i have a question regarding a degree theory argument and an apparent contradiction. Let me point out that I am a complete novice with degree theory and really i am just pushing some symbols with no real under standing of what they mean. So the equation I want to consider is

$$-\Delta u(x) - \lambda u(x)=| \nabla u(x)|^p$$ in $ \Omega$ with $u=0$ on $ \partial \Omega$. Here $0<\lambda<\lambda_1$ ($\lambda_1$ is the first eigenvalue of $-\Delta$) is fixed and $\Omega$ is a bounded domain in $ R^N$ with smooth boundary. We are taking $ 1<p$ (but as close to $1$ as one wishes).

So I would like to try and show there are no positive classical solutions of the above pde. Now given $u \in C^{0,1}$ (bounded gradient) and $0 \le t$ define $K_t(u)=v$ where $v$ satisfies $$-\Delta v - \lambda v = | \nabla u|^p + t $$ in $ \Omega$ with $v=0$ on $ \partial \Omega$. Let $ X=C^{0,1}$ with norm $ \| \nabla u \|_{L^\infty}$ (functions in $X$ are zero on the boundary). Then $K_t$ is compact on $X$. Using some other arguments I can show for large enough $T>0$ we have there is no solution of $$ -\Delta u -\lambda u = | \nabla u|^p +T$$ in $ \Omega$ (classical solution with $u=0$ on $ \partial \Omega$. This shows that $deg(I-K_T,B_R,0)=0$ for all $ R>0$ (here $B_R$ is the open ball centered at the origin in $X$).

Now we claim that for $R>0$ large enough we have $$0=deg(I-K_T,B_R,0)=deg(I-K_0,B_R,0).$$ We suppose not, then there is some $R_m \rightarrow \infty$ and $ 0 \le t_m \le T$ and $\| \nabla u_m \|_{L^\infty}=R_m$ such that $u_m-K_{t_m}(u_m)=0$ and so $u_m$ satisfies $$ -\Delta u_m -\lambda u_m = | \nabla u_m|^p + t_m$$ in $\Omega$ with $u_m=0$ on $ \partial \Omega$. But I now claim that i can prove some estimates that in fact show there is some $C>0$ such that $ \| \nabla u_m \|_{L^\infty} \le C$ (independent of $m$) (lets assume this claim is valid).

We now show for $ \epsilon>0$ small enough we have $deg(I-K_0,B_\epsilon,0)=deg(I,B_\epsilon,0) $ and this equals $1$. If the result is false there is some $ \epsilon_m \searrow 0$ and $ 0 \le t_m \le 1$ and $\| \nabla u_m \|_{L^\infty}=\epsilon_m$ with $ u_m-t_m K_0(u_m)=0$ which gives $$-\Delta u_m - \lambda u_m = t_m | \nabla u_m|^p$$ and we now consider $v_m(x)= \frac{u_m(x)}{\epsilon_m}$ and note that $\| \nabla v_m \|_{L^\infty}=1$. But $v_m$ satisfies $$-\Delta v_m - \lambda v_m = t_m \epsilon_m^{p-1} | \nabla v_m|^p$$ and since the right hand side goes to zero in $L^\infty$ we see that $\| \nabla v_m\|_{L^\infty} \rightarrow 0$; a contradiction. We now use for large enough $R>0$ and small enough $\epsilon>0$ that $$ 0=deg(I-K_0,B_R,0)= deg(I-K_0, B_R \backslash B_\epsilon,0)+ deg(I-K_0, B_\epsilon,0) $$ and hence $deg(I-K_0, B_R \backslash B_\epsilon,0)=-1$ and hence there is a non-zero classical solution of the original pde. This same argument works for $ \lambda=0$ but we know this equation $ -\Delta u = | \nabla u|^p$ in $\Omega$ with $u=0$ on $ \partial \Omega$ does not have a classical nonzero solution (by the maximum principle).

The degree theory arguments were taken from: W. Reichel and T. Weth ``Existence of solutions to nonlinear subcritical higher order elliptic Dirichlet problems'' J. Differential Equations 248 (2010), 1866–1878.