Solve this Dirichlet problem Show that the Dirichlet problem
$$ \left\{ \begin{array}{l} u_{xx}+ u_{yy}=u^3 \ \text{in} \ x^2+y^2 \lt 1 \\ u=0 \ \text{on} \ x^2+y^2 = 1 \end{array} \right.$$
where $u=u(x,y)$, has only the trivial solution $u \equiv 0$
Thanks
 A: We have 
$$\Delta u=u^3\mbox{ on }D$$
where $D=\{(x,y):x^2+y^2< 1\}$. Multipying $u$ on both sides and integrating it over $D$, we get
$$\tag{1}\int_Du\Delta u=\int_D u^4\geq 0.$$
By Stokes' Theorem, we have 
$$\tag{2}\int_Du\Delta u=-\int_D|\nabla u|^2+\int_{\partial D}u\frac{\partial u}{\partial v}
=-\int_D|\nabla u|^2$$
where $\frac{\partial u}{\partial v}$ is the normal derivative of $u$ on $\partial D$, and the last equality follows from the boundary condition that $u=0$ on $\partial D=\{(x,y):x^2+y^2= 1\}$.  Combining $(1)$ and $(2)$, we obtain 
$$\int_D|\nabla u|^2\leq 0$$
which implies that $\nabla u\equiv 0$ in $D$, or equivalently, $u$ is a constant in $D$. Since $u=0$ on $\partial D$, we must have $u=0$ in $D$. 
A: A somewhat different take on it:
Let $u(x, y)$ be a solution to 
$u_{xx} + u_{yy} = u^3 \tag{1}$
in the open unit disk $B(0, 1) = \{ (x, y) \in \Bbb R^2 \mid x^2 + y^2 < 1 \}$, with
$u(x, y) = 0 \tag{2}$
on $\partial B(0, 1) = \{ (x, y) \in \Bbb R^2 \mid x^2 + y^2 = 1 \}$.  Without going into further details of the analysis, I am going to assume $u(x, y) \in C^2(\bar B(0, 1), \Bbb R)$, where $\bar B(0, 1) = \{(x, y) \in \Bbb R^2 \mid x^2 + y^2 \le 1 \}$;  note that $\bar B(0, 1) = B(0, 1) \cup \partial B(0, 0)$.  Then $u(x, y)$, being a continuous, real-valued function on the compact set $\bar B(0, 1)$, actually takes on its maximum and minimum values at some points in $\bar B(0, 1)$; let $M$ be the maximum, and $m$ be the minimum, values of $u(x, y)$ on the set $\bar B(0, 1)$so that
$m \le u(x, y) \le M$ for $(x, y) \in \bar B(0, 1)$.  Then we have
$m \le 0 \le M. \tag{3}$
(3) follows from the boundary conditions imposed upon $u(x, y)$:  if we had $M < 0$, then $u(x, y) \le M < 0$ on $\bar B(0, 1)$, contradicting (2); a similar argument shows $m \le 0$.  Now if $m = M = 0$, then $u(x, y) = 0$ and there is nothing left to prove; so we can assume at least one of $m, M \ne 0$.  If $M > 0$, there must be a point $P_M \in B(0, 1)$ with $u(P_M) = M > 0$; $P_M \notin \partial B(0, 1)$ since $u(x, y) = 0$ for $(x, y) \in \partial B(0, 1)$.  We have
$\nabla u(P_M) = 0, \tag{4}$
since $P_M$ is a local (at least) maximum of the function $u(x, y)$.  Furthermore, at $P_M$, (1) becomes
$u_{xx}(P_M) + u_{yy}(P_M) = u^3(P_M) = M^3 > 0. \tag{5}$
Next, we examine the Hessian, or matrix of second derivatives, of $u$ at $P_M$:
$Hess(u)(P_M) = \begin{bmatrix} u_{xx}(P_M) & u_{xy}(P_M) \\ u_{yx}(P_M) & u_{yy}(P_M) \end{bmatrix}; \tag{6}$
we see from (5) that the trace of $Hess(u)(P_M)$ is
$\text{Tr}(Hess (u)(P_M)) = u_{xx}(P_M) + u_{yy}(P_M) = M^3 > 0, \tag{7}$
which implies that at least one eigenvalue of the symmetric matrix $Hess(u)(P_M)$ is positive!  Let $\lambda$ be such a positive eigenvalue of $Hess(u)(P_M)$, and let $\mathbf e$ be the corresponding unit eigenvector; we know such $\lambda, \mathbf e$ exist since $Hess(u)(P_M)$ is symmetric.  Then
$Hess(u)(P_M) \mathbf e = \lambda \mathbf e, \tag{8}$
and we can compute the first and second derivatives of $u(x, y)$ along the line $P_M + t \mathbf e$ at the point $P_M$ where $t = 0$ as follows:
$(du(P_M + t \mathbf e) / dt)_{t = 0} =  \nabla u(P_M) \cdot \mathbf e = 0 \tag{9}$
by (4); 
$(d^2u(P_M + t \mathbf e) / dt^2)_{t = 0} = \mathbf e \cdot Hess(u)(P_M) \mathbf e = \mathbf e \cdot \lambda \mathbf e = \lambda > 0. \tag{10}$
Equation (9) is well-known and (10) follows from the formula
$du(P_M + t \mathbf e) / dt =  \nabla u(P_M + t\mathbf e) \cdot \mathbf e \tag{11}$
by performing a second differentiation with respect to $t$, using the chain rule, and finally evaluating at $t = 0$.  The details are easy to execute.  Taken together, (9) and (10) show that, for sufficiently small $t$, $u(x, y)$ is in fact increasing along the line $P_M + t \mathbf e$, so that for such small $t \ne 0$ $u(P_M + t\mathbf e) > u(P_M)$, which contradicts the assumption that $u(x, y)$ takes on its maximum value $M > 0$ at $P_M$.  A corresponding, similar argument disposes of the case $m < 0$; then we have $\text{Tr}(Hess (u)(P_m)) = m^3 < 0$ so that one of the eigenvalues $\lambda < 0$ and so forth.  In any event, these deliberations indicate that the only way to avoid a contradiction is to assume $m = M = 0$, in which case $u(x, y) = 0$ on 
$\bar B(0, 1)$ as we have seen.  QED.
Hope this helps.  Cheerio,
and as always, 
Fiat Lux!!!
