Chandrasekhar–Kendall function

Chandrasekhar–Kendall functions are the eigenfunctions of the curl operator derived by Subrahmanyan Chandrasekhar and P. C. Kendall in 1957 while attempting to solve the force-free magnetic fields. The functions were independently derived by both, and the two decided to publish their findings in the same paper.

If the force-free magnetic field equation is written as $$\nabla\times\mathbf{H}=\lambda\mathbf{H}$$, where $$\mathbf{H}$$ is the magnetic field and $$\lambda$$ is the force-free parameter, with the assumption of divergence free field, $$\nabla\cdot\mathbf{H}=0$$, then the most general solution for the axisymmetric case is
 * $$\mathbf{H} = \frac{1}{\lambda}\nabla\times(\nabla\times\psi\mathbf{\hat n}) + \nabla \times \psi \mathbf{\hat n} $$

where $$\mathbf{\hat n}$$ is a unit vector and the scalar function $$\psi$$ satisfies the Helmholtz equation, i.e.,
 * $$\nabla^2\psi + \lambda^2\psi=0.$$

The same equation also appears in Beltrami flows from fluid dynamics where, the vorticity vector is parallel to the velocity vector, i.e., $$\nabla\times\mathbf{v}=\lambda\mathbf{v}$$.

Derivation
Taking curl of the equation $$\nabla\times\mathbf{H}=\lambda\mathbf{H}$$ and using this same equation, we get


 * $$\nabla\times(\nabla\times\mathbf{H}) = \lambda^2\mathbf{H}$$.

In the vector identity $$ \nabla \times \left( \nabla \times \mathbf{H} \right) = \nabla(\nabla \cdot \mathbf{H}) - \nabla^{2}\mathbf{H}$$, we can set $$\nabla\cdot\mathbf{H}=0$$ since it is solenoidal, which leads to a vector Helmholtz equation,


 * $$\nabla^2\mathbf{H}+\lambda^2\mathbf{H}=0$$.

Every solution of above equation is not the solution of original equation, but the converse is true. If $$\psi$$ is a scalar function which satisfies the equation $$\nabla^2\psi + \lambda^2\psi=0$$, then the three linearly independent solutions of the vector Helmholtz equation are given by


 * $$\mathbf{L} = \nabla\psi,\quad \mathbf{T} = \nabla\times\psi\mathbf{\hat n}, \quad \mathbf{S} = \frac{1}{\lambda}\nabla\times\mathbf{T}$$

where $$\mathbf{\hat n}$$ is a fixed unit vector. Since $$\nabla\times\mathbf{S} =\lambda\mathbf{T}$$, it can be found that $$\nabla\times(\mathbf{S}+\mathbf{T})=\lambda(\mathbf{S}+\mathbf{T})$$. But this is same as the original equation, therefore $$\mathbf{H}=\mathbf{S}+\mathbf{T}$$, where $$\mathbf{S}$$ is the poloidal field and $$\mathbf{T}$$ is the toroidal field. Thus, substituting $$\mathbf{T}$$ in $$\mathbf{S}$$, we get the most general solution as


 * $$\mathbf{H} = \frac{1}{\lambda}\nabla\times(\nabla\times\psi\mathbf{\hat n}) + \nabla \times \psi \mathbf{\hat n}. $$

Cylindrical polar coordinates
Taking the unit vector in the $$z$$ direction, i.e., $$\mathbf{\hat n}=\mathbf{e}_z$$, with a periodicity $$L$$ in the $$z$$ direction with vanishing boundary conditions at $$r=a$$, the solution is given by


 * $$\psi = J_m(\mu_jr)e^{im\theta+ikz}, \quad \lambda =\pm(\mu_j^2+k^2)^{1/2}$$

where $$J_m$$ is the Bessel function, $$k=\pm 2\pi n/L, \ n = 0,1,2,\ldots$$, the integers $$m =0,\pm 1,\pm 2,\ldots$$ and $$\mu_j$$ is determined by the boundary condition $$a k\mu_j J_m'(\mu_j a)+m \lambda J_m(\mu_j a) =0.$$ The eigenvalues for $$m=n=0$$ has to be dealt separately. Since here $$\mathbf{\hat n}=\mathbf{e}_z$$, we can think of $$z$$ direction to be toroidal and $$\theta$$ direction to be poloidal, consistent with the convention.