Weak maximum principle for coercive divergence-form equations

Homogeneous maximum bound. Put $t:={\sup }_{\mathrm{\partial }\mathrm{\Omega }}{u}^{+}$. If $t=+\mathrm{\infty }$ the bound is immediate. Otherwise $t\ge 0$ and $w:=(u-t{)}^{+}\in H_0^1(\mathrm{\Omega })$ by [F1]. Since the equation is homogeneous, boundedness of the form and density extend its inequality from nonnegative compactly supported smooth tests to all nonnegative $H_0^1$ tests. For the weak sign condition, choose real ${\varphi }_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ with ${\varphi }_j\to w$ in $H_0^1$. Then ${\varphi }_j^2\ge 0$ and ${\varphi }_j^2\to w^2$ in $W^{1,1}$, since Cauchy--Schwarz gives convergence of both the functions and their gradients. Thus $w^2\in W_0^{1,1}$ with $D(w^2)=2wDw$, and boundedness of $b,c$ makes $\zeta \mapsto \int (c\zeta +b^i{D}_i\zeta )$ continuous on $W^{1,1}$; the sign condition therefore holds on $w^2$ without asserting $w^2\in H_0^1$. Testing with $w$ and using $u=w+t$ on $\{w>0\}$ gives $0\ge a(u,w)={\int }_{\mathrm{\Omega }}{a}^{ij}{D}_{j}w{D}_{i}w+{\int }_{\mathrm{\Omega }}(c{w}^2+b^i{D}_{i}ww)+t{\int }_{\mathrm{\Omega }}cw\mathrm{.}$ The lower-order quadratic term is ${\int }_{\mathrm{\Omega }}(c{w}^2+b^i{D}_{i}ww)=\frac{1}{2}{\int }_{\mathrm{\Omega }}c{w}^2+\frac{1}{2}{\int }_{\mathrm{\Omega }}(c{w}^2+b^i{D}_i(w^2))\ge 0,$ by $c\ge 0$ and the extended weak sign condition; the boundary-shift term $t\int cw$ is nonnegative as well. Hence $\theta \mathrm{\parallel }Dw{\mathrm{\parallel }}_2^2\le 0$, and Poincare gives $w=0$. Therefore ${\mathrm{ess\hspace{0.17em}sup}}_{\mathrm{\Omega }}u\le t$.

Forcing energy bound. Assume $b\equiv 0$, $c\ge 0$, and $f\in L^q(\mathrm{\Omega })$ with the stated exponent. If $t:={\sup }_{\mathrm{\partial }\mathrm{\Omega }}{u}^{+}=+\mathrm{\infty }$, the claim is immediate; otherwise set $v:=(u-t{)}^{+}\in H_0^1(\mathrm{\Omega })$. Since $q>n\mathrm{/}2$, Sobolev and Holder show that $f$ defines a continuous functional on $H_0^1(\mathrm{\Omega })$, so the local subsolution inequality extends to this test. On $\{v>0\}$, $u=v+t$, and testing gives $\theta \mathrm{\parallel }Dv{\mathrm{\parallel }}_2^2\le {\int }_{\mathrm{\Omega }}{f}^{+}v\le \mathrm{\parallel }{f}^{+}{\mathrm{\parallel }}_q\mathrm{\parallel }v{\mathrm{\parallel }}_{q^{\mathrm{\prime }}}\mathrm{.}$ For $n\ge 3$ take $\kappa =2^{\ast }$; for $n=2$ take any finite $\kappa >q^{\mathrm{\prime }}$. Holder, Poincare and the available Sobolev embedding imply $\mathrm{\parallel }v{\mathrm{\parallel }}_{q^{\mathrm{\prime }}}\le C\mathrm{\parallel }Dv{\mathrm{\parallel }}_2$. Thus $\mathrm{\parallel }v{\mathrm{\parallel }}_{H_0^1}+\mathrm{\parallel }v{\mathrm{\parallel }}_2\le C_E\mathrm{\parallel }{f}^{+}{\mathrm{\parallel }}_q$, where constants depend only on the parameters in the Statement.

