This question was originally posted in ME: https://math.stackexchange.com/questions/4725157/what-is-an-explicit-subset-of-mathbbz3-that-makes-bigl-sinn-cdot-x
but more and more I think about it, this problem looks nontrivial. So, I ask for help here.
Basically I would like to find an explicit orthonormal basis of $L^2([0,1]^3, \mathbb{R})$ as eigenfunctions of $-\Delta$. Here, we assume periodic boundary conditions on $[0,1]^3$.
Let $n \in \mathbb{Z}^3$ and $x \in [0,1]^3$. Then, it is clear that the collection
\begin{equation}
\Bigl( \sqrt{2}\sin(2\pi n \cdot x),\sqrt{2}\cos(2\pi n \cdot x) \Bigr)_{n \in \mathbb{Z}^3}
\end{equation}
are eigenfunctions of $-\Delta$ on $[0,1]^3$ with the eigenvalues $4\pi^2 \lvert n \rvert^2$. Moreover this collection clearly generates the Hilbert space $L^2([0,1]^3, \mathbb{R})$ and each element is of unit norm.
However, I would like to extract some "orthonormal basis" from this generating set. For example, the sine and cosine functions corresponding to $n=(1,1,1)$ and $n=(-1,-1,-1)$ are NOT linearly independent.
If I regard $L^2([0,1]^3, \mathbb{R})$ as $3$-fold tensor product of $L^2([0,1], \mathbb{R})$ and construct its orthonormal basis as tensor products of sines and cosines such as $\sin(n_1 x_1) \cos(n_2x_2) \sin(n_3 x_3)$ then, it would suffice to set $n_1, n_2, n_3 \geq 0$.
On the other hand, the situation becomes much more complicated if I seek an orthonormal basis of the form mentioned above..
Could anyone please help me?
Best Answer
One can use, for example, $n \in \mathbb{Z}^3$ with the following restrictions (edited upon comment by Alexei Kulikov to be careful about cases with $n_i =0 $): $n_1 >0 \ \ \lor \ (n_1 =0 \land n_2 > 0) \ \lor \ (n_1 = n_2 =0 \land n_3 \geq 0) $.
To abbreviate things, introduce the notation $$ sss = \sin 2\pi |n_1 | x_1 \sin 2\pi |n_2 | x_2 \sin 2\pi |n_3 | x_3 \\ ssc = \sin 2\pi |n_1 | x_1 \sin 2\pi |n_2 | x_2 \cos 2\pi |n_3 | x_3 \\ \ldots \hspace{8.2cm} $$ Then, for given $n$ with $n_1 , n_2 , n_3 \geq 0$, your preferred functions are $$ \sin (2\pi n\cdot x) = scc-sss+csc+ccs \\ \cos (2\pi n\cdot x) = ccc-css-ssc-scs $$ Note that the right hand sides contain all 8 functions that would (as you note) constitute a basis, namely, $sss,css,scs,ssc,ccs,csc,scc,ccc$ (reduced in number in cases where a $n_i $ vanishes, stipulating that the zero function is not a basis vector and shall simply be discarded whenever it arises); however, of those 8 functions (or a reduced number), we only have 2 linear combinations.
Let's start by considering $n_1 > 0$. To disentangle the 8 individual functions (or a reduced number), it is sufficient to supplement with $n_2 < 0$, yielding the combinations $$ scc+sss-csc+ccs \\ ccc+css+ssc-scs \ ; $$ with $n_3 < 0$, yielding the combinations $$ scc+sss+csc-ccs \\ ccc+css-ssc+scs \ ; $$ and with $n_2 , n_3 < 0$, yielding the combinations $$ scc-sss-csc-ccs \\ ccc-css+ssc+scs \ ; $$ Adding and subtracting the first two sets yields the combinations $$ scc+ccs \\ sss-csc \\ ccc-scs \\ css+ssc $$ and adding and subtracting the last two sets yields the combinations $$ scc-ccs \\ sss+csc \\ ccc+scs \\ css-ssc $$ Finally, adding and subtracting these new sets of 4 functions each fully disentangles the 8 functions (or a reduced number) that we know to be part of a basis.
Now, in the case $n_1 =0$, consider to begin with $n_2 > 0$. In this case, we need only the 4 functions $ccc,csc,ccs,css$ (or a reduced number if $n_3 =0$), of which we already have 2 linear combinations; to disentangle the functions, we only need to supplement with $n_3 <0$, i.e., we again have disentangled the functions that we know to be part of a basis.
Finally, for $n_1 =n_2 =0$, we only need the 2 functions $ccc,ccs$, which are already disentangled, i.e., we are left with the restriction $n_1 =n_2 =0$ and $n_3 \geq 0$.
Altogether, we thus have a one-to-one mapping between the described cases and a known basis; therefore, the described set of your preferred functions forms a basis (again, the zero function is discarded whenever it arises).