Can I approximate $\rho_n$ somehow so it satisfies this Bessel function conidition?

Question

While solving a PDE using the eigenfunction expansion method, I came across the following condition for $\rho _n$:

$$ J_0(\rho_n r_{max}) = J_2(\rho_n r_{max}) \tag 1$$

where $r_{max}$ is known. Is there any approximation I can use for $\rho_n$ so it approximately satisfies the above condition? I know for instance that the roots of the Bessel function of the first kind, order zero, can be approximated with $ n \pi - \pi / 4 + 1 / (8 n \pi - \pi / 4)$. I found this out in this paper. I was hoping there was something similar I can use for this problem.

Answer 1

I prefer to add a separate answer.

More or less, we want to find the zero's of function

$$f(t)=J_0(t)- J_2(t)$$

which are closer and closer to $(n\pi-1)$ which is normal since these zero's are also those of $$g(t)=t\, J_0(t)-J_1(t)$$

So, let $$\large \color{blue}{ x_n=\rho_n\, r_{max}\,j_{0,n}}$$ and, emprically, use $$x_n=1+\sum_{k=1}^p \frac{a_k}{(n\pi-1)^k}$$

With $t=\frac 1 {n\pi -1}$

$$x_n=1-\frac{t}{35768}-\frac{3765 t^2}{3776}+\frac{1371 t^3}{3877}-\frac{1092 t^4}{3167}-\frac{5258 t^5}{3191}$$

The results using the exact values of $j_{0,n}$ $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 1.841183798 & 1.841183781 \\ 2 & 5.331442543 & 5.331442771 \\ 3 & 8.536320547 & 8.536316357 \\ 4 & 11.70599581 & 11.70600490 \\ 5 & 14.86358308 & 14.86358863 \\ 6 & 18.01552771 & 18.01552786 \\ 7 & 21.16437382 & 21.16436986 \\ 8 & 24.31133342 & 24.31132686 \\ 9 & 27.45705857 & 27.45705057 \\ 10 & 30.60193159 & 30.60192297 \\ \end{array} \right)$$

Now, look at this paper.

Answer 2

Let $x=\rho_n \,r_{max}$ and you want to find the zero's of function $$f(x)=J_0(x)- J_2(x) $$

If you compute the very first ones (or plot the function), you will observe that they are closer and closer to $$x_0^{(n)}=n\pi-1$$

If you want to improve it, use the first iteration of a Newton-like method ; this will give (using $c=\cos(1)$ and $s=\sin(1)$) $$x_1^{(n)}=x_0^{(n)}+2\frac{J_0\big(x_0^{(n)}\big)-J_2\big(x_0^{(n)}\big)}{3 J_1\big(x_0^{(n)}\big)-J_3\big(x_0^{(n)}\big)}$$ Expanding for large values of $n$ and continuing with Taylor expansions $$x_1^{(n)}=(n\pi-1)+\frac{s-c}{c+s}-\frac{9-4 c s}{4 \pi n (c+s)^2}+O\left(\frac{1}{n^2}\right)$$that is to say $$x_1^{(n)}=(n\pi-1)-\frac {4 \pi n \cos (2)+(9-2 \sin (2)) }{4 \pi n (1+\sin (2)) }+O\left(\frac{1}{n^2}\right)$$

A few numbers for illustration

$$\left( \begin{array}{cccc} n & x_0^{(n)} & x_1^{(n)} & \text{solution} \\ 1 & 2.14159 & 2.06024 & 1.84118 \\ 2 & 5.28319 & 5.35149 & 5.33144 \\ 3 & 8.42478 & 8.54296 & 8.53632 \\ 4 & 11.5664 & 11.7095 & 11.7060 \\ 5 & 14.7080 & 14.8661 & 14.8636 \\ 6 & 17.8496 & 18.0176 & 18.0155 \\ 7 & 20.9911 & 21.1663 & 21.1644 \\ 8 & 24.1327 & 24.3133 & 24.3113 \\ 9 & 27.2743 & 27.4590 & 27.4571 \\ 10 & 30.4159 & 30.6040 & 30.6019 \\ \end{array} \right)$$

Be sure that we could so much better. For example, the next term of the expansion would be $$\frac{-67 \sin (1)+2 \sin (3)+195 \cos (1)+18 \cos (3)}{32 \pi ^2 n^2 (\sin (1)+\cos (1))^3}$$ and for $n=5$, it would give $14.8645$