Operator corresponding to the form $int_a^b f'(x)overlineg'(x)dx$ with $f,g in H^1(a,b)$

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There's a theorem that states if $s$ is a densely defined, semi-bounded, closed, symmetric sesquilinear form then there exists a unique self-adjoint operator $T$ such that $s(f,g)=langle Tf, grangle$. The form $s[f,g]=int_a^b f'(x)overlineg'(x)dx$ satisfy all those requirements with $f,g in H^1(a,b)$. The corresponding operator is $Tf = -f''$, but my question is what's the domain?

Intuitively, $f$ should be in in $H^2(a,b)$. The problem is with the boundaries, since the partial integration term can vanish in several different ways for this operator (for example $f(a)=0=f(b)$ and the theorem states that there exists a UNIQUE Self-adjoint operator. Any thoughts?

real-analysis functional-analysis operator-theory

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asked Jul 28 at 20:21

ProShitposter

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There's a theorem that states if $s$ is a densely defined, semi-bounded, closed, symmetric sesquilinear form then there exists a unique self-adjoint operator $T$ such that $s(f,g)=langle Tf, grangle$. The form $s[f,g]=int_a^b f'(x)overlineg'(x)dx$ satisfy all those requirements with $f,g in H^1(a,b)$. The corresponding operator is $Tf = -f''$, but my question is what's the domain?

Intuitively, $f$ should be in in $H^2(a,b)$. The problem is with the boundaries, since the partial integration term can vanish in several different ways for this operator (for example $f(a)=0=f(b)$ and the theorem states that there exists a UNIQUE Self-adjoint operator. Any thoughts?

real-analysis functional-analysis operator-theory

share|cite|improve this question

asked Jul 28 at 20:21

ProShitposter

737

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

There's a theorem that states if $s$ is a densely defined, semi-bounded, closed, symmetric sesquilinear form then there exists a unique self-adjoint operator $T$ such that $s(f,g)=langle Tf, grangle$. The form $s[f,g]=int_a^b f'(x)overlineg'(x)dx$ satisfy all those requirements with $f,g in H^1(a,b)$. The corresponding operator is $Tf = -f''$, but my question is what's the domain?

Intuitively, $f$ should be in in $H^2(a,b)$. The problem is with the boundaries, since the partial integration term can vanish in several different ways for this operator (for example $f(a)=0=f(b)$ and the theorem states that there exists a UNIQUE Self-adjoint operator. Any thoughts?

real-analysis functional-analysis operator-theory

share|cite|improve this question

asked Jul 28 at 20:21

ProShitposter

737

There's a theorem that states if $s$ is a densely defined, semi-bounded, closed, symmetric sesquilinear form then there exists a unique self-adjoint operator $T$ such that $s(f,g)=langle Tf, grangle$. The form $s[f,g]=int_a^b f'(x)overlineg'(x)dx$ satisfy all those requirements with $f,g in H^1(a,b)$. The corresponding operator is $Tf = -f''$, but my question is what's the domain?

Intuitively, $f$ should be in in $H^2(a,b)$. The problem is with the boundaries, since the partial integration term can vanish in several different ways for this operator (for example $f(a)=0=f(b)$ and the theorem states that there exists a UNIQUE Self-adjoint operator. Any thoughts?

real-analysis functional-analysis operator-theory

share|cite|improve this question

asked Jul 28 at 20:21

ProShitposter

737

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 20:21

ProShitposter

737

asked Jul 28 at 20:21

ProShitposter

737

asked Jul 28 at 20:21

ProShitposter

737

737

add a comment | 

add a comment | 

1 Answer
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accepted

The following computation shows that if we don't impose any boundary data then $s$ does not correspond to the operator $-fracd^2dx^2$. For ease of typsetting I consider real-valued $f$ and $g$ and for ease of discussion I assume $f$ and $g$ are smooth.
begineqnarray*
s[f,g]
& = &
-int_a^b f''(x) g(x); rm d x + f'(x) g(x)bigg|_a^b
\
& = &
int_a^bleft(-f''(x) + mathscr B(f)right)g(x); rm dx,
endeqnarray*
where $mathscr B(f)$ is the boundary operator whose action on $g$ is
beginequation*
int_a^b mathscr B(f)(x) g(x); rm dx
=
f'(x) g(x)bigg|_a^b.
endequation*
According to this computation, for all smooth, $mathbb R$-valued functions $f$ and $g$ we have $s[f,g] = langle Tf, grangle$, where $T= -fracd^2dx^2 + mathscr B$.

