\pi}} \lim_{\sigma_1\rightarrow 0+}\frac{1}{\sigma_1\sqrt{2\pi}} \int_{0}^s \exp{-[\frac{(t-t_1)^2}{2\sigma_1^2}}+\frac{(t-t_2)^2}{2\sigma_2^2}] dt$

From now on we will use only one "small parameter" $\sigma=\sigma_1=\sigma_2$

$\mu(r,s,u,v)=\frac{1}{|w_1w_2|}\lim_{\sigma\rightarrow 0+}\frac{1}{\sigma^22\pi}\int_{0}^s \exp{-[\frac{(t-t_1)^2+(t-t_2)^2}{2\sigma^2}]} dt$

The integral has primitive,

$-\frac{1}{2}\sqrt{\pi}\sigma\exp{-\frac{(t_1-t_2)^2}{4 \sigma^2}}[\mathrm{erf}(\frac{t_1+t_2-2t}{2\sigma})]_0^s$

Hence,

$\mu(r,s,u,v)=\frac{1}{|w_1w_2|}\lim_{\sigma\rightarrow 0+}\frac{\sqrt 2}{4}\frac{1}{\sigma\sqrt{2\pi}}\exp{-\frac{(\frac{t_1-t_2}{\sqrt2})^2}{2 \sigma^2}}[\mathrm{erf}(\frac{t_1+t_2-2t}{2\sigma})]_s^0=\frac{1}{|w_1w_2|}\delta(t_1-t_2) \lim_{\sigma\rightarrow 0+}\frac{1}{2}[\mathrm{erf}(\frac{t_1+t_2-2t}{2\sigma})]_s^0 $

Now we can discuss the limit of the error function in the primitive.

• $t_1+t_2\neq 0\ \wedge\ t_1+t_2-2s\neq 0$ then $\lim_{\sigma\rightarrow 0+}[\mathrm{erf}(\frac{t_1+t_2-2t}{2\sigma})]_s^0 = 1 - 1 = 0$

Then $\mu(r,s,u,v)=0$.

• $t_1+t_2 =0\ \wedge\ t_1+t_2-2s\neq 0$ then $\lim_{\sigma\rightarrow 0+}[\mathrm{erf}(\frac{t_1+t_2-2t}{2\sigma})]_s^0 = 0-1=-1$

Then $\mu(r,s,u,v)=-\frac{1}{2|w_1w_2|}\delta(t_1-t_2)$.

• $t_1+t_2 \neq 0\ \wedge\ t_1+t_2-2s= 0$ then $\lim_{\sigma\rightarrow 0+}[\mathrm{erf}(\frac{t_1+t_2-2t}{2\sigma})]_s^0 = 1-0= 1$

Then $\mu(r,s,u,v)=\frac{1}{2|w_1w_2|}\delta(t_1-t_2)$.

• $t_1+t_2 = 0\ \wedge\ t_1+t_2-2s= 0$ then $\lim_{\sigma\rightarrow 0+}[\mathrm{erf}(\frac{t_1+t_2-2t}{2\sigma})]_s^0 = 0-0= 0$

Then $\mu(r,s,u,v)=0$.

You can back-substitute then $t_1,t_2$, $w_1$, $w_2$ in the solutions.

Hope this helps.