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



$$int_0^1 u(x)^2 dx le int_0^1 u'(x)^2 dx quad textrm{?}$$










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  • To be clear, there are $many$ versions of that inequality .
    – DaveNine
    Nov 17 at 0:24















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



$$int_0^1 u(x)^2 dx le int_0^1 u'(x)^2 dx quad textrm{?}$$










share|cite|improve this question






















  • To be clear, there are $many$ versions of that inequality .
    – DaveNine
    Nov 17 at 0:24













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



$$int_0^1 u(x)^2 dx le int_0^1 u'(x)^2 dx quad textrm{?}$$










share|cite|improve this question













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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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


















  • 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










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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
$$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$






share|cite|improve this 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
    $$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$






    share|cite|improve this answer

























      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
      $$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$






      share|cite|improve this answer























        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
        $$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$






        share|cite|improve this answer












        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
        $$int_0^1 |u(x)|^2 , dx le int_0^1 int_0^1 |u'(t)|^2 , dt , dx = int_0^1 |u'(t)|^2 , dt.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 16 at 17:49









        Umberto P.

        38.2k13063




        38.2k13063






























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