Let $\;\Delta\;$ be the Laplacian operator in $\;\mathbb R^n\;$ and consider two functions $\;{\zeta}_0 \in C^{\infty}_{0}(\mathbb R^n),{\zeta}_1 \in C^{\infty}(\mathbb R^n)\;$ such that $\;{ {\zeta}_0}^2+{ {\zeta}_1}^2=1\;$.
Then a simple computation shows: $\;\Delta = {\zeta}_{0}\Delta{\zeta}_{0}+{\vert \nabla {\zeta}_{0} \vert}^2+{\zeta}_{1}\Delta{\zeta}_{1}+{\vert \nabla {\zeta}_{1} \vert}^2\;$ where the functions here are taken as multiplication operators.
The above is a part of a proof I'm reading at the moment. Although the computation is a simple one, I have trouble deducing the above equality. What I see is:
$\;\Delta = \Delta(1)=\Delta({ {\zeta}_0}^2+{ {\zeta}_1}^2)=2({\zeta}_{0}\Delta{\zeta}_{0}+{\vert \nabla {\zeta}_{0} \vert}^2+{\zeta}_{1}\Delta{\zeta}_{1}+{\vert \nabla {\zeta}_{1} \vert}^2)\;$
What am I missing here? Any help would be valuable.
Thanks in advance!
For $n=1$ it is probably easier to see. In this case $$\Delta:=\frac{d^2}{dx^2}$$ Thus $$\Delta (\zeta_1^2+\zeta_2^2)=\frac{d^2}{dx^2}(\zeta_1^2+\zeta_2^2)=\frac{d}{dx}(2\zeta_1\zeta_1'+2\zeta_2\zeta_2')=2(\zeta_1\zeta''_1+(\zeta'_1)^2+\zeta_2\zeta_2''+(\zeta_2')^2)$$ On the other hand $$\Delta(1)=\frac{d^2}{dx^2}(1)=0$$ Therefore $$\zeta_1\zeta''_1+(\zeta'_1)^2+\zeta_2\zeta_2''+(\zeta_2')^2=0$$ For general $n>1$ it holds $$\zeta_1\Delta\zeta_1+|\nabla\zeta_1|^2+\zeta_2\Delta\zeta_2+|\nabla\zeta_2|^2=0$$ where $$\Delta:=\frac{d^2}{dx_1^2}+...+\frac{d^2}{dx_n^2}$$ and $\nabla$ is the gradient.