Weak Formulation's Well-Posedness with Poincare-Friedrichs Inequality
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I have arrived at the weak form for a PDE which has the following form.
Let $V={ vin H^1(0,1):v(1)=0 }$, $F$ is a bounded linear functional on V.
$$left{begin{array}{ll}
textrm{Given } Fin V^* textrm{, find } uin V textrm{ such that}\
int_{0}^{1}{u'(x)v'(x) dx} =F(v) \
forall vin V
end{array}
right.$$
I understand that the Poincare-Friedrichs inequality can be used to show that when $V=H_0^1 (0,1)$ the coercitivity of the Lax-Milgram holds, but what about for when I have only one Dirichlet boundary condition and I don't have completely defined Dirichlet conditions?
Does the Poincare-Friedrichs inequality apply to my space $V$, like so
$$int_0^1 u(x)^2 dx le int_0^1 u'(x)^2 dx quad textrm{?}$$
functional-analysis pde sobolev-spaces
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up vote
0
down vote
favorite
I have arrived at the weak form for a PDE which has the following form.
Let $V={ vin H^1(0,1):v(1)=0 }$, $F$ is a bounded linear functional on V.
$$left{begin{array}{ll}
textrm{Given } Fin V^* textrm{, find } uin V textrm{ such that}\
int_{0}^{1}{u'(x)v'(x) dx} =F(v) \
forall vin V
end{array}
right.$$
I understand that the Poincare-Friedrichs inequality can be used to show that when $V=H_0^1 (0,1)$ the coercitivity of the Lax-Milgram holds, but what about for when I have only one Dirichlet boundary condition and I don't have completely defined Dirichlet conditions?
Does the Poincare-Friedrichs inequality apply to my space $V$, like so
$$int_0^1 u(x)^2 dx le int_0^1 u'(x)^2 dx quad textrm{?}$$
functional-analysis pde sobolev-spaces
To be clear, there are $many$ versions of that inequality .
– DaveNine
Nov 17 at 0:24
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have arrived at the weak form for a PDE which has the following form.
Let $V={ vin H^1(0,1):v(1)=0 }$, $F$ is a bounded linear functional on V.
$$left{begin{array}{ll}
textrm{Given } Fin V^* textrm{, find } uin V textrm{ such that}\
int_{0}^{1}{u'(x)v'(x) dx} =F(v) \
forall vin V
end{array}
right.$$
I understand that the Poincare-Friedrichs inequality can be used to show that when $V=H_0^1 (0,1)$ the coercitivity of the Lax-Milgram holds, but what about for when I have only one Dirichlet boundary condition and I don't have completely defined Dirichlet conditions?
Does the Poincare-Friedrichs inequality apply to my space $V$, like so
$$int_0^1 u(x)^2 dx le int_0^1 u'(x)^2 dx quad textrm{?}$$
functional-analysis pde sobolev-spaces
I have arrived at the weak form for a PDE which has the following form.
Let $V={ vin H^1(0,1):v(1)=0 }$, $F$ is a bounded linear functional on V.
$$left{begin{array}{ll}
textrm{Given } Fin V^* textrm{, find } uin V textrm{ such that}\
int_{0}^{1}{u'(x)v'(x) dx} =F(v) \
forall vin V
end{array}
right.$$
I understand that the Poincare-Friedrichs inequality can be used to show that when $V=H_0^1 (0,1)$ the coercitivity of the Lax-Milgram holds, but what about for when I have only one Dirichlet boundary condition and I don't have completely defined Dirichlet conditions?
Does the Poincare-Friedrichs inequality apply to my space $V$, like so
$$int_0^1 u(x)^2 dx le int_0^1 u'(x)^2 dx quad textrm{?}$$
functional-analysis pde sobolev-spaces
functional-analysis pde sobolev-spaces
asked Nov 16 at 14:42
Joel Biffin
515
515
To be clear, there are $many$ versions of that inequality .
– DaveNine
Nov 17 at 0:24
add a comment |
To be clear, there are $many$ versions of that inequality .
– DaveNine
Nov 17 at 0:24
To be clear, there are $many$ versions of that inequality .
– DaveNine
Nov 17 at 0:24
To be clear, there are $many$ versions of that inequality .
– DaveNine
Nov 17 at 0:24
add a comment |
1 Answer
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If $u in V$ then
$$u(x) = u(x) - u(1) = - int_x^1 u'(t) , dt$$
so that $$|u(x)| le int_x^1 |u'(t)| , dt le left( int_0^1 |u'(t)|^2 right)^{1/2}$$
and consequently
$$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If $u in V$ then
$$u(x) = u(x) - u(1) = - int_x^1 u'(t) , dt$$
so that $$|u(x)| le int_x^1 |u'(t)| , dt le left( int_0^1 |u'(t)|^2 right)^{1/2}$$
and consequently
$$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$
add a comment |
up vote
1
down vote
If $u in V$ then
$$u(x) = u(x) - u(1) = - int_x^1 u'(t) , dt$$
so that $$|u(x)| le int_x^1 |u'(t)| , dt le left( int_0^1 |u'(t)|^2 right)^{1/2}$$
and consequently
$$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$
add a comment |
up vote
1
down vote
up vote
1
down vote
If $u in V$ then
$$u(x) = u(x) - u(1) = - int_x^1 u'(t) , dt$$
so that $$|u(x)| le int_x^1 |u'(t)| , dt le left( int_0^1 |u'(t)|^2 right)^{1/2}$$
and consequently
$$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$
If $u in V$ then
$$u(x) = u(x) - u(1) = - int_x^1 u'(t) , dt$$
so that $$|u(x)| le int_x^1 |u'(t)| , dt le left( int_0^1 |u'(t)|^2 right)^{1/2}$$
and consequently
$$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$
answered Nov 16 at 17:49
Umberto P.
38.2k13063
38.2k13063
add a comment |
add a comment |
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To be clear, there are $many$ versions of that inequality .
– DaveNine
Nov 17 at 0:24