1.1.The Maximum Principle 11 (1.1.2).Suppose that uC(B)C2(B)satisfies LuinB,u(<u() for any x∈B,andu(xo)≥0.Then, w>0, where v is the exterior unit normal to B at xo. Proof.Without los of generality,we assume B=Bg for someR.By the continuity of u up to aBr,we have,for any BR, u(x)≤u(xo) For positive constants u and e to be determined,we set w(z)=e-ulzP2-e-uR2 and v(z)=u(z)-u(xo)+Ew(x). We consider w and v in D=BR\BR/2. A direct calculation yields Lw =e[4uazj-2uaijg-2b+c)-ce-R ≥e4e{4u2ax)-2μ(a6+biz)+c, where we used c in BR.By the strict ellipticity (1.1.2),we have a,ez≥AzP≥AR2inD Hence, Luw≥e4eP{u2R2-2u(a46+bz）+c}≥0inD if we chooseμsufficiently large.Byc≤0andu(xo)≥O,we obtain,for any e>0, Lv=Lu+eLw -cu(xo)0 in D. Next,we discussvon in two cases.First,on we have u(< 0,and hence u-u(zo)<-e for some>0,by the continuity of u on Bg/2. Note that w<1 on aBR/2.Then for such an e,we obtain v<on BR/ Second,on aBR,we have w=0 and u<u(o).Hence,v0 on BR and v(xo)=0.Therefore,v<0on 8D. In conclusion,Lv >0 in D andv0 on OD.By the maximum principle, we have v≤0inD 由扫描全能王扫描创建 ▣ 由 扫描全能王 扫描创建