How to choose the Testfunction-Space?











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Every time I want to transform a boundary-value-problem of the form $-u'' + ... = f$ to it's weak formulation, I have problems choosing the testfunction-space.



I have the feeling that in scripts the standard testfct-space is $H_0^1(Omega)$ (of course just if $u(x)=0 forall x in partialOmega$).



Maybe it is a silly question, but why do we choose $H_0^1(Omega)$ and not just $W_0^{1,1}$?



Is it because we want to use the great properties of Hilbert-Spaces, especially of $L^2$ and also because we have reflexiveness then, ...?



I hope some of you can give me some answers and help me understanding the theory of differential equations better!










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    Every time I want to transform a boundary-value-problem of the form $-u'' + ... = f$ to it's weak formulation, I have problems choosing the testfunction-space.



    I have the feeling that in scripts the standard testfct-space is $H_0^1(Omega)$ (of course just if $u(x)=0 forall x in partialOmega$).



    Maybe it is a silly question, but why do we choose $H_0^1(Omega)$ and not just $W_0^{1,1}$?



    Is it because we want to use the great properties of Hilbert-Spaces, especially of $L^2$ and also because we have reflexiveness then, ...?



    I hope some of you can give me some answers and help me understanding the theory of differential equations better!










    share|cite|improve this question
























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      Every time I want to transform a boundary-value-problem of the form $-u'' + ... = f$ to it's weak formulation, I have problems choosing the testfunction-space.



      I have the feeling that in scripts the standard testfct-space is $H_0^1(Omega)$ (of course just if $u(x)=0 forall x in partialOmega$).



      Maybe it is a silly question, but why do we choose $H_0^1(Omega)$ and not just $W_0^{1,1}$?



      Is it because we want to use the great properties of Hilbert-Spaces, especially of $L^2$ and also because we have reflexiveness then, ...?



      I hope some of you can give me some answers and help me understanding the theory of differential equations better!










      share|cite|improve this question













      Every time I want to transform a boundary-value-problem of the form $-u'' + ... = f$ to it's weak formulation, I have problems choosing the testfunction-space.



      I have the feeling that in scripts the standard testfct-space is $H_0^1(Omega)$ (of course just if $u(x)=0 forall x in partialOmega$).



      Maybe it is a silly question, but why do we choose $H_0^1(Omega)$ and not just $W_0^{1,1}$?



      Is it because we want to use the great properties of Hilbert-Spaces, especially of $L^2$ and also because we have reflexiveness then, ...?



      I hope some of you can give me some answers and help me understanding the theory of differential equations better!







      differential-equations hilbert-spaces sobolev-spaces lp-spaces boundary-value-problem






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      asked Nov 21 at 17:21









      Spinnennetz

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          Working on a Hilbert space is nice of course, but arguably the most important property is reflexiveness, which $W_0^{1,1}$ lacks.



          Indeed, reflexive normed spaces are characterized by the property that the closed unit ball is weakly compact. This is known as Kakutani's theorem. Moreover, by the Eberlein-Smulian theorem, weak compactness and sequential weak compactness are equivalent on a Banach space, so it also follows that on a reflexive Banach space bounded sequences admit a weakly converging subsequence.



          A typical approach in proving existence results for (weak) solutions to boundary value problems is to construct a functional $J$ on some function space $X$ whose minimizers are the (weak) solutions to the boundary value problem. You then consider a minimizing sequence $left{u_nright}$, i.e. such that
          $$lim_{nto +infty}J(u_n)=inf_XJ $$
          and prove that it is bounded. If $X$ is reflexive, then by the above property $left{u_nright}$ has a weakly converging subsequence $left{u_{n_k}right}to u_0in X$. Then, if you also prove that $J$ is weakly lower semicontinuous (which is not particularly restrictive - for instance the norm is always weakly lower semicontinuous), you get that
          $$inf_X J=liminf_{kto +infty}J(u_{n_k})geq J(u_0) $$
          which implies that $u_0$ is a minimizer for $J$, and hence a (weak) solution to the associated boundary value problem.






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            Working on a Hilbert space is nice of course, but arguably the most important property is reflexiveness, which $W_0^{1,1}$ lacks.



            Indeed, reflexive normed spaces are characterized by the property that the closed unit ball is weakly compact. This is known as Kakutani's theorem. Moreover, by the Eberlein-Smulian theorem, weak compactness and sequential weak compactness are equivalent on a Banach space, so it also follows that on a reflexive Banach space bounded sequences admit a weakly converging subsequence.



