How to apply maximum/minimum principle to $\Delta u=u^3$ to prove uniqueness? I was told to use the maximum principle $$\max_{\bar{\Omega}}u=\max_{\partial\Omega}u$$ for subharmonic functions to prove that the problem $\Delta u=u^3$ in $\Omega=B(0, 1)$, $u=0$ in $\partial \Omega$, has unique solution ($u=0$). So assume that $v$ is another solution. Then define $w=u-v$. Then $w=0$ in $\partial \Omega$ and $$\Delta w=\Delta u-\Delta v=(u^2+uv+v^2)w,$$ where $u^2+uv+v^2\geq 0$. I should somehow conclude that $\Delta w\geq 0$ to be able to apply the maximum principle, and I believe I must also use the minimum principle. So I should end with $w\leq 0$ and $w\geq 0$ in $\Omega$ and hence $w=0\Leftrightarrow u=v$. Any help is appreciated.
 A: Suppose $u$ has a minimum at some point $p \in \Omega$.  Then
$u(p) \ge 0; \tag 1$
otherwise, we have
$u(p) < 0, \tag 2$
whence,
$\nabla^2 u(p) = \Delta u(p) = u^3(p) < 0; \tag 3$
(3) asserts that the trace of the Jacobean matrix  $J_u$ of $u$, which is
$J_u = \left [ \dfrac{\partial^2 u}{\partial x_i \partial x_j} \right ], \; 1 \le i, j \le n, \tag 4$
obeys
$\text{Tr} J_u(p) < 0, \tag 5$
since
$\text{Tr} J_u(p) = \displaystyle \sum_1^n \dfrac{\partial^2 u}{\partial x_i^2} = \Delta u(p) < 0; \tag 6$
here the $x_i$ are the usual cartesian coordinates on $\Bbb R^n$.  By virtue of (6), some eigenvalue $\lambda$ of $J_u(p)$ must satisfy
$\lambda < 0; \tag 7$
this in turn implies $u$ is decreasing in the direction of an eigenvector $\vec v \ne 0$ corresponding to $\lambda$, 
$J_u(p) \vec v = \lambda \vec v, \tag 8$
and hence we cannot have
$u(p) < 0 \tag 9$
at a minimum $p \in \Omega$ of $u$.  So
$u(p) \ge 0, \tag {10}$
which in turn implies
$u(x) \ge 0, \; \forall x \in \Omega. \tag{11}$
Now by virtue of (11), 
$\Delta u(x) = u^3(x) \ge 0 \tag{12}$
for $x \in \Omega$, so $u$ is subharmonic.  The fact that $u$ is subharmonic in turn implies that 
$\max_{\bar{\Omega}}u=\max_{\partial\Omega}u, \tag{13}$
as pointed out by our OP Infinitebig in the text of the question.  Thus, since $u(x) = 0$ for $x \in \partial \Omega$, we have
$u(x) \le 0 \tag{14}$
for $x \in \Omega$.  (11) and (14) together show that
$u(x) = 0, \; \forall x \in \Omega, \tag{15}$
and thus the uniqueness of the solution $u(x) = 0$ is established.