The forcing iteration. Write $a:=\mathrm{\parallel }{f}^{+}{\mathrm{\parallel }}_{L^q(\mathrm{\Omega })}$. If $a=0$, step 1.2 gives $v=0$. Otherwise fix $T>0$ and define $k_j=t+T(1-2^{-j})$, $E_j:=\{u>k_j\}$, $Y_j:={\int }_{E_j}(u-k_j{)}^2$, and $B_j:={\int }_{\mathrm{\Omega }}\mathrm{\mid }D(u-k_j{)}^{+}{\mathrm{\mid }}^2$. Let $q^{\mathrm{\prime }}$ be conjugate to $q$, and choose $\kappa =2^{\ast }$ for $n\ge 3$; for $n=2$ choose finite $\kappa >2q^{\mathrm{\prime }}$. Set $\delta :=1-2\mathrm{/}\kappa >0$, $\gamma :=1\mathrm{/}{q}^{\mathrm{\prime }}-1\mathrm{/}\kappa >0$, and $\beta :=\delta +2\gamma$. For $n\ge 3$, $\delta =2\mathrm{/}n$ and $q>n\mathrm{/}2$ gives $\beta =1+4\mathrm{/}n-2\mathrm{/}q>1$; for $n=2$, $\beta =1+2\mathrm{/}{q}^{\mathrm{\prime }}-4\mathrm{/}\kappa >1$ by the choice of $\kappa$. Testing with $(u-k_j{)}^{+}$ and using $c\ge 0$, Holder on $E_j$, and Sobolev gives $B_j^{1\mathrm{/}2}\le Ca\mathrm{\mid }{E}_j{\mathrm{\mid }}^{\gamma }$ (if $B_j=0$, Poincare gives $(u-k_j{)}^{+}=0$). Also $\mathrm{\mid }{E}_{j+1}\mathrm{\mid }\le (T{2}^{-j-1}{)}^{-2}{Y}_j$ and Sobolev gives $Y_{j+1}\le C\mathrm{\mid }{E}_{j+1}{\mathrm{\mid }}^{\delta }{B}_{j+1}$. Consequently $Y_{j+1}\le C_0{a}^2{T}^{-2\beta }{2}^{2\beta (j+1)}{Y}_j^{\beta }\mathrm{.}$ Set $B:=2^{2\beta }$ and $Z_j:=Y_j\mathrm{/}{T}^2$. Choose $T=C_{1}a$ with $C_1^2\ge C_{0}B$ and $C_1^2\ge C_E(2B{)}^{1\mathrm{/}(\beta -1{)}^2}$, where $Y_0\le C_E{a}^2$ by step 1.2. Then $Z_{j+1}\le (C_{0}B\mathrm{/}{C}_1^2)B^j{Z}_j^{\beta }\le B^j{Z}_j^{1+(\beta -1)}$ and $Z_0\le (2B{)}^{-1\mathrm{/}(\beta -1{)}^2}$. The nonlinear iteration [F5] gives $Z_j\to 0$, hence $Y_j\to 0$. Since $(u-t-T{)}^{+}\le (u-k_j{)}^{+}$ and $(u-t-T{)}^{+}\in L^2(\mathrm{\Omega })$, this forces $(u-t-T{)}^{+}=0$ a.e. on $\mathrm{\Omega }$, proving the forcing bound. The finite $\kappa$ choice in dimension two uses the full open range of the critical Sobolev embedding.

Supersolutions and equality. If $u$ is a weak supersolution of $Lu=f$, then $-u$ is a weak subsolution of the same operator with coefficients $(a,b,c)$ and source $-f$, by linearity of the form; applying step 1.1 or step 2.1 yields the stated lower-bound versions with $u^{-}=(-u{)}^{+}$ and $f^{-}$. If $u$ is a weak solution of $Lu=0$ with $b=c=0$, let $s:={\sup }_{\mathrm{\partial }\mathrm{\Omega }}u$. For every finite a.e. upper bound $t$ on $u$, $(u-t{)}^{+}=0$, so [F1] implies $Tu\le t$ a.e.; taking infima gives $s\le {\mathrm{ess\hspace{0.17em}sup}}_{\mathrm{\Omega }}u$. If $s=+\mathrm{\infty }$, this forces ${\mathrm{ess\hspace{0.17em}sup}}_{\mathrm{\Omega }}u=+\mathrm{\infty }=s$. If $s$ is finite, $(u-s{)}^{+}\in H_0^1(\mathrm{\Omega })$, and density extends the weak identity to this test. Since $b=c=0$, it gives $0=a(u,(u-s{)}^{+})=\int a^{ij}{D}_j(u-s{)}^{+}{D}_i(u-s{)}^{+}$, so $u\le s$ a.e. The reverse trace bound just proved gives $s\le {\mathrm{ess\hspace{0.17em}sup}}_{\mathrm{\Omega }}u$, and hence ${\mathrm{ess\hspace{0.17em}sup}}_{\mathrm{\Omega }}u=s$.