Proof of coercivity Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)When does non-negativity of the integral of a function imply that the function itself is non-negative?An existence TheoremShowing boundedness and a coercivity condition for a bilinear formAn integral of a sequence of functionsCoercivity of a bilinear formShow that form is coerciveProve that $lim_n rightarrow infty int_Omega |f_n - f|^p dm=0$.Schwarz symmetrization is equimeasurableBoundedness of a function implies coercivity of certain functional
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Proof of coercivity
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)When does non-negativity of the integral of a function imply that the function itself is non-negative?An existence TheoremShowing boundedness and a coercivity condition for a bilinear formAn integral of a sequence of functionsCoercivity of a bilinear formShow that form is coerciveProve that $lim_n rightarrow infty int_Omega |f_n - f|^p dm=0$.Schwarz symmetrization is equimeasurableBoundedness of a function implies coercivity of certain functional
$begingroup$
I have the intution that the functional $$A(u)=int_Omega |nabla u(x)|^2 dx + 1/2cdotint_Omega (1-|u(x)|^2)^2 dx quad forall u in H^1(Omega) $$ where $Omega $ is a bounded domain of $mathbbR^N$, is coercive, that is $$ lim_Vert u Vertrightarrow inftyA(u)=infty$$ where $Vert uVert^2= int_Omega|nabla u|^2 +int_Omega | u |^2 $. I dont know how to prove it. I am stucked. Is there any theorial result that can be helpful in order to prove it that I should know?
This is my attempt:
$$ A(u)geq int_Omega |nabla u(x)|^2 dx +1/2 int_u (1-|u(x)|^2)^2 dx geq int_Omega |nabla u(x)|^2 dx +1/2 int_u |u(x)|^2 dx = Vert u Vert^2- int_ |u(x)|^2 dx = Vert u Vert^2-Crightarrow infty$$ am I right?
real-analysis functional-analysis
asked Apr 1 at 17:32
R. N. MarleyR. N. Marley
1259
$endgroup$
add a comment |
$begingroup$
I have the intution that the functional $$A(u)=int_Omega |nabla u(x)|^2 dx + 1/2cdotint_Omega (1-|u(x)|^2)^2 dx quad forall u in H^1(Omega) $$ where $Omega $ is a bounded domain of $mathbbR^N$, is coercive, that is $$ lim_Vert u Vertrightarrow inftyA(u)=infty$$ where $Vert uVert^2= int_Omega|nabla u|^2 +int_Omega | u |^2 $. I dont know how to prove it. I am stucked. Is there any theorial result that can be helpful in order to prove it that I should know?
This is my attempt:
$$ A(u)geq int_Omega |nabla u(x)|^2 dx +1/2 int_u (1-|u(x)|^2)^2 dx geq int_Omega |nabla u(x)|^2 dx +1/2 int_u |u(x)|^2 dx = Vert u Vert^2- int_ |u(x)|^2 dx = Vert u Vert^2-Crightarrow infty$$ am I right?
real-analysis functional-analysis
asked Apr 1 at 17:32
R. N. MarleyR. N. Marley
1259
$endgroup$
add a comment |
$begingroup$
I have the intution that the functional $$A(u)=int_Omega |nabla u(x)|^2 dx + 1/2cdotint_Omega (1-|u(x)|^2)^2 dx quad forall u in H^1(Omega) $$ where $Omega $ is a bounded domain of $mathbbR^N$, is coercive, that is $$ lim_Vert u Vertrightarrow inftyA(u)=infty$$ where $Vert uVert^2= int_Omega|nabla u|^2 +int_Omega | u |^2 $. I dont know how to prove it. I am stucked. Is there any theorial result that can be helpful in order to prove it that I should know?