            A typical approach in proving existence results for (weak) solutions to boundary value problems is to construct a functional $J$ on some function space $X$ whose minimizers are the (weak) solutions to the boundary value problem. You then consider a minimizing sequence $left{u_nright}$, i.e. such that
            $$lim_{nto +infty}J(u_n)=inf_XJ $$
            and prove that it is bounded. If $X$ is reflexive, then by the above property $left{u_nright}$ has a weakly converging subsequence $left{u_{n_k}right}to u_0in X$. Then, if you also prove that $J$ is weakly lower semicontinuous (which is not particularly restrictive - for instance the norm is always weakly lower semicontinuous), you get that
            $$inf_X J=liminf_{kto +infty}J(u_{n_k})geq J(u_0) $$
            which implies that $u_0$ is a minimizer for $J$, and hence a (weak) solution to the associated boundary value problem.






            share|cite|improve this answer



























              up vote
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              down vote













              Working on a Hilbert space is nice of course, but arguably the most important property is reflexiveness, which $W_0^{1,1}$ lacks.



              Indeed, reflexive normed spaces are characterized by the property that the closed unit ball is weakly compact. This is known as Kakutani's theorem. Moreover, by the Eberlein-Smulian theorem, weak compactness and sequential weak compactness are equivalent on a Banach space, so it also follows that on a reflexive Banach space bounded sequences admit a weakly converging subsequence.



              A typical approach in proving existence results for (weak) solutions to boundary value problems is to construct a functional $J$ on some function space $X$ whose minimizers are the (weak) solutions to the boundary value problem. You then consider a minimizing sequence $left{u_nright}$, i.e. such that
              $$lim_{nto +infty}J(u_n)=inf_XJ $$
              and prove that it is bounded. If $X$ is reflexive, then by the above property $left{u_nright}$ has a weakly converging subsequence $left{u_{n_k}right}to u_0in X$. Then, if you also prove that $J$ is weakly lower semicontinuous (which is not particularly restrictive - for instance the norm is always weakly lower semicontinuous), you get that
              $$inf_X J=liminf_{kto +infty}J(u_{n_k})geq J(u_0) $$
              which implies that $u_0$ is a minimizer for $J$, and hence a (weak) solution to the associated boundary value problem.






              share|cite|improve this answer

























                up vote
                1
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                up vote
                1
                down vote









                Working on a Hilbert space is nice of course, but arguably the most important property is reflexiveness, which $W_0^{1,1}$ lacks.



                Indeed, reflexive normed spaces are characterized by the property that the closed unit ball is weakly compact. This is known as Kakutani's theorem. Moreover, by the Eberlein-Smulian theorem, weak compactness and sequential weak compactness are equivalent on a Banach space, so it also follows that on a reflexive Banach space bounded sequences admit a weakly converging subsequence.



                A typical approach in proving existence results for (weak) solutions to boundary value problems is to construct a functional $J$ on some function space $X$ whose minimizers are the (weak) solutions to the boundary value problem. You then consider a minimizing sequence $left{u_nright}$, i.e. such that
                $$lim_{nto +infty}J(u_n)=inf_XJ $$
                and prove that it is bounded. If $X$ is reflexive, then by the above property $left{u_nright}$ has a weakly converging subsequence $left{u_{n_k}right}to u_0in X$. Then, if you also prove that $J$ is weakly lower semicontinuous (which is not particularly restrictive - for instance the norm is always weakly lower semicontinuous), you get that
                $$inf_X J=liminf_{kto +infty}J(u_{n_k})geq J(u_0) $$
                which implies that $u_0$ is a minimizer for $J$, and hence a (weak) solution to the associated boundary value problem.






                share|cite|improve this answer














                Working on a Hilbert space is nice of course, but arguably the most important property is reflexiveness, which $W_0^{1,1}$ lacks.



                Indeed, reflexive normed spaces are characterized by the property that the closed unit ball is weakly compact. This is known as Kakutani's theorem. Moreover, by the Eberlein-Smulian theorem, weak compactness and sequential weak compactness are equivalent on a Banach space, so it also follows that on a reflexive Banach space bounded sequences admit a weakly converging subsequence.



                A typical approach in proving existence results for (weak) solutions to boundary value problems is to construct a functional $J$ on some function space $X$ whose minimizers are the (weak) solutions to the boundary value problem. You then consider a minimizing sequence $left{u_nright}$, i.e. such that
                $$lim_{nto +infty}J(u_n)=inf_XJ $$
                and prove that it is bounded. If $X$ is reflexive, then by the above property $left{u_nright}$ has a weakly converging subsequence $left{u_{n_k}right}to u_0in X$. Then, if you also prove that $J$ is weakly lower semicontinuous (which is not particularly restrictive - for instance the norm is always weakly lower semicontinuous), you get that
                $$inf_X J=liminf_{kto +infty}J(u_{n_k})geq J(u_0) $$
                which implies that $u_0$ is a minimizer for $J$, and hence a (weak) solution to the associated boundary value problem.







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                edited Nov 21 at 19:21

























                answered Nov 21 at 19:15









                Lorenzo Quarisa

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                3,021316






























                     

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