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
asked Aug 28 at 15:33
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61
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up vote
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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
asked Aug 28 at 15:33
dittì
61
2
One counterexample should be enough in my opinion.
– supinf
Aug 28 at 15:36
add a comment |Â
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
asked Aug 28 at 15:33
dittì
61
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
asked Aug 28 at 15:33
dittì
61
asked Aug 28 at 15:33
dittì
61
asked Aug 28 at 15:33
dittì
61
asked Aug 28 at 15:33
dittì
61
61
2
One counterexample should be enough in my opinion.
– supinf
Aug 28 at 15:36
add a comment |Â
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
add a comment |Â
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2
One counterexample should be enough in my opinion.
– supinf
Aug 28 at 15:36