This is my attempt:
$$ A(u)geq int_Omega |nabla u(x)|^2 dx +1/2 int_u (1-|u(x)|^2)^2 dx geq int_Omega |nabla u(x)|^2 dx +1/2 int_u |u(x)|^2 dx = Vert u Vert^2- int_ |u(x)|^2 dx = Vert u Vert^2-Crightarrow infty$$ am I right?
real-analysis functional-analysis
asked Apr 1 at 17:32
R. N. MarleyR. N. Marley
1259
$endgroup$
I have the intution that the functional $$A(u)=int_Omega |nabla u(x)|^2 dx + 1/2cdotint_Omega (1-|u(x)|^2)^2 dx quad forall u in H^1(Omega) $$ where $Omega $ is a bounded domain of $mathbbR^N$, is coercive, that is $$ lim_Vert u Vertrightarrow inftyA(u)=infty$$ where $Vert uVert^2= int_Omega|nabla u|^2 +int_Omega | u |^2 $. I dont know how to prove it. I am stucked. Is there any theorial result that can be helpful in order to prove it that I should know?
This is my attempt:
$$ A(u)geq int_Omega |nabla u(x)|^2 dx +1/2 int_u (1-|u(x)|^2)^2 dx geq int_Omega |nabla u(x)|^2 dx +1/2 int_u |u(x)|^2 dx = Vert u Vert^2- int_ |u(x)|^2 dx = Vert u Vert^2-Crightarrow infty$$ am I right?
real-analysis functional-analysis
real-analysis functional-analysis
asked Apr 1 at 17:32
R. N. MarleyR. N. Marley
1259
asked Apr 1 at 17:32
R. N. MarleyR. N. Marley
1259
edited Apr 1 at 18:00
R. N. Marley
asked Apr 1 at 17:32
R. N. MarleyR. N. Marley
1259
asked Apr 1 at 17:32
R. N. MarleyR. N. Marley
1259
asked Apr 1 at 17:32
R. N. MarleyR. N. Marley
1259
1259
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your functional is not coercive. For example, fix $Omega = [0,1] subseteq mathbbR$ and consider $u_n(x) = x^n+1$. Then $nabla u(x) = (n+1)x^n$ and so
$$|u|^2 = (n+1)^2 int_0^1 x^2n dx + int_0^1 x^2(n+1) dx = frac(n+1)^22n + frac12(n+1) to infty$$ as $n to infty$. However
$$Au_n = (n+1) int_0^1 x^n dx + int_0^1 (1- x^2(n+1))^2 dx = 2 + frac14n+1 - frac1n+1$$
is a bounded sequence.
Edit: After you've edited, your new functional is coercive by basically your attempt, except for the fact a factor of a half mysteriously goes missing at some point. This is easily dealt with however since you only want a lower bound that doesn't care about multiplicative factors.
answered Apr 1 at 17:50
Rhys SteeleRhys Steele
7,9101931
$endgroup$
$begingroup$
Another question, why $1-|u|^2in L^2(Omega)$ if $uin H^1(Omega)$??
$endgroup$
– R. N. Marley
Apr 1 at 19:24
$begingroup$
It needn't be in general (unless you assume enough to get this from a sobolev embedding). I assume the correct interpretation of your problem is that $A$ takes values in $[0,infty]$ and if $Au = infty$ then you have no problem with coercivity.
$endgroup$
– Rhys Steele
Apr 1 at 20:33
$begingroup$
I see... so my functional is not well defined on $H^1(Omega )$.
$endgroup$
– R. N. Marley
Apr 1 at 20:50
1
$begingroup$
Well, it's well defined as a functional to the extended real line. Note that all the integrands are non-negative and so the integrals still all make sense.
$endgroup$
– Rhys Steele
Apr 1 at 20:51
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your functional is not coercive. For example, fix $Omega = [0,1] subseteq mathbbR$ and consider $u_n(x) = x^n+1$. Then $nabla u(x) = (n+1)x^n$ and so
$$|u|^2 = (n+1)^2 int_0^1 x^2n dx + int_0^1 x^2(n+1) dx = frac(n+1)^22n + frac12(n+1) to infty$$ as $n to infty$. However
$$Au_n = (n+1) int_0^1 x^n dx + int_0^1 (1- x^2(n+1))^2 dx = 2 + frac14n+1 - frac1n+1$$
is a bounded sequence.
Edit: After you've edited, your new functional is coercive by basically your attempt, except for the fact a factor of a half mysteriously goes missing at some point. This is easily dealt with however since you only want a lower bound that doesn't care about multiplicative factors.
answered Apr 1 at 17:50
Rhys SteeleRhys Steele
7,9101931
$endgroup$
$begingroup$
Another question, why $1-|u|^2in L^2(Omega)$ if $uin H^1(Omega)$??