If you want to have $T = -fracd^2dx^2$ then you first restrict the domain of $s$ to some subspace $mathscr Hsubset H(a,b)$ on which $mathcal B = 0$ (in the sense that $langle mathscr B(f), grangle = 0$ for all $f, gin mathscr H$). Once you've chosen $mathscr H$, your theorem guarantees that $T$ is unique. The point here is that $T$ depends on your choice of $mathscr H$. Common choices of $mathscr H$ are $mathscr H = H_0^1(a,b) = fin H(a,b): f(a) = 0 = f(b)$ and $mathscr H = fin H(a,b): f'(a) = 0 = f'(b)$.

share|cite|improve this answer

edited Jul 29 at 15:40

answered Jul 28 at 23:41

BindersFull

498110

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for your response. I was just wondering why $s$ isn't symmetric? By the definition $overlines[g,f]= overlineint_a^b g'(x) overlinef'(x)dx = int_a^b f'(x)overlineg'(x)dx=s[f,g]$.
  – ProShitposter
  Jul 29 at 8:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ProShitposter You are correct, $s$ is certainly symmetric. I have edited my answer to provide more useful information.
  – BindersFull
  Jul 29 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I have just one additional question: If we choose $s$ on $H_0^1 (a,b)$ for example, then all the requirements of the theorem are satisfied. So $T = -f''$ is the self-adjoint operator, but is the domain then $H^2(a,b)$ with $f(a)=f(b)=0$?
  – ProShitposter
  Aug 1 at 16:54
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The following computation shows that if we don't impose any boundary data then $s$ does not correspond to the operator $-fracd^2dx^2$. For ease of typsetting I consider real-valued $f$ and $g$ and for ease of discussion I assume $f$ and $g$ are smooth.
begineqnarray*
s[f,g]
& = &
-int_a^b f''(x) g(x); rm d x + f'(x) g(x)bigg|_a^b
\
& = &
int_a^bleft(-f''(x) + mathscr B(f)right)g(x); rm dx,
endeqnarray*
where $mathscr B(f)$ is the boundary operator whose action on $g$ is
beginequation*
int_a^b mathscr B(f)(x) g(x); rm dx
=
f'(x) g(x)bigg|_a^b.
endequation*
According to this computation, for all smooth, $mathbb R$-valued functions $f$ and $g$ we have $s[f,g] = langle Tf, grangle$, where $T= -fracd^2dx^2 + mathscr B$.

If you want to have $T = -fracd^2dx^2$ then you first restrict the domain of $s$ to some subspace $mathscr Hsubset H(a,b)$ on which $mathcal B = 0$ (in the sense that $langle mathscr B(f), grangle = 0$ for all $f, gin mathscr H$). Once you've chosen $mathscr H$, your theorem guarantees that $T$ is unique. The point here is that $T$ depends on your choice of $mathscr H$. Common choices of $mathscr H$ are $mathscr H = H_0^1(a,b) = fin H(a,b): f(a) = 0 = f(b)$ and $mathscr H = fin H(a,b): f'(a) = 0 = f'(b)$.

share|cite|improve this answer

edited Jul 29 at 15:40

answered Jul 28 at 23:41

BindersFull

498110

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for your response. I was just wondering why $s$ isn't symmetric? By the definition $overlines[g,f]= overlineint_a^b g'(x) overlinef'(x)dx = int_a^b f'(x)overlineg'(x)dx=s[f,g]$.
  – ProShitposter
  Jul 29 at 8:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ProShitposter You are correct, $s$ is certainly symmetric. I have edited my answer to provide more useful information.
  – BindersFull
  Jul 29 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I have just one additional question: If we choose $s$ on $H_0^1 (a,b)$ for example, then all the requirements of the theorem are satisfied. So $T = -f''$ is the self-adjoint operator, but is the domain then $H^2(a,b)$ with $f(a)=f(b)=0$?
  – ProShitposter
  Aug 1 at 16:54
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

The following computation shows that if we don't impose any boundary data then $s$ does not correspond to the operator $-fracd^2dx^2$. For ease of typsetting I consider real-valued $f$ and $g$ and for ease of discussion I assume $f$ and $g$ are smooth.
begineqnarray*
s[f,g]
& = &
-int_a^b f''(x) g(x); rm d x + f'(x) g(x)bigg|_a^b
\
& = &
int_a^bleft(-f''(x) + mathscr B(f)right)g(x); rm dx,
endeqnarray*
where $mathscr B(f)$ is the boundary operator whose action on $g$ is
beginequation*
int_a^b mathscr B(f)(x) g(x); rm dx
=
f'(x) g(x)bigg|_a^b.
endequation*
According to this computation, for all smooth, $mathbb R$-valued functions $f$ and $g$ we have $s[f,g] = langle Tf, grangle$, where $T= -fracd^2dx^2 + mathscr B$.

