Biharmonic boundary condition

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I try to solve

$$\Delta^2u=f$$ on unit square. with $f=4\sin(\pi x)\sin(\pi y)$

Using $v=-\Delta u,$ leads to $$v+\Delta u=0,$$ $$-\Delta v=f.$$

By Dirichlet boundary condition on $u$.

What is the another boundary condition? Could I use Dirichlet boundary condition for $v$?When I use this condition the answer is true. Otherwise it is false.

How could I determine the boundary condition. Is there any refrence to help me about that?

Please help me.Thanks a lot.