Let $D$ be a bounded region. For the problem
$\Delta u+au-bu^2 = 0$ in $D$
$u=0$ on the boundary of $D$,
show that there are no positive solutions if $a<\lambda_1$ (which is the smallest eigenvalue of $\Delta u + \lambda u = 0$ on $D$, with $u=0$ on the boundary of $D$).
I know to multiply by $u$, then integrate over $D$. I have then been given that I should use Green's formula to answer the question. However, I'm not really sure how to link the two.
Any help is very much appreciated!
I am assuming $b>0$. Let's first examine the smallest eigenvalue of $\Delta$ for $u=0$ on boundary: if $$ -\Delta u = \lambda u $$ Multiply both side by $u$, integrate: $$ \int_D -u\Delta u \,dx= \int_D \lambda u^2\,dx \tag{1} $$ Green's formula is using divergence theorem $$ \int_D \nabla \cdot \boldsymbol{F} \,dx = \int_{\partial D} \boldsymbol{F}\cdot\boldsymbol{n} \,dS\tag{$\dagger$} $$ on $\boldsymbol{F} = \psi \nabla \varphi$, thus $\nabla \cdot \boldsymbol{F} = \psi \Delta \varphi + \nabla \varphi \cdot \nabla \psi$ and $(\dagger)$ becomes: $$\int_D \psi \Delta \varphi \,dx = -\int \nabla \varphi \cdot \nabla \psi\, dx +\int_{\partial D} \psi ( \nabla \varphi \cdot \boldsymbol{n} )\, dS $$ Plugging this formula back to (1), and using $u=0$ on boundary: $$ \int_D \nabla u\cdot\nabla u \,dx - \int_{\partial D} u ( \nabla u \cdot \boldsymbol{n} )\, dS= \int_D |\nabla u|^2 \,dx= \int_D \lambda u^2\,dx $$ if $a<\lambda_1$, the smallest possible eigenvalue, this implies for ANY permissible u in some space (in this case $u$ solves the original PDE) $$ \int_D |\nabla u|^2 \,dx \geq \int_D \lambda_1 u^2\,dx > \int_D a u^2\,dx\tag{2} $$
Now multiply both sides of $$ \Delta u+au-bu^2 = 0 $$ by $u$ and integrate over $D$: $$ \int_D(u\Delta u + au^2 -bu^3) dx = 0 $$ again by Green's identity $$ \int_D(-\nabla u\cdot \nabla u + au^2 -bu^3) dx + \int_{\partial D} u ( \nabla u \cdot \boldsymbol{n} )\, dS = 0 $$ By $u=0$ on boundary, above integral is: $$ \int_D(-|\nabla u|^2 + au^2 -bu^3) dx = 0 $$ which is: $$ \int_D bu^3 \,dx = \int_D(-|\nabla u|^2 + au^2) dx < 0 $$ by (2). Therefore there does not exist a positive $u$ on $D$, otherwise $\displaystyle\int_D bu^3 \,dx >0$.
