How to prove $u=0$ in the unit ball?

Question

Given that $u\in C^4(B_1)$, ${\triangle}^2{u}=\triangle{\triangle{u}}=0$ in the unit ball, And $u=\left| \nabla{u} \right|=0$ on the boundary of the unit ball,then how to deduce that $u=0$ in the unit ball? This is one of my ended midterm problems, to which the solution I missed thanks for any clues!

Answer 1

Take integration by part:

$0=\int_{B_1} u {\triangle}^2{u} dy=-\int_{B_1} \nabla u \nabla ({\triangle}u) dy+\int_{\partial B_1} u \frac{\partial{\triangle}u}{\partial n}dS=-\int_{B_1} \nabla u \nabla ({\triangle}u) dy$,

as u and its gradient are both null on boundary.

$\int_{B_1} \nabla u\nabla ({\triangle}u) dy=-\int_{B_1}{\left|{\triangle}{u}\right|}^2dy+\int_{\partial B_1} u \frac{\partial{\triangle}u}{\partial n}dS=-\int_{B_1}{\left|{\triangle}{u}\right|}^2dy=0$

Therefore ${\triangle}u=0$ in $B_1$ (As u is $C^4$-continuous.)

Then by maxmum principle, we have $u=0$ in $B_1$.