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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.

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  • $\begingroup$ Let me be a bit more clear. I am attempting to find out what is wrong, either this degree theory argument is false (i am sure me taking the ideas from the quoted paper and applying them here introduced errors...) or else the result that i claim: that i can prove a sequence of smooth solutions $u_m$ has bounded gradient. $\endgroup$ – Math604 Mar 6 '17 at 1:27
  • $\begingroup$ Could you remind us of your notion of degree, please? $\endgroup$ – Sebastian Goette Mar 6 '17 at 12:57
  • $\begingroup$ I guess i am using the Leray–Schauder degree. $\endgroup$ – Math604 Mar 7 '17 at 0:22

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