operator not satisfying weak maximum principle

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I have to prove that the operator $Lu=Delta u +u$ does not satisfy the weak maximum principle, i.e. if $u$ is such that $Lu=0$, then $u$ can attain its maximum in the interior of the bounded domain $Omega$ that is considered for $u$.



In the monodimensional case one can simply consider $u(t)=sin (t)$ in $Omega=(0,pi)$ so I tried to adapt it to the general case using for example $u(x_1dots x_n)=sum_i=1^n sin (x_i)$ and prove that its maximum is inside the interior of, e.g. $Omega=(0,pi)^n$.



Is there a more general way to prove it? Thanks in advance.




functional-analysis pde elliptic-operators




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    One counterexample should be enough in my opinion.
    – supinf
    Aug 28 at 15:36














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I have to prove that the operator $Lu=Delta u +u$ does not satisfy the weak maximum principle, i.e. if $u$ is such that $Lu=0$, then $u$ can attain its maximum in the interior of the bounded domain $Omega$ that is considered for $u$.



In the monodimensional case one can simply consider $u(t)=sin (t)$ in $Omega=(0,pi)$ so I tried to adapt it to the general case using for example $u(x_1dots x_n)=sum_i=1^n sin (x_i)$ and prove that its maximum is inside the interior of, e.g. $Omega=(0,pi)^n$.



Is there a more general way to prove it? Thanks in advance.




functional-analysis pde elliptic-operators




share|cite|improve this question
















  • 2




    One counterexample should be enough in my opinion.
    – supinf
    Aug 28 at 15:36












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have to prove that the operator $Lu=Delta u +u$ does not satisfy the weak maximum principle, i.e. if $u$ is such that $Lu=0$, then $u$ can attain its maximum in the interior of the bounded domain $Omega$ that is considered for $u$.



In the monodimensional case one can simply consider $u(t)=sin (t)$ in $Omega=(0,pi)$ so I tried to adapt it to the general case using for example $u(x_1dots x_n)=sum_i=1^n sin (x_i)$ and prove that its maximum is inside the interior of, e.g. $Omega=(0,pi)^n$.



Is there a more general way to prove it? Thanks in advance.




functional-analysis pde elliptic-operators




share|cite|improve this question












I have to prove that the operator $Lu=Delta u +u$ does not satisfy the weak maximum principle, i.e. if $u$ is such that $Lu=0$, then $u$ can attain its maximum in the interior of the bounded domain $Omega$ that is considered for $u$.



In the monodimensional case one can simply consider $u(t)=sin (t)$ in $Omega=(0,pi)$ so I tried to adapt it to the general case using for example $u(x_1dots x_n)=sum_i=1^n sin (x_i)$ and prove that its maximum is inside the interior of, e.g. $Omega=(0,pi)^n$.



Is there a more general way to prove it? Thanks in advance.





functional-analysis pde elliptic-operators





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 28 at 15:33









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  • 2




    One counterexample should be enough in my opinion.
    – supinf
    Aug 28 at 15:36












  • 2




    One counterexample should be enough in my opinion.
    – supinf
    Aug 28 at 15:36







2




2




One counterexample should be enough in my opinion.
– supinf
Aug 28 at 15:36




One counterexample should be enough in my opinion.
– supinf
Aug 28 at 15:36















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