Some multivalued problems - Yosida approximation
$begingroup$
I study multi-valued problems on the basis of M. Chipot's book and I have a seemingly simple problem.
Set for a.e. $x in Omega$, $lambda > 0$, $forall t in mathbb{R}$:
$$ J_lambda (x,t) = (I +
lambda beta(x,cdot))^{-1}(t) $$
$$ beta_lambda (x,t) = (t-J_lambda(x,t))/lambda .$$
So $J_lambda$ is resolvent and $beta_lambda$ is Yosida approximation of $beta(x,t)$. Then there is said that it is easy to show that
$$ lambda_1 geq lambda_2 Longrightarrow beta_{lambda_1}(t) leq beta_{lambda_2}(t) $$
and for positive constant $c>0$:
$$ (c beta)_lambda = cbeta_{c lambda}.$$
I can't see it and and maybe some of you will be able to help?
Thanks in advance!
ordinary-differential-equations pde approximation multivalued-functions
asked Jan 10 at 0:26
WawMathematicianWawMathematician
13
$endgroup$
add a comment |
$begingroup$
I study multi-valued problems on the basis of M. Chipot's book and I have a seemingly simple problem.
Set for a.e. $x in Omega$, $lambda > 0$, $forall t in mathbb{R}$:
$$ J_lambda (x,t) = (I +
lambda beta(x,cdot))^{-1}(t) $$
$$ beta_lambda (x,t) = (t-J_lambda(x,t))/lambda .$$
So $J_lambda$ is resolvent and $beta_lambda$ is Yosida approximation of $beta(x,t)$. Then there is said that it is easy to show that
$$ lambda_1 geq lambda_2 Longrightarrow beta_{lambda_1}(t) leq beta_{lambda_2}(t) $$
and for positive constant $c>0$:
$$ (c beta)_lambda = cbeta_{c lambda}.$$
I can't see it and and maybe some of you will be able to help?
Thanks in advance!
ordinary-differential-equations pde approximation multivalued-functions
asked Jan 10 at 0:26
WawMathematicianWawMathematician
13
$endgroup$
add a comment |
$begingroup$
I study multi-valued problems on the basis of M. Chipot's book and I have a seemingly simple problem.
Set for a.e. $x in Omega$, $lambda > 0$, $forall t in mathbb{R}$:
$$ J_lambda (x,t) = (I +
lambda beta(x,cdot))^{-1}(t) $$
$$ beta_lambda (x,t) = (t-J_lambda(x,t))/lambda .$$
So $J_lambda$ is resolvent and $beta_lambda$ is Yosida approximation of $beta(x,t)$. Then there is said that it is easy to show that
$$ lambda_1 geq lambda_2 Longrightarrow beta_{lambda_1}(t) leq beta_{lambda_2}(t) $$
and for positive constant $c>0$:
$$ (c beta)_lambda = cbeta_{c lambda}.$$
I can't see it and and maybe some of you will be able to help?
Thanks in advance!
ordinary-differential-equations pde approximation multivalued-functions
asked Jan 10 at 0:26
WawMathematicianWawMathematician
13
$endgroup$
I study multi-valued problems on the basis of M. Chipot's book and I have a seemingly simple problem.
Set for a.e. $x in Omega$, $lambda > 0$, $forall t in mathbb{R}$:
$$ J_lambda (x,t) = (I +
lambda beta(x,cdot))^{-1}(t) $$
$$ beta_lambda (x,t) = (t-J_lambda(x,t))/lambda .$$
So $J_lambda$ is resolvent and $beta_lambda$ is Yosida approximation of $beta(x,t)$. Then there is said that it is easy to show that
$$ lambda_1 geq lambda_2 Longrightarrow beta_{lambda_1}(t) leq beta_{lambda_2}(t) $$
and for positive constant $c>0$:
$$ (c beta)_lambda = cbeta_{c lambda}.$$
I can't see it and and maybe some of you will be able to help?
Thanks in advance!
ordinary-differential-equations pde approximation multivalued-functions
ordinary-differential-equations pde approximation multivalued-functions
asked Jan 10 at 0:26
WawMathematicianWawMathematician
13
asked Jan 10 at 0:26
WawMathematicianWawMathematician
13
edited Jan 10 at 19:00
WawMathematician
asked Jan 10 at 0:26
WawMathematicianWawMathematician
13
asked Jan 10 at 0:26
WawMathematicianWawMathematician
13
asked Jan 10 at 0:26
WawMathematicianWawMathematician
13
13
add a comment |
add a comment |
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