Let $u \in C^4(\mathbb R^n)$ be a harmonic function
Prove: $\Delta^2 \vert x \vert^2 u =0 \;$
I calculated the partial derviates: $\frac{d^2}{d^2x_i}\vert x \vert^2 u $
and got to the following:
$\Delta (\vert x \vert^2 u) = 2u+4x \nabla u+x^2\Delta u = 4x \nabla u + 2u $
Applying $\Delta $ again: $\Delta^2 (\vert x \vert^2 u )= 10\Delta u+\sum_{i=1}^nx_i\frac{d^3}{d^3x_i}u$
The first term vanishes, but why does the second ? Maybe I miscalculated
Would appreciate any help
$\nabla^2(fg) = \nabla \cdot \nabla(fg) = \nabla \cdot (g \nabla f + f \nabla g) = g\nabla^2 f + 2\nabla f \cdot \nabla g + f \nabla^2 g; \tag 1$
$f = \vert x \vert^2 = \displaystyle \sum_1^n x_i^2, \; g = u; \tag 2$
$\nabla f = (2x_1, 2x_2, \ldots, 2x_n); \nabla^2 f = \nabla \cdot \nabla f = 2n; \tag 3$
$\nabla^2(fg) = 2n u + f \nabla^2 u + \displaystyle \sum_1^n 2x_i u_i = 2n u + \sum_1^n 2x_i u_i, \tag 4$
where
$u_i = u_{x_i} = \dfrac{\partial u}{\partial x_i}; \tag 5$
$\nabla^2(\nabla^2 (fg)) = 2n \nabla^2 u + \nabla^2 \left ( \displaystyle \sum_1^n 2x_i u_i \right ) = \nabla^2 \left ( \displaystyle \sum_1^n 2x_i u_i \right ); \tag 5$
$\nabla^2 \left ( \displaystyle \sum_1^n 2x_i u_i \right ) = \displaystyle \sum_1^n \nabla^2(2x_i u_i); \tag 6$
$\nabla^2 (x_i u_i) = u_i \nabla^2 x_i + 2\nabla x_i \cdot \nabla u_i + x_i \nabla^2 u_i; \tag 7$
$\nabla^2 (x_i) = 0; \; \nabla^2 u_i = (\nabla^2 u)_i = 0; \tag 8$
$\nabla^2 (x_i u_i) = 2\nabla x_i \cdot \nabla u_i = 2 u_{ii}; \tag 9$
thus,
$\nabla^2 \left ( \displaystyle \sum_1^n 2x_i u_i \right ) = \displaystyle 2 \sum_1^n \nabla^2 (x_i \cdot u_i) = 2\sum_1^n 2u_{ii} = 4 \nabla^2 u = 0, \tag{10}$
whence
$\nabla^2 \nabla^2 (\vert x \vert^2 u) = \nabla^2 \nabla^2 (fg) = 4 \nabla^2 u = 0. \tag{11}$