$endgroup$
– R. N. Marley
Apr 1 at 19:24
$begingroup$
It needn't be in general (unless you assume enough to get this from a sobolev embedding). I assume the correct interpretation of your problem is that $A$ takes values in $[0,infty]$ and if $Au = infty$ then you have no problem with coercivity.
$endgroup$
– Rhys Steele
Apr 1 at 20:33
$begingroup$
I see... so my functional is not well defined on $H^1(Omega )$.
$endgroup$
– R. N. Marley
Apr 1 at 20:50
1
$begingroup$
Well, it's well defined as a functional to the extended real line. Note that all the integrands are non-negative and so the integrals still all make sense.
$endgroup$
– Rhys Steele
Apr 1 at 20:51
add a comment |
$begingroup$
Your functional is not coercive. For example, fix $Omega = [0,1] subseteq mathbbR$ and consider $u_n(x) = x^n+1$. Then $nabla u(x) = (n+1)x^n$ and so
$$|u|^2 = (n+1)^2 int_0^1 x^2n dx + int_0^1 x^2(n+1) dx = frac(n+1)^22n + frac12(n+1) to infty$$ as $n to infty$. However
$$Au_n = (n+1) int_0^1 x^n dx + int_0^1 (1- x^2(n+1))^2 dx = 2 + frac14n+1 - frac1n+1$$
is a bounded sequence.
Edit: After you've edited, your new functional is coercive by basically your attempt, except for the fact a factor of a half mysteriously goes missing at some point. This is easily dealt with however since you only want a lower bound that doesn't care about multiplicative factors.
answered Apr 1 at 17:50
Rhys SteeleRhys Steele
7,9101931
$endgroup$
$begingroup$
Another question, why $1-|u|^2in L^2(Omega)$ if $uin H^1(Omega)$??
$endgroup$
– R. N. Marley
Apr 1 at 19:24
$begingroup$
It needn't be in general (unless you assume enough to get this from a sobolev embedding). I assume the correct interpretation of your problem is that $A$ takes values in $[0,infty]$ and if $Au = infty$ then you have no problem with coercivity.
$endgroup$
– Rhys Steele
Apr 1 at 20:33
$begingroup$
I see... so my functional is not well defined on $H^1(Omega )$.
$endgroup$
– R. N. Marley
Apr 1 at 20:50
1
$begingroup$
Well, it's well defined as a functional to the extended real line. Note that all the integrands are non-negative and so the integrals still all make sense.
$endgroup$
– Rhys Steele
Apr 1 at 20:51
add a comment |
$begingroup$
Your functional is not coercive. For example, fix $Omega = [0,1] subseteq mathbbR$ and consider $u_n(x) = x^n+1$. Then $nabla u(x) = (n+1)x^n$ and so
$$|u|^2 = (n+1)^2 int_0^1 x^2n dx + int_0^1 x^2(n+1) dx = frac(n+1)^22n + frac12(n+1) to infty$$ as $n to infty$. However
$$Au_n = (n+1) int_0^1 x^n dx + int_0^1 (1- x^2(n+1))^2 dx = 2 + frac14n+1 - frac1n+1$$
is a bounded sequence.
Edit: After you've edited, your new functional is coercive by basically your attempt, except for the fact a factor of a half mysteriously goes missing at some point. This is easily dealt with however since you only want a lower bound that doesn't care about multiplicative factors.
answered Apr 1 at 17:50
Rhys SteeleRhys Steele
7,9101931
$endgroup$
Your functional is not coercive. For example, fix $Omega = [0,1] subseteq mathbbR$ and consider $u_n(x) = x^n+1$. Then $nabla u(x) = (n+1)x^n$ and so
$$|u|^2 = (n+1)^2 int_0^1 x^2n dx + int_0^1 x^2(n+1) dx = frac(n+1)^22n + frac12(n+1) to infty$$ as $n to infty$. However
$$Au_n = (n+1) int_0^1 x^n dx + int_0^1 (1- x^2(n+1))^2 dx = 2 + frac14n+1 - frac1n+1$$
is a bounded sequence.
