If $u=\log{\big(x^3+y^3+z^3-3xyz\big)}$.Find $\Big(\frac{\partial}{\partial{x}}+\frac{\partial}{\partial{y}}+\frac{\partial}{\partial{z}}\Big)^{2}{u}$

Question

I just started learning functions of several variables and found the above problem in my book.

The solution given in my book was by finding the value of $\Big(\frac{\partial{u}}{\partial{x}}+\frac{\partial{u}}{\partial{y}}+\frac{\partial{u}}{\partial{z}}\Big)$ and squaring it which gives: $$\Big(\frac{\partial}{\partial{x}}+\frac{\partial}{\partial{y}}+\frac{\partial}{\partial{z}}\Big)^{2}{u}=\dfrac{9}{(x+y+z)^2}$$

I also found the same question here and the solution given here by @Ivo Terek seems to be perfect. But here: $$\Big(\frac{\partial}{\partial{x}}+\frac{\partial}{\partial{y}}+\frac{\partial}{\partial{z}}\Big)^{2}{u}=\dfrac{-9}{(x+y+z)^2}$$

My Doubt:

Is $\Big(\frac{\partial{u}}{\partial{x}}+\frac{\partial{u}}{\partial{y}}+\frac{\partial{u}}{\partial{z}}\Big)^{2} = \Big(\frac{\partial}{\partial{x}}+\frac{\partial}{\partial{y}}+\frac{\partial}{\partial{z}}\Big)^{2}{u}$ and which of the above solution is correct ?

Also it will be great help if someone can attach the source to learn this specific concept as my book doesn't have much clear explanation. Thanks!

Answer 1

Note that $$ \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)^2u = \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)(u_x+u_y+u_z) = u_{xx}+u_{yy}+u_{zz} +2u_{xy}+2u_{yz}+2u_{zx} $$ but $$ \left(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}\right)^2 = (u_x)^2+(u_y)^2+(u_z)^2+2u_xu_y+2u_yu_z+2u_zu_x $$ which are clearly not equal in general.

The problem comes from the fact that the square in the first expression is a squared composition, i.e. the composition of an operator with itself, whereas the square in the second expression is the "standard" algebraic square, i.e. multiplication with itself.

You should follow IvoTerek's resolution $-$ which saves a lot of time by the way $-$, which produces a negative sign (it was expected, since $(\ln x)'' = -1/x^2$); however, beware of his solution still different from yours.