If you want to have $T = -fracd^2dx^2$ then you first restrict the domain of $s$ to some subspace $mathscr Hsubset H(a,b)$ on which $mathcal B = 0$ (in the sense that $langle mathscr B(f), grangle = 0$ for all $f, gin mathscr H$). Once you've chosen $mathscr H$, your theorem guarantees that $T$ is unique. The point here is that $T$ depends on your choice of $mathscr H$. Common choices of $mathscr H$ are $mathscr H = H_0^1(a,b) = fin H(a,b): f(a) = 0 = f(b)$ and $mathscr H = fin H(a,b): f'(a) = 0 = f'(b)$.

share|cite|improve this answer

edited Jul 29 at 15:40

answered Jul 28 at 23:41

BindersFull

498110

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for your response. I was just wondering why $s$ isn't symmetric? By the definition $overlines[g,f]= overlineint_a^b g'(x) overlinef'(x)dx = int_a^b f'(x)overlineg'(x)dx=s[f,g]$.
  – ProShitposter
  Jul 29 at 8:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ProShitposter You are correct, $s$ is certainly symmetric. I have edited my answer to provide more useful information.
  – BindersFull
  Jul 29 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I have just one additional question: If we choose $s$ on $H_0^1 (a,b)$ for example, then all the requirements of the theorem are satisfied. So $T = -f''$ is the self-adjoint operator, but is the domain then $H^2(a,b)$ with $f(a)=f(b)=0$?
  – ProShitposter
  Aug 1 at 16:54
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The following computation shows that if we don't impose any boundary data then $s$ does not correspond to the operator $-fracd^2dx^2$. For ease of typsetting I consider real-valued $f$ and $g$ and for ease of discussion I assume $f$ and $g$ are smooth.
begineqnarray*
s[f,g]
& = &
-int_a^b f''(x) g(x); rm d x + f'(x) g(x)bigg|_a^b
\
& = &
int_a^bleft(-f''(x) + mathscr B(f)right)g(x); rm dx,
endeqnarray*
where $mathscr B(f)$ is the boundary operator whose action on $g$ is
beginequation*
int_a^b mathscr B(f)(x) g(x); rm dx
=
f'(x) g(x)bigg|_a^b.
endequation*
According to this computation, for all smooth, $mathbb R$-valued functions $f$ and $g$ we have $s[f,g] = langle Tf, grangle$, where $T= -fracd^2dx^2 + mathscr B$.

If you want to have $T = -fracd^2dx^2$ then you first restrict the domain of $s$ to some subspace $mathscr Hsubset H(a,b)$ on which $mathcal B = 0$ (in the sense that $langle mathscr B(f), grangle = 0$ for all $f, gin mathscr H$). Once you've chosen $mathscr H$, your theorem guarantees that $T$ is unique. The point here is that $T$ depends on your choice of $mathscr H$. Common choices of $mathscr H$ are $mathscr H = H_0^1(a,b) = fin H(a,b): f(a) = 0 = f(b)$ and $mathscr H = fin H(a,b): f'(a) = 0 = f'(b)$.

share|cite|improve this answer

edited Jul 29 at 15:40

answered Jul 28 at 23:41

BindersFull

498110

The following computation shows that if we don't impose any boundary data then $s$ does not correspond to the operator $-fracd^2dx^2$. For ease of typsetting I consider real-valued $f$ and $g$ and for ease of discussion I assume $f$ and $g$ are smooth.
begineqnarray*
s[f,g]
& = &
-int_a^b f''(x) g(x); rm d x + f'(x) g(x)bigg|_a^b
\
& = &
int_a^bleft(-f''(x) + mathscr B(f)right)g(x); rm dx,
endeqnarray*
where $mathscr B(f)$ is the boundary operator whose action on $g$ is
beginequation*
int_a^b mathscr B(f)(x) g(x); rm dx
=
f'(x) g(x)bigg|_a^b.
endequation*
According to this computation, for all smooth, $mathbb R$-valued functions $f$ and $g$ we have $s[f,g] = langle Tf, grangle$, where $T= -fracd^2dx^2 + mathscr B$.

