Divergence form of the fractional Laplacian

Question

Can I write the fractional Laplacian $$(-\Delta)^{\alpha/2} u(x) : = c_{\alpha,d} \mathrm{P.V.}\int_{\mathbb{R}^2} \frac{u(x) - u(y)}{|x-y|^{d+\alpha}}dy$$ in the divergence form $$(-\Delta)^{\alpha/2} u(x) = \nabla \cdot J(x)$$ for some function $J$? Or something else?

Answer 1

The answer is yes, under some regularity assumptions for $u$.

I will show two such representations:

1. Since $\Delta=\operatorname{div}\nabla=\nabla\cdot\nabla$ we have:

$$ (-\Delta)^{\alpha/2}u=(-\Delta)(-\Delta)^{(\alpha-2)/2}u= \nabla\cdot(-\nabla((-\Delta)^{(\alpha-2)/2}u). $$

2. Another representation uses Riesz transforms. With the Fourier transform defined by $$ \hat{u}(\xi) = \int_{\mathbb{R}^n} u(x)e^{-2\pi i x\cdot\xi}\, dx $$ we can express the derivatives and the fractional Laplacian by $$ \frac{\partial u}{\partial x_j} = (2\pi i\xi_j\hat{u})^\vee, \qquad (-\Delta)^{\alpha/2}u=((4\pi^2|\xi|^2)^{\alpha/2}\hat{u})^\vee. $$ The Riesz transforms are defined by $$ R_j u = \left(-\frac{i\xi_j}{|\xi|}\hat{u}\right)^\vee \quad \text{for $j=1,2,\ldots,n$,} $$ and the vector valued Riesz transform is $$ Ru=\langle R_1 u,\ldots,R_n u\rangle $$ Then it is easy to check that

$$ (-\Delta)^{\alpha/2}u = \nabla\cdot(R(-\Delta)^{\frac{\alpha-1}{2}}u). $$

Indeed, by taking the Fourier transform this equality reduces to $$ (4\pi^2|\xi|^2)^{\alpha/2}= \sum_{j=1}^n (2\pi i\xi_j)\left(-\frac{i\xi_j}{|\xi|}\right)(4\pi^2|\xi|^2)^{(\alpha-1)/2}. $$

The Riesz transform can be expressed as the singular integral operator.