0

The biharmonic operator is $\nabla^4 \phi \equiv \nabla^2 (\nabla^2 \phi)$. Are there any identities for it? I need to find $\phi$ such that

$~\\$ $\nabla^4 \phi = \frac{1}{3}\nabla^4 u^3 - u \nabla^4 u^2$,

where we know $\nabla^2 u = 0$. Alternatively, also $\psi$ such that $\nabla^2 \psi = u \nabla^4 u^2$ would be great. Any help is greatly appreciated.

user251041
  • 21
  • Look at the simplest cases: In 1D any function that satisfy $\nabla^4\phi = 0$ suffices (i.e. a qubic polynomial) since both $u$-terms vanish. In 2D and higher it becomes much harder. For 3D spherical symmetry we get the horrible equation $\frac{d^2}{dz^2}\left[z^4\frac{d^2\phi}{dz^2}\right] = 16z^5\left(\frac{du}{dz}\right)^3$ (with $z=-1/r$). The non-linearity + the '$u$' outside the differation operator means that usual tricks do not work (transforms). I doubt there is a 'nice' general formula for $\phi$ in terms of $u$ here. – Winther Jun 29 '15 at 07:27
  • Thanks! Yeah, I guess one has to make educated guesses for $\phi$. – user251041 Jun 30 '15 at 02:36

0 Answers0