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!
differential-equations hilbert-spaces sobolev-spaces lp-spaces boundary-value-problem
asked Nov 21 at 17:21
Spinnennetz
214
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up vote
1
down vote
favorite
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
asked Nov 21 at 17:21
Spinnennetz
214
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 21 at 17:21
Spinnennetz
214
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
differential-equations hilbert-spaces sobolev-spaces lp-spaces boundary-value-problem
asked Nov 21 at 17:21
Spinnennetz
214
asked Nov 21 at 17:21
Spinnennetz
214
asked Nov 21 at 17:21
Spinnennetz
214
asked Nov 21 at 17:21
Spinnennetz
214
asked Nov 21 at 17:21
Spinnennetz
214
214
add a comment |
add a comment |
1 Answer
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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.
answered Nov 21 at 19:15
Lorenzo Quarisa
3,146316
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1 Answer
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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.
answered Nov 21 at 19:15
Lorenzo Quarisa
3,146316
add a comment |
up vote
2
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.
answered Nov 21 at 19:15
Lorenzo Quarisa
3,146316
add a comment |
up vote
2
down vote
up vote
2
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.
answered Nov 21 at 19:15
Lorenzo Quarisa
3,146316
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.
answered Nov 21 at 19:15
Lorenzo Quarisa
3,146316
edited Nov 21 at 19:21
answered Nov 21 at 19:15
Lorenzo Quarisa
3,146316
answered Nov 21 at 19:15
Lorenzo Quarisa
3,146316
answered Nov 21 at 19:15
Lorenzo Quarisa
3,146316
3,146316
add a comment |
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