Edit: After you've edited, your new functional is coercive by basically your attempt, except for the fact a factor of a half mysteriously goes missing at some point. This is easily dealt with however since you only want a lower bound that doesn't care about multiplicative factors.
answered Apr 1 at 17:50
Rhys SteeleRhys Steele
7,9101931
edited Apr 1 at 18:10
answered Apr 1 at 17:50
Rhys SteeleRhys Steele
7,9101931
answered Apr 1 at 17:50
Rhys SteeleRhys Steele
7,9101931
answered Apr 1 at 17:50
Rhys SteeleRhys Steele
7,9101931
7,9101931
$begingroup$
Another question, why $1-|u|^2in L^2(Omega)$ if $uin H^1(Omega)$??
$endgroup$
– R. N. Marley
Apr 1 at 19:24
$begingroup$
It needn't be in general (unless you assume enough to get this from a sobolev embedding). I assume the correct interpretation of your problem is that $A$ takes values in $[0,infty]$ and if $Au = infty$ then you have no problem with coercivity.
$endgroup$
– Rhys Steele
Apr 1 at 20:33
$begingroup$
I see... so my functional is not well defined on $H^1(Omega )$.
$endgroup$
– R. N. Marley
Apr 1 at 20:50
1
$begingroup$
Well, it's well defined as a functional to the extended real line. Note that all the integrands are non-negative and so the integrals still all make sense.
$endgroup$
– Rhys Steele
Apr 1 at 20:51
add a comment |
$begingroup$
Another question, why $1-|u|^2in L^2(Omega)$ if $uin H^1(Omega)$??
$endgroup$
– R. N. Marley
Apr 1 at 19:24
$begingroup$
It needn't be in general (unless you assume enough to get this from a sobolev embedding). I assume the correct interpretation of your problem is that $A$ takes values in $[0,infty]$ and if $Au = infty$ then you have no problem with coercivity.
$endgroup$
– Rhys Steele
Apr 1 at 20:33
$begingroup$
I see... so my functional is not well defined on $H^1(Omega )$.
$endgroup$
– R. N. Marley
Apr 1 at 20:50
1
$begingroup$
Well, it's well defined as a functional to the extended real line. Note that all the integrands are non-negative and so the integrals still all make sense.
$endgroup$
– Rhys Steele
Apr 1 at 20:51
$begingroup$
Another question, why $1-|u|^2in L^2(Omega)$ if $uin H^1(Omega)$??
$endgroup$
– R. N. Marley
Apr 1 at 19:24
$begingroup$
Another question, why $1-|u|^2in L^2(Omega)$ if $uin H^1(Omega)$??
$endgroup$
– R. N. Marley
Apr 1 at 19:24
$begingroup$
It needn't be in general (unless you assume enough to get this from a sobolev embedding). I assume the correct interpretation of your problem is that $A$ takes values in $[0,infty]$ and if $Au = infty$ then you have no problem with coercivity.
$endgroup$
– Rhys Steele
Apr 1 at 20:33
$begingroup$
It needn't be in general (unless you assume enough to get this from a sobolev embedding). I assume the correct interpretation of your problem is that $A$ takes values in $[0,infty]$ and if $Au = infty$ then you have no problem with coercivity.
$endgroup$
– Rhys Steele
Apr 1 at 20:33
$begingroup$
I see... so my functional is not well defined on $H^1(Omega )$.
$endgroup$
– R. N. Marley
Apr 1 at 20:50
$begingroup$
I see... so my functional is not well defined on $H^1(Omega )$.
$endgroup$
– R. N. Marley
Apr 1 at 20:50
1
1
$begingroup$
Well, it's well defined as a functional to the extended real line. Note that all the integrands are non-negative and so the integrals still all make sense.
$endgroup$
– Rhys Steele
Apr 1 at 20:51
$begingroup$
Well, it's well defined as a functional to the extended real line. Note that all the integrands are non-negative and so the integrals still all make sense.
$endgroup$
– Rhys Steele
Apr 1 at 20:51
add a comment |
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