2
$\begingroup$

How can I solve this equation?

I tried the following:

I take the Fourier transform on both sides and obtain

$$-(k_xk_y + k_xk_z + k_yk_z)\hat{u}(k_x, k_y, k_z) = 1$$ i.e. $$\hat{u}(k_x,k_y, k_z) = \frac{-1}{(k_xk_y + k_xk_z + k_yk_z)}$$ and taking the inverse Fourier transform I obtain $$u(x,y,z) = \frac{-1}{(2\pi)^3} \int_{-\infty}^{\infty} dk_x \int_{-\infty}^{\infty} dk_y \int_{-\infty}^{\infty} dk_z \frac{e^{-i(k_x x + k_y y + k_z z)}}{k_z(k_x +k_y) + k_xk_y}$$ and by the residue theorem I have a single pole at $k_z = -\frac{k_xk_y}{(k_x + k_y)}$ and thus I obtain $$\frac{-i}{(2\pi)^2} \int_{-\infty}^{\infty} dk_x \int_{-\infty}^{\infty} dk_y \frac{e^{-i(k_x x + k_y y - \frac{k_xk_yz}{k_x+k_y})}}{k_x + k_y}$$ but this I can not evaluate since it has an essential singularity at $k_y = -k_x$.

$\endgroup$
  • $\begingroup$ I don't know how to conclude. However it may be useful to note that there is a scaling symmetry to this problem, because if $u$ is a solution then $u_\lambda(x,y,z)=\lambda u(\lambda x, \lambda y, \lambda z)$ is a solution too. You may try searching for a solution that is invariant under this scaling. $\endgroup$ – Giuseppe Negro Mar 8 '18 at 12:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.