If you want to have $T = -fracd^2dx^2$ then you first restrict the domain of $s$ to some subspace $mathscr Hsubset H(a,b)$ on which $mathcal B = 0$ (in the sense that $langle mathscr B(f), grangle = 0$ for all $f, gin mathscr H$). Once you've chosen $mathscr H$, your theorem guarantees that $T$ is unique. The point here is that $T$ depends on your choice of $mathscr H$. Common choices of $mathscr H$ are $mathscr H = H_0^1(a,b) = fin H(a,b): f(a) = 0 = f(b)$ and $mathscr H = fin H(a,b): f'(a) = 0 = f'(b)$.

share|cite|improve this answer

edited Jul 29 at 15:40

answered Jul 28 at 23:41

BindersFull

498110

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 15:40

edited Jul 29 at 15:40

edited Jul 29 at 15:40

answered Jul 28 at 23:41

BindersFull

498110

answered Jul 28 at 23:41

BindersFull

498110

answered Jul 28 at 23:41

BindersFull

498110

498110

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for your response. I was just wondering why $s$ isn't symmetric? By the definition $overlines[g,f]= overlineint_a^b g'(x) overlinef'(x)dx = int_a^b f'(x)overlineg'(x)dx=s[f,g]$.
  – ProShitposter
  Jul 29 at 8:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ProShitposter You are correct, $s$ is certainly symmetric. I have edited my answer to provide more useful information.
  – BindersFull
  Jul 29 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I have just one additional question: If we choose $s$ on $H_0^1 (a,b)$ for example, then all the requirements of the theorem are satisfied. So $T = -f''$ is the self-adjoint operator, but is the domain then $H^2(a,b)$ with $f(a)=f(b)=0$?
  – ProShitposter
  Aug 1 at 16:54
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank you for your response. I was just wondering why $s$ isn't symmetric? By the definition $overlines[g,f]= overlineint_a^b g'(x) overlinef'(x)dx = int_a^b f'(x)overlineg'(x)dx=s[f,g]$.
  – ProShitposter
  Jul 29 at 8:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ProShitposter You are correct, $s$ is certainly symmetric. I have edited my answer to provide more useful information.
  – BindersFull
  Jul 29 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you. I have just one additional question: If we choose $s$ on $H_0^1 (a,b)$ for example, then all the requirements of the theorem are satisfied. So $T = -f''$ is the self-adjoint operator, but is the domain then $H^2(a,b)$ with $f(a)=f(b)=0$?
  – ProShitposter
  Aug 1 at 16:54
  
  

  
  
  
  

1

1

Thank you for your response. I was just wondering why $s$ isn't symmetric? By the definition $overlines[g,f]= overlineint_a^b g'(x) overlinef'(x)dx = int_a^b f'(x)overlineg'(x)dx=s[f,g]$.
– ProShitposter
Jul 29 at 8:23

Thank you for your response. I was just wondering why $s$ isn't symmetric? By the definition $overlines[g,f]= overlineint_a^b g'(x) overlinef'(x)dx = int_a^b f'(x)overlineg'(x)dx=s[f,g]$.
– ProShitposter
Jul 29 at 8:23

1

1

@ProShitposter You are correct, $s$ is certainly symmetric. I have edited my answer to provide more useful information.
– BindersFull
Jul 29 at 15:42

@ProShitposter You are correct, $s$ is certainly symmetric. I have edited my answer to provide more useful information.
– BindersFull
Jul 29 at 15:42

Thank you. I have just one additional question: If we choose $s$ on $H_0^1 (a,b)$ for example, then all the requirements of the theorem are satisfied. So $T = -f''$ is the self-adjoint operator, but is the domain then $H^2(a,b)$ with $f(a)=f(b)=0$?
– ProShitposter
Aug 1 at 16:54

Thank you. I have just one additional question: If we choose $s$ on $H_0^1 (a,b)$ for example, then all the requirements of the theorem are satisfied. So $T = -f''$ is the self-adjoint operator, but is the domain then $H^2(a,b)$ with $f(a)=f(b)=0$?
– ProShitposter
Aug 1 at 16:54

add a comment | 

 

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