Evaluate: $\left(\frac{\partial }{\partial x}+\frac{\partial }{\partial y}+\frac{\partial }{\partial z}\right)^2u$

Question

If $u=\ln(x^3+y^3+z^3-3xyz)$, show that $$\bigg(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}+\dfrac{\partial }{\partial z}\bigg)^2u=\dfrac{-9}{(x+y+z)^2}$$

I don't know how to interpret $\bigg(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}+\dfrac{\partial }{\partial z}\bigg)^2u$.

If $$\bigg(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}+\dfrac{\partial }{\partial z}\bigg)^2u=\dfrac{\partial ^2u}{\partial x^2}+\dfrac{\partial ^2u}{\partial y^2}+\dfrac{\partial ^2u}{\partial z^2}+2\dfrac{\partial ^2u}{\partial x\partial y}+2\dfrac{\partial ^2u}{\partial y\partial z}+2\dfrac{\partial ^2u}{\partial z\partial x}$$

then is there any form of theory or properties regarding this, because I don't want to calculate every term and plug it into the giant expression. I also observe that I can made the given problem into the homogenous function of degree $3$. Is that useful?

Please help.

Answer 1

Your interpretation of $\bigg(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}+\dfrac{\partial }{\partial z}\bigg)^2u$ is correct. In order compute it in a convenient way, just use symmetry and the factorization $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xz-yz-xz).$$ It follows that $$\frac{\partial u}{\partial x}=\frac{1}{x+y+z}+\frac{2x-z-y}{x^2+y^2+z^2-xz-yz-xz}.$$ Hence $$\bigg(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}+\dfrac{\partial }{\partial z}\bigg)u=\frac{3}{x+y+z}+0.$$ Now it remains to compute $$\bigg(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}+\dfrac{\partial }{\partial z}\bigg)^2u=\bigg(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}+\dfrac{\partial }{\partial z}\bigg)\left(\frac{3}{x+y+z}\right)=-\dfrac{3\cdot 3}{(x+y+z)^2}.$$

Answer 2

First, write the argument of $\ln$ in terms of the elementary symmetric polynomials in three variables: $$ x^3+y^3+z^3 - 3xyz = (x+y+z)^3 -3(x+y+z)(xy+yz+xz) = E_1^3 - 3E_1 E_2 $$ Now note that the operator $D = \partial /\partial x+\partial/\partial y + \partial/\partial z$ acts on the elementary symmetric polynomials by lowering their order: $D[E_2]= 2E_1$ and $D[E_1] = 3$. So $$ D[\ln(E_1^3 - 3E_1E_2)] = \frac{9E_1^2-9E_2-6E_1^2}{E_1^3-3E_1E_2} = 3\frac{E_1^2-3E_2}{E_1(E_1^2-3E_2)} = \frac{3}{E_1}, $$ and $$ D^2[\ln(E_1^3 -3E_1E_2)] = D\left[\frac{3}{E_1}\right] = -\frac{9}{E_1^2}. $$