I'm trying to find an upper bound of the first zero (not including $0$) of Bessel's function of orden zero, $J_0$.

The method proposed is using the Rayleigh quotient evaluated at a simple function.

First of all, the associated Sturm-Liouville problem of the Bessel equation of order zero is:


With boundary conditions $y(1)=0$ and $y(0)=0$. The solution is $J_0(\alpha_n x)$.

Since by the boundary condition $J_0(\alpha_n)=0$, we would just have to find the eigenvalue $\lambda_0=\alpha_0^2$, in order to find the first zero.

The eigenvalue is given by minimizing Rayleigh's quotient. So any other value will be an upper bound of $\lambda_0$.

The problem is that I've only seen this produce on equations with boundary conditions $y(0)=y(1)=0$. Will it work if $y(0)=1$?

An alternative I though of would be to calculate an upper bound of the first zero of $J_1$, which seems a weaker procedure, but at least the boundary conditions are $y(0)=y(1)=0$.

Anyway, the Sturm-Lioville problem in this case is:


In general is the equation is of the type $(ry')'+(\lambda p +q)=$, the Rayleigh quotiente is:

$$R[y]=\frac{\int_a^b ry'^2-qy^2dx}{\langle y,y\rangle}=\frac{\int_0^1 xy'^2 +y^2/x dx}{\int_0^1 y^2 dx}$$

I evaluated $R$ at $y(x)=x^2-x$. Which yields $\lambda\geq7.5$. Taking the square root, $\alpha \geq 2.74$ this is a good bound for the zero of $J_0$, but it should be a bound of the zero of $J_1$, which is not, because the it's approximately $3.83$.

Does anyone see why this contradicts the hyphothesis?

  • $\begingroup$ Your boundary condition should not be $y(0)=1$, it should instead be Neumann condition of $y'(0)=0$ to make it a sensible eigenproblem. $\endgroup$ – Ruslan Jan 6 '15 at 11:40

First, an eigenproblem needs homogeneous boundary conditions. So your $y(0)=1$ is no good. In general, the correct boundary condition for Bessel eigenproblem at left boundary is that of regularity. In the particular case of $J_0(x)$ it simplifies to Neumann condition of $y'(0)=0$.

The Rayleigh's quotient would be then computed just in the same way as in the case of $J_1(x)$:

$$R(L,y)=\frac{\langle y,Ly\rangle}{\langle y,y\rangle}.$$

But you have to note that the inner product in Sturm—Liouville problem is not just an integral of $y_1^*(x)y_2(x)$. It's a weighted inner product:

$$\langle y_1,y_2\rangle=\int_a^b w(x)y_1^*(x)y_2(x)dx,$$

where in your case of Bessel equation the weight function $w(x)=x$.

Now taking a test function of $y(x)=1-x^2$, I compute the Raleigh quotient equal to $6$. Taking square root of it, I have


while the true first zero of $J_0(x)$ is


which seems not too far off.


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