Proving that 2D eigenfunctions of a biharmonic PDE span the space Let $U=(0,1)\times(0,1)$. Consider the elliptic boundary value problem on this domain:
$$
\Delta^2u = f,
$$
where $u: U\rightarrow \mathbb{R}$ and $f\in L^2(U)$. The boundary conditions are: $u(x,y) = \Delta u(x,y) = 0$ for $(x,y)\in\partial U$. 
I was supposed to solve this problem by first solving the corresponding eigenvalue equation, i.e. $\Delta^2 \phi = \lambda \phi$, and then perform an expansion of u in these eigenfunctions. By separation of variables I found the following eigenfunctions: $\phi = \sin(n\pi x)\sin(m\pi y)$ for $n,m\in\mathbb{Z}$, which my teacher confirmed. The expansion in these eigenfunctions is quite easy, but the problem for me is proving that these eigenfunctions form a basis for $L^2(U)$. One can easily prove that they are orthogonal, but proving that they span the space is my biggest problem. My teacher was talking about defining a function on $(0,1/2)\times(0,1/2)$ and then mirroring it some lines to obtain u and thus proving that they span the space. This explanation was beyond me, so I hope you guys have a better one!
 A: First, some eigenfunctions are missing (per the answer by Shuhao Cao). Second, it is a little easier to prove completeness working over the complex numbers. This is because the linear span of the functions $$\phi_{m,n}(x,y)=e^{i\pi m x}e^{i\pi n y}, \ m,n\in\mathbb Z^2$$ is an algebra which contains constants and separates the points of $[0,1]\times [0,1]$. By the Stone-Weierstrass theorem, every continuous function on $[0,1]\times [0,1]$ is a uniform limit of such linear combinations. Since the closed linear span of $\{\phi_{m,n}\}$ contains all continuous functions, it is all of $L^2$. 
You can translate the above into the language of real-valued functions by considering $\operatorname{Re}\phi_{m,n}$ and $\operatorname{Im}\phi_{m,n}$, but you'll need some trigonometric identities to verify that the linear span is an   algebra. 
A: HINT: I believe the eigenfunction set you found is not complete.
The basis set of $L^2(0,1)$ is $\{e^{i k\pi x}\}_{k\in \mathbb{Z}}$. Any $f\in L^2(0,1)$ can be written as:
$$f(x) = \sum_{k = -\infty}^{\infty} f_k \,e^{ik\pi x},$$
where $f_k$ is the Fourier coefficient. 
In 2 spacial dimension case, $U = (0,1)\times(0,1)$, the basis set is $\{e^{im\pi x}e^{in\pi y}\}_{m,n\in \mathbb{Z}}$.
Using Euler's formula gives $$\{\sin(m\pi  x)\sin(n\pi  y), \cos(m\pi x)\sin(n\pi  y), \sin(m\pi x)\cos(n\pi y), \cos(m\pi x)\cos(n\pi y)\}_{m,n\in \mathbb{Z}}$$
as the possible basis set for $L^2(U)$. Now for any function in $L^2(U)$, try to find its Fourier series expansion using these functions. If you could represent any function in $L^2(U)$ as an expansion using these functions, then these functions form a set of basis for $L^2(U)$.
