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Norm estimate of $(-lambda A+ I)^-1$ for strictly elliptic operator

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1

$begingroup$

Let $Omega$ be a smooth domain in $mathbbR^n$, and $A$ be a strictly elliptic operator
$$
Au=partial_i(a^ij(x)partial_j)u,
$$
where $a^ij$ are bounded functions satisfying
$$
a^ij(x)xi_ixi_jge alpha |xi|^2,quad forall xiin mathbbR^n,,xin mathbbR^n.
$$
Then for any $fin L^2(Omega)$ and $lambda>0$, there exists $uin H^2(Omega)$ such that
$$
-lambda Au+ u=f,quad xin Omega; quad x=0,quad xin partialOmega.
$$
This implies $(-lambda A+I)^-1:L^2(Omega)to H^2(Omega)$ is a bounded operator (we assume zero boundary condition). Is it possible to get a precise estimate for the operator norm $|(-lambda A+I)^-1|_mathcalL(L^2(Omega),H^2(Omega))$ in terms of $lambda$? In particular, I would like to know the behavior of the norm as $lambdato 0$.

---

Let $Dom(A)=H^2(Omega)cap H^1_0(Omega)$, we know $A$ is a maximal monotone operator on $L^2(Omega)$, hence $ |(-lambda A+I)^-1|_mathcalL(L^2(Omega))le 1$ for all $lambda$. But it does not give the estimate in terms of $H^2$ norm.

functional-analysis sobolev-spaces regularity-theory-of-pdes

share|cite|improve this question

edited 2 days ago

John

asked 2 days ago

JohnJohn

9,50411336

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know some form of the spectral theorem? That makes it relatively easy to compute such norms.
  $endgroup$
  – MaoWao
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MaoWao Thanks. I will check the theorem.
  $endgroup$
  – John
  yesterday
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $Omega$ be a smooth domain in $mathbbR^n$, and $A$ be a strictly elliptic operator
$$
Au=partial_i(a^ij(x)partial_j)u,
$$
where $a^ij$ are bounded functions satisfying
$$
a^ij(x)xi_ixi_jge alpha |xi|^2,quad forall xiin mathbbR^n,,xin mathbbR^n.
$$
Then for any $fin L^2(Omega)$ and $lambda>0$, there exists $uin H^2(Omega)$ such that
$$
-lambda Au+ u=f,quad xin Omega; quad x=0,quad xin partialOmega.
$$
This implies $(-lambda A+I)^-1:L^2(Omega)to H^2(Omega)$ is a bounded operator (we assume zero boundary condition). Is it possible to get a precise estimate for the operator norm $|(-lambda A+I)^-1|_mathcalL(L^2(Omega),H^2(Omega))$ in terms of $lambda$? In particular, I would like to know the behavior of the norm as $lambdato 0$.

---

Let $Dom(A)=H^2(Omega)cap H^1_0(Omega)$, we know $A$ is a maximal monotone operator on $L^2(Omega)$, hence $ |(-lambda A+I)^-1|_mathcalL(L^2(Omega))le 1$ for all $lambda$. But it does not give the estimate in terms of $H^2$ norm.

functional-analysis sobolev-spaces regularity-theory-of-pdes

share|cite|improve this question

edited 2 days ago

John

asked 2 days ago

JohnJohn

9,50411336

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know some form of the spectral theorem? That makes it relatively easy to compute such norms.
  $endgroup$
  – MaoWao
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MaoWao Thanks. I will check the theorem.
  $endgroup$
  – John
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Let $Omega$ be a smooth domain in $mathbbR^n$, and $A$ be a strictly elliptic operator
$$
Au=partial_i(a^ij(x)partial_j)u,
$$
where $a^ij$ are bounded functions satisfying
$$
a^ij(x)xi_ixi_jge alpha |xi|^2,quad forall xiin mathbbR^n,,xin mathbbR^n.
$$
Then for any $fin L^2(Omega)$ and $lambda>0$, there exists $uin H^2(Omega)$ such that
$$
-lambda Au+ u=f,quad xin Omega; quad x=0,quad xin partialOmega.
$$
This implies $(-lambda A+I)^-1:L^2(Omega)to H^2(Omega)$ is a bounded operator (we assume zero boundary condition). Is it possible to get a precise estimate for the operator norm $|(-lambda A+I)^-1|_mathcalL(L^2(Omega),H^2(Omega))$ in terms of $lambda$? In particular, I would like to know the behavior of the norm as $lambdato 0$.

---

Let $Dom(A)=H^2(Omega)cap H^1_0(Omega)$, we know $A$ is a maximal monotone operator on $L^2(Omega)$, hence $ |(-lambda A+I)^-1|_mathcalL(L^2(Omega))le 1$ for all $lambda$. But it does not give the estimate in terms of $H^2$ norm.

functional-analysis sobolev-spaces regularity-theory-of-pdes

share|cite|improve this question

edited 2 days ago

John

asked 2 days ago

JohnJohn

9,50411336

$endgroup$

share|cite|improve this question

edited 2 days ago

John

asked 2 days ago

JohnJohn

9,50411336

share|cite|improve this question

edited 2 days ago

John

asked 2 days ago

JohnJohn

9,50411336

asked 2 days ago

JohnJohn

9,50411336

asked 2 days ago

JohnJohn

9,50411336