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Finding the eigenvalue of the Bessel function. By applying the right boundary condition, we have: $$ A J_{0} (\lambda a) = 0 $$ Here, we require that $A \neq 0$ to avoid a trivial solution and we let $J_{0} (\lambda a) = 0$. How do you derive the value of $\lambda$ from this equality. In the heat equation with planar symmetry, we use the fact that $\sin (n \pi ) = 0$ and equate it with for example $\sin (\lambda x/L)$ to derive the eigenvalue $\lambda = \frac{n \pi L}{x}$.

Could someone show a resource such as a table containing the roots of $J_{0} (\lambda a)$?

Stephanie
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  • I don't believe there is a closed form solution for this. You can find these roots numerically. Many mathematical softwares have these zeros. – Gregory Jan 17 '21 at 03:22
  • You can google the zeros of the Bessel function. If accuracy is not key, you can appeal to large $x$ asymptotic: $J_0(x) \sim \cos \left( x - \frac{\pi}{4} \right)$. – Gregory Jan 17 '21 at 03:48
  • I believe Mathematica has an inbuilt function that returns these roots. BesselJZero[Bessel function order, index of zero] – K.defaoite Jan 17 '21 at 13:55

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I give you below the list of the first zero's of $J_0(x)$ $$\{2.404825558,5.520078112,8.653727913,11.79153444,14.93091771,18.07106397,21.211636 63,24.35247153,27.49347913,30.63460647,33.77582021,36.91709835,40.05842576,43.19 979171,46.34118837,49.48260990,52.62405184,55.76551076,58.90698393,62.04846920,6 5.18996481,68.33146933,71.47298161,74.61450065,77.75602563,80.89755587,84.039090 78,87.18062985,90.32217264,93.46371878,96.60526795,99.74681986,102.8883743,106.0 299309,109.1714897,112.3130503,115.4546127,118.5961766,121.7377421,124.8793089,1 28.0208770,131.1624463,134.3040166,137.4455880,140.5871604,143.7287336,146.87030 76,150.0118825,153.1534580,156.2950343,159.4366112,162.5781887,165.7197667,168.8 613454\}$$

If you need more than these $54$, just tell.

For large values on $n$, you can use $$t+\frac{1}{8 t}-\frac{31}{384 t^3}+\frac{3779}{15360 t^5}-\frac{6277237}{3440640 t^7}+\frac{2092163573}{82575360 t^9}+O\left(\frac{1}{t^{11}}\right)$$ where $t=\pi \left(n-\frac{1}{4}\right)$.

For $n=50$, this would give $$156.29503426853352381954991$$ while the exact value is

$$156.29503426853352381954950$$

Claude Leibovici
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