Let's take two Weibull distributions with parameters $(\alpha,\beta)$ and $(k,\lambda)$, respectively.
We know the general form of the Weibull density is:
$$f(x)=\bigg(\frac{k}\lambda\bigg)\bigg(\frac{x}\lambda\bigg)^{k-1}{e^{-(x/\lambda)}}^{k}$$
Obviously, we can see that for any $k>1$, $\lambda>0$ that there is an intersection at
$$x=0$$
In fact, the number of intersection points is dependent on $\alpha,k$. We can summarise the number of intersection points below:
$$\begin{array}{cc|lll}
&&&k&&\\
&& <1 & =1 & >1 \\
\hline
& <1 & 2 & 2 & 2 \\
\alpha & =1 & 2 & 1^{1} & 2 \\
& >1 & 2 & 2 & 3^{2}
\end{array}$$
$^1$This is the special case of the intersection of two exponential density functions. In this case, the intersection point is given by $x=\beta\cdot\lambda\cdot\log(\beta/\lambda)/(\beta-\lambda)$.
$^{2}$Where one of the intersection points is always at $(x,y)=(0,0)$.
If we equate the two density functions, we get:
$$(k-\alpha)\log(x)-\bigg(\frac{x}{\lambda}\bigg)^{k}+\bigg(\frac{x}{\beta}\bigg)^{\alpha}-\log\bigg(\frac{\alpha\cdot\lambda^{k}}{k\cdot\beta^{\alpha}}\bigg)=0$$
Unfortunately, there isn't a simple solution to this. We can solve for the remaining roots numerically:
a=1.7
b=2
k=2
l=1
fun = function(x, pars) {
return((pars[3]-pars[1])*log(x)-(x/pars[4])^pars[3]+(x/pars[2])^pars[1]+log(pars[3])-pars[3]*log(pars[4])- log(pars[1])+pars[1]*log(pars[2]))
}
ss=uniroot(f = fun, pars = c(a,b,k,l), interval = c(3, 10), extendInt = "yes")
> ss$root
[1] 1.414019
You may need to try different starting points to find all of the roots.
The two curves will always converge in the tails:
$$\lim\limits_{x\to\infty}f(x;\alpha,\beta)-f(x;k,\lambda)=0$$
We can visualize an example of these intersections below for each of